1. Explanations: Fundamentals of Logical Reasoning …media.careerlauncher.com.s3.amazonaws.com/mba/mba_e-lib...Hence, margin = 20%. Profit amount = 20% of sales = 0.2 × 22.5 = Rs

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1. Explanations: Fundamentals of Logical Reasoning and Data Interpretation 1

2. Explanations: Fundamentals of Ratio & Proportion 71

3. Explanations: Fundamentals of Time, Speed & Distance 93

4. Explanations: Fundamentals of Grammar 119

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Explanations: Fundamentals ofLogical Reasoning & Data Interpretation

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Data Interpretation

Practice exercise – A1

From the pie charts, the sales figures in rupees (in crores)are:

Company 2002 2003

HLL 45 150

P & G 37.5 75

Henkel 22.5 75

Nirma 30 56.25

Others 15 18.75

Above table has been made to explain the things. You neednot make it. Also, in pie-charts of this sorts, it is veryadvantageous to identify the ratio and percentage increaseof total size of pie in the year after year. In this case the ratioof the total size of market is 6 : 15 i.e. 2 : 5

1. b %150100150

150375 =×−

Since one would already have found the ratio ofthe pies, before starting, as 2 : 5, one could also

have mentally just calculated − = =5 2 3

150%2 2

2. e Growth in sales of P & G = (0.2 × 375 – 0.25 × 150)= Rs. 37.5 crore.HLL = (0.4 × 375 – 0.25 × 150) = Rs.112.5 crore.Henkel = (0.2 × 375 – 0.15 × 150) = Rs.52.5 crore.Others = (0.05 × 375 – 0.1 × 150) = Rs.3.75 crore.

Alternative method:Since growth in total sales has been 150% from2002 to 2003.So we can change the pie chart of 2003 asHLL = 40 × 2.5 = 100%P & G = 20 × 2.5 = 50%Henkel = 20 × 2.5 = 50%Nirma = 15 × 2.5 = 37.5%Other = 5 × 2.5 = 12.5%These percentages are of market size in 2002, i.e.Rs. 150 crore.Now, visually, one can figure out the minimum growthin sales is in others.

3. b Growth rate of HLL = 150 45

100 233.3%45− × =

4. e Sale of P & G in 2002 = 25% of 150.Sale of P & G in 2003 = 20% of 375.

Ratio of sales in 2002 : 2003 = 0.25 150 10.2 375 2

× =× .

Alternately, since we know the ratio of the piesis 2 : 5, one just needs to do the following 25% of2 : 20% of 5 i.e. 50 : 100

5. a Detergent market grows by 10% of 150= Rs. 15 crore.

6. a It is equivalent to that of the whole market = 10%.

7. a 10% annual growth for 2 years means 21% growth

(using the formula, ab

a + b + 100

).

Sales in 2005 = 1.21 × 375 = Rs. 453.75 crore.

8. d Costs of HLL in 2002 = 30 × 1.1 = Rs. 33 crore.Profit = 45 – 33 = Rs. 12 crore.

Profit percentage = %4.361003312 =×

Margin = 100Sales

ofitPr × = %67.261004512 =×

9. a Henkel = %3.2331005.22

5.2275 =×−

P & G = %1001005.37

5.3775 =×−

⇒ Difference = 133.33%

Alternative method:If we keep base for 2002 and 2003, then (As shownin the solution of question 2)

Henkel =50 15 7

15 3− = × 100 = 233%

P & G = 50 25

25−

= 100%

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10. c The ratio will not change because if the total salesare doubled, the sales of Nirma will also double forboth the years.

Hence, the ratio will be 56.25 15

= 30 8

.

Again one could have used the ratio of pie 2 : 5effectively and answer would have been 15% of5 : 20% of 2 i.e. 75 : 40 i.e. 15 : 8

11. c Surf Excel = 30% of HLL.HLL = 30% of detergent market.Surf Excel = 30% of (30% of detergent market)= 9% of detergent market.

12. c Profit = 25%. Hence, margin = 20%.Profit amount = 20% of sales = 0.2 × 22.5= Rs. 4.5 crore.

13. a Since all the companies have the same expenditure,the company with the maximum sales will have thehighest profit as well as profit percentage.

14. d Cost =Sale

100 Pr ofit percentage+ × 100

Cost of Henkel in 2002 =150 15% 22.5

1.25 1.25× = = 18 crore.

Henkel’s cost in 2003 = 19.8.Henkel’s sale in 2003 = 75.

Profit percentage =55

19.8 × 100 = 278%.

15. c P & G’s profit in 2003 = S 0.8S

100 25%0.8S− × = .

If both sales and costs increase by 10%, then saleswill be 1.1S and cost will be 1.1(0.8S) and hencethe profit percentage will remain at 25%.


Index is a representation of actual value. A year is designatedas the base year and the index for the base year is taken as100, e.g. in the given graph for sales, cost and profit, thebase year is 1993 (index = 100). For 1994, the sales indexis 120, which means that sales in 1994 is 20% more thanthe sales in 1993 and sales in 1995 (index 131) is 31% morethan in the base year (1993) sales. Similarly, the cost indexof 98 in 1994 indicates that the cost has gone down by 2%over the base year of 1993 and in 1996 the cost has goneup by 10% over cost in 1993. (Index = 110)

Index is used to compare data over a large number of yearsand to determine trends. Because the base year value isalways taken as 100, it is easy to determine the percentage

change in any year with respect to the base year. An indexcannot give the actual value for any year unless the actualvalue and the index of any one year is given to us.e.g. if, for the given graph, actual sales in 1994 (sales index120) is Rs. 144 crore, then we can say that:Sales index of 120 = Sales of 144 crore.

Sales index of 1 = Sales of 144

120

= Rs. 1.2 crore.

Sales index of 100 = Sales of Rs. 120 crore.

Thus, we now have the conversion factors, i.e. to convertsales index to sales : Multiply index with 1.2 and to convertsales into sales index : Divide sales by 1.2.

Thus, now we can determine the index of sales for anyyear if the sales (rupees in crores) is known and also thesales (rupees in crores) of any year if the sales index isknown.

In the solutions below the abbreviations being used are:

SV = Sales value SI = Sales indexCV = Cost value CI = Cost indexPV = Profit value PI = Profit index

1. e Index does not give the actual value. The profitindex is 110.

2. a Sales in 1993 = Rs. 500 crore = Index of 100.In 1998, SI = 154, which is 54% above the index of1993. Thus, sales of 1998 should be 54% morethan the sales in 1993.Thus, sales in 1998 = 500 × 1.54 = Rs. 770 croreORIndex of 100 = Sales of Rs. 500 crore.Index of 1 = Sales of Rs. 5 crore.Index of 154 = Sales of 154 × 5 = Rs. 770 crore.

3. c In 1996, CI = 110 and CV = Rs. 550.

In 2002, CI = 190. Thus, CV = 550

190 Rs. 950110

× = .

4. e It is not possible to find the value across indices.With one value of cost we can find out all the valuesof cost but none of the values of sales or profits.

5. e Gross profit = (Sales – Cost), since we do not haveSV and CV for various years, the gross profit cannotbe calculated.

6. b In 2003, PI = 130 and PV = Rs. 780.

In 2001, PI = 110. Thus, PV = 780

110 Rs. 660130

× = .


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7. a In 1993, SI = CI = 100,SV = Rs. 300 andCV = Rs.120. Thus, SV in 1998 = 154 × 3 = Rs. 462and CV in 1998 = 153 × 1.2 = Rs. 183.6.Gross profit = 462 – 183.6 = Rs. 278.4.In the given question, we cannot determine the profitbecause it is defined as (Sales – Cost – Tax) andwe do not have the tax for 1998.

8. b In 1993, SI = CI = 100,SV = Rs. 300 andCV = Rs.120. Thus, SV in 2003 = 188 × 3 = Rs. 564and CV in 2003 = 166 × 1.2 = Rs. 199.2.Gross profit = 564 – 199.2 = Rs. 364.8.

9. c In 1993, SI = 100 and SV = Rs. 700.Thus, SV in 1998 = 154 × 7 = Rs. 1078.In 1996, CI = 110 and CV = 600.

Thus, CV in 1998 = 153 × 600

110= 834.54.

Gross profit = 1078 – 834.54 = Rs. 243.45.

10. b Current SI in 1993 = 100 and SI in 1994 = 120.New SI in 1994 = 120.

Thus, new SI in 1993 = 100

120

× 100 = 83.33.

11. a Current PI in 1996 = 125 and PI in 2003 = 130.New PI in 1996 = 100.

Thus, new PI in 2003 = 100

125

× 130 = 104.

12. b Only 3 years — 2000, 2001 and 2002 — have theirCI more than the CI for 2003. So for these years,CI will be more than 100.

13. b Total Sales Value for the period 1993-98= 1200 + (1.2 × 1200) + (1.31 × 1200) +(1.52 × 1200) + (1.62 × 1200) + (1.54 × 1200)= Rs. 9828

or (Sum of SI from 1993 to 1998) × 1200

100.

14. a Sales Value for the period 1993-98 = 9828.(Refer question 13)Total Cost Value for the period 1993-98

= (100 + 98 + 95 + 110 + 121 + 153) × 800

100= 5416.Difference = 9828 – 5416 = Rs. 4412.

15. c Average of Cost Index = 1537

11 = 139.72

Average of Sales Index – Average of Cost Index= 147.27 – 139.72 = 7.54


1. e We need the total number of students opting forfinance in three institutes to arrive at the average.

2. c Average salary at IIML

(60 × 550000) + (40 × 420000) + (50 × 725000)

150=

= 574000.Rather than multiplying by 60,40 and 50 you canalso multiply by 6,4,5 and then divide by 6 + 4 + 5= 15.

3. a Calculating as above, average salary at IIMK= 75900.Average salary of IIMK – IIML = 759000 – 574000= 185000.

4. b Average salary at IIMB

(50 × 540000) + (45 × 453000) + (110 × 775000)

= 205

= 647000.

5. d The average salary for finance in IIMB is significantlylower than that for systems. Since the number infinance has gone up, the average salary will belower than what we calculated for question 4.Thus, the answer is 626. (Check by calculation)

6. c Now since the number in systems has increased,the average salary will be more than the averagesalary in question 5 by a large margin.So the answer is 667.

7. c Average salary for finance in IIMA, IIMB and IIMC

(540000 × 50) + (520000 × 90) + (725000 × 45)=

185= 624000

8. b Average salary of the remaining students of

systems = (500 × 850000 – 1000000 × 200)

300= 750000.Instead of the above calculation, one could alsohave done the following mentally : Of 5 studentswith average salary 850000, 2 are in dotcomcompanies with average 1000000. Thus theycontribute 300000 more than overall average andthus on an average each of the 3 non-dotcomstudents would have a salary of (850000 – 100000)= 750000.


9. b Average salary of IIMA finance students placed in

India = (1 × 620 – 0.6 × 800)

0.4 = 350.

Alternately, as explained in the oral calculationsfor solutions to 17, in this question 60% could havebeen taken as the fraction 3/5 and thus 5 studentshave average of 620. Of these 3 have average of800 and thus contribute a total of 540 more to thesum than the average. Thus the remaining 2 studentson an average has 620 – 270 = 350

10. d Rather than taking number as 150 and 50, to reducecalculations, we can take their ratio 3 : 1.

Hence, average = (3 × 453 + 1 × 575)

4= 483.

Practice exercise –A4

1. a Land used for paddy in 2001-02 = 12.71 Mn Ha.Total land used in 2001-02 = 29.93 Mn Ha.Percentage of land used for paddy

12.71100 42.5%

29.93= × =

or %3.43~1003013 −× approximately.

2. c Productivity = Production/Area = Yield.Yield of ragi in 2002-03 = 1327 kg/Ha.Yield of ragi in 2001-02 = 1378 kg/Ha.Percentage change in yield

= 1378 1327

100 3.72%1378

− × �

Calculation technique: One needs to calculate51/1348. 10% of 1378 is 137.8 and hence 5% is68.9. Thus 51 will be less than 5% and only oneoption is left.

3. b Barley. We do not need to solve it. Mere inspectionreveals the answer.

4. c Production = Area × Yield.If yield increases by 10% and the area remains thesame, then the production will also go up by 10%.Hence production in 2003-04= Production in 2002-03 × 1.1 = 2.43 × 1.1= 2.43 + 0.243 = 2.67 Mn T.

5. d Production = Area × Yield.Production = 521 × 12.71/1000 = 6.62 Mn T.

Alternatively, we know 12.5% = 18

.

So 521 ×1 100

8 1000×× = 6.5 Mn T.

It would be higher than 6.5 Mn T as 12.717 > 12.5

6. a Rephrase the question as, “Which crop has thehighest yield over the years?” Answer is barley.

7. b Percentage increase in production of maize in2000-01 = (9.99 – 8.06) × 100/8.06 = 24%.Production in 2002-03 = 9.99 × (1.24)2

= 15.36 Mn T.

Yield = Production/Area.Thus, yield = 15.36 × 1000/6.1 = 2518 kg/Ha.

Alternative method:We can approximate, though some risk is involved.But whenever you approximate, you need tounderstand how much error you are introducing andwhether your actual answer would be more than orless than the calculated answer.Percentage increase in production of maize in

2000-01 =10 8

8−

× 100 = 25%.

Production in 2002-03 = 10 × (1.25)2 = 10 × 1.5625= 15.62 Mn T.But actual answer would be less than this.Let us say → 15.30 Mn T

Now yield =15.3 1000

6.1×

= 2,508 kg/ma.

8. c It can be visually figured out that the year would be2000-01 therefore total production would be 35.52Mn T.

9. a In the year 2000-01, the production should be2.53 + 2.58 i.e. the production should be a little morethan double the original production. Thus yield alsoshould be a little more than double of the given yieldi.e. a little more than 2 × 1329 = 2658. Only oneoption is in that range.

10. d Production of paddy in 2001-02 = 1602 × 12.71= 20.36 Mn Ha.Production of paddy in 2000-01 = 12.81 Mn T.Increase in production

= %5910081.12

81.1236.20 =×−.

Alternatively, we can approximate as well.Production of paddy in 2001-02 = 1600 × 12.71

= 1600 ×100

8 100× = 20 Mn Ha as 1

12.58

=

Production of paddy in 2000-01= 12.81 Mn Ha ~ 12.5 Mn Ha.

Increase in production =20 12.5

12.5−

× 100 ~ 60%.


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11. c In such questions use options. Start with an intelligentchoice of option e.g. as maize appears in 3 of thefour choices, check trend for maize. As trend formaize does not satisfy the required condition, safelymark the fourth option i.e. (c).

12. e The given data pertains to production and not sales.We cannot assume that 100% of paddy produced,was sold. Thus, data is inadequate to solve thegiven problem.

13. c Cost of production of barley = Rs. 1,250 per quintal.If there is no loss of produce, then selling price for25% profit = 1250 × 1.25 = Rs. 1562.5.Loss in produce is 10%.Thus, for every 100 kg produced, only 90 kg can besold. So for 25% profit, SP = (1562.5/0.9)= Rs. 1,736 per quintal.

14. b By rule of alligation, ratio of quantity available toPDS to open market = 2 : 5.Therefore, 2/7th or 28.57% is allocated for PDS.

15. d Yield required = 1518 kg/Ha.Amount of production = 8.06 Mn T.

Thus, land required =8.06 1000

5.311518

× = Mn Ha.

Land should be less by (5.86 – 5.31) = 0.55 Mn Ha.


1. b Number of Esteems = 10% of 40% of 2.5 lakh = 4% of 2.5 lakh = 0.1 lakh.

2. c 60% of 40% of total car sales = 48000, i.e. 24% oftotal sales = 48000.Thus, total sales = 48000/0.24 = 2 lakh.

3. a HM market share = 15%.Zen market share = 20% of 40% = 8%.Therefore, difference = (15 – 8)% = 7%.= 7% of 2.5 lakh = 17500.

4. c Number of Maruti cars sold = 40% of 2 lakh = 80000.Sale of Maruti 800 = 48000.Increase of Maruti 800 sold = 25% of 48000 = 12000.New total sales of Maruti = 92000 = 0.92 lakh.New total car sales = 200000 + 12000 = 2.12 lakh.

Percentage share of Maruti = %4.4310012.292.0 =× .

Thus, percentage points by which Maruti share hasincreased = 43.4 – 40 = 3.4 points.

5. c The number accounted for by Hyundai and Daewoo= (10 + 25)% of 4 lakh = 35% of 4 lakh = 1.4 lakh.

6. a Number of Santros = 75% of 25% of 4 lakh = 75000.

7. d Number of Santros = 75000 (from question 6).Number of Zen = 20% of 40% of 4 lakh = 32000.Difference between Santro and Zen sales = 43000.

8. c Sales of Maruti 800 in Mumbai= 60% of 40% of 5% of total India sales= 0.012 × 24 lakh = 28800.

9. b Sales in 2004 = 1.15 × 4 lakh = 4.6 lakh.Thus, HM sales = 15% of 4.6 lakh = 69000.

10. c Maruti sales in 2003 = 40% of 4 lakh = 1.6 lakh.Increase in Maruti sale in 2004 = 20% of 1.6 lakh= 0.32 lakh.Thus, Maruti sales in = 1.6 + 0.32 = 1.92 lakh andtotal sales = 4 + 0.32 = 4.32 lakh.Percentage share of Maruti = (1.92/4.32) × 100

= 44.44%.

11. d Hyundai’s sales in 2003 = 25% of 4 lakh = 1 lakh.Hyundai’s sales in 2004 = 90% of 1 lakh = 0.9 lakh.Total sales in 2004 = 130% of 4 lakhs = 5.2 lakh.Hyundai’s share of sales = (0.9/5.2) × 100 = 17.3%.

12. a Total sales in 2004 = 5.2 lakh. (Refer question 11)Daewoo’s sales in 2003 = 10% of 4 lakh = 0.4 lakh.Daewoo’s share in 2004 = (0.4/5.2) × 100 = 7.7%.

13. b Sales of Maruti in 2003 = 40% of 4 lakh = 1.6 lakh.Increase in sales of Esteem in 2004 = 40000.Thus, sales of Maruti in 2004 = 2 lakhs.Total sales in 2004 = 1.2 × 4 lakh = 4.8 lakh.Percentage share of Maruti in 2004= (2/4.8) × 100 = 41.67%.

14. c Sales of Esteem in 2003 = 16000.Sales of Esteem in 2004 = 40000 + 16000 = 56000.Sales of Maruti in 2004 = 2 lakh. (Refer question 13)Esteem’s share in Maruti sales = (0.56/2) × 100

= 28%.

15. d Sales of Opel in 2003 = 25% of 10% of 4 lakh = 10000.



We are making table for convenience sake. You can avoidmaking it. The following figures, based on the graph, havebeen used for the calculations.

YearShareholders

wealthTotal debt

Net fixed

assets

Net working capital

1999 125 410 180 350

2000 280 350 250 375

2001 475 430 335 550

2002 750 675 550 745

2003 1150 800 760 100

Following abbreviations have been used.

SW = Shareholder’s wealth TD = Total debtNFA = Net fixed assets NWC = Net working capital

1. a Average of the total debt is 556.Hence, the total debt is above the average for2 years (2002 and 2003).

2. d NWC in 1999 = 350, NWC in 2003 = 1000.If NWC in 1999 = 100, then

NWC in 2003 = 1000

100 285.6350

× =

3. d NFA growth rate in 2000 = 250 – 180

180 = 38.9%.

NFA growth rate in 2002 = 550 – 335

335 = 64.2%.

Hence, the difference is 25.3%.

4. c Total growth = 1150 – 125 = 1025.

Thus, average annual growth = 1025

4 = 256.

Thus, SW in 2004 = SW in 2003 + 256 = 1150 + 256 = 1406.

5. e Average annual growth rate

× =1150 –125100 205%

125(Refer question 4)Growth rate is less than 205% for all the choices.

6. d The word compounded has no relevance and theitem which has grown the maximum will have thehighest CAGR and average annual growth rate.So shareholder’s wealth has grown the maximum.

7. b NWC growth rate in 2002 = 745 – 550

550 = 35.45%.

SW growth rate in 2003 = 1150 – 750

750 = 53.33%.

Thus, NWC is less than SW by 53.33 –35.45

3.33= 33%.

Alternatively, you can approximate, since choicesare far apart.

NWC growth rate in 2002 ≈750 550 200 4

550 550 11− = =

= 36.36% 1

9.09%11

=

SW growth rate in 2003 ≈ 1150 750 400 8750 750 15

− = =

= 53%.

NWC is less than SW by = 53 36 17 1

53 53 3− = ≈ = 33.3%.

8. e Growth rate of NWC is less than that of SW for allthe years.

9. c Growth rate of:NFA in 2000-2002 = 120% and TD in 2000-2002= 92%.SW in 1999-2000 = 124% and SW in 2002-2003= 53%.TD in 2001-2002 = 56% and NWC in 2001-2002= 35%.TD in 2002-2003 = 23% and SW 2001-2002 = 58%.

10. a While doing this visually, the only doubt is whetherfor the year 2001 the NWC is above average or not.Assume the average as the value in 2001 and usethe method of deviation. If sum of deviation is +,average is above the value of 2001 or else otherway round. In this case the sum of deviation wouldbe –200 – 175 + 225 + 450 which will obviously bepositive. Thus average will be greater than that foryear 2001. Hence, (a) is correct.


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For questions 1 to 10:Again we are making the table, but you can avoid that.The following figures, based on the graph, have been usedfor the calculations. The table gives the percentage changeover the previous year or the growth rates.

Year General Mining Mfg. Elect. FD GDP

1998-99 8.5 7 8.5 8.5 10 4.2

1999-2000 13 9.5 14 8 8 5

2000-01 5.5 -2 6.5 4 7.5 4

2001-02 6.5 6 6.5 6.5 5.5 6

2002-03 4 -2 4.5 6.3 5.8 5

1. b Growth rate of manufacturing in 2001-02 = 6.5%.Thus, growth in manufacturing in 2001-02 = 6.5%of 80 = 5.2Mining data is redundant over here.

2. d Average of GDP growth rate

4.2 5 4 6 5.55%

5+ + + += =

3. b Add the growth rate of 5 years for all the items, thissum will give the relative ranking of the growthrates. This sum for mining and electricity is lessthan that of fiscal deficit.In question of this type where one has to eliminateusing options, choose which sector to start withvery intelligently. Choose that one which appears intwo options and thus with just one calculations,atleast you can shortlist your choice to two options.

4. e The chart gives us the percentage change from theprevious year. With the GDP given, Fiscal deficitcan not be calculated because the data does notgive the relation between GDP and fiscal deficit.

5. c Add the percentage change for the four items from1999-2000 to 2002-03, the item which has thehighest sum will end up with the highest figure.

6. e Cannot be determined because the data gives usthe percentage change only and we do not havethe actual values for any of the years.

7. b GDP 1999-2000 = 3,09,000 × 1.05= Rs. 3,24,450 croreFiscal deficit 1999-2000 = 11,200 × 1.10 × 1.08= Rs. 13,305 croreFiscal deficit as percentage of GDP in 1999-2000

= 13305

324450 =

13324

= 4.1%

8. c If initially the amount of electricity = 1, then at theend of period amount of electricity= 1.085 × 1.08 × 1.04 × 1.065 × 1.063 = 1.3796 orelectricity will grow by 37.96%If growth rate of electricity in 2000-01 is replacedwith growth rate of mining 1999-2000 (9.5%), thenelectricity = 1.085 × 1.08 × 1.095 × 1.065 × 1.063

= 1.4526or electricity will grow by 45.26%Hence, difference = 45.26 – 37.96 = 7.3%

9. d The highest total production would be in 2001-02,because in the last year 2002-03, there is a negativegrowth and thus is less than the figure in 2001-02.

10. c The highest growth rate for the various years are:1998-99-10%, 1999-2000-14%, 2000-01-7.5%2001-02-6.5% and 2002-03-6.3%Thus, figure at the end of the period= 1.1 × 1.14 × 1.075 × 1.065 × 1.063 = 1.526Hence, overall growth = 52.6%


We have made the table for convenience sake, but youshould not make it.Let production of :A in the various years be given by A2000, A01, A02 and A03

B in the various years be given by B2000, B01, B02 and B03

C in the various years be given by C2000, C01, C02 and C03

D in the various years be given by D2000, D01, D02 and D03

The following table, based on the graph, gives the values ofthe percentage share of production for A, B, C and D for thefour years

Year A B C D

2000 23% 37% 20% 20%

2001 23% 35% 22% 20%

2002 20% 30% 30% 20%

2003 10% 30% 20% 40%

Let total production in 2001 = p, thus total production in2003 = 1.21pGiven C03 – A01 = 1320 thus 0.2 x 1.21p – 0.23p = 1320thus p = 110,000 MT, and total production in MT in2000 = 1,00,000 2001 = 1,10,0002002 = 1,21,000 2003 = 1,33,100

1. a C2000 = 20% of 1,00,000 = 20,000 MT.C02 = 30% of 1,21,000 = 36,300 MT

Thus, 236300 20000(1 r /100) r 35%= + ⇒ =One could also have approximated it as follows:The production increases from 20k to 36k i.e. aincrease of 80%. Thus CAGR has to be less than40% and once could check with 35%


2. e Given data pertains to production and not sales.

3. a Product B

4. d Cumulative production of B will still be the highestbecause the ratio of production for the variousproducts for the various years is constant.

5. e Cannot be determined because the price of A is notknown for the four years.

6. c The total revenue from selling B and D= 3k × 150 + 2k × 120 = 690k.Thus total cost in producing B and D

= 690k

1.05 = 657k

Cost in producing D = 2k × 100 = 200k.

Rs.120{ cost of one unit Rs.100

1.20= =� }

⇒ Cost in producing B = 457k

⇒ CP of 1 unit of B = 457

3 = 152.3

⇒ Thus loss % on selling B = 2.3

150×100 = 1.53%

7. b Remember that in this question you cannot use thetotal production in 2003 as 133,100 because thiswas arrived at when there is a 10% increase inproduction in each year. So all calculations have tobe done again for this question.

C03 – A01 = 1320

20% × Total03 – 23% × Total01 = 1320

20% × 1.5625 × Total01 – 23% × Total01 = 1320

(0.3125 – 0.23) × Total01 = 1320

Total01 = 16,000Total production in 2003 = 16,000 × 1.5625 = 25000Thus required difference = 10% of 25000 = 2500

8. c All this problems asks is that what is the value ofx% such that -5% and x% successively amountsto 21%. Thus 0.95 × k = 1.21k = 1.2736Thus growth in 2002 should be 27.36%Though data of growth rate given originally ischanged, but since this question has nothing to dowith actual production amounts and we do not haveto use 1320 MT, the question is easy.

9. e This is very similar to question number 7. The originaldata of 10% increase every year is changed andhence we cannot use the production figures foundearlier. With the new growth rates we have,C03 – A01 = 1320

20% × Total03 – 23% × Total01 = 132020% × 0.95 × 1.1 × Total01 – 23% × Total01 = 132020% × 1.045 × Total01 – 23% × Total01 = 1320This gives a negative production figure for 2001and hence data is inconsistent.

10. e Total production of B in 4 years= 37000 + 38500 + 36300 + 39930 = 1,50,630Total sale of B in 4 years= 37000 + 38500 + 36300 + 39930 × 0.5 = 1,31765Sales Value of B = 131765 × 400 = Rs. 5,27,6000Total cost of B = 100 + 150630 × 200 = Rs. 30126100Profit = 74.9%


1. b Percentage change in the average price for2000 = 50%2001 = 16.66%2002 = 14.28%2003 = 37.5%One does not need to calculate this.One can just check it visually. The changes are 1,0.5, 0.5, –1.5 and the base are 2, 3, 3.5, 4.5. Theyear where change is maximum and base is leastwill have the highest growth rate.

2. b Production of Maa-roti in 2000 = 30,000 andProduction of Maa-roti in 2003 = 60,000Average annual growth rate for 2000-2003

= %3.3331

100000,30

000,30000,60 =××−

3. b Average price per car

40 2 30 3 50 3.5 40 4 60 2.5220

Rs. 2.97 lakh

× + × + × + × + ×=

=

4. e Cannot be determined because total cars producedby the car industry is not known for any year.

5. b Production of Maa-roti in 2001 = 50,000

Percentage share of Maa-roti = 50,000

100200,000

×

= 25%

6. a Production 2001 = Production 1996 × 1.08 × 1.15,

Production in 1999 =200000

1610301.08 1.15

=×

7. e The assumption that production = sales is not valid unless it is specifically stated.

8. a Cars sold in 2000 = 80% of 30,000 = 24,000Price per car in 2000 = Rs. 3 lakhRevenue in 2000 = 24,000 × 3 lakh = 72,000 lakh


Page: 9��

9. c Revenue in 2000 = 72,000 lakh (Refer question 8)In 2001, the entire production of 2001 and 20%2000 production is sold = 56,000Revenue in 2001 = 56000 × 3.5 = 196000 lakhRevenue 2000 + 2001 = 2,68,000 lakh

10. c Revenue for each year is given below:1999 : 40 × 2 = Rs. 80,000 lakh2000 : 30 × 3 = Rs. 90,000 lakh2001 : 50 × 3.5 = Rs.1,75,000 lakh2002 : 40 × 4 = Rs.1,60,000 lakh2003 : 60 × 2.5 = Rs.1,50,000 lakhThus, in 2001 Maa-roti has the highest growth.

11. a 2001. Calculate from question 10.

12. e Again, we cannot assume that production = sales.

13. c Average price per car for Maa-roti = Rs. 4 lakhAverage price per car of car industry = Rs. 4.5 lakhMarket share of Maa-roti = 62.5%Let average price of rest of the car industry = p,then 0.625 × 4 + 0.375 p = 4.5Thus, p = Rs. 5.33 lakhRecollect that in case of averages, oral calculationsare far better as follows : 62.5% is 5/8 and thus 8items have an average price of 4.5 and 5 of themhave an average of 4. Thus these 5 were “given” atotal of 0.5 × 5 = 2.5 by rest 3 and thus rest 3 will

have an average of 4.5 + 2.5

3 = 4.5 + 0.83 = 5.33

14. e FC = Fixed cost VC = Variable cost/unitSP = Selling price N = Number of units sold

Break Even Point = BEP is the production level atwhich the company is in a no profit or no losssituation.

BEP = FC / (SP – VC), since the question does notgive us all the required data, we cannot determinethe BEP.

15. c 40,000 cars are produced in 1999If 10% gets rejected, net production = 36,000Turnover = 36,000 × 2 lakh = 72,000 lakh= 720 crore

16. b Number of Maa-roti Omni produced in 2001= 15% of 50,000 = 7500

17. e Since we do not have the average price per car forMaa-roti 800 for the two years, we cannot determinethe ratio of sales revenue.

18. c Ratio of Maa-roti Gypsy produced in 2002 to 2001

= 5:2500200

000,50of%10000,40of%5 ==

19. b Maa-roti EsteemProduction 2001 = 20% of 50,000 = 10,000Production 2002 = 22% of 40,000 = 8,800

Growth rate = 8800 10000

100 12%10000

− × = −

20. a Refer to question 10 for data, simple annual growthrate

= 150000 80000 1

100 21.8%80000 4

− × × =

21. a Refer to question 9 for data. For CAGR apply formulafor CI, A = P (1 + r/100)n

A = 1,50,000 lakh, P = 80,000 lakh and n = 4 timeperiods, 150 = 80 (1 + r/100)4 thus r = 17%

22. e Only the average price per car is given. The numberof cars produced is not given thus the averageprice per car for 2001-2004 cannot be determined.

23. a In 1999 Maa-roti produced 40,000 cars, which is20% more than in 1998, thus production in 1998

= 40000

1.2 = 33333

Practice Exercise – A10

The following figures, based on the graph, have been usedfor the calculations. Abbreviations used are :

Int. = Interest Depn. = Depreciation

NP = Net Profit OI = Other Income

raeY TIDBPO .tnI .npeD PN xaT IO

9991 011 03 5 57 0 1

0002 582 08 02 561 02 2

1002 593 08 05 502 06 4

2002 025 011 08 582 58 01

3002 083 541 021 01 501 6

1. d Operating profit after interest and taxes but beforedepreciation = Net profit + Depreciation, this ishighest for 2002.

2. c Refer to the table above, tax and depreciation havewitnessed a constant growth across the years.


3. b NP has grown from 75 in 1999 to 165 in 2000,hence percentage increase in NP for 1999-2000= 120%.OPBDIT has grown from 395 in 2001 to 560 in 2002,hence percentage increase in OPBDIT for2001-2002 = 41.8%Thus, difference = 120 – 41.8 = 78.2%.

4. b Interest grows from 30 in 1999 to 145 in 2003, thussimple annual growth rate

145 – 30 195.8%

30 4 = × =

5. e For CAGR, use the formula for compound interest145 = 30(1 + r/100)4 thus r = 48.3%

Alternative method:Total growth from 1999 to 2003 is 400%.Now, let us take a convenient value, say 50%(4 year)

Use a + b + ab

100

50 + 50 +50 50

100×

= 125

125 + 125 +125 125

100×

~ 400%

So CAGR is closer to 50%, hence the answer is 48%.

6. b The company would have been placed in the MATcategory in 1999 because its tax for the year iszero.

7. a OPBDIT 2002 = 520 and OPBDIT 2003 = 380OPBDIT 2003 as a percentage of OPBDIT 2002

=380520

= 73.1%

8.b NP and OPBDIT increase till 2002, after which bothof them decrease.

9. a Interest in 1999 = 30, then interest in 2003 = 145,thus if interest in 1999 = 100, then interest in 2003

= 145 × 100

30 = 483

10. c The maximum depreciation was in 2003, hencemaximum assets were acquired in 2002.


This set is DI type reasoning set, where the reasoning isprimarily used in solving individual questions.

1. d Total bill amount:(150 × 1 + 50 × 3.5 + 25 × 7 + 14 × 5 + 15 × 2) + 99= 699

2. b Total bill amount:(150 × 1+ 50 × 3.5 + 25 × 7 + 14 × 5 + 15 × 2) + 99+ 12% on (25 × 7) = 720

3. a Since, in STD calls it’s not mentioned that the callhas been made to which operator, (like: Airtel, WLLetc.) one may get tempted to mark option (d).But upon solving, and considering all the possiblecases, we get

Rohan spent:On STD: 18 × 1.5(Airtel, GSM) = Rs 27

Or 18 × 2 (WLL) = Rs 36

Local: 60% of 120 = 72 min (because Airtel andGSM have same rates) @ 1 Re. min.∴ Amount spent = 72 × 1 = Rs 7240% of 120 = 48 min @ Rs. 2/min.∴ Amount spent = 48 × 2 = Rs 96Total Bill = (72 + 96 + 27 or 36)= Rs 195 or Rs 204

Similarly for Mohan:On STD: 30 × 1.5 = Rs 45 or 30 × 2 = 60Local: 70 × 1 = Rs 70

30 × 2 = Rs 60Minimum and Maximum amounts that Mohan wouldhave spent are Rs.175 and Rs.1907.The maximum possible difference is(Rs.204 – Rs.125) = Rs.59.

4. a Normal tariff for 40 SMS/month would have been= 40 × 1.5 = Rs 60With SMS scheme: 40 × 0.60 + 35 = Rs.59In fact, in order to benefit from the scheme, oneneeds to send a minimum of 39 SMS.So, option (a) is the correct choice.

5. d Option (a): 30 × 1.5 + 55 + 99 (Price of Plan)= Rs 199 (assuming local calls are made @ Re. 1)

Option (b): 16 × 1.5 + 76 + 99 (Price of Plan)= Rs 199 (assuming local calls are made @ Re. 1)

Option (c): 10 × 1.5 + 10 × 2.5 + 10 × 2.5 = Rs 65Now local calls (since not mentioned) may bedistributed like 5 × 2 (WLL/ Landline) + 25 × 1 (GSM/Airtel) = Rs. 35Total 65 + 35 + 99 (Price of Plan) = Rs 199Therefore, bill amount of Rs 199 per month ispossible in each of the three options (a), (b) and(c).


Page: 11��

∴ Option (d)Total bill = 8 × 1.5 + 4 × 3.0 + 7 × 3.5 + (55 × 1 or 55× 2) + 99 = Rs.202.5 or Rs.257.5Hence (d) is correct.

6. a Checking optionsOption (a): 8 ISD calls can be distributed as:4 calls to Rest of the world and another 4 calls toGulf.Bill amount = 4 × 40 + 4 × 10 + 99 = Rs 299

Option (b): Checking with the maximum rate ofboth GSM and WLL, maximum possible bill amount= 35 × 3 + 10 × 3.5 + 99 = Rs 239

7. b Rates in Delhi-Mathura segment:

Outgoing calls to Earlie r Ne w Diffe re nce

Airtel 1.5 1 0.5

GSM/CDMA 1.5 1 0.5

WLL/Landline 2 1 1

Therefore, to cover the additional charge ofRs.20/month

One has to make =

200.5 40 calls to Airtel/GSM

and =

201 20 calls to WLL/Landline.

Hence, inOption (a): No profitOption (b): profit = (35 × 0.5 + 10 × 1) – 20 = Rs. 7.50Option (c): profit = (50 × 0.5) – 20 = Rs. 5

8. c In order to have no loss from the plan, user have tobalance out additional charge of Rs. 250 by makingcalls.For to be definitely sure, we will take that existingSTD rate which will give us minimum differencewhen compared with new scheme∴ STD of Airtel = Rs. 1.5/minDifference = 1.5 – 1 = 0.50/min

∴ Minimum calls = 2500.5

= 500

Minimum 500 calls has to be made such that I amdefinitely sure of having no loss from opting for thisplan.

9. b Checking options:(a) Difference in the rates

In STD = 1.5 – 0.5 = Rs 1/minAmount saved = 1 × 12 = Rs 12In local = 1.0 – 0.5 = Rs 0.5/minAmount saved = 0.5 × 8 = Rs 4Total saved amount = 12 + 4 = Rs 16Additional Charge levied @ Rs. 5/day for 3 days= Rs 15∴ One can gain Re. 1 in this calling pattern.

(b) Difference in the rates:In STD = 1.5 – 0.5 = Rs 1/minAmount saved = 1 × 14 = Rs 14In local = 1.0 – 0.5 = 0.5/minAmount saved = 0.5 × 18 = Rs 9Total amount saved = 9 + 14 = Rs 23Additional charge paid= 5 × 5 = Rs 25∴ Loss = Rs (25 – 23) = Rs 2Option (b) is the correct choice.Rest of the options need not be checked.

10. b Difference in amount = 1 (Airtel local) – 0.5 = Rs 0.5/ call is saved

Additional Rent paid = Rs. 5 dailyTo balance the additional charge, minimum no. of

calls that should be made = 5

0.5 = 10 calls

∴ In order to GAIN one should make atleast10 + 1 = 11 calls

11. e This was a tricky one.There is no need to calculate anything.Absolute profit will always remain the same.Profit will always be :[100 × 1 (free calls) + 100 × 1.5 (free SMS) – Rs. 200]= Rs. 250 – Rs. 200 = Rs. 50Note: Has it been said which will have “maximumprofit percentage”, then one has to calculate.Be careful while reading the question.

12. a This one is also a tricky question.Total ISD calls made 12 min/month and 80% of thebill came from the Rest of the world.Let, calls to USA, Canada, Europe (Fixed Lines) beof ‘a’ minutes duration, calls to GULF, Europe (Mobile),SAARC be of ‘b’ minutes duration and calls to Restof the World be of ‘c’ minutes duration.

⇒ 7a 10b 40 c+ + = Total Bill (in Rs.)

Or7a + 10b = 20% of (7a + 10b + 40c)

7a + 10b = 15

(7a + 10b + 40c)

⇒ 4(7a + 10b) = 40c ...(i)and also,a + b + c = 12 (total duration) ...(ii)Solving equations (i) & (ii), we get only one set aspossible solution. (0, 6, 6)i.e.a = 0 minuteb = 6 minutesc = 6 minutes∴ calls to USA, Canada, Europe (Fixed Lines) werefor minimum duration.


For questions 13 and 14: The outgoing call and SMScharges are reduced to half.

13. b Since, 20% drop in the total bill was because ofdrop in the rates by 50%.

∴ In order to get 20% drop, he needs to spend 200.5

= 40% of the call-time during the Happy Hours.⇒ He made 60% of the calls during the day and40% during Happy Hours

Ratio = 60 340 2

=

14. e Since, details of local calls made and SMS sent isnot available, percentage of local calls made duringthe day cannot be determined.


The set can be solved in two ways:A. Adding an extra column for points in Set A and

tabulating all points before proceeding. Thisapproach will involve repeating this step for atleast60% of the teams again after question 6. Thisapproach, though time consuming, is likely to giveguaranteed results. For anyone who finds this settough to conceptualize, this may be the preferredapproach

B. Take each question as and when it comes and tackleit on merit.

For questions 1 to 15:

1. c (32 teams) × (6 Games Played) / 2 = 96 (it takes twoteams to play a match) matches will be played inround 1, 8 matches in round 2, 4 in Quarter-finals,2 in Semi-finals & 1 in the finalTotal = 96 + 8 + 4 + 2 + 1 = 111

Short-cut: Answer has to be > 96 eliminating optionsa & b instantly

Common-error – counting each match twiceforgetting that two teams are playing and hence notdividing by 2

2. e Will be either 3 or 4 depending on whether thesecond round was a ‘home’ match or not.

3. b All teams play atleast 3 away matches in the firstround and another 3 away matches in the quarter-final, semi-final and final.Remember, ‘neutral’ venue is also an away matchfor any team.

4. c The fact that the question exists means that only 1of the 4 options can be right. Check each one till wefind a team which, based on its score after 5 games,played can still reach the 2nd round.After tabulating the points, we can see that onlyBenfica has a chance to qualify for the next round.Other teams given in the options will never qualifyfor the next round.

Sure-shot Approach: Ideally the points for thegroups to which each option pertains should becalculated and noted for future use. (1 at a time:and not all at the start). People trying this questionmentally stand a good chance at getting it wrong.

Common Speed Breaker: Checking all 4 optionsinspite of having found the right option, just to bedoubly sure, often costs students extra time in theexam. Be sure of yourself!

5. b Easy, the team which has the most money at thisstage, which is the team which has won all thegames it has played - Arsenal. This question is arare example of where scanning the data ratherthan reviewing the options would give a fastersolution.

6. c Add results / points for new instructions given andthen use options. Need to be done only for the 4options, not for entire data set.Anderlecht is yet to score a point so far in thetournament and playing “away” in the last match ofthe 1st round will result in a loss for them, thusearning ‘zero’ point from the competition.

7. a All home terms win their matches in their last firstround matches. Lyon , Inter Milan, Barcelona,Liverpool and Arsenal have 16, 12, 13, 11 and 18points respectively.

8. e Cannot be determined because the scores from thelast set of matches of round 1 are not available

9. c Can be counted and is equal to 77.Short-cut: The total number of games won in thelast round was 16 and this was an exception asthere were no draws in this set, hence, the totalnumber of game won in all 6 rounds has to be< 6*16 = 96. This rules out Option (d).Option (a) is less than 16 (last set wins) and has tobe wrong.Simple observation shows that 38 cannot be theanswer, so it has to be 77

10. c Review the options.Option (a): Even after losing the last match of theirgroup, Juventas and Bayern Munich, with 12 pointseach will reach the next round, leaving behind clubBrugge with 9 points.


Page: 13��

Option (b): Rangers by virtue of winning the lastmatch of the group, will end with 9 points, Thus willqualify from group H as a ‘runners-up’ behind InterMilan.Option (c): Lille, after losing the last match of thegroup (because Lille is playing away in the lastmatch) will end up with only 6 points. So, fromgroup D, Benfica and Villareal will quality with 8 and10 points respectively.So, option (c) is the right answer choice.

11 a From the table given in the explanation of theprevious question, it is obvious that Juventas is theonly surviving team among the options given in thisquestion.So option (a) is the correct answer.

12. c As Arsenal has won the most number of games (6)till this stage they are ahead of all teams on themoney earners list. Juventus have won 5 & have 0draws, Liverpool have won 4 & have 2 draws andare hence also behind. Rangers with 2 wins and 3draw’s is not in the reckoning. The key in thisquestion is that the actual values of amount earnedneed not be calculated.

13. c Again doesn’t need to be calculated, the two teamsrecords (wins, draws & losses) need to becompared and are found to be identical

14. a Options (c) & (d) are out as Arsenal has beenknocked out as seen in the last question, leavingJuventus as an automatic choice and just a checkrequired between Liverpool & Rangers.Since all ‘Draws’ took place in 1st round, comparingthe number of draws by Liverpool and Rangers,we get Rangers had 3 draws, while Liverpool had2 draws. So option (a) is the correct choice.

15. d We have no data on quarter-final results, hencethis question only checks on the teams which havedefinitely been knocked out and hence cannot be inthe final. AC Milan & Lyon have both been knockedout, making this an easy choice.


If N is the number of LPS students who took standard Xexamination, then the number of students who passed theexamination is 0.8 N.Number who joined the various streams:Science = 0.2 × 0.8N = 0.16NArts = 0.3 × 0.8N = 0.24NCommerce = 0.5 × 0.8N = 0.4NNumber of students joining from outside = 0.2 × 0.8N =0.16N Ratio of these students in = Science : Arts : Commerce = 2 : 3 : 5, thus number of these students inScience = 0.2 × 0.16N = 0.032NArts = 0.3 × 0.16N = 0.048NCommerce = 0.5 × 0.16N = 0.08N

Number of students in XI Arts = 720 (Given), thus0.24N + 0.048N = 0.288N = 720, thus N = 2500We will have the following:

LPS Outside Total

Total standard X 2500

Total standard XI 2000 400 2400

XI science 400 80 480

XI commerce 1000 200 1200

XI arts 600 120 720

1. b As N = 2500.

2. a 2000 (solved above)

3. c (0.4 + 0.08) × 2500 = 1200

4. b Percentage of arts students in LJC standard XI

= 720

2400 = 30%

5. c 480 (solved above)

6. a Number of science students in XI = 480Number of science students in XI from LPS = 400Percentage of science students in LJC standard XI

who are from LPS = 400

480 = 83.33%

7. e We do not know what percentage of the girls tookarts though we can find out the total number of girls.

8. c 10% of commerce stream = 10% of 1200 = 1205% of the arts stream = 5% of 720 = 36, thus numberof students in XII arts = 720 + 120 – 36 = 804

9. a 5% of arts stream = 5% of 720 = 366% of commerce stream = 6% of 1200 = 72, thusnumber of students in XII LJC = 2400 – 36 – 72= 2292

10. c 8% of commerce stream = 8% of 1200 = 96,10% of science stream = 10% of 480 = 48, thuschange in number of commerce students = –48Percentage change in commerce students= –48/ 1200 = –4%

11. e Though we know the ratio, we do not have the ratioof number of boys and girls in any stream or in total.

12. b Number of girls in science = 50% of boys in scienceThus, girls form 33.33% of science = 480/3 = 160Thus, number of girls in arts = 3/2 × 160 = 240because the ratios Arts : Science is 3 : 2.


13. b Number that failed in XI class = 10% 2400 = 240

14. b Number of boys in science stream = 320(from question 12)Number of boys in arts stream = 720 – 240 = 480Thus, ratio = 320 : 480 = 2 : 3

15. b Total number of students in standard XI afterintroduction of home science = 2400 + 300 = 2700Percentage of home science students in standardXI = 300 / 2700 = 11.11%

Logical Reasoning

Practice Exercise – B1

For questions 1 to 5: Note that from statements I, IV and VIwe can make out that Chandra lives in New Delhi. Now fromV, we can make out that Aparna is not the dancer; and fromII, since Chandra lives is New Delhi, she is also not thedancer. Therefore, Bharti is the dancer and lives in Kolkata.Also from VIII, Bharti is married to Ramsingh. Now sinceChandra lives in New Delhi and Bharti in Kolkata, it is obviousthat Aparna lives in Mumbai, and from IX, she (Aparna) ismarried to Mansingh. Hence, the result is as follows.

atakloK ihleDweN iabmuM

sretsiS itrahB)recnaD(

ardnahC anrapA

srehtorB hgniSmaR hgniSmihB hgniSnaM

1. b

2. a

3. a

4. c

5. a


According to the question from VI, we get__ __ 20 pounds __ __ ... (1)

American

From VIII, we get

Trunk25 pounds __ 20 pounds __ __ ... (2) American

From V, we get

Trunk Box25 pounds 5 pounds 20 pounds __ __ ... (3)

American

From II, we get

Trunk Box25 pounds 5 pounds 20 pounds __ 10 pounds ... (4) American

Logically, the left place in (4) will be occupied by 15 poundsbelonging to Indian as per IV

Trunk Box25 pounds 5 pounds 20 pounds 15 pounds 10 pounds

American Indian ... (5)

From III, Swede can occupy the second spot from left only

Trunk Box25 pounds 5 pounds 20 pounds 15 pounds 10 pounds ... (6)

Swede American Indian

From VII, we get

Trunk Box Crate/Suitcase Crate/Suitcase Carton25 pounds 5 pounds 20 pounds 15 pounds 10 poundsGerman Swede American Indian Belgian

Here 10 pounds weight will be owned by Belgian becauseit is the one which is not owned by German as it is a carton.Therefore, owned by Belgian and consequently 25 poundswill be owned by German.

6. c

7. d

8. e

9. b

10. e


Page: 15��

For questions 11 to 16: According to the question, wehave

Persons Campsites Lakes States

A E I M

B F J N

C G K O

D H L P

If we represent the data in the following way, we get

– , H , J, N

A, – , – , O B, – , K , –

From (ii) (1 )

(4 ) (2 )

(3 )

From (i) From (v)

From (iv)D , F, – , –

We have I camping on P ... from (iii) ... (3)Now, from (3), I and P have to be together. These I and Pcannot be with (1), (2) or (4) as they are already occupied.Therefore, I and P will be there with (3).So we have

– H J N

B K A O

D F I P

Logically, as all the states are occupied except M,M will come along with B and K. Therefore, wehave the final arrangement depicted by the tablegiven below.

nosreP A B C D

etispmaC H F

ekaL L K J I

etatS O M N P

And on the basis of this table we can solve rest ofthe questions very easily.

11. b 12. a 13. d 14. a 15. d 16. b

For questions 17 and 18: As given that the names ofbrothers and sisters do not begin with the same letter andPinku and Gaurav are not Saroj or Sangeeta’s brothers,Pinku cannot be the brother of Pooja and hence he is thebrother of Rakhi.

Now we have that Gaurav cannot be the brother of Saroj,Sangeeta or Rakhi. Therefore, Gaurav is the brother ofPooja. As given that Saroj is not Ratan’s sister and Rakhiand Pooja can also not be the sister’s of Ratan (from aboveconclusions), Ratan is the brother of Sangeeta. Anil willhave to be the brother of Saroj as this is the only validcombination left. Therefore, we have this table finally.

rehtorB retsiS

ukniP ihkaR

varuaG ajooP

nataR ateegnaS

linA joraS

17. d

18. b

19. a Relative speed of approach of the cyclists= (15 + 15) = 30 miles per hour.

Therefore, they will meet after 3030

= 1 hr.

Since the fly has travelled throughout this 1 hr(regardless of direction) at a speed of 20 miles perhour, total distance travelled by it is (1 × 20)= 20 miles.

20. b He has got = 749

= 7 cigarettes.

∴ The duration of time he will take to smoke these

7 cigarettes = 43

7 × hr = 5.25 hr (i.e. 5 hr and

15 min). Now note that after he has smoked these 7cigarettes, he will collect 7 more stubs (one fromeach), from which he will be able to make another

cigarette. This will take him another 43

hr (45 min)

to smoke. Therefore, total time taken = 6 hr.




Note: Jayesh is to the right of Alok, i.e. J, A. Pramod isbetween Bhagat and Subodh, i.e. B, P, S. Subodh isbetween Jayesh and Pramod. So the sequence is:

Bhagat Pramod Subodh Jayesh Alok

1. a Alok is at the extreme left end.

2. d Subodh is in the middle.

3. a Statement I is superfluous.

For questions 4 to 8: According to the question from II, weget

DC

From III, we get_ B_ E_ __ __ _E or _B _

From IV, we get_ B

_ E_ GF FG or _E _B _

From V, we get Bthe final sequence Eas G

FD ... (1)CA

Now on the basis of the sequence given in (1), wecan solve rest of the questions very easily.

4. e

5. a

6. c

7. d

8. e

For questions 9 to 13: On the basis of the given information,we arrive at the following sitting plan that does not violateany of the given conditions.

H

K

F

G

J L

I E

And on the basis of the above figure rest of thequestions are solved as follows:

9. e K is seated between E and H.

10. c Three persons H, L and J or G, I and E are seatedbetween K and F.

11. b The three lady members are E, H and G.

12. c J is to the immediate left of F.

13. a Clearly, J is a male member.

For questions 14 and 15: According to the given questionfrom II, we get

H is to ry

C iv ics ... (1)

From III, we get

G eography

Eng lish ... (2)


Page: 17��

From IV (1) and (2), we get

H isto ry

C iv ics

G eography

English

Econom ics

... (3)

Since history and civics cannot be at any otherplace than this, according to the given conditions.On the basis of this very arrangement, rest of thequestions can be solved very easily.

14. c Clearly, C gives us the clue that the science book isplaced at the bottom. Thus, we know that there arethree books between the civics and science books.

15. d Clearly, history, civics and geography are the threebooks kept above the English book. To deduce this,no additional information is required.

For questions 16 to 20:From the given information in the question:From II, we get Dr Choudhary teaches zoology in MumbaiUniversity.From III, we get Dr Natrajan is neither in Osmania nor in DelhiUniversity. Therefore, he will be either at Mumbai or GujaratUniversity. Similarly, as he teaches neither geology norhistory, therefore, he must be teaching physics or botany.

... (1)From IV,Dr Zia → Physics but as he is not teaching in either Mumbaior Osmania University, he must be teaching either in Delhi orGujarat University ... (2)

From V, we get Dr Joshi teaches history in Delhi University.... (3)

From (1) and (2), we conclude that Dr Natarajan teachesbotany.

And from (1), (2) and VI, we get both Natarajan and Ziateach in Gujarat University. Finally, on summarisation wecan prepare the following table.

Names University Subject

Dr Joshi Delhi History

Dr Davar Osmania Geology

Dr Natrajan Gujarat Botany

Dr Choudhary Mumbai Zoology

Dr Zia Gujarat Physics

On the basis of the above table, rest of the questionscan be solved very easily.

16. c 17. a 18. d 19. b 20. d

Practice exercise – B3

For questions 1 and 2:Check out the numbers that have four factors and try tounderstand, e.g. 6 – 1, 2, 3, 6.Note that in case of 1, 2, 4, 8, x = 2 (not 4), which is prime.Note that x and y will always be prime.Therefore, x . y = NTherefore, x . x . y = x . N

1. a

2. d


8 8

9

9

r

r

r

r

r – 8

r – 9

9

r

r

8 – r

9 – r

8

(r – 8)2 + (r – 9)2 = r2

∴ r = 29, 5

3. b 4. c 5. b


For questions 6 and 7:1000009 = 293 × 3413You could get the factors using options.

6. e

7. a

For questions 8 to 11:The prime numbers less than 45 are as follows:2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43.So there are 14 prime numbers. Now, since the average ofthe goals scored is also a prime number, for question (10),we can readily eliminate choice (a). Now the sum of all theprime numbers listed above is 281. So the sum of total goalsscored must be less than 281. Note that only choice (b), i.e.23 gives a total number of goals of (23 × 11) = 253 which isless than 281. Hence, average is 23.Thus, 23 is not the number of goals scored by any player.Of the remaining 13 numbers, whose sum is (281 – 23)= 258, we have to delete two numbers, such that the sum ofthe other 11 numbers is 253. That means the sum of thosetwo numbers, (which are to be deleted) is 5. Obviously,those two numbers must be 2 and 3.

8. b 9. a 10. a 11. b

12. e 222221 = 619 * 359. However, we do not knowwhich is the number of compartments, and whichis the number of soldiers in each compartment.Therefore, the answer cannot be determined dueto insufficient data.

13. b Now, since we know that the number ofcompartments is less than the number of soldiers ineach compartment, there have to be 359compartments.

For questions 14 and 15:Obviously, the maximum number of such points possiblewould be 3, where the points are the vertices of anequilateral triangle.If the points do not have to lie in a plane, then we can havefour such points, where they form the vertices of a trapezoid.

14. d

15. b

For questions 16 to 18:Let the length be x, and the breadth y.Since area = Perimeter, xy = 2(x + y).The only values of x and y which satisfy this equation are(x, y) = (4, 4) and (x, y) = (3, 6).Therefore, there are two such rectangles, and only onesuch square.Therefore, if the rectangle is not a square, its dimensionswill be 6 by 3.

16. b 17. a 18. b

Practice exercise – B4

For questions 1 to 4: The given information can betabulated as follows.

Male Female Male Female Male Female

Professor A E H, I L O, P S

Lecturer B, C, D F, G J, K M, N Q, R T

Department

Civil Mechanical Electrical

Further, following constraints are added.

(i) Male lecturer in Civil Department

× Female lecturer in Electrical Department

∴ (B, C, D) × T… (Not allowed)

(ii) Female professor × Male lecturer

∴ (E, L, S) × B, C, D, J, K, Q, R (Not allowed)

As given that BCD cannot be with T and ELS cannotbe with BCDJKQR, then using these conditions wecan solve rest of the questions very easily.

(1) Choice (a) is not valid as no lecturer fromelectrical department is there. Choice (b) is notvalid as no lecturer from civil department isthere. Choice (c) is not valid as no lecturer frommechanical department is there. Therefore,choice (d) is the only valid choice as it justifiesall the given conditions.

(2) According to the question. HIL cannot go withBCDFG. ... (1)The delegation must have minimum threeprofessors. ... (2)Representation from every department mustalso be there. ... (3)

Checking from the above conditions, we get(a) is not valid as it violates (1).(b) is not valid as A, H, D are males.(c) is again not valid as it violates (1).(d) obviously, it is the valid choice because it fulfills

all the conditions.

(3) According to the question, only civil andelectrical departments must be used. ..(1)Committee must have two professors and twolecturers. ...(2)


For questions 17 and 18:Doctor cannot be a woman. (As the two women,Sujith’s mother and his wife are not blood relatives.) Soncannot be the doctor because he is the youngest of themand all people are his blood relatives.⇒ Sujith is the doctor. Lawyer can be anyone exceptSujith and his mother (Sujith’s mother also because ofcondition I).

17. a 18. c

For questions 19 and 20:The joker can be with Ajay or Sujith, since Sanjay cannothave it.Three queens can only be with Sanjay, as 3 queens plus aminimum of 1 king add up to a minimum of 4 cards.

19. d 20. b


For questions 1 and 2:Here, the best method to solve these questions is going bychoices. In (1) choice (b) is only valid choice as it is between70 and 79 and is not a multiple of 6, whereas other choicesare not valid as they do not follow the given criteria.

In (2) obviously, none of the information is superfluous.Therefore, choice (e) is the answer.

1. b 2. e

For questions 3 to 6:According to the given conditions in the question.

From III and IV, we get that Puneet arrived at the mansionafter midnight and because the detective arrived at themansion at midnight, therefore, Puneet is not the detective ... (i)Now the detective can be either Aditya or Vijay.

From II, we get that the detective came at the second number(then only II is justified) .From III and V, we can conclude that Vijay was the earlierarriver between Puneet and Vijay, therefore, he is not thedetective. Similarly, from (i) Puneet is not the detective.Therefore, logically, it is Aditya who is the detective andarrived at the mansion at the second number. ... (ii)From IV, I and VI, we conclude that Vijay came first. Aditya(the detective) came second and Puneet came third. Nowfrom VI Puneet was not the murderer and from I and IV itwas Aditya, the detective, who was the murderer . ... (iii)Using (i), (ii) and (ii) rest of the questions can be solvedvery easily.

3. c 4. b 5. c 6. a

For questions 7 to 9: Simple! You are being asked aboutthe product of which two consecutive integers is equal toproduct of three consecutive integers. You have two suchsets.(i) 1 × 2 × 3 = 2 × 3 (This is not possible in this case.)(ii) 5 × 6 × 7 = 14 × 15

7. e 8. c 9. e

10. d2r

2r

2r2r

In fact you will get a hexagon joining the centres of all theexternal circles.

For questions 11 to 13:We haveP + Q + R + S = 45 …(i)

and (P + 2) + (Q – 2) + 2R + 12

S = 45 …(ii)

2(P + Q) = R + S ... (iii)

Solving the equations, we have their ages as follows.

P Q R S

? ? 01 02

11. b 12. c 13. a

For questions 14 to 18: If a room has an odd number ofvisitors, it will be closed.Any room number that is twice a perfect square will havean odd number of visitors. The room with the largest suchnumber (twice a perfect square) will be the last room tohave an odd number of visitors.Note that the 38th room with an open door will be the 38throom whose number is not twice a perfect square, whichhappens to be 88.Note that if only occupants of 2 to 1000 do the task, thenonly 1 person, i.e. the occupant of number 1000 will go toroom number 2000, (since 2000 is a multiple of 1000), andthis room will therefore be closed. (Since it is initially open.)Number of room closed upto 1000 = 22Since there are 22 twice a perfect square numbers.Number of remaining room = 500Number of twice a perfect number between 1000 and 2000= 9.All rooms with twice a perfect square between 1000 and2000 will be open.


Page: 21��

∴ Number of room closed = 22 + 500 – 9 = 513As regards question 18, anyone who lives in a room with anumber greater than 1000 will obviously have to visit onlythat particular room, as it will not have any multiple which isnot greater than 2000.

14. c 15. b 16. a 17. a 18. b

19. e The only such number below 200 is 153.

Therefore, Q is 5.

20. e You have two such numbers: 370 or 371.Therefore, R may be 0 or 1.

Practice Exercise – B6For questions 1 to 5:According to the given information in the question,we getO is a bank manager and is a male (as he is married to a ladyprofessor) ... (X)Q is a medical representative and is a male (as he is the sonof M ... (Y)N is a chartered accountant and is a female (as she is thedaughter-in-law of L) ... (Z)As given in VI that businessman is married to the charteredaccountant, this would mean that the businessman cannotbe L (as N, chartered accountant, is the daughter-in-law ofL), O, Q and P (as they are having different professions,other than business). Therefore, logically M is thebusinessman with whom N is married. So M is businessman(male). ... (A)Given that P is an unmarried engineer.

Here as we know the profession of all the members exceptL and only profession unclaimed is professorship, therefore,L is a professor and because L is the grandmother of Q,therefore, L is female ... (B)From III, II and (B), we can very clearly conclude that O, themale bank manager is married to lady professor L.Using (A) and (V), we get that M is the son of L and O.From (IV), we get that Q and P are kids of M ... (C)So, all the conclusions using (X), (Y), (Z), (A), (B) and (C)can be depicted with the help of this family tree with thehelp of which rest of the question can be solved very easily.

(Ma le / Bank M anager) (Fema le/Pro fessor)

M

O

N(Ma le /Bus inessman) Son (D aughte r-in-law /C hartered accountan t)

P Q

(U nm arried Engineer) (Son /Medical R epresen tative)

L

1. d 2. c 3. a 4. e 5. b

For question 6 to 10:The best method to solve this question is to make a table andfill the places according to the information given logically.On analysing the information given in the question, wearrive at the following table.

tnedutS tcejbuSyroslupmoC tcejbuSlanoitpO

M )nevig(yhpargoeG )nevig(hsilgnE

N )nevig(yhpargoeG )nevig(ygoloiB

O )nevig(yhpargoeG )nevig(scisyhP

P )nevig(hsilgnEyhpargoeG

elameF)nevig()nevig(tneduts

Q )nevig(yrtsimehC )nevig(scisyhP

R )nevig(scisyhP )nevig(yrtsimehC

6. d 7. a 8. e 9. a 10. d

For questions 11 to 15:The given information can be tabulated as follows.

seciohCstnedutS

A B C D E F

1 R M R M M R

2 B P B C P B

3 C B T T B P

R ⇒ Reading B ⇒ BadmintonC ⇒ Cricket M ⇒ MusicP ⇒ Painting T ⇒ Tennis

Note: The best way to solve this problem isto f i l l the informat ion given logical ly andcomplete the table. On the basis of the giveninformation the third choice for both C and Dwould be tennis.

11. c 12. d 13. c 14. c 15. b


For questions 16 to 20: The given information canbe tabulated as follows:

emaN tcejbuS ytiC

AyhposolihP

)nevig(dabaredyH)i(morf...

BscitamehtaM

)nevig(erolagnaB)ii(morf...

C

sa(scimonocEtonseodD

hcaetscimonoceC,erofereht).tisehcaet

)nevig(ihleD

Dylnosa(yrotsiH

)tfelsiyrotsih)nevig(rupiaJ

EyhpargoeG

)nevig(wonkcuL

)vi(morf...

16. b 17. d 18. c 19. d 20. b


For questions 1 and 2:Five such pairs are possible. (10x + y) – (10y + x) = 36⇒ x – y = 4∴ Numbers are 15 and 51, 26 and 62, 37 and 73, 48 and 84,and 59 and 95.

1. e 2. b

For questions 3 and 4:x = 10a + b, x2 = 100a2 + 20ab + b2

y = 10 b + a, y2 = 100b2 + 20ab + a2

⇒ In the square, middle number (ten’s place) will alwaysbe the same.122 = 144, 212 = 441132 = 169, 312 = 961

3. b 4. b

For questions 5 and 6:

Black cat White cat Initially: x y When the white cat gives z ladoos to black cat

x + z y – z ⇒ (y – z) = 3 (x + z)

Black cat gives z ladoos to white cat

x – z y + z ⇒ (y + z) = 5 (x – z)

Solving the two equations, we getx = 5z and y = 19z.Therefore, the ratio y : x = 19 : 5.19 + 5 = 24 which means that there must be at least 24ladoos as the cats have an integral number of ladoos. Ifthere are 30 ladoos, the monkey must have cheated thecats of 6 ladoos (i.e. 30 – 6).

5. b 6. c

For questions 7 and 8: From the given data, we can makethe following table with the help of which rest of thequestions can be solved very easily.

)04(elaM )03(elameF

52evobA

deirraM 7 21

deirramnU 5 0

52woleB

deirraM 8 3

deirramnU 02 51

latoT 04 03

7. b 8. d

For questions 9 and 10:For every plate of ice cream, there are 2 employees.For every plate of sambar, there are 3 employees.For every plate of chicken, there are 4 employees.∴ Number of employees must be a multiple of theLCM of 2, 3 and 4, i.e. 12.Only choice (d) of question 9 satisfies this condition.Another way of looking at this is as follows.Let the number of employees be x.

Therefore, the number of plates of ice cream = 2x

.

Therefore, the number of plates of sambar = 3x .

Therefore, the number of plates of chicken = 4x

.

Therefore, 6512

x134x

3x

2x ==++

Therefore, x = 60 (employees).

Number of ice cream dishes = 2

60 = 30.

9. d 10. a


Page: 23��

For questions 11 and 12:21 can be factorized as 3 × 7. Thus, to have 21 handshakes the combinations for number of males in the twofamilies can be as follows.

P R

1esaC 7 3

2esaC 3 7

3esaC 1 12

4esaC 12 1

Let there be x females in P family and y females in R family.Then for the first case, number of kisses will be7y + 3x + xy = 34.Note that y = 2 and x = 4 satisfy the equation. So the totalnumber of females will be 6. This will be true for the secondcase also. For case 3, total number of kisses will be equal to1y + 21x + xy = 34.

Note: That x = 0, y= 34 satisfy the equation. Therefore, thenumber of females will be 34. This will be true for case 4also. Thus, the number of females can be 34.

11. e 12. e

For questions 13 and 14:1 × 2 × 3 = 2 × 35 × 6 × 7 = 14 × 15The choices in question 14 give the answer.

13. b 14. c

For questions 15 and 16ABCABC is divisible by 2 since it is even. Also, according tothe divisibility rule of 7, ABC – ABC= 0 → The number is divisible by 7.

⇒ It is divisible by 14.For 16, data is insufficient.

15. a 16. a

For questions 17 and 18:1001 = 7 × 11 × 13

I am obviously more than 21

95 -year-old. Therefore, my

only possible age is 11 x 13 = 143 years. One of my great-grandsons is 7-year-old and one of them is 1-year-old.

17. c 18. c

For questions 19 and 20:Suppose the holiday lasted for x days.∴ There were 11 nice morning.∴ On x – 11 days it rained in the morning.Similarly, on x –12 days it rained in the afternoon.It never rained in the morning as well as afternoon.∴ We have x – 11 + x – 12 = 13 or x = 18.Therefore, the holiday lasted for 18 days.It obviously never rained in the morning as well as in theafternoon.

19. a 20. e


1. a Shikha is to the left of Reena and Manju is to herright. Rita is between Reena and Manju. So theorder is: Shikha, Reena, Rita, Manju. In thephotograph, Rita will be second from left.

2. a B is to the right of D. A is to the right of B. E is to theright of A and left of C. So the order is: D, B, A, E, C.Clearly, A is in the middle.

3. b Q is to the left of R and to the right of P, i.e. P, Q, R.O is to the right of N and to the left of P, i.e. N, O, P.S is to the right of R and to the left of T, i.e. R, S, T.So the order is: N, O, P, Q, R, S, T.Clearly, Q is in the middle.

4. a S is sitting next to P. So the order S, P or P, S isfollowed. K is sitting next to R. So the order R, K isfollowed because R is on the extreme left. T issitting not next to K or S (as given). So thearrangement will be R, K, S, P, T. Clearly, P and K aresitting adjacent to S.

5. a Clearly, the order is Anuradha, Rashi, Monika,Sulekha, Abha because Rashi is not adjacent toSulekha or Abha and Anuradha is not adjacent toSulekha. Therefore, Anuradha is adjacent to Rashi.

6. c Putting the conditions given in the question, theposition of swimmers can be:

1 2 3 4 5

A D/C B C/D E

7. e1 2 3 4 5

A D C B E

Positions of B and E violate rule 1.


8. a There is only one possible arrangement, whichsatisfies the rule, i.e.

1 2 3 4 5

B D A C E

9. d D is to the left of C, i.e. C, D. B is to the right of E, i.e.B, E. A is to the right of C, i.e. A, C. B is to the left ofD, i.e. D, B. From the above statements, the correctorder is: A, C, D, B, E. Clearly, D is sitting in thecentre.

Note: It is given that A, B, C, D and E are sittingfacing you. So your right and left will be consideredas left and right respectively.

For questions 10 to 14: According to the question

From III, Indian is wearing a green cap and a Jacket ... (1)From VI, Kurta is worn alongwith red cap and sits next toJapanese ... (2)From VIII, T-shirt with white cap combination is seated atone end.So from (1), (2), (3), VII and I we conclude that the Japanesewear a shirt of yellow colour.From IV, V, VI and VII, we conclude that the placement ofpeople will be like(1) (2) (3) (4)German American Japanese Indian

From (2) and IV, we arrive at the following table with thehelp of which rest of the questions can be solved veryeasily.

ytilanoitaN namreG naciremA esenapaJ naidnI

sehtolC trihs-T atruK trihS tekcaJ

spaC pacetihW pacdeRwolleY

pacneerG

pac

10. d 11. c 12. c 13. c 14. c

15. a From I, we get France, America, India ... (i)From (II), we get India, Australia, Japan, China ... (ii)Combining (i) and (ii), we get the correct sequenceas: France, America, India, Australia, Japan, China.The two flags in the centre are of India and Australia.

For questions 16 to 20: According to the given question,

From IFargoCad illac ... (i)

From II

Fia t

Fa rgo

Cad illac

... (ii)

From III, we getFiat

Am bassador

Fargo

Bedford

M aru ti

Cadillac

M ercedes [logica lly i ]t has to be here on ly

From IV

Fiat

Am bassador

Fargo

Hence, the sequence of cars is as follows:Fiat, Bedford, Maruti, Ambassador, Fargo, Cadillac,Mercedes.

16. d Clearly, Maruti is in the third place and Mercedes inthe seventh, i.e. Mercedes is fourth to the right ofMaruti.

17. b Clearly, Cadillac is in the sixth place, to the immediateleft of Mercedes, which is in the seventh place(from the top).

18. d On the sides of the Cadillac are the Fargo and theMercedes.

19. a Clearly, Maruti is in third place (from top), and is tothe immediate left of the Ambassador, which is inthe fourth place.

20. c To the right of Ambassador are Fargo, Cadillac andMercedes.


Page: 25��


For questions 1 to 5: On the basis of the given informationfollowing graph can be constructed and from the graph, wecan easily find the answers of the rest of the questions.

I

1 km

1 km

0.5 km

0.5 km

M

J

K

L H

G

D B0.5 km

E C A

1 km

1 km1 km

1. d

2. a

3. d

4. c

5. d

6. a According to the given question, we have

Total audience Male Female

At 12 noon (Waiting) 180 108 72After announcement

(Waiting) 108 72 36

Left audience 72 36 36

At 12.50 p.m. (108 + 18)

126 90 36

Required percentage = 90 250

100 % 83.3%108 3

× = =

7. c Let A, M, P denote the sets of people liking apples,mangoes and pineapples respectively.n(M) = 20% of 10,000 = 2000

( )n(A M) 5% of 10,000 500, n M P∩ = = ∩= 3%of 10,000 = 300.

n(A M P) 2% of 10,000 200∩ ∩ = =We have to find number of people belonging to theshaded region in the following Venn diagram.

M P

ANumber of people who likes only mangoes

= ( )n A M P′ ′∩ ∩

( )( )n M A P ′= ∩ ∪

= ( ) ( )( )n M – n M A P∩ ∪

= ( ) ( ) ( )n M – n M A M P ∩ ∪ ∩

= ( ) ( ) ( ) ( ) ( )( )n M – n M A n M P – n M A M P ∩ + ∩ ∩ ∩ ∩

= ( ) ( ) ( ) ( )n M – n M A n M P – n M A P ∩ + ∩ ∩ ∩ = 200 – [500 + 300 – 200] = 1,400

8. d Using the data given, the seven friends are sitting inthe following order.E A G B D F C

Therefore, E was sitting fourth to the left of D.

9. c Given information can be rewritten as(Nidhi – Pavbhaji) __ × ... (i)(Dhruv – Kriti) __ �� ... (ii)

(Parul – Movie) __ × ... (iii)(If beach – Paanipuri) __�� ... (iv)(Nidhi – Beach) __ × ... (v)(Circus – Pavbhaji) __ × ... (vi)(Harsh – Parul) __ × ... (vii)Clearly (ii) and (vii) implies Dhruv – Kriti, Harsh –Nidhi, and Amit – Parul.Also (iv) and (v) implies that Nidhi did not eat paanipuri.Also using (i), we conclude that Nidhi and Harshate chaat. Now clearly only the followingcombinations of place to visit and food items arepossible.


They are Movie – Pavbhaji or Chaat (using iv)Beach – PaanipuriCircus – Chaat, PaanipuriApplying elimination method and using (iv) we getbeach – paanipuri, and using (vi) we get circus –chaat as the true combination

and finally Movie – pavbhajiThus we haveHarsh – Nidhi – Circus – Chaat as one correctcombination.Then the other combinations are:Using (iii), we get Amit – Parul – Beach – Paanipuriand obviously Dhruv – Kriti – Movie – Pavbhaji.

10. c We get the following table directly from theinformation given.

2 3 4 5 6 7 8 9

–×

yadnuSscisyhP – – – –

lacisyhPnoitacude

Now, using (VIII) there is a gap of two days betweencomputers and biology. This means that computersexamination is either on 2nd or 5th and biology iseither on 5th or 8th. If we take the second case, i.e.computer 5th and biology 8th, then using (VII)mathematics will be on 6th but this does not satisfythe (V) condition given.Now we consider the other case, i.e. computer 2ndand biology 5th. Using (vii), we have mathematics7th and using (v) chemistry 6th. The remainingengineering drawing is on 8th, the only day left. So,in tabular form we have

2 Computer Saturday

3 Holiday Sunday

4 Physics Monday

5 Biology Tuesday

6 Chemistry Wednesday

7 Mathematics Thursday

8 Engineering draw ing Friday

9 Physical education Saturday

11. b Y is intelligent in mathematics and physics. Also Yisintelligent in chemistry and physics.

12. a (d) is not possible as rubber and watches cannotbe exported together. (b) and (c) are not possibleas timber and garlic both are exported. So (a) is theanswer.

13. c On the basis of the analysis of the information given,the combinations of the movies and their type andthe number of prizes they bagged is as follows.

D Love story 3 (As it has won maximum awards)

C Horror 2

A Action 1

B Kids None

So (c) is the answer.

14. d The combinations are as follows.

A G Beach

B F Rose

C E Movie

D H Chocolate

15. c The combinations are as follows.

A G Beach

B F Rose

C E Movie

D H Chocolate

16. b If we take first statement of I as false, then discountis with saree and wristwatch is free with it. Theonly condition where the logic of one statement isfalse and other true is justified, is possible whensecond statement of II is false and first statement ofIII is false. On this basis, we conclude that cap isfree with bedsheet and T-shirt is free with shirt.Hence, choice (b) is the correct answer.

17. d The given information can be depicted with the helpof following diagram.

880430

190

460 345

690

675

Sw im ming

B illiards

Tennis

460 + (430 – 190) + (345 – 190)= 460 + 240 + 155 = 855


Page: 27��

18. a Members playing only tennis= 880 – (430 + (460 – 190)) = 880 – (430 + 270)= 880 – 700 = 180Members playing only billiards= 675 – (460 + (345 – 190)) = 675 – (460 + 155)= 675 – 615 = 60So 180 + 60 = 240

19. a Option (a) has D and G both which is not allowed.All others are allowed. Therefore, choice (a) is thecorrect answer.

20. a Both B and C cannot be selected. So the other doorpaint has to be A. So A is always selected.


1. d All of D’s children are in Z. So (d) is the answer.

2. c From the question, we get that_ Sapna _ _ _ _ _ _ _1 2 3 4 5 6 7 8 9 ... (A)Given that(i) Megha, Sapna and Riya cannot sit at 1 or 9.(ii) Beena and Megha does not have anybody sittingadjacent to them.(iii) There is only one empty chair between Meghaand Riya.(iv) Charu is adjacent to both Jiya and Riya.From (iv) it is very clear that Jiya - Charu - Riya orRiya - Charu - Jiya will be the sitting arrangement

... (v)From (v) and (iii), we getMegha — Riya - Charu - Jiya ... (vi)From (i) and (ii), we conclude that Beena and Meghamust have nobody adjacent to them. It means theymust have at least one place empty adjacement tothem. Now based on all the conditions (i), (ii), (iii),(iv), (v) and (vi), we get the only possible and validarrangement as

1 2 3 4 5 6 7 8 9 X Sapna Jiya Charu Riya X Megha X Beena

... (vii)From (vi), we get Megha will sit at chair number 7.

3. e One of the possible sitting arrangements based ongiven information is depicted by the diagram givenbelow.

Shehul / M ilind

Shehul / M ilind

Anil Rajesh

Kiran

Vinay

Since the given information is not sufficient (becausenothing is mentioned about the positioning of Vinay)to definitely depict the exact seating arrangement,‘none of these’ is the correct choice.

4. d On the basis of the analysis of the given information,we arrive at the following diagram with the help ofwhich question can be solved easily.

Pallav i

Kom olika

Anu

Mand ira

R am ola

N atasha

5. b Applying the rules given in the question, only twocombinations are possible.

A Cricket or A Cricket

B Football B Cricket

C Football C Cricket

So A always play cricket.

6. b (a) is not feasible as it does not satisfy condition (i).(c) is not feasible as it does not satisfy condition(iii). Again option (d) does not satisfy condition (i).The only feasible group is given by option (b).

7. c Clearly, using (II) and (III) Poonam and Poornima donot stay on 1st floor and Priya and Priyanka do notstay on top floor.Using (I), Priya does not stay on 1st floor.So the only option for the girl staying on 1st floor isPriyanka. Thus, Poornima stays on 2nd floor, soPriya stays on 3rd floor and Poonam on the 4thfloor.

8. a Using (II) and (VII), their full names are SunilSachdeva, Rohit Sehwag and Sandeep Sharma.Using (III) Mr. Sachdeva – PurseNow, using (V) Mr. Sharma – Cosmetics andMr. Sehwag – Saree.Thus, (IV) implies that Himani is Mrs. Sehwag.Using (I) Dhwani is Mrs. Sachdeva and thus Vidhi isMrs. Sharma.Thus can be tabulated as follows.

Name Surname Gift Wife’s name

Sunil Sachdeva Purse Dhwani

Rohit Sehwag Saree Himani

Sandeep Sharma Cosmetics Vidhi


9. e Let x be the number of employees at branch A.Then, 320 – x is the number of employees at branchB.

Total number of absentees = 10

%100

of 320 = 32

At branch A, number of absentees = 20% of x

At branch B, number of absentees = 9

7 %13

of

(320 – x)

Now, 20 100

x (320 – x) 32100 13 100

+ =×

x 1(320 – x) 32

5 13⇒ + =

x 60⇒ =

⇒ 320 – x = 260But the number of females present on the day is30% of the total employees present. But it is notgiven that how many of those who were absentare females. So, we cannot find the total number ofmales at branch B.

10. d Let us assume that they carried N chapattis.Let N = 3x + 1 ... (i)He eats his share.Number of chapattis left = 2xLet 2x = 3y + 1 ... (ii)The second friend eats his share.Number of chapattis left = 2yLet 2y = 3z + 1 ... (iii)The third friend eats his share.Number of chapattis left = 2z.Let 2z = 3w + 1 ... (iv)All three of them get up and eat their shares.

Using (i), (ii), (iii) and (iv) we get 81 65

N W8 8

= +

For even W – N will not be an integral number for

W = 1, 73N

4=

W = 3, 77N

2=

W = 5, 235

N4

=

W = 7, N = 79∴ Minimum number of chapattis = 79

11. c Gunjan and Piya have to be together.∴ Sameer would not be there according to (ii) andhence Priyanka would not be there according to(v). So the four girls are Shikha, Aastha, Gunjanand Piya. Hence, Anand and Vineet would not bethere according to (iv) and (iii). Hence one of eitherBiswas and John will be there according to (vii).Since Biswas and Aastha have to be together, themember of the team are the ones given by choice(c).

12. c Clearly, using (III) the only salaries that satisfy thisstatement is Rs. 6,000.

36000 10,000

5= ×�

∴ Kishan’s salary is Rs. 6,000 ... (1)Using (IV) and (V) Ganesh’s salary is not Rs. 10,000

... (2)Also Ganesh – EXCELConvergys – Rs. 10, 000 ... (3)Then (VI) and (I) implies that Shiv’s salary is eitherRs. 12,000 or Rs. 15,000 and Shyam’s salary iseither Rs. 6,000or Rs. 7,500 ... (4)Using (VII) person working in Daksha getsRs. 15,000 ... (5)Using (1) and (4), Shyam’s salary is Rs. 7,500∴ Shiv’s salary is Rs. 15,000 and hence he worksin Daksha [Using (5)].Thus, Ravi’s salary is Rs. 10,000 and he works forConvergys (Using 3)Using (III), Kishan works in Global Vantedge andShyam in GE Capital.

emaN emans'ertneCllaC ).sRni(yralaS

maR sygrevnoC 00001

mayhS latipaCEG 0057

nahsiK egdetnaVlabolG 0006

vihS ahskaD 00051

hsenaG LECXE 00021

13. c The information given can be written in acompressed from as follows.(i) Aman, Algebra, 2(ii) Ankur, __ , 4, Scientist(iii) __, Botany, 1, Teacher(iv) Ankit, __, __, Businessman(v) Amrita, Geometry, __, Architect(vi) Mathematician and Physics expert stay on samefloor 2.


Page: 29��

Using (i), (ii) and (iii). Aman is not a scientist orteacher and Ankur is neither good at algebra nor atbotany.Again (i), (iv) and (v) imply that Aman is not abusinessman. So, by elimination method, Aman isan architect. Similarly, Ankur is either good at zoologyor physics. But using (vi), Ankur is a zoology expert.One of the mathematician stays at 2nd floor andthat is Aman. So, Amrita is not the one staying on2nd floor. Hence, by elimination she stays at 3rdfloor. Thus the other person staying on 2nd floor isAnkit and he is the physics expert.So using (iv), Ankit, Physics, 2, Businessman is thecorrect combination.

DS Practice Exercises

Practice exercise – C1

1. d Statement I is very tempting as 4 = 4 × 1, but boththese numbers can be negative also. Hence, I is notenough. Statement II says that both A and B arepositive. So both statements taken together solvethe problem.

2. a Since all the 3 are odd and z – x = 4, they have to beconsecutive.

3. c By statement I, a + 1 = a + a + 2∴ a = –1By statement II,a(a + 2) = –1∴ a = –1∴ Answer is (c), i.e. it can be found out using eitherstatements alone.

4. d From statement I, we get

a – 2 a + 2a

–1 1 3

–3 –1 1

–5 –3 –1

All 3 cases we ge t ‘–1 ’

So we do not know what the average will be fromstatement I only.Using only II we can say that the numbers are–1, 1, 3 or 1, 3, 5.Using both the only possibility is –1, 1, 3 and hencewe can find the average.∴ Answer is (d).

5. e Neither statements I and II alone nor taken togethercan give remainder as remainder will vary accordingto the numbers.∴ Answer is (e).

6. e Statement I does not have data to find the cost ofeach album. Statement II gives the total amount spentbut we cannot find out the percentage of sales taxper copy.

7. d Combine both the statements to find the answer.

8. e Since nothing about expenditure and the rate of taxis mentioned, the salary cannot be calculated.

9. d Let the total number of students appear in theexamination be x.Statement I will give the information about passedstudents.Statement II is also not complete in itself.Combining the statements I and II, we get

x10051

10x10050 =+

Hence, x can be calculated.

10. b Statement II clearly gives that a can of beer is costlier.Thus, (b).

11. b CP =23

SP ⇒ SP = 1.5 CP

Profit percentage is 50.

12. d Only statement I or II alone is not sufficient.Combining statements I and II, we can calculate thelist price as (Rs. 1485 × 12) = Rs. 17820.Hence, discount is Rs. 2,320.

13. a From statement I, cost price = Rs. 1512

and

selling price = Rs. 64

Hence, profit can be worked out. So statement I issufficient. From statement II, we have no idea aboutthe selling price and hence the profit. So statementII alone is insufficient.

14. e Knowing the highest and the lowest scores tells usnothing about the other scores. So statement I isnot enough. Statement II is very tempting but studentsmust realize that suppose 2 people take the GRE,one scores 2300 and the other scores 1000. Surelythe average is not 2000. So statement II cannotgive us the answer. The two statements takentogether also cannot answer our question.


15. c Statement I indicates that the number of 50-paisacoins is 2 and the number of one-rupee coins is 3.Statement II independently gives the same result.

16. d A relation between a and c can be found using both

statements I and II, i.e. 1ca

cca +=+

17. a From statement I, it can be found that the total of thedeposits = 75 × 5 = $375

18. c d = Kt2, u = 0 from statement I, 1205 = K.

⇒ 25 K = 25

1205

We need to determine D10

– D9

D10 = 10025

5.120 × , D9 = 8125

5.120 ×

D10 – D9 = )19(25

5.120)81–100(

255.120 =

Note:You do not have to calculate the actual answer.From statement II, 490.4 = Kt

2

K . (102) ⇒ K =

1004.490

Proceeding as above, D10 – D9 can be determined.

19. e There are more unknowns than the number ofequations. Hence, both statements I and II are notsufficient.

20. d Both statements are required.


1. a Using only statement I, x32

182x =+ , where x is

the initial amount of money he took to the mall.

2. a a b a a1 : 1

a b b b+ = + − −

3. e Both the statements give the same information.

4. e 10 students in statement II may not be the remainingbeyond the 30 given in statement I.

5. e Using both statements I and II, we still cannot sayanything about the number of students in the class.

6. e Statement II repeats statement I. Hence, the questioncannot be solved.

7. e Let the amount of money Prem has be P, and theamount that Jagdish has be Q.Statement I gives the inequality P ≥ J + 100.Statement II gives the inequality P + J ≤ 500.It is obvious that we cannot get the answer fromstatement I alone or statement II alone or even fromstatements I and II together.

8. e Both the statements do not give any informationabout the speed limit. Thus, (e).

9. e It is not specified which tap is opened and whichone is closed, and what part of the tank was initiallyfull.

10. d From statement I, we cannot find the hours.

From statement II, we get x y

15 3

+ = which enables

us to get the answer. By using both the statementstogether we get the answer.

11. b Statement I gives only the relative efficiencies andno idea of time taken is there.So statement I is insufficient.

Statement II gives time required by B as 1

102

×

= 5 days. Hence, statement II alone is sufficient.

12. e The statements given do not give any informationabout the number of rooms to be vacuumed.

13. b From statement I, we can only get an equation interms of x and y, but not the value of x and y.From statement II, we can get the value of x as 0.

14. a q2pq

pqpq2 2+=+

15. b Statement (I) gives us an inequality which is notenough to answer the question. Statement IIindicates that g is greater than h becauseirrespective of the sign of the integer, the integerwhose cube is greater will obviously be the greaterone. Therefore, statement II alone is enough.

16. d From the given condition, a6 = b6

But since the power is even we do not knowwhether a = b. But from I and II we can concludethat both ‘a’ and ‘b’ are +ve. Thus, a = b∴ a3 – b3 = 0∴ Answer is (d).


Page: 31��

17. d From I, we have r = 0 or r = 3. So it is not sufficient.From II, we have r = (–1) or r = 3.So it is also not enough. The two statements takentogether give r = 3.

18. d Statement I says that there could be 25 or morebooks. Statement II says that there could be 25 orless books. The two statements taken together giveus the answer as 25 books.

19. a x2 + 2xy + y2 = (x + y)(x + y)

20. e From I and II, we get 2 equation in two variable. Butthey are dependent. Thus, eventually we get oneequation in two variables. Which has infinite solutionset. Thus, (e).


1. b Statement I can be factorized as [x – 3][x – 1] = 0giving two possibilities, statement II is also a quadraticequation but when factorised it yields (x – 1)2 = 0

2. b Statement I does not give any information, since if‘x’ is –ve and ‘y’ is +ve, then also x2y is +ve and if xis +ve and y is +ve, then also x2y is +ve.Statement II tells us either both x and y are +ve or

both –ve. In either case xy is +ve.

3. d From statement I, we have x² – 7x ≥ 8, meansx² – 7x – 8 ≥ 0. So (x – 8)(x + 1) ≥ 0. So x ≤ –1.And x ≥ 8. And combining with statement II, we getx = 8.

4. e Both the conditions are not sufficient to say ifxy > 1.Since if x = 0.25 and y = 10∴ xy = 2.5 > 1 and if x = 0.025y = 1.2xy < 1∴ Answer is (e).

5. b From statement I, we have 1x

< 1

Consider x = 4

∴ 14

= 0.25 < 1; for x > 1

Now consider x = –4

10.25 1

4∴ = − <

−; for x < 1

So even if ‘x’ is > 1 or x < 1, statement I is satisfied.Thus, we cannot say from that if x > 1 or x < 1

From statement II, we get x2 < 1x2 – 1 < 0(x + 1)(x – 1) < 0

–1 +1Thus, we can say that x lies between +1 and –1.Thus, x is not greater than 1. Thus, (b).

6. e 2x > 6 implies x > 3, while 3x < 10.9 impliesx < 3.6333.Hence, the answer cannot be conclusivelydetermined.

7. d From statement II, we get x as positive.Hence, using this in the first statement we can inferthat x > y.

8. d Since the value of x2 – 7x + 12 either increases ordecreases depending upon the domain of x, bothstatements I and II are required to answer thequestion.

9. a Statement I implies that x can be between 1 and 2 or–1 and –2.Statement II implies that x is between 1 and 2.

10. a Statement I is true only if a = b = c = 0

11. d From statement I, we can say1

4 10.25

= >

12

= 0.5 < 1

Thus, statement II is needed where x < 1∴ Answer is (d).

12. e From statement I, nm

= n + p cannot be said if n > p.

From statement II, n > m, i.e. n = km, where k > 1

Again it cannot be said if n > p. Combining,nm

= pn

∴ x1

= n + p, i.e. n + p < 1

It cannot be determined if n > p.

13. a From statement I,a = 40°∴ a + b + c = 180°∴ b + c = 140°And x + b + c + y = 360°∴ x + y = 360 – (b + c) = 360 – 140 = 220∴ The answer is (a).


��14. b From statement I alone we do not get the answer.

From statement II alone we get the height as 10 ft.

From the figure below.

5

5 ft 5 ft

Thus, statement I is of no importance.Thus, (b).

15. c From statement I, we get as below

12

0°

4 3√ Thus, we can find each side and hence the area.From statement II, we get120 °8Thus, area can be found out independently.Thus, (c).

16. d Statement I is not enough to answer the question asthe line could be the Y-axis or parallel to theY-axis. Statement II is not enough as there could beinfinite lines passing through (6, 0).However, statements I and II together indicate thatline I does not pass through the origin. So thequestion is answered using both the statements.

17. e Considering both statements carefully, it is notpossible to prove the congruency of the twotriangles, as even though both are right-angled andhave the same perimeter. Their corresponding sidesmight measure differently and they might still add upto the same perimeter.

18. b In ∆PQR,let angles be ∠P = p, ∠Q = q and ∠R = r.From statement I, q = r∠PRS = 180 – r

∴ ∠TRS =

21[180 – r]

This is not sufficient to deduce that TR || PQ.

From statement II, 2r902 r180

qp −=−==And ∠TRS = 21[180 – r] = 2r

90 −∴ TR || PQ using corresponding angles are equal.

19. c Sum of the radii of the two smaller circles is equal tothe radius of the larger circle also the two smallercircles have equal radius. (Both the circles touchthe centre and circumference of the larger circle atone point each.) Thus answer is (b) as we can findout ‘R’ independently from 2 sentences.

20. c From statement I the angles are 30, 90 and 60 whichmeans the triangle is a right-angled triangle.Hence, area can be found.Statement II is also sufficient since the sides of the

triangle are in the ratio

23

:21

:1, i.e. the triangle is

a right-triangled.Practice exercise – C41. b From hint I, 3 2 3a–a 9a –1 2+ ≥ ⇒3 2 3a–a 9a –3 0+ ≥ ⇒ 2(3a–1)(a 3) 0+ ≥ Since ()3a 3+ is always greater than 0,(3a–1) 0≥or,

1a

3≥From hint II, 22a 6a≥( ) 26a 2a 0 2a 3a 1 0⇒ − ≤ ⇒ − ≤ or,

10 a

3≤ ≤ From hint III, a ≥ 0It is to be noticed that hint III can be deduced byeither hint II or hint I. Combining hints I and II, we get

the value of a as

13.

So hint III is redundant


Page: 33


For questions 4 to 6:Another new type of question set. Basically, instructionswould have given you the idea that the question is not to beanswered but the information required to find the answer tothe question must be singled out from the sea of data.We analyze the facts one by one in a tabular format.

Fact # Height Weight (i) C > D C > D (ii) D > B B > C (iii) A > B B > A

(iv) - C

DA

>

(v)

A

C B

D

> -

SUMMING UP

AC B

D> >

CB D

A> >

4. a Here, we are interested in the weights of B and Drelative to each other. We see that the first twofacts tell us that B > C and C > D. Thus, both of themcombined have told us that B > C > D and hence, B> D. Note that even (iii) and (iv) combined wouldhave directed us to the result(B > A > D). But in the options, we hada. (i), (ii) B > C > Db. (ii), (iii) No information about Dc. (i), (iv) All information (about weights) in fact

(i) is already contained in fact (iv).d. (ii), (v) Fact (v) does not tell anything about

weights and (ii) alone is insufficient.So, only option (a) contains all the facts required toanswer the question. The answer to the questionasked is "YES", but we have nothing to do with thatanswer.

5. e Here, a sequential procedure was to be followed.First, we had to find who is the shortest individual.Then, we had to find who is the heaviest individual.Finding both of them can confirm if the same individualis referred or not. Hence, from facts (i), (ii), (iii) and(iv) we get to know that B is the shortest as well asheaviest. So, option (e) is correct.

6. c From option (a), it becomes clear that B or C cannotbe the lightest (because they are heavier thansomebody). But lightest could be either A or D.From option (b), information revealed is that C or Ais not the lightest. But still, we cannot be sure aboutD because information about B is still concealed.From option (c), we come to know that neither Bnor C nor A is the lightest. Hence, D has to be thelightest.

From option (d), it can be seen that fact (v) isworthless as far as the data about weights go.Also, facts (i) and (ii) do not confirm D as the lightestamong all.Therefore, option (c) consists of sufficient factsand hence, is our answer.

In questions 7 to 11, the task was not to identify the optioncontaining sufficient facts, but to tell the minimum sufficientnumber of facts required. In this scenario, redundancy andintersection of the information in the facts must also be keptin mind while marking the answer. In these questions, theultimate result acquired through the synthesis of all theinformation given by all the hints taken together would beemployed.

7. a In the table, we can see that C is the tallest. Heightorder of C can be computed by fact (v) alone, whichdirectly tells that C is the tallest. Minimum 1 factrequired. Option (a).

8. c From the table, B is the heaviest. To know that B isthe heaviest, we must make sure that there is atleast one person heavier than each of A, C and D.For that purpose, we require facts[(ii) AND (iii) AND { (i) or (iv) } ]. Minimum 3 factsrequired. Option (c).

9. a We can see in the table that D is not the shortest. Toconfirm this, we need just one information that couldtell us that there is at least one person shorter thanD. Fact (ii) tells us just that. Only one fact required.Option (a).

10. b From the table, D is the lightest. To know that D isthe lightest, we must make sure that there is at leastone person lighter than each of A, B and C. For thatpurpose, we require facts[(iv) AND { (ii) or (iii) } ]. Minimum 2 facts required.Option (b).

11. c B is the shortest. To know that B is the shortest, wemust make sure that there is at least one personshorter than each of A, C and D. Only facts [(ii) AND(iii) AND { (i) or (v) } ] can confirm the premise that Bis the first in ascending order of heights. Minimum 3facts required. Option (c).


Page: 35��

For questions 12 to 20:This is a typical kind of set similar to that featured in CAT2005. The questions asked you to decipher the data andthen ascertain the dependency of the two statementsspecific to each question. Hence, it must be understood thatthese kinds of questions ask for a bit more than formingcases or inferring a pictorial / tabular simplification of theinformation. The data can be synthesized as:

Five ladies Colour of sarees

Mrs. Laali Red

Mrs. Neelima Blue

Mrs. Hariyaali Green

Mrs. Kesari Orange

Mrs. Shyama Black

However, the colour of the saree worn by an individuallady in particular is not known.Only possible source of information to come to any kind ofconclusion regarding the colour of saree each lady wearsis the following table along with three additional informationgiven in the question.

Colony

Red Blue Green Orange Black

Affection 2 4 2 3 1

Adorable 4 2 4 2 3

Fondness 1 3 4 1 3

Total 7 9 10 6 7

Number of extra-marital affairs

Colour of the saree

Since time is what we fall short of while attempting suchcrucial questions in any exam, we must turn as economicalfor that resource as possible. Therefore, Mrs. Laali will beaddressed as L, Mrs. Neelima as N, Mrs. Hariyaali as H, Mrs.Kesari as K and Mrs. Shyama as S from now on.

Information 1: K is having 1 affair more than NInterpretation: Number of affairs of K has to be exactly onemore than that of N.The possibilities from the table above areN = 6 and K = 7 orN = 9 and K = 10There is no other possibility. So it is clear that: -(i) N wears the orange saree or blue saree(ii) K wears the red or green or black saree

Also,N (blue) ⇒ K (green) and N (orange) => K (black / red)

Information 2: S has lesser affairs than LInterpretation: S < L. First things first. S cannot bemaximum(Green) and L cannot be minimum(Orange). Thatis because number of affairs of S is less than someoneelse’s. Hence, it cannot be maximum (10 for green).

Similarly, number of affairs of L is greater than that ofsomeone else. So, it can never be minimum (6 for orange).

Information 3: In Adorable colony, H has the maximum numberof affairs.Interpretation: From the table, we can see that the maximumnumber of affairs inAdorable colony is 4, which are those of the lady in Red andthe lady in Green. Hence, H is wearing either Red or Green.Now, we must move on to the questions.

12. c Straight from the information 3, we deduced that His in red or green. So, none of the two statements isdefinitely true. However, exactly one of them istrue. Hence, option (c)

13. a Option (a): Statement I tells that green is the colourof the saree worn by K. And since, from informationI, if total affairs of K = 10, total affairs of N will be 9,that means N is wearing blue coloured saree. Sostatement I confirms statement II. Note that weassumed statement I to be true and concluded thetruth of statement II. Hence, this option is correct.Option (b): If statement II is assumed to be true, thenaccording to information 1, i.e. K = 1+N, K has 10affairs. Hence, K wears green saree which makesstatement I true. So option (b) is not correct, becauseit concludes statement I to be false.Option (c): If statement I is assumed to be false,then K doesn’t wear green saree. Hence, she wearseither red or black saree. That means, number ofher affairs = 7. Thus, number of affairs of N = 6(orange saree). So statement II is also false. Hence,option (c) is not correct, because it concludesstatement II to be true.Option (d): None of the two statements is definitelytrue because there are other cases possible.

14. a 2 affairs in Affection colony means either red orgreen. And Mrs. Hariyaali is in one of red or greensarees (from information 3). Hence, statement I istrue. 2 affairs in Adorable colony means either blueor orange. And from information 1, we have seenthat K cannot wear blue or orange. Hence, statementII is also true.

15. d S cannot be green and L cannot be orange frominformation 2. Hence, only statement II is true.

16. c We know that K wears the red or green or blackcoloured saree only. Hence, blue is not possible.Green coloured saree for K is possible. So from theoptions, we see that option (c), which says thatstatement II is true and statement I could be false, isthe correct option.Now information 1, 2 and 3 must be combined fornecessary deductions. If N is not wearing orange,she must be wearing blue. And N(blue) => K(green).This way, H (green or red) has to wear red. Andsince S < L, S (orange) and L(black) is the onlycombination possible for N wearing blue.


Now, we assume that N is wearing orange. Hence,K will be wearing black or red. When K wears red,H (green or red) wears green and hence S wearsblack and L wears blue (because S < L). When Kwears black, H wears green or red and accordingly,S wears red or blue obeying S < L.The exhaustive list of possibilities

Red (7) Blue (9) Green (10) Orange (6) Black (7)

H N K S L

K L H N S

S L H N K

H S L N K

17. a Minimum number of affairs = 6 (orange). Neelimanot wearing orange means she wears blue. In thatcase, Shyama will wear orange (see table) andshe would be having only 1 and not 3 affairs inFondness colony. Neelima wearing orange meansShyama could be wearing red or blue or black. 3affairs in Fondness colony is true for lady wearingblue saree and lady wearing black saree. Shyamacould have 1 affair in Fondness colony if shehappens to wear the red saree. Hence, statement IIcould be true if statement I is false.

18. c We know that S cannot wear green. And K cannotbe orange because K can be red or green or blackonly. Hence irrespective of the conditions, bothstatement I and statement II are definitely false.

19. d Option (a): If we take statement II to be true, L hasmaximum (4) affairs in Affection colony. So shewears blue saree. In that case, H has the maximumnumber of affairs (green saree). Hence if statementII is true, statement I is also true. This option claimsthat statement I is false. Hence, not the right option.Option (b): If statement I is taken to be false, Hdoesn’t have maximum affairs and therefore, shedoesn’t wear green. So she wears red. It impliesthat L could wear black or green. Therefore, L cannever have maximum affairs in Affection colony(which is true for lady wearing blue only). Sostatement II is certainly false. Hence, this option isnot correct.Option (c): Both the statements are notindependently true, as can be seen from the table.We do have cases when H does not have maximumnumber of affairs. Hence, this option is not valid.Option (d): H having maximum affairs means H iswearing green (10 affairs). In that case, L definitelywears blue (as can be seen from the table) andhence she has the maximum affairs in Affectioncolony (4 affairs).So, statement II is necessarily true if, statement I istrue.So, this is the right option choice.

20. b Option (a): If statement I is true, H doesn’t weargreen. That means, H wears red. Correspondingly,S could wear orange or blue, and not necessarilyorange.Option (b): H not wearing green means H wearsred. In that case, S can wear blue or orange. Butwhen it is given that S wears orange, we can seefrom the table that H wears red for sure (and hencenot green). So given that statement II is true, wecan infer that statement I must be true. So, this isthe right option choice.Option (c): If statement I is false, it means that Hwears green. Then S wears black or red but neverorange. Hence if statement I is false, statement IIwould necessarily be false and not true.Option (d): If statement II is false, S wears black orred or blue saree. Hence, H may or may not wearthe green saree.

LRDI Practice Exercises

Practice exercise – D1

1. b There are just two hotels having grade ‘A’ andexceeding 13,450 customers, viz. Taj Residencyand Mandar International.

2. c Just count the hotels with grade ‘A’ or ‘B’ and havingmore than 11,500 customers per year.

3. b There are just five hotels with their names startingwith alphabet ‘M’ or ‘R’. Of these Madhuban Deluxehas the highest customers per year and which isgreater than 1,500, the second highest. So evenwith increased customers for other hotels,Madhuban Deluxe will have the highest customersper year.

4. b Same explanation as in question 3.

5. b In metro, valuation is > 10 times the equity. In Acircles, valuation is more than 4 times but less than5 times. In B and C circles, valuation is more than 5times, so valuation is least for A circles.

6. c Combined share of metro and A circles = Rs. 1,48,350 millionTotal valuation of all circles = Rs. 2,02,900 millionThus, percentage share of metro + A circles

= 148350202900

< 74%, i.e. 73.18%

7. c Total losses written off is Rs. 1,000 crore = Rs.10,000 millions. Thus, the increase in valuation forMetro and A circles is Rs. 4,444.4 million and Rs.5,555.5 million respectively. Valuation to equity ratiohas to be equal for A and B circles by reducing the


Page: 37��

equity of A circles, thus 67100 5556+

x =

445007875

or

x = 72656 7875

44500×

= Rs.12,858 million.

Thus, percentage decrease in equity of A

= 12858

115300

− = 16%

8. e Total valuation forDelhi : 81250 × 0.42 = 34,125Maharashtra : 67100 × 0.25 = 16,775Karnataka : 67100 × 0.22 = 14,762⇒ Delhi : Maharashtra : Karnataka = 34.1 : 16.7 : 14.7

9. d By observation, we can say that E has registeredmaximum increase.

10. b (i) Price of share A on day 1 = Rs. 198Price of 100 shares on day 1 = 198 × 100= Rs. 19,800If we encash 50 shares of A on day 5, we get201 × 5 = Rs. 10,050And on day 6 = 203 × 50 = Rs. 10,150Gain in case of share A is (10050 + 10150) –19800 = 20200 – 19800 = Rs. 400

(ii) Investment of share B on day 1 = 1012 × 50Total investment = Rs. 50,600Total investment on day 1 on share C = 52 × 50= Rs. 2,600

(iii) Investment of share B on day 1 = 1012 × 100If 50 shares of B and C are encashed on Days5 and 6, then amount encashed = (1034 + 1067)× 50 – 1012 × 100 = Rs. 3850

Amount of share B on day 6 = 1067 × 50 = Rs. 3,350Amount of share C on day 6 = 56 × 50 = Rs. 2,800Total return = 53350 + 2800 = Rs. 56,150Gain = 56150 – 53200 = Rs. 2,950So, (ii) is better than (i).

11. e We can observe from the table that the percentageincrease for stock E is more than 20%.

12. e Stock C has shown the least increase in absoluteterms but in percentage terms stock D is the least.Volatility is not defined whether it is in percentageterms or in absolute or something else.So none of these is the correct choice.

13. b Day 4 = 199 × 20 + 1030 × 30 + 54 × 25 + 414 × 15+ 62 × 10 = 3980 + 30900 + 1350 + 6210 + 620Day 4 = 43060Day 5 = 201 × 20 + 1034 × 30 + 55 × 25 + 417 × 15+ 63 × 10 = 43300

Percentage change

430604306043300

4Day4Day5Day −=−=

%55.01004306028050 =×=

14. e Percentage increase in gross assets from 2001-02

to 2005-06 = %4.5151001130800 =×

−

15. b Total disbursements from 2002-03 to 2004-05= 560 – 65 = Rs. 495 crore

16. b Maximum percentage increase in gross profit is for

1995-96 = %1.5710012133 =×

−


17. b

Route CarrierNo. of

ticketsFare Fuel (Rs.)

I Jet Airw ays 3 Rs. 14100 2820

II Indian Airlines 4 Rs. 13640 2728

III Air Deccan 2 Rs. 3510 912

Rs. 6460

18. b Total expenses = Fare + Time not used in workingWage = Rs. 200 / hr⇒ By air = Rs. 2,375 + Rs. 200 × 2 = Rs. 2,775 andBy train = Rs. 2,286 + Rs. 200 × 9 = Rs. 4,086So, profit = Rs. 4,086 – Rs. 2,775 = Rs. 1,311 (Sincetotal expenses are lower for sending him by AirDeccan).

19. d Neither remuneration nor number of employees isknown.

20. c Check the options, (c) is maximum.

21. d The last digit can be found only if we know whichnumber is higher. Hence, both the statements areneeded.

22. d Overall winning percentage is the weighted averageof percentage won in the 2 years, with the numberof matches being the weights.Thus, we need both the statements.


23. c If he was dot on time for the class as per his watch,

he would have been 2221

min late.

24. c We can write the equation as 3rd (r + d) = 0, r can

be 3− or r can be 0. Hence using either of the

statements alone, one can say that r 3.= −

25. d Combining (I) and (II):Time taken by Shyam = 2 hrRatio of speed = 2 : 3

So time taken by Ram 2 203 t

=

t = 45 minSo Ram reaches at 11.40 a.m. + 45 min = 12.25 p.m.


1. a Part of body made of neither bones nor skin norMuscles

1 1 1 10 3 5 181 1 1

3 10 6 30 30+ + = − + + = − = −

6 4 21

10 10 5= − = =

2. e Quantity of water in a body of 50 kg = 70% of 50 kg

7050 35 kg

100= × =

3. d Skin 110%

10= =

Proteins = 16%

10100 62.5%

16⇒ × =

4. c Percentage requirement of proteins and other dryelements = 30%

Therefore, the required angle30

360100

= ×

�

= 108°

5. b In value term, Godrej appreciation has been highest= 200

6. a Very clearly, Dabur, which has become almost 3times in 2008 from 2006.

7. dEP

ratio has not decreased by more than 20%. So

no share is bought. So (0, 0)

8. d Shares are bought and not sold, so no tax is paid,since tax is paid only when shares are sold.

9. a Shastri 1436

7 =×=

Akram = 2036

10 =×

Razzaq = 7216

12 =×

Binny = 2726

9 =×

10. c Akram = 10332 >

Azad = 332

331 <

Shastri 43

12 ==

Binny = 162

32 =

11. d It seems, either Maninder or Shastri is the answer.

= =21Maninder 1.9

11, 7.1

712

Shastri ==

12. e We cannot find the runs scored as we do not knowthe extras.

13. e Number of literate females in India in 1991= 39.4% of 430 million = 169.4 million

14. e Cannot be determined because we do not have thepopulation (the weights for average) for the states.

15. e Number of illiterates in Maharashtra in 2001 = 36.9% of 19 million = 7.01 million

16. e Because the infant population of states is not given.Hence, the data is insufficient.


17. e Some people can be lawyers as well as womenAlso passage of the bill is not defined.∴ We cannot find out the answer.

18. a63 550 58 250

Average age550 250

× + ×=

+ = 61.43.


Page: 39��

19. c Minimum case is when all the Rajya Sabha femalemembers are Hindus, i.e. 42.So, minimum number of Hindu males is 64.

20. b Percentage of Congress(I) members = 267 112

550 250

++

37947.37%

800= =

21. b From statement I, we have Y > – |X|, which meansY is less than or greater than X. So statement I byitself cannot answer the question. From statementII, we have Y > |X| which means Y is greater than X.So it is sufficient.

22. e The sum of a, b and c cannot be found fromstatements I and II individually or together.

23. d Statement I alone is not sufficient as the total numberof guests is unknown.Statement II alone is not sufficient as the proportionof single-scoop and double-scoop is not known.Combining statements I and II, we can find thesolution.

24. c Statement I indicates that n is divisible by 1, the 2prime numbers and their product. Hence, there are4 different positive integers. From statement II, thefactors of 8 are 1, 2, 4 and 8 itself, again 4 differentpositive integers.

25. a Using statement (I):a × b = 16 ⇒ a = 4, b = 4a = 2, b = 8a = 8, b = 2In all the above cases, the last digit of (5 × ab)ab willbe zero.So statement (I) alone is sufficient.Statement (II) alone is not sufficient.


1. e356

Percentage 10 10%3694

= × ≈

2. b Population of Hungary = 123

5120.24

=

Population of Pakistan = 285

1092.62

=

So difference = 403

3. a Japan

4. b Aggregate deposits for the year ended March 1999= (717271 – 111861) = Rs. 6,05,410.Hence, percentage increase in deposits

111861605410

= 18.5%=

5. b

=× %51100

717271366003

6. e

=×−

%5.11100717271

717271800000

7. b For a credit deposit ratio of 60%, if deposit isRs. 8,00,000 crore, then deposit should be Rs.4,80,000 crore.

=×−

%31100366003

366003480000

8. c Hindu population in 1971= 1.2 × 0.78 × 665.3 = 622.72

Percentage increase = 622.72 –549.8

100549.8

×

13.3%=

9. d Increase in total population = 548 – 439 = 109millionIncrease in Hindu population = 453 – 366 = 87millionIncrease in Hindu population in total population

increased 87

80%109

= ≈

10. a Population of Hindu, Jain and others in 1941= 424 millionPopulation of Hindu, Jain and others in 1961= 641 million.

424 2Ratio

641 3= ≈

11. e Hindu in 1971=( )2550549.8

549.8 660 million453.4 450

× ≈ ≈

Others in 1971 = 275.5 75.5 (75)

94 million61.4 60

× ≈ =

Ratio = 660 794 1

≈


12. a Clear from Data.

13. a Clear from Data.

14. a R.D. = 604 604

37.75 30.216 20

− = − = 7.55

J.K. = 503 503

25.15 22.8620 22

− = − = 2.29

M.K. = 902 902

43 39.2121 23

− = − = 3.79

A.S. = 832 832

39.62 34.6720 27

− = − ≈ 5

Clearly R.D. has the highest difference.∴ Option (a) is the correct choice.

15. d Calculating the strike rate of the given bowlers weget(a) 3.2 for Harmission(b) 4.68 for Shane Warne(c) 4.26 for M. Muralidharan(d) 2.9 for Pollock


16. c Except Reliance Growth, Birla Dividend Yield Plusand Templeton India Growth all others fit in thecondition. Hence option (c) is the correct choice.

17. a Reliance Growth is maximum at 0.58.

18. a Templeton India Growth is 2nd best at 0.52.

19. a HSBC Equity – 0.09Franklin India Blue-chip – 0.14

20. c Barring top – 3, all others are eligible.

21. d Combining both the statements, we can determinethat S is standing at the middle position.

22. d Combining (I) and (II), we get the following sittingarrangement.

R itu

S u kan ta

M and ar

B a sis th

S h ikh a

Stu ti

23. e Combining (I) and (II):The oldest institute could be any one of the three,i.e. Career Launcher, Erudite, or IMS.Hence, it cannot be determined.

24. a The cyclicity of 2 is 4.Using statement (I), xy can be 43, 49, 83, or 89.So last digit of (72)xy could be 2 or 8.Using statement (II), xy can be 46 or 86.So the last digit of (72)xy has to be 4.

25. d Combining (I) and (II):c = xa = x – 1b = 2xd = 2x + 2

So a x 1 c x

andb 2x d 2x 2

−= =+

ad = (x – 1)(2x + 2) = 2x2 – 2bc = 2x (x) = 2x2

As ad < bc, a cb d

<


1. b Total amount spent on ad= Ad spent on (Print + TV + Others)In 1998 = 6824, in 2000 = 8988Average annual growth rate of ad spent

%8.152

1001

68248988 =×

−

=

2. b Growth rate of ad spent is the highest for TV asshown:

Print = %2.24138204745 =−

TV = %44124603542 =−

Others = %8.281544701 =−

3. e Amount spent on TV as a percentage of total ad

spent for 1999 = %4.3876872952 =

4. b Decreases continuously.

5. c %6.2110023.236502.512 =×


Page: 41��

6. b Exports of pearls in September 1998 have fallenby US $0.26 million (0.54-0.28) from the value inSeptember 1997.

Percentage decrease = %1.4854.026.0 =

7. e %68.110063.1628.0 =×

8. e 2365100 82.7%

2860 × =

9. e (210 – 220) + (225 – 230) + (232 – 215) +(215 – 205) + (215 – 248) + (226 – 247) +(245 – 239) + (250 – 223) + (217 – 248) = – 40

10. c Export earnings of jewellery + leather = 32% of230food grain imports = 46% of 225.Percentage of food grain imports that could not be

paid for %88.28100)22546.0()23032.0(

1 =×

××−=

11. b In June, cotton exports = 27% of (100 – 7.25)% of215 = 0.27 × 0.9275 × 215 = Rs. 53.8 million

12. e Chemical import = 0.23 × 225 = Rs. 51.75 millionExport of jewellery = 0.18 × 230 = Rs. 41.4 millionDifference to be covered by spices = Rs.10.35millionSpice exports = 0.07 × 230 = Rs.16.1 millionThus, percentage of export of spices spent in

chemical imports = %28.641001.16

35.10 =×

13. c Export earning from cotton in April in US dollar

41.1$42

)22027.0( =×= million

Import of chemicals would cost 0.23 × 210= Rs. 48.3 millionFor this to match $1.41 million, the exchange rate

has to be 25.3441.13.48 =


Total matches played = 50Total number of Losses = 44 + Y

Total number of Ties = 11 Z

2

+

Total matches played = number of loss + number of tiesTotal number of wins = Total number of losses

So, 50 = 44 + Y + 11 Z

2

+

⇒ 2Y + Z = 1Since Y and Z are non-negative integers.Z = 1, Y = 0So, X = 19.

14. c

15. a

16. e


17. b Total oranges produced in May = 31 × 5 = 155Checking the options,(1) 26th JuneTotal oranges needed by Kitto = 1 + 2 + 3 + ... + 26= 351Total oranges produced till (and on) 26th June= 155 + 26 × 8 = 155 + 208 = 363So, 12 oranges will be left.(2) 27th June, Kitto will be in need of 27 oranges.Taking a clue from option (1), maximum availableorange= 12 + 8 = 20.So, it will run short of oranges on 27th June.

18. b Total oranges produced in May = 31 × 8 = 248Checking the options,(1) 1st JulyTotal oranges needed till (and on) 1st July= 1 + 2 + ... + 31 = 496Total oranges produced till (and on) 1st July= 62 × 8 = 496On 2nd July, kitto will obviously run start of oranges.So, option (b) is the answer.

19. e Total oranges needed till 1st July= 1 + 2 + … + 15 + 15 × 15 = 345Total oranges produced till 1st July = 31 × 5 + 30 × 8 = 395So oranges left on 1st July = 50.

20. d Combining I and II: There are 25 students, one girland 24 boys in the class. Santosh’s rank is 24th inthe class and since the girl is not 25th, she has tobe above Santosh. Therefore, Santosh’s rankamong the 24 boys is 23rd. So the answer is (d).


21. d Combining I and II, the code for ‘love you’ is ‘mikesika’ (from I). If we combine I with II, ‘you’ has to be‘mika’. So ‘love’ has to be ‘sika’. The answer is (d).

22. e All that we can get after combining the informationgiven in the two statements is that ‘God is’ coded as‘mau pau’, but we cannot find out exact code ofGod.

23. c Statement I:Z has to be second.X has to be third.Y has to be first.Statement II:Y finished second.Z has to be third.X has to be first.Thus, both the statements individually give theanswer.

24. e Combining I and II: We known that A and C are males,and B and D are females. But we do not know thegender of E who could be a male or female.

25. e Combining I and II:Even after combining both the statements, we cannotdetermine who is the tallest. It could be either Rohitor Manish. When we say that Rohit is not taller thanManish, it means Rohit could be of the same orlesser height than Manish.


1. d In 1985, production of rice and wheat is 393 lakhtonne and 331 lakh tonnes respectively.Hence, the ratio of production of rice to wheat

= 393 : 331 = 1.2 : 1

2. d Among all the crops grown in Bihar, MP andMaharashtra, Jowar has the highest production.

3. c Production of wheat by Punjab as a percentage of

total production of India = %1.31331103 =

4. c Total number of OBC students= 10 + 12 + 11 + 10 + 11 + 13 = 67Total number of industry-sponsored students= 2 + 1 + 1 + 2 + 1 + 2 = 9Total number of SC/ST students= 15 + 16 + 18 + 19 + 11 + 10 = 89

5. a Total number of outside university students= 3 + 2 + 4 + 1 + 2 + 3 = 15Total number of outside state students= 2 + 2 + 4 + 1 + 2 + 3 = 14Total number of OBC students= 10 + 12 + 11 + 10 + 11 + 13 = 67Total number of college III students:General category = 35% of 500 = 175Others = 3 + 4 + 4 + 5 + 11 + 18 + 1 = 46

Hence, %4.43100)46175(

)671415( =×+

++

6. d In college III, total number of handicapped, outsideuniversity, NRI and OBC students = 23 and totalnumber of outside state, SC/ST and industry-sponsored students = 23.

7. b In college IV, students admitted through quotas= 38.Students admitted through general quota= 14% of 500 = 70

Hence, %28.541007038 =×

8. d None of the above: the figures given are for increasein GDP.

9. a 1 × (1.025)4 × (1.018)4 = 1.1854

10. b It is clear from the table.

11. e Since we cannot take average value of GDPbecause of possibility of different GDP figures fordifferent regions within a group.

12. e Cannot determined because data of each region isnot given.

13. e Required cost= (240 × 1.5) + (1200 × 3) + (840 × 1.8)= Rs. 5,472

14. a Required average

)4848484848(1482484.2486.1485.148

++++×+×+×+×+×=

or 5

124.26.15.1 ++++ = Rs. 1.7

15. d The cost of transportation for each alternativeare:(1) 300 × 2.8 = Rs. 840(2) 200 × 3 = Rs. 600(3) 300 × 1.5 = Rs. 450(4) 400 × 1.1 = Rs. 440Obviously, the cost of transportation ofalternative (d) is the least.


Page: 43��

16. d Required percentage

= %400%100240

2401200 =×−.

Alternative method:In stead of using the formula for percentage growth

Final Initial100;

Initial−= × we can use;

Final100 1 ;

Initial × −

as it saves one mathematical operation ofsubtraction.

So %400100)15(1001240

1200 =×−=×

−

17. b 1997-98, nearly 50% increase. (Can be figured outby looking at the slope)

18. a Company turnover in 1997-98 = Rs. 250 crore.

Total industry turnover in 1997-98 = 250 × 100

7

Total industry turnover in 1998-99 = 1.17

100250 ××

Company’s share in 1998-99 = Rs. 220 crore

∴ Company’s share = 1001.1250

1007220××

××=

2514

= 5.6%

19. e 200% increase in 4 years. (From 3 crore to 9 croreor 50% average annual growth rate.)

20. c Ratio = 553

165 = . (Do not write data for every

year, try to do it mentally)This is maximum in 1996-97.

21. d Statements I and II alone are not sufficient todetermine Ravi’s age. From statement I, we getRavi’s age but from I, we get Ravi + 10 = 2(Ram + 10) and from statement II, Ram = 5 years old.Therefore, Ravi = 30 –10 = 20-year-old.

22. c From statement I, 4 boys and 3 girls,i.e. every boy has 3 brothers and 3 sisters.From statement II,every girl has 3 brothers and 3 sisters∴ The answer is (c).

23. e ‘x’, ‘y’, ‘z’ be the lengths of the ropes in the increasingorder of length.Then from statement I, y + z = 10and from statement II, x + y = 92 equations and 3 variables. Thus, cannot get ‘x’.∴ The answer is (e).

24. d Combining I and II, we must have 6 silver and 2bronze coins. No other possibility is there.

25. e Combining I and II, the only information we get is thatC has four sisters. Now R has four siblings buthow many brothers and sisters is not known.Hence, the number of uncles for K cannot bedetermined. So the answer is (e).


1. e ≈ 9.2 : 28.2 ≈ 1 : 3

2. c Total sales = crore4.176.710032.1 =×

3. b Answer is ‘The Decan Herald in 1999’.‘Others’ is not included because it is not anewspaper. ‘Other ’ here includes newspaperswhich are not mentioned.

4. b Deccan Herald gained Rs. 5.34 crore.

5. a Jadeja’s total = 164.Tendulkar’s average = 57.5Hence, the ratio is 1 : 2.86.

6. a Tendulkar’s points are 8 + 3 + 4 + 4 + 7 = 26.Dravid’s points are 4 + 10 + 10 = 24.Others are below this.

7. a Since the total of 5 batsmen is more than 226 runs.

8. d Against England = 382

= 19

Africa = 18

181

=

Australia = 31

7.753

=

Sri Lanka =�

143.5

4=

9. e Sales = 200971× crore1600~2008~ −×−

10. b From a figure of Rs. 90 crore in 1996-97, salestouch Rs. 175 crore in 1999-2000. Since sales haveless than doubled, so percentage increase < 100%.It has to be around 95%.

11. e We do not have data on market shares in 1997-98.


12. a 40% of others = 40% of 5% = 2%, which is 1

4 5. of

Zenith home PC market share.

∴ Wipro home PC sales = 1

4 5. of 200 = 200

92 ×

= Rs. 44.44 crore

13. d Among all the products, E has maximum sale aswell profitability. So obviously E will be having thehighest profit.Sale = 60 lakhProfit = 20%

Cost = lakh502.1

60 =

Profit = 50 × 20% = 10 lakh

14. c Profit earned by B = lakh1140

10010

1.140 =×

Profit earned by E = 10 lakh

So %36.3611

400101110040

100)ofit(Pr)ofit(Pr

E

B ==××=×

15. c Profit earned by A = lakh510020

2.130 =×

Profit earned by B = lakh6.3lakh1140 =

Profit earned by lakh115300

10015

15.120

C =×=

= 2.6 lakh

Profit earned by lakh86300

1005.7

075.150

D =×=

= 3.5 lakhProfit earned by E = 10 lakhTotal profit = 24.7 lakh

16. c The percentage share of the profits of A and

E = %63.6010074.24

15 =×


17. e It can be observed that shares give a flat 11% return.Now, making all the possible investment from Rs.10,000 – Rs. 1,00,000, return in dairy does notbecome equal to the return obtained by shares.Short cut: Check it through options.

18. c Since after the investment of Rs. 30,000, for everyinvestment of Rs. 10,000, return from dairy is Rs.900 whereas for the same amount, return fromshares is Rs. 1,100.

19. b If equal amount is invested, than net return =Rs. 4,200, which is maximum in all the cases.

20. e It can be observed that if he investsRs. 20,000 — AgricultureRs. 10,000 — Dairy andRs. 20,000 — PoultryTotal return = Rs. 2,600 + Rs. 1,400 + Rs. 2,750= Rs. 6,750So, the percentage return = 13.5%

21. b Statement I gives us superfluous data. To find outthe volume of the glass, we need its height anddiameter. Hence, statement II alone is sufficient.

22. e No information has been given about the area orany sides of the square base of pyramid, we areonly provided with distance from the square basefrom IInd statement. Hence we cannot determinethe attitude of the pyramid by using the statements.

23. e You do not know the shape of the bathroom floor.Thus, the exact number of tiles cannot bedetermined.

24. e We still do not know the height of the room.

25. a Using statement I, we get

( ) t)r5.4(k10070

t5.4k 222 ×−×π×=×××π×

Solving this we can find r.k is constant of proportionality to convert volumeinto weight. Thus answer is (a).


1. a Percentage change in exports of automobilesexcluding HCVs in 1993-94 to 1990-91

= 1004234

44–20–58370 ×++

= %2.2100423494 =×

2. e We know the total number of two-wheelers in1990-91 and 1993-94. But we do not know thenumber of two-wheelers which were exported toEU in 1990-91 and 1993-94.Hence, data is insufficient.


Page: 45��

3. d The total revenue of government through duty on

exports 72 10285107212.133000 ×=×××== Rs. 285 crore

4. d Revenue earned by LCVs = Rs. 1012 × 5.5 crore= Rs. 5,566 croreRevenue earned by LMCs = Rs. 823 × 4.2 crore= Rs. 3,456.6 croreTotal revenue in US dollars

= crore5.42

6.34565566 + = $ 21.2 million

5. d Cost of production (or) wage bill= Man-hour worked × Hourly wages × Days workedNow it is very clear from the table that year 2000has the highest value for all the three parameters.

6. a It is clear that the least man-hours worked is in1970 = 1300 × 12 = 15600 man-hours × 250 days

7. c Percentage increase in production is always higherthan the increase in number of workers.

8. b 1,300 workers by working 12 hr per day canproduce 550 tonnes. Hence, by working 24 hr perday they can produce a maximum of 1,100 tonnes.Hence, they can meet targets of 1980 but not of1985.


mreT ecirplaicepSeussirep

esaercedegatnecrePmretsuoiverpmorf

1 22 —

2 02 %9.xorppA

3 81 %01

4 61 %11.11

5 5.31 %6.51

9. e

10. b Effective price (EP) = 480 + 48 – 0 = 528 for 24 issues

IssueEP

= Rs. 22

EP for 5 years = 810 + 0 – (12 × 25) = 510 for 60issues.

IssueEP

= Rs. 8.5

Difference = Rs. 13.5

11. dComputed special price

News s tand price for each−

12. c EP (3 years) = 648 + (24 × 3) – 60 = 660

issueEP

=66036 = Rs. 18.33

13. e 1250 = 18%, then

100% crore700010070~18

1001250 =×−×=

14. e Cannot be determined as only market share isgiven.

15. d HP’s sales of peripherals = 7

2500

= Rs. 357 croreHP’s share of peripheral market = 30.4%

Hence, total market = 4.30100357 ×

= Rs. 1174 crore

16. a Revenues from consultancy services= 8% of 6945 crore = 555.6

Revenues from PC servers = 9.30100357 ×

= 1,155

Hence, consultancy services as percentage of

PC servers =555.61155

× 100 = 48.1%.

For questions 17 to 20: In the pie diagram, the expensesare given in total amount of Rs. 3,600.

17. b Family A yields = 960036018 × = Rs. 480

B yields = 1440036015 × = Rs. 600

⇒ Ratio = 480 : 600 = 4 : 5

18. e Family A spends = 480036048 × = Rs. 640

B spends = 720036039 × = Rs. 780

⇒ Required percentage = %82100780640 =×


19. b For family A, light = 240480036018 =×

For family B, light = 30072003615 =×

⇒ Difference = 300 – 240 = Rs. 60 (least)

20. d Percentage of expenditure of A on food = 10036096 ×

Percentage of expenditure of B on food = 100360110 ×

⇒ Ratio = 96 : 110 = 48 : 55

21. b The factors of 630 = 2 × 3 × 3 × 5 × 7. By using thestatement II, the sum of the 2 numbers is 153.Therefore, out of the factors of 630,(3 × 3 × 7 = 63) + (3 × 3 × 2 × 5 = 90) add up to give153. Hence, the 2 numbers are 63, 90 and we canfind the absolute difference between them.Hence, by using statement II only we get the answer.Hence, (b).

22. c From statement I, 110

X Y100

=

Required ratio = 100110

10090

X100Y90 ×=

Statement I alone is sufficient.

From statement II, Y10010

1110

100X10 ×=

Similarly, statement II alone is sufficient.

23. d Combining I and II, we must have 6 silver and2 bronze coins. No other possibility is there.

24. d Statement I is necessary and statement II is alsoneeded as if we take:

x 11

y 2= <

c = 98

x 98 97 97 1y 98 96 96 2

− −∴ = = >− −

x c 1y c 2

−∴ >−

This defeats the given case. Thus, both statementsI and II are needed. ∴ Answer is (d).

25. e The IInd statement tells us that triangle QPR is a rightangled triangle but the information is given only ofperpendicular distance QR, no information is givenabout the base, so even by combining both thestatements we cannot get the length of PR.It cannot be answered using both the statements.∴ Answer is (e)


1. e If the Republicans win 97 seats, percentage of

swing 84

)6597( =−=

Democrats win 90 – 8(3) = 66 seats.

2. e No relation between voters turnout (votes polled)and percentage of swing are given.

3. c (iii) adds up to more than 200 seats.(ii) gives negative value for Socialists.

4. d For republicans to get 100 seats the vote swingmust be 8.75% which is not an integer.

5. d The maximum observed value of ‘s’ is 10. Sam musthave answered 10 questions at least on one day.The minimum he answers on any day is 5. FromMonday to Thursday there are 14 queries from theprevious week and 10 from that week. Hence in allthere are 24 queries. So there is no way he couldhave answered 10 queries on any day from Mondayto Thursday. So he must have answered themaximum (10) number of queries either on Friday oron Saturday.

6. b The queries received on Saturday have to be allanswered on Saturday. On Thursday Sam has toanswer atleast 1 query that was received that dayand on Friday Sam has to answer at least threequeries that were received on that day. All the otherqueries can definitely be answered with atleast aday’s delay.Hence the minimum number that has to be answeredon the day of receipt is(4) + (5) + (1) + (7 + 3 + 1) = 21.For Example, they can plan their schedule in the

following manner:

Mon Tue Wed Thu Fri Sat

DI8 by Anu

18 by Anu

18 by Anu

10 by Anu

4 by Anu(4)* + 2 by Anu

LR6 by Dudi

6 by Dudi

6 by Dudi

6 by Dudi

1 by Dudi(5)* + 7 by dudi

QA 2 by Tiru

15 by Tiru

15 by Tiru

15 by Tiru

6 by Tiru, 5 by Dudi and 4 by Anu

(1)* + 9 by Tiru

VA 5 by Sam

5 by Sam

5 by Sam

4 + (1)* by Sam

(3)* + 4 by Sam

(7)* + 3 by Sam

( )* indicates queries received on the same day.


Page: 47��

7. d If all conditions have to be satisfied on each dayatleast 21 queries have to be answered.The number of incoming queries on M, W, Th, Fri,Sat are all less than 21, hence there must definitelybe some carry over from the previous day in theweek for each of these days.

8. e Number of queries of the beginning of the week= 103.Number of queries received during the week = 107.Hence number of queries answered= (103 + 107) – 20 = 190

9. c In all DI has (24 + 40) = 64 queries to be handled.Similarly LR has 37 queries to be handled.QA has 72 queries to be handled.VA has 37 queries to be handled.

Now range for dudi is ≤ ≤6 d 12; so at least on one

day he answered 12 queries, so minimum numberof queries answered by Dudi in the week= 6 × 5 + 12 = 42.

42 ≤ Number of queries that can be answered by

Dudi in a week ≤ 66.As explained above, minimum number of queriesanswered by Tiru in the week = 25.

25 ≤ Number of queries that can be answered by

Tiru in a week ≤ 77

Similarly, 58 ≤ Anu ≤ 98 and 35 ≤ Sam ≤ 55.If Dudi answers the LR queries than he has to moveto other sections viz. either DI or QA.So, atmost queries of three sections could behandled by one single respondent.


10. d The three friends whose houses are at a minimumpossible distance from the house of A are C, D andE.There are 6 ways by which A can visit C, D and E.

Route Distance (kms) AH-EH-CH-DH 48 AH-CH-DH-EH 44 AH-CH-EH-DH 40 AH-DH-CH-EH 45 AH-DH-EH-CH 39 AH-EH-DH-CH 46

Now, DH – E0 < EH – E0 and CH – E0DH – C0 < EH – C0 and CH – C0DH – D0 < EH – D0 and CH – D0and for AH – CH – EH – DH the distance travelled is40 Kms.

Now from D there are again 6 ways by which A canreach his office Ao

Route Distance (kms) DH-EO-CO-DO-AO 54 DH-EO-DO-CO-AO 53 DH-CO-EO-DO-AO 52 DH-CO-DO-EO-AO 59 DH-DO-CO-EO-AO 60 DH-DO-EO-CO-AO 52

Therefore minimum distance that A can travel is40 + 52 = 92 kms.

11. b AH-AO 12 kms BH-BO 13 kms CH-CO 18 kms DH-DO 12 kms EH-EO 17 kms FH-FO 19 kms

Now combining each of the above distances withthe distances given in table 2. i.e. Distance of AHfrom BH, CH, DH, EH and FH and so on, we find thatfor C, E and F the distances of their house fromtheir respective offices is not less than the distancefrom the houses of any of his friends.

12. a

B D D BH H O O15km 12km [email protected]/km @Rs.3/km @Rs.2/km

8Rs. 115 36 30

→ → →

+ + =

13. b

The shortest route can be found through iteration

12km 13km 14km 16km 14km 13kmD E F B C A D → → → → → →

In all it takes 82 kms of travelling (or) Rs. 82/-

For questions 14 to 17: The following values have beentaken from the graphs for calculation

raeY 5991 6991 7991 8991 9991 0002

)rCsR(eulaV 004 013 051 002 095 057

)gknM(ytitnauQ 502 571 031 051 562 024

14. c Value per kilogram for:1995 = Rs. 19.50 1996 = Rs. 17.711997 = Rs. 11.53 1999 = Rs. 22.26Thus, it is the lowest for 1997.


15. c Increase in value of jute exports for1996 = –22.5% 1998 = 33.3%1999 = 195% 2000 = 27.11%The graph shows that the steepest incline is from1998 to 1999.Hence, percentage change is maximum in 1999.

16. e Refer solution to question 14.2000 = 17.851998 = 13.33

Alternative method:Average Price =

19.50 17.71 11.53 13.33 22.26 17.856

+ + + + +

= 17.75

17. b The price per kilogram has decreased in 1996, 1997and 2000, has been worked out in questions14 and16.

Alternative method:For 1995 cannot be determined, but otherwisegrowth is the highest in 1999 from 13.33 to 22.26.

18. e 1.25 × 10–3 times the world’s fresh cut flower exports= Rs. 100 lakhSo the value of the world export of fresh cut flowers

= 31025.1

100−×

lakh; 25.1

10100 3×lakh = 8 × 104 lakh

19. c 20% of ‘Others’ = 4% of world’s export of cutflowers in 1990-91= 17 lakh. So 4% of x = 17 lakh.

⇒ x = 1004

17 × = 4.25 crore

20. e Netherlands may be exporting flowers other thancut flowers.

21. c Using statement I and II we can independently findout the unknown side of rectangle which comesout to be ‘6’. So perimeter can be calculated bymaking use of both statements independently.∴ Ans (c)

22. b From statement I, a + b = 3k but nothing can be saidabout C. Therefore, statement I is not sufficient.From statement II, c = 3p⇒ 3(a + b) + c = 3(a + b) + 3p = 3(a + b + p) = 3q⇒ 3(a + b) + c is divisible by 3.Hence, statement II alone is sufficient.

23. e The volume is unknown and the cost of eachindividual element except zinc is unknown. Cost ofcopper cannot be found.

24. d Statement I indicates the time taken by Ravi to eat24 eggs. Statement II indicates the time taken byHarish to finish eating 24 eggs. Together they givethe answer.

25. e From statement I, we get that there are balls of2 colours. This gives the probability of selection of

the ball of other colour as 114

. However, it has no

information about the total number of balls.From statement II, we do not get any informationabout the total number of balls.


1. c Required percentage growth

= (68718 – 42137) × 42137100

Students please note that to calculate the exactvalue of this expression, we need calculator. Sinceoptions given are not very close to each other, wecan approximate the values.And using approximations, we get the value of the

required ratio = (68600 – 42000) × 42000100

= 42

2650

= 63%

2. c Books 1975 1980

Percentage growth

Primary 42137 68718 66%

Secondary 8820 20177 128%

Highersecondary

65303 82175 26%

G raduatelevel

25343 36697 44%

Hence, percentage growth is least for highersecondary books, viz. 26%.

3. b Again referring to the above table, we can see thatthe percentage growth rate is maximum forsecondary level books, viz. 125%.

4. d It can be seen from the given table that thoughprimary level books have shown a consistentgrowth, it has declined in 1978. On the other hand,even secondary and higher secondary level bookshave shown a consistent increase except for 1977when it had declined. But the graduate level bookshave shown a consistent growth over the period.


Page: 49��

For questions 5 to 8: Let us refer these hawkers as J, R,S and P respectively in place of Jayram, Rajaram, Sitaramand Peariram. From information (II), we know that P is sellingChatpata. From information (I), we gather that R do not selleither Fruits or Plastic Toys. So R must be sellingNewspapers. So J and S are selling Fruits and Plastic Toys,not in any particular order.

Regarding their revenue earned, we know thatJ + R + S + P = 1200.

But from information (IV), S = 250.Therefore J + R + P = 950, or R + P = (950 – J) …(i)

From information (III), we can say that

R P50 J

2+ + = …(ii)

Solving (i) and (ii), we get that J = 350 and (R + P) = 600From information (I), we know that R > 300. But if, sayR = 301, then P = 299, which is greater than 250. But frominformation (II), P earned the minimum revenue. So, P shouldearn less than S (Rs. 250).So, we can say that R > 350 and P < 250. We can compilethe following table with this derived conclusion.

Hawkers Jayram Rajaram Sitaram Peariram

ItemsFruits/ Plastic Toys

News papers

Fruits/ Plastic Toys

Chatpata

Revenue earned (Rs.)

350 >350 250 <250

5. b According to the question, S + P = R.We know that S = 250 and P = (600 – R)Solving we get, R = 425. Since Rajaram soldNewspapers, the answer is (b).

6. e We know that either J or S is the Fruit-seller. J solditems worth Rs. 350 and S sold items worth Rs.250. If one orange was sold at Rs. 2.50 per piece,then either 140 or 100 oranges were sold on thatday by one of these hawkers.

7. d The amount to be paid to the hooligans by J and Sare Rs. 35 and Rs. 62.50 respectively. In order tomaximise their total ‘take-home’ revenue, the amountto be paid by R and P should be considered. We canobserve that P has to pay a lesser percentage ascompared to R. So we maximise the revenue earnedby P at Rs. 249, (corresponding revenue earned byR is Rs. 351), to get the minimum amount to be paidto the hooligans. We can compile the result in thefollowing table.

Hawkers Jayram Rajaram Sitaram Peariram TotalRevenue

earned (Rs.)350 351 250 249 1200

Amount to be paid to the hooligans

(Rs.)

35 70.2 62.5 37.35 205.05

Take-home' revenue (Rs.)

315 280.8 187.5 211.65 994.95

So, (d) is the right answer.

8. c As integral number of Chatpatas’ were sold so (600- Rajaram’s salary) should be divisible by 7.Option (c) i.e. (600 –494) = Rs.106 is not divisibleby 7. So Rs.494 cannot be the revenue earned byRajaram on that day.

9. b6

140180 − = 6.67 million tonnes (Approximately)

10. a In 1997, kharif crop production is

11099 108.9

100× = million tonnes

And rabi crop production in the same year is

11681 93.96

100× = million tonnes

∴ Total production increases by

108.9 93.96 180100 12.7%

180+ − × =

11. c Given that 83% equals 110. So 100% should

equal 110100 132.5

83× = million tonnes(Approx.)

So quantity of food grains India should have imported= 132.5 – 110 = 22.5 million tonnes (approximately)



Ferrari was on the podium in 2

rd3

of the races i.e. 12 races.

Hence in (9 + 9) – 12 = 6 races it held both the 1st and 3rdpositions.

McLaren was on the podium in 2

rd3

of the races i.e. 12

races.Hence since 5 + 5 + 2 = 12 in no race did it hold more than 1position.

Renault was on the podium in 2

th9

of the races i.e. 4 races.

Hence it could have held the position (1, 3); (1, 3); (1); (3).Based on this the following chart could be drawn:

Race Ist IInd IIIrd

18 R R

17 R R

16 R M

15 Others M

14 M R

13 M Others

12 M F

11 M F

10 M F

9 F F

8 F F

7 F F

6 F F

5 F F

4 F F

3 F M

2 F M

1 F M

The above chart represents one of the options in the races.Every other option would also satisfy the same condition.Ferrari in the 2nd position can range anywhere from race 1to race 12 but not beyond it (since Ferrari held the podium inonly 12 races)

12. c From the table we can see that there would be 6races when both Ferrari and McLaren woulddefinitely be on the podium.

13. b Renault had podium positions in 4 races.It won 6 podium positions.Hence it is possible only in the following cases:(1, 3), (1, 3), (1), (3). Hence the answer is (b).

14. c The maximum number of times when all the three onthe podium can be from Ferrari is 6. For example, inall races from 4 to 9 the 2nd position is occupied byFerrari.The minimum could be 0, when Ferrari occupies2nd position in races say 1 to 3 and 10 to 12.

15. c Total number of points obtained by Ferrari team= 9 × 10 + 6 × 9 + 9 × 8 = 216McLaren team = 5 × 10 + 2 × 9 + 5 × 8 = 108Renault team = 3 × 10 + 0 × 9 + 3 × 8 = 54Let’s assume the remaining podium positions werewon by a single team; Y, thenY = [18 – (9 + 5 + 3)] × 10 + [18 –(6 + 2 + 0)] × 9 +[18 – (9 + 5 + 3)] × 8= 1 × 10 + 10 × 9 + 1 × 8 = 108∴ The minimum difference in points between thetop two teams = 216 – 108 = 108

16. a From the table it can be seen that all the three cannever be simultaneously on the podium.

For questions 17 to 20: Suppose that the selling price isRs. 100.

17. a 25% (of 30% (of 20%)) = 1.5%.

18. c 6 lacs = 0.6 million

19. a Appraisal cost = 6 million⇒ Product Testing cost = 3 million.

20. b Error cost = 20% of quality cost

= ( )×120% of total cos t

5

= × =1 14%

5 5 of total cost

21. c Statements I and II both individually give a relationbetween the capacities of water barrels A and B.

22. c From statement I, we easily get the answer.Since we know both the distance and speed.

From statement II, x + 50 t2

→

x → t

xt = (x + 50)t2

= 45

∴ xt = 45

4525t 45

2+ =

∴ t = 54 minThus, we can find time from statement II alsoindependently. Thus, (b).


Page: 51��

23. e From statement I, 1 1 1A C 8

+ =

From statement II, 1 1 1B C 6

+ = . Here we have two

equations and three unknowns. Hence, we needone more equation.

24. a Statement I clearly shows that x is either zero ornegative, i.e. not positive. We do not get anyinformation from statement II. Thus, (a).

25. c From statement I, the SP of a pair can be obtainedwhile the CP can be obtained from statement II.


1. d All are true.

2. e New demand is 114% of 175000 = 199500 pairs.So demand supply gap= 199500 – 175000 = 24500 pairs

3. a Actual demand in 1995-96 is 110% of 190= 209000 pairsActual production in 1995-96 is 85% of 190= 161.5 thousand pairsSo (Demand – Supply) = 209 – 161.5 = 47.5 × 1000 = 47500 pairs

For Questions 4 to 7:

4. c If B’s speed is 20 mph.Then A’s speed = 30 mph C’s speed = 20 mph D’s speed = 80 mphRatio of their speeds = 3 : 2 : 2 : 8

5. a At that instant A, B, C and D passed through thepoints P, Q, R and S. As they passed these points,all of them were moving parallel to each other. Lettheir speeds be Va, Vb, Vc and Vd respectively.At that instant A & B were moving in the samedirections (both clockwise), so we must have.Va + Vb = 50 …(i)Similarly C and D also moved in the same directions(either both in the clockwise direction or both in theanticlockwise direction) and their speeds withrespect to B are 40 mph and 60 mph respectively.

Case I:

R

Q

P

S

V c

V a

V b

V d

F ig . C & D , b oth anti-c lock w ise

If both C & D are moving in the anticlockwisedirection, then

Vc – Vb = ± 40 … (ii)Vb + Vd = 60 … (iii)Adding (i) and (ii)Va + Vc = 90 or 10 … (iv)Adding (iii) and (iv)Va + Vb + Vc + Vd = 150 or 70Therefore, the average of the speeds of these cars

is: 150

4 0r

704

i.e. 37.5 mph or 17.5 mph

Case II:

R

Q

P

S

V c

V a

V b

V d

If both C & D are moving in the clockwise direction,thenVd – Vb = +60 …(v)Vb + Vc = 40 …(vi)

Solving, we get the value of the average of thespeeds of the four cars i.e. 37.5 mph.Hence (a) is the only possible choice.


6. b C observes that D is moving at a speed of 20mph.As C and D are moving in the same direction wemust have:Vc + Vd = 20From this equation, the maximum values of eitherVc OR Vd can be 20mph only irrespective of thedirection of C & D ...(i)Again, there are two cases:

Case I: Both C and D drive in the clockwise direction.Vb – Vc = +40 …(ii)Vb + Vd = 60 …(iii)

Case II: Both C and D drive in the anticlockwisedirection.Vb – Vd = 60 …(iv)Vb + Vc = 40 …(v)Case II is impossible.

As all the speeds are either positive or zero,equations (ii) and (iii) must be changed as following:Vb – Vc = 40 or Vb = Vc + 40 …(vi){since Vc ≤ 20}Vb + Vd = 60 or Vb = 60 – Vd …(vii){since Vd ≤ 20}From both equations, we get40 ≤ Vb ≤ 60.Out of the given options, only option (ii) lies in thegiven range. Hence (b) is the correct choice.

7. d As E is stationary, the speeds observed by E, arethe actual speeds of the cars. Let both the driversC and D were driving with a speed of V mph. As Cand D have the same speed, we need not find thesolution under different cases as we did in theprevious problems. We can write the followingequations:

Va — Vb = ± 50 …(i)Vb + V = 60 …(ii)V — Vb = ± 40 …(iii)Adding (ii) and (iii) we get:V = Vc = Vd = 50 or 10 …(iv)Subtracting (iii) from (ii) we get:Vb = 10 or 50 …(v)Putting both these values of Vb in (i), we get

Va = 50 ± 50 or 10 ± 50 i.e.Va = 100, 0 or 60, – 40As Va represents speed, it cannot be negative,hence Va = –40 is discarded.Va = 100, 0 or 60 …(vi).From (iv), (v) and (vi) we observe that Vb and Vado not have unique values rather they have morethan one allowed value. As a result (Va, Vb, Vc,Vd.) can have any one of the following combinationsof values:

Case I: (Va, Vb, Vc, Vd.) have values (0,50, 10,10)ORCase II: (Va, Vb, Vc, Vd.) have values(100,50, 10,10)ORCase III: (Va, Vb, Vc, Vd.) have values(60, 10, 50,50)Options (a), (b) and (c) are not correct. If we checkthe options, only option (d) is correct. As V

b = 10

and Va = 60 (more than double) is a possible

combination of speeds.

8. b IT has the lowest acquisition value US $21.6 m.

9. b For 2001 - 03, average acquisition

cost $ 1.6 b

$ 13.3 m120

= =

So 3 companies were acquired at lower thanaverage cost - Expert Information Services,Alpharma, Dashiqiao.

10. d The Australian companies were acquired at a highervalue as compared to the others,Total = US $ 59.5 mNext is USA = US $ 39. 8 m, then UK and then China.

11. e London Stock Exchange has more than 15companies listed. But how many exactly is notknown. Also, a company may be listed in more thanone international stock exchange.So we have no data to confirm that these more than25 companies are all different.

12. c Top 3 acquisition value = US $ 386.4 mThe total acquisition value for top 9 acquisitions inthe period = US $ 461.5 m.Hence ratio = 5 : 6

For questions 13 to 15:The following table can be made:

Revenue Profit % Profit Revenue Profit % Profit

A 4600 15 600 6720 12 720

B 3900 30 900 5000 25 1000

C 2700 35 700 3840 28 840

D 6600 10 600 5000 25 1000

E 7700 10 700 8800 10 800

2004 2005

(all figures in rupee crores)

By the given information, A is Dixons and D is Eletropaulo.

13. a Here B is Coleco. The second highest profit in 2004is Rs.700 crore. So both Alta Vista and Bultaco isthe answer.


Page: 53��

14. b Here C is Alta Vista. The highest profit earned byany Company in 2005 is Rs. 1000 crore.So, from the choices, Coleco and Electropaulo isthe answer.

15. d % increase in profit over 2004 of A:

= × =120100 20%

600% increase in profit over 2004 of B:

= × =100100 11.11%

900% increase in profit over 2004 of C:

= × =140100 20%

700% increase in profit over 2004 of D:

= × =400100 66.67%

600% increase in profit over 2004 of E:

× =100100 14.28%

700Here E is Bultaco.So C is either Coleco or Alta Vista.Hence, the data is insufficient.

For questions 16 and 17:A total of 5 different arrangements is possible

1 3 2 5 4 6

1 4 2 5 3 6

1 4 2 6 3 5

1 5 2 4 3 6

1 6 2 4 3 5

X X X X X X

X X X X X X

X X X X X X

X X X X X X

X X X X X X

− − −− − −− − −− − −− − −

16. b From the various possible arrangement shownabove, there are two possible arrangements for(X1, X4).

17. b If the pairs (X1, X3), (X1, X5), (X1, X6), (X2, X6)or (X4, X6) are given, then all other pairs can bedetermined,while if the pairs (X1, X4), (X2,X4), (X2,X5), (X3,X5)or (X3, X6) are given, then all other pairs can’t bedetermined.Therefore required probability

Total number of favourable cases

Total number of possible cases= 5

10=


Designation Manager Assistant Manager

Executive

Salary 15x 12x 8x

Name Anny Bobby Citra

Expenditure 5y 4y 3y

18. d Anny does not draw the maximum and the Executivedoes not spend the minimum. Hence, Anny is notthe Manager and Citra is not the Executive.

Anny Bobby Citra

Manager ×Assistant Manager

Executive ×

Now if we analyse the options,Option (a) is wrong, since Citra cannot be theExecutive. Hence, our conclusion can never bereached.Option (b) says that Bobby is the Manager. SinceCitra cannot be the Executive, she has to be theAssistant Manager and therefore our conclusion iscontradicted.Option (c) says that Citra is not the AssistantManager. That means Citra is the Manager. Now,either of Anny or Bobby can be the AssistantManager. Hence, the information in this option is notsufficient for us.Option (d) tells us that Bobby is the Executive. Since,Anny cannot be the manager, she must be theAssistant Manager. So this option is the correctchoice.

19. c Manager spends the maximum. Hence, Anny is theManager.Since Bobby is the executive, hence Citra must bethe Assistant ManagerConclusion says that Citra saves the maximum.Savings = Salary – Expenditure


Executive

Name Anny Citra Bobby

Salary 15x 12x 8x


Savings 15x – 5y 12x – 3y 8x – 4y

Bobby’s savings = 8x – 4yBobby earns less than Citra and spends more thanher.


So, her savings are certainly lesser than that ofCitra. Since, according to the conclusion drawn,Citra saves the maximum, we have

(12x – 3y) > (15x – 5y) ⇔ 2y > 3xTherefore, if with the help of certain additionalinformation, we can establish 2y > 3x, that will bethe right answer choice.We have logically deduced that Anny is the Managerand Citra is the Assistant Manager. So, options (b)and (d) can be out rightly rejected.In option (a), Anny’s savings is twice herexpenditure. Therefore, Anny’s salary must be threetimes her expenditure.⇒ 15x = 3 × 5y⇒ x = y and hence Citra is not saving the maximum.In option (c), Citra’s salary is twice her savings.Hence, her salary is also twice her expenditure.⇒ 12x = 2 × 3y⇒ y = 2x⇒ 2y > 3x and hence this information is correct aswell as sufficient to lead us to the stated conclusion.So, option (c) is the right choice.

20. a Citra’s savings = Citra’s expenditure = 12

× Citra’s

salaryCitra’s savings is 25% less than that of AssistantManager. Therefore, Citra is not the AssistantManager.


Executive

Salary 15x 12x 8x

Citra’s expenditure = 3y

Case I: Case II:Citra is the Manager Citra is the Executive

⇒ 3y = 12

× 15x 3y = 12

× 8x

⇒ 2y = 5x 3y = 4x

If Citra happens to be the Manager,Then, Executive’s savings is either equal to(8x – 4y) or (8x – 5y), which can never be negative.But, if Citra is the Manager, then 2y = 5x.Therefore, both (8x – 4y) and (8x – 5y) becomenegative quantity, which is not possible.Hence, Citra has to be the Executive.∴ Citra’s savings = 8x – 3y = 6y – 3y = 3yThis saving is 25% less than the savings of theAssistant Manager.Hence, Assistant Manager must be saving 4y.⇒ 12x – Assistant Manager’s expenditure = 4y⇒ 9y – Assistant Manager’s expenditure = 4y⇒ Assistant Manager’s expenditure = 5y

Therefore, Anny is the Assistant Manager and thus,Bobby is the Manager.

Designation Manager Assistant Manager Executive

Name Bobby Anny Citra

Salary 15x 12x 8x


29 29 16Savings x or y x or 4y 4x or 3y

3 4 3

Conclusion says that savings of Executive is Rs.12,000 per month. Hence, 4x = 3y = 12000. Assumingthe stated conclusion to be true, we can compilethe following table.

Designation ManagerAssistant Manager

Executive

Name Bobby Anny Citra

Salary 45000 36000 24000

Expenditure 16000 20000 12000

Savings 29000 16000 12000

Option (a) says that Executive spends Rs. 12000per month. This enables us to find the statedconclusion.Option (b):Bobby is the Manager but monthly salaryof Citra, if Rs. 36000 per month would mean that 8x= 36000 and hence, x = 4500 (misleading)Option (c): Anny is the Assistant Manager but if shesaves Rs. 20000 per month, 4y = 20000 and hence,y = 5000 (misleading)Option (d): Manager’s salary = 15xAnny’s expenditure = 5yIt gives x = 1600 (misleading)So, option (a) is the correct choice.

21. d From the first condition the number can be977988

From the second condition we get112233 :

99∴ From I and II combined we get number as 88.∴ Answer is (d).

22. a Statement I implies that y is odd since the only wayfor the average of x and y to be a whole number isthat x + y must be even. Statement II is not sufficientsince either y or y + 1 must be even.Hence, x + y + 1 must be even.

23. e Since the ball that has been picked up is notidentified, the probability cannot be found.


Page: 55��

24. e Statement I gives probability of A winning but herewe do not know the number of contestants.Statement II provides the information that there areonly 2 contestants but we do not know the probabilityof a tie. Thus, answer is (e).

25. e In statement I, 2500 includes families who haveneither of the devices and no data is given aboutsuch families. Hence, even statement II is not useful.


1. b The most obvious one is rank number 5. Since Hhas scored less than each of A, B, C and J in all thethree sections, its total also has to be less thanthese four students. By the same reasoning, totalmarks of H have to be more than D, E, F, G and I.Thus H is ranked 5th.

D’s is definitely more than the total of F and I (as Dhas scored more in each of the sections comparedto these students). Also D cannot be less than G(worst case scenario for D, as compared to E, isQuant -5, EU +4 and DI +2). Similarly D cannot beless than E. Thus 6th rank will be D.

The other close option will be rank 10. Howeverone cannot conclude that F will come last. E and Fboth could be tied for the rank 9 or 10. Considerworst case scenario for E and best case scenariofor F. Thus E will score 8 more than F in EU and 2more than F in DI. But F can make up for thisdifference by scoring 10 more than E in quant. Thusfor rank 10, one cannot identify the student.

Among, A, B, C and J, J could be at any of the ranks1 to 4 i.e. J could be greater or less than each of A,B and C. Hence one cannot identify who ranks 1 to4.

By same logic between E, F, G and I, student Ecould be at any position from rank 7 to 10. Thus thestudent at rank 7 to 10 also cannot be identified.

2. a The minimum difference between marks scored byF and C in quant can be 10 and this will happen onlywhen the difference between consecutively rankedstudents (F & D; D & G; G & H; H & B; B & C) is 2.

3. e The minimum marks that A can score is 19 in Quant,20 in EU and 22 in DI i.e. a total of 61. The maximummarks that A can score is 25 in Quant, 23 in EU and25 in DI i.e. a total of 73. Thus the required differenceis 12 marks.

4. e As explained in answer explanation to first questionof this set, the student at rank number 7 cannot befound as E could be ranked anything from 7 to 10.

5. a The total sales over the period shown for differentmanufactures are given as follows:(a) For A = 440 + 480 + 470 + 500 + 520 + 510 = 2920 × 1000 units(b) For B = 400 + 410 + 415 + 415 + 420 + 430 = 2490 × 1000 units(c) For C = 380 + 390 + 390 + 400 + 420 + 495

= 2475 × 1000 units(d) For D = 360 + 380 + 400 + 415 + 440 + 500

= 2495 × 1000 units(e) For E = 480 + 440 + 440 + 420 + 425 + 435

= 2640 × 1000 unitsThus, the sales of the manufacturer A is the highest.

Alternative method:‘A’ sales has been more than all other manufacturerin all the years except 1995.

6. d Required share

= %5.19100440380390410480

410 ≈×

++++

(Observe that the sum total of the numbers in thedenominator would be higher than ‘five times 410’and hence, the required percentage would be alittle less than 20%.)

7. c It can be easily observed that, since the differenceis the highest in case of manufacturer C, the largestpercentage growth would naturally occur for him,as the base is the smallest.

8. e Required ratio = (400 ÷ 500) = 0.8

9. d The figure for the highest sales of scooters overthe period shown is 520 units which occurs in 1999in case of the manufacturer A.

For questions 10 to 12:Only those cities that have three roads emanating fromthem can be start/end city.

10. b

11. b Three routes starting from 2 and 4 each.

12. d

13. c Total number of line employees with 3 to 5 years ofemployment = 140Out of the above, the number that attended trainingon financial management = 80∴ The number of line employees with 3 to 5 yearsemployment who did not attend training on financialmanagement is (140 – 80), i.e. 60.


14. b Number of employees with less than 3 years ofemployment who attended training on decision-making alone = (40 – 10) + (30 – 15) = 45Similarly, the number of employees with less than3 years of employment who attended training onfinancial management alone= (30 – 10) + (20 – 15) = 25Answer = 45 + 25 = 70.

15. a Number of line employees with more than 5 yearsof employment who attended at least oneprogramme = 50 + 40 – 30 = 60Number of staff employees with more than 5 yearsof employment who attended at least oneprogramme = 40 + 50 – 20 = 70Total number of employees with more than 5 yearsof employment who attended at least oneprogramme = 60 + 70 = 130Percentage of employment with more than 5 yearsof employment who did not attend either workshop

= 64100360230

100160200

130)160200( =×=×+

−+

16. e (50 + 40) + (40 + 50) – (30 + 20) = 130

17. d Middle management use highest percentage ofspread sheet, which is 53 out of 150

%3.3510015053 =×=

18. e Option (e). Answer is 202.

19. b Top management have second highest proficiencyin spread sheet. Just check the second highest

value from the four of those given in the table.

20. a 42 + 23 = 65

21. d From statement (A), he cannot watch a newschannel at night.From statement (B), he can watch an entertainmentchannel only in the evening and night (combiningwith statement A).Also note, he watches only one type of channel ineach of 4 parts of the day. Hence (d).

22. b Statement (A) is irrelevant. It only says Pakistan isthe eventual winner. But nothing is said about points.Statement (B), explains the point system.Based on this, the maximum points possible to bescored by the winning team is (3 × 3 matches) = 9points. Hence (b)

23. b Statement (A) indicates that the chief guest spokelonger than 25 min but does not say how muchlonger than 25 min. So statement (B) indicates thatthe chief guest spoke for lesser than 35 min.Hence, we can infer from statement (B) alone thatthe chief guest did not speak for more than 45 min.

24. e It is not possible to answer the question. Although(c) might seem to be a tempting offer, it is not,because we do not know whether Rashmi will goshopping if Ranjana goes.

25. d Both the statements are required, since from thestatement (A), we get

A B CA

3+ +> or 3A A B C> + + or B C

A2+>

From the statement (B), we get

A BB

2+> or 2B > A + B or B > A

Combining both the statements, we get

B CA B

2+ < <

or B C 2B+ < or B > C∴ ‘B’ weighs most.Alternately: Statement (A) says there has to be atleast one of B or C or both B and C less than A.Statement (B) says B is more than A.So combining we can say the order has to beC, A, B.


1. b75.68

75.6837.79 − 80 70 114.28%

70 7−− = =�

2. e Demand = 1.15 × 102 = 117 MMTGap = 117 – 98 = 19 MMT

3. e It can be checked by multiplying the All India supplyfigures by 0.3 and comparing it with southern Indiasupply figures. No year satisfies that.

4. b Supply = 22 × 1.3 = 28.6 MMTDemand = 28 × 1.4 = 39.2 MMTDeficit = 10.6 MMT


Page: 57��


Each of the participants received at least one vote in round1. If the minimum number of votes received is 2, then 4 of the11 votes are accounted for, since two of the contestantswere tied for the last place in Round 1. Payal has received4 votes. Taking these votes into consideration, 8 votes areaccounted for. The remaining 3 votes can be divided amongthe other two participants as either (3 + 0) or (2 + 1), both ofwhich are not possible. (3 + 0) is not possible becauseeach participant has received at least 1 vote and (2 + 1) isnot possible because we have considered the lowest num-ber of votes as 2.

If the minimum number of votes received by two partici-pants is 3 each, then including Payal’s 4 votes, 10 out of 11votes would have been accounted for and therefore theremaining two participants cannot receive at least 1 voteeach.

Therefore, the only possible combination is when twoparticipants receive 1 vote each (the minimum), Payalreceives 4 votes while the other 2 participants receive 3and 2 votes respectively.

Further, one of the participants has received ‘0’ votes inround 2.i. That participant cannot be Priti because she has re-

ceived 1 vote in round 2.ii. That participant cannot be Priyanka because Mr. Biyani

has voted for her in round 2.iii. That participant cannot be Payal because the judge

who voted for Poonam in round 1 voted for her inround 2

iv. That participant cannot be Pooja because 50% of thejudges who voted for Payal in round 1 voted forPooja in round 2.

v. Therefore, it is Poonam who got ‘0’ votes in round 2.

Further, it is given that the judge who voted for Poonam inround 1 voted for Payal in round 2. Therefore, Poonamwould have got 1 vote in round 1. Also, Payal would havegot 3 votes in round 2. (50% of votes from earlier round and1 vote of the judge who voted for Poonam in round 1).

The remaining 8 votes were divided between Pooja andPriyanka. For this to be possible and Priyanka to be jointsecond with another person, the only possible combinationin round 2 can be:Priti : 1 vote, Payal : 3 votes, Priyanka : 3 votes, Pooja : 5votesFrom condition II, Pooja got 2 additional votes in round 2.Therefore Pooja would have got 3 votes in round 1. Priyankagot 1 additional vote of Mr. Biyani in round 2 and ended with3 votes. Therefore Priyanka would have got 2 votes inround 1.

The number of votes received after the first 2 rounds wereas follows:

Pooja Payal Priti Priyanka Poonam

B'lore Delhi B'lore

Round 1 3 4 1 2 1

Round 2 5 3 1 3 0

Since Poonam was did not contest in round 3, Priyankamust be the other girl from Delhi. Further, the total number ofvotes in Round 3 is 13. (Poonam will also vote). Fromcondition IV, the total votes won by Pooja and Priti (two girlsfrom Bangalore) will be 7 while Payal and Priyanka (twogirls from Delhi) together secured 6 votes in Round 3.

5. b If Priyanka received 2 votes in round 3, then Payalwould have received 4 votes in round 3, becauseboth of them together received 6 votes in round 3.

6. e Pooja was the person with the highest votes at theend of round 2.

7. d If Priti received 3 votes in round 3 and Payal re-ceived 50% of the remaining votes which is 5, thenPooja received 4 votes in round 3 while Priyankareceived 1 vote in round 3. Therefore, option (d) isdefinitely true i.e. Priyanka received the minimumnumber of votes in round 3.

8. d Priti and Poonam are the 2 contestants who re-ceived the minimum number of votes (1 each) inround 1.

For questions 9 to 11:Iqbal is watched by Bimal only. Beckham watches 3 movies— Hence, does not watch Iqbal which is screened at plaza.

Movie Salaam Namaste Iqbal

Movie Hall Satyam Chanakya Priya Plaza

Bunty � � � �

Babli � � � �

Bonny ��

Bimal � � � �

Beckham � � � �

9. c

10. e

11. e


12. a Net revenue by the sale of scooters

= 48

100237841× . = 11416.368 million

Number of scooter units sold = 60

1307211100

×

= 784326.6

Revenue per scooter6.7843261037.11416 6×=

= Rs. 14,555.63 ≈ Rs. 14,500.

Alternative method: Net income of scooters

= 24000 × 50

100 ~ 12000

but it has to be less than 12,000, because both theapproximate values are higher. So let us take11,500 million.

Total scooters sold = 1300000 × 60

100 = 780000

Net realisation = 611500 10

780000×

~ 14500

13. a Take total sales volume as 100, then motorcyclesales volume = 2225% increase = (25 × 22) ÷ 100 = 5.5This 5.5 is the decrease in scooter sales volume⇒ Percentage decrease in sales volume= (5.5 × 100) ÷ 60 = 9.1%

14. d If others is excluded, then of a total of 100, netrevenue left = 89Of this, 3 is from mopedsHence, percentage of mopeds

= 3

100 3.4 % approximately89

× =

15. c Net profit

= 20 48 17 25 21 3

23784.1million100 100 100 100

+ + + × = (0.13 + 0.06) × 23,784.1 = Rs. 4,500 million

16. a Net revenue per vehicle would be the highest forthree-wheelers. (The ratio of percentage revenueand percentage volume sales is greater than 1,only for three-wheelers.)

For questions 17 to 20: From the data that is given, wecan find the following data (the explanation of how thefollowing values were arrived at is given after the table):

Item 1999-2000 2000-01

Food (percentage) 22% 23%

Food (value) 4928 5934

Manufactured articles 11648 11352

Raw material 5824 8514

Total value of exports Rs. in crore 22400 25800

17. d Food related exports in 2000-01= 0.23 × 25800 = 5934So food related exports in 1999-00= (5934 – 1006) = 4928Hence, percentage of food related exports in

1999-2000 = 22100224004928 =×

18. b In 1999-2000, the value of manufactured articlesand raw materials export = Rs. (22400 – 4928) = Rs. 17472 croreSince export in manufactured goods is twice thatof raw materials, Rs. 17,472 has to be divided in theratio 2 : 1, viz. export of manufactured goods = Rs.11,648 crore and raw materials = Rs. 5,824 crore.Hence, the difference between raw materials andfood = Rs. (5824 – 4928) crore = Rs. 896 crore.

19. d In 2000-01, the combined percentage ofmanufactured articles and raw materials = 77 andthis is in the ratio 4 : 3.Hence, percentage of manufactured articles export= 44% and that of raw materials export = 33%Hence, the value of manufactured = 0.44 × 25800= Rs. 11352 crore and the value of raw materials= Rs. 8514 crore.Hence, percentage of difference between the valueof raw materials in-between 1999-2000 and2000-01

= Rs. (8514 5824)

8514−

= 31.6%

20. a The change in the value of exports from 1999-2000to 2000-01 = Rs. (11648 – 11352) crore= Rs. 296 crore.

21. e Combining (I) and (II):The only information we have is about the sittingpositions of the highest and the lowest agedpersons. Nothing can be said about theoccupation of chairs 2, 3 and 4. Hence, wecannot find the age of C.

22. a Statement (I) alone:A → QC → SB →E →D → TB does not visit T, and E obviously cannot visit T.So D will surely visit city T.

23. e Statement A is not sufficient to answer the givenquestion. Statement B indicates the days on whichthe club needs to be run but no information isprovided regarding saturday and sunday.


Page: 59��

24. e Statement A gives the number of users in Decemberbut does not indicate the volume. Statement Bindicates some idea on the volume reduction andthe number of maximum users. But, even together,they do not give any answer to the volume of waterin the pool in the particular month of December.

25. b From statement A,

R ahul

Kamal

G aurav

4

2

∴ Obviously, Rahul is older than Gaurav by 2 years.Statement B does not give any information.∴ The answer is (b).


1. a Interest earned = 7201 – 12 × 5 × 100 = Rs. 1,201

2. d 2:9~12015409

600072011200017409

year5onInterestyear10onInterest −=

−−=

3. b Interest on 250 monthly investment for 20 year= 130991 – (12 × 20 × 250) = Rs. 70,991Interest on 500 monthly investment for 10 year= 87047 – 60000 = Rs. 27,047How much less would be earned= 70991 – 27047 = Rs. 43,944

Alternative method:The total amount invested in both the cases will besame as half the amount is paid for twice the time.Thus difference in earnings will be the difference inamounts, i.e. 130991 – 87047 = Rs. 43,944

4. b Very clearly the growth has been minimum in

1993-94 = 732

= 2.74%

5. b Profit of banks in 1991-92 = 3% of 630 = 18.9Profit of mutual fund in 1992-93 = 5% of 710 = 35.5Difference = 16.6

6. d Profit to sales is maximum in 1994-95. It is 10%,otherwise in all the year, it is less than that.

For questions 7 to 10: The following table can be drawnas per the information given.

P Q R S T 4 2 A ��

B ��

C 1 D ��

E ��

Since S cannot be supplied by 4 vendors (Q is being suppliedby 4 vendors and each component is supplied by differentnumber of vendors), hence S has to be supplied by all 5.Thus C who supplies just 1 component, supplies S. Furthersince Q is supplied by 4 vendors, A and B have to supply itas well. Thus table now looks like:

P Q R S T (2+) 4 5 2 A ��

B ��

C 1 �� D ��

E ��

Since P is supplied by at least 2, R has to be supplied by just1 vendor and P has to be supplied by 3 vendors. Nothingrow wise, D is the only vendor who can supply 2 parts (asall other supply more than 2). Thus table now looks as:

P Q R S T 3 4 1 5 2 A ��

B ��

C 1 �� D 2 �� E ��

Now it is obvious that P is supplied by E and thus E supplies4 components. Only A can supply all 5 components andhence, B supplies 3. Thus final table is:

P Q R S T 3 4 1 5 2 A 5 ��

B 3 �� C 1 �� D 2 �� E 4 ��

Therefore, 7. a 8. c 9. d 10. a


For questions 11 and 12:

11. b Given: B > L …(i)To be concluded: I > RIn option (a), B > I and L > R. Combining it with (i), B> L > R and B > I. Therefore, conclusion cannot beconfirmed for sure.In option (b), B < I and L > R. Combining it with (i),I > B > L > R. Therefore, conclusion has beenconfirmed.In option (c), B > I and L < R. Even if we combine itwith (i), our conclusion cannot be confirmed.In option (d), B < I and L < R.Combining it with (i), L < B < I and L < R.Therefore, conclusion cannot be confirmed.

12. b Here, x2 > x⇒ x2 – x > 0⇒ x(x – 1) > 0⇒ either x > 1 or x < 0Our conclusion, which states that x < 0 would betrue only when we come to know that otherpossibility, i.e. x > 1 is not the case.For that matter, if we know that x is a real numberless than 1, we can say that x is not greater than 1and hence, x < 0. So option (b) would be the correctchoice.

13. b 1970-71 = 4.11.3

59.09.29.16

42.3+

−+

~ 0.17 – 0.13

= 0.04

1973-74 = 1.6

73.04.23

22.4 − ~ 0.18 – 0.12 = 0.06

1974-75 = 2.7

78.01.27

2.5 − ~ 0.19 – 0.11 = 0.08

1975-76 = 1.7

84.02849.5 − ~ 0.19 – 0.12 = 0.07

So, it is highest in 1974-75.

14. c Or ratio of gross cropped area to consumption offertilizers should be highest

1970-71 = 174.8 174.8

1.11 3.42 0.59 5.12=

+ +

1971-72 = 51.5

1.173; = 177

1972-736.06

1974-75 = 71.7

4.180; 1975-76 =

22.88.187

So, obviously, 1970-71 is answer.

15. e In 1972-73 = 23.20 + 32.77 = 55.97In 1973-74 = 24.00 + 34.20 = 58.20More = 2.23 (Remember these figures are cumulativefigures)

16. c If you look at the trend, area under irrigation hasbeen increasing over the year in minor as well asmajor. Only in 1974-75, in case of minor, it hasdecreased which suggests that some minor areahas came under major.

For questions 17 to 20: Australia won all 5 and Hollandlost all 5. For India or Germany, 7 points can be scored as{W, W, D, L, L} or {W, D, D, D, D}. But since Australia defeatedboth of them, the only possibility is {W, W, D, L, L}.For Pakistan also, there has to be at least one W and at leastone L⇒ Possibility = {W, D, L, L, L}Total W = Total L⇒ South Korea won 3 matches and lost one.Table is:

P W L D Points

India (Ind) 5 2 2 1 7

Germany (Ger) 5 2 2 1 7

Australia (Aus) 5 5 0 0 15

Holland (Holl) 5 0 5 0 0

South Korea (SK) 5 3 1 1 10

Pakistan (Pak) 5 1 3 1 4

Win Loss Draw

Aus � � �

Holl � � �

Pak Holl Ind + SK + Aus Ger

SK Holl + Pak + Ger Aus Ind

Ger Ind + Holl SK + Aus Pak

Ind Pak + Holl Ger + Aus SK

The matches between India and Germany as well as SouthKorea and Germany did not end in a draw. Hence, Pakistanand Germany match must have been a draw.⇒ India defeated Holland and Pakistan only.

17. b The result between India and Germany must be awin in favour of Germany.

18. a 10

19. e 7 + 7 + 15 + 10 + 4 = 43

20. c There were 2 matches which added in a draw inthe whole tournament.

21. c Statement A indicates Bunty gets the middle mostsequence. Statement B indicates Bunty is the lastone in sequence.Hence each statement individually is sufficient.

22. d Statement A gives the number of hours Mr. Sharmawalks on weekdays and Statement B indicates thenumber of hours he walks on Sunday.


Page: 61��

23. d From statement (A),since ‘D’ always lies, B > W.From statement (B),since ‘c’ always speak truth, c = White hat.∴ From statements (A) and (B), ‘A’ and ‘B’ have towear black hats.∴ The answer is (d).

Persons A B C D

Hat B B W W

24. e Both the statements are not sufficient since ‘A’, ‘B’,‘C’ positions can be any of these.

C

25

A B

15

B

A C15

25

15

CAB

25

And many such positions are possible.The answer is (e).

25. d Both the statements together give the lengthXY = 10 cm.

Practice exercise – D14For questions 1 to 5: It is given that each member represents one of the five schools. Hence, the only logic which needsto be understood is that if a particular school has won 1 red medal, the individual member winning 2 red medals cannotbelong to that school. On this inference, it could be seen that Qadir, Rishi and Yousuf have 2 red medals each. Thus, theycannot be from Lucknow school or Ahmedabad school, because their respective cumulative red medal count is less than2. On a similar reasoning, we can limit the possibilitiesThe following table illustrate the possible linkage between the individual and the schools the represent.

Red Blue White Green De lhi Indore Lucknow Kolkata Ahm edabad

Puneet 1 1 1 1 ×

Qadir 2 0 0 1 × ×

Rishi 2 0 0 0 × ×

Satyam 0 1 3 0 × × ×

Tarun 0 1 1 2 × ×

Umesh 0 1 2 2 × × ×

Vipul 0 1 0 1

Wasim 0 0 2 1 ×

Xavier 0 2 1 0 × × ×

Yousuf 2 0 1 0 × ×

Zaheer 1 1 1 0 ×

M em bers

M edals Poss ible School

Till here, it can be seen that Satyam could be from Delhi school or Indore school. Let us assume that Satyam is from Indoreschool. Thus, his blue and white medals account for the entire tally of blue (1) and white (3) medals of Indore school. Theremaining 2 Green and 2 Red medals of Indore school must have been won by other member(s) representing Indoreschool. These other member(s) surely cannot win any Blue or White medal.

For any other member of Indore school, blue and white medal count should be 0 (only Qadir and Rishi). But even both ofthem combined cannot equal the green medal count of Indore school. Hence, this case is not possible and our assumptionthat Satyam is from Indore school is wrong. Therefore, Satyam must be from Delhi school.


Putting Satyam to Delhi school, the remaining medal tally of all the schools is as follows

Red Blue White Green

Delhi School* 3 1 1 1

Indore School 2 1 3 2

Lucknow School 1 3 2 1

Kolkata School 2 1 1 2

Ahmedabad School 0 1 2 2

SchoolsMedals

*(without the medals won by Satyam)

Now Xavier, who has won 2 blue medals, must be from Lucknow school. The remaining medal tally is


Delhi School 3 1 1 1


Lucknow School* 1 1 1 1



SchoolsMedals

*(without the medal won by Xavier)

Now, Lucknow’s one remaining red medal could come from Puneet or Zaheer. But Zaheer has won 1 each of red, blue andwhite medals and hasn’t won any green medal. If Zaheer is from Lucknow school, it won’t get just one green medal fromsome other member, because none of the members has won just one Green medal.Hence, Puneet is from Lucknow school. This completes the member list of Lucknow school.

The remaining medal tally:




Lucknow School* 0 0 0 0



SchoolsMedals

*(without medal won by Xavier & Puneet)

Delhi school, which has got 3 red medals, must have got 1 red medal from a member and 2 red medals from anothermember. Only remaining member having 1 red medal is Zaheer. He must be from Delhi school. Remaining medal tally:


Delhi School* 2 0 0 1





SchoolsMedals

*(without medal won by Satyam & Zaheer)


Page: 63��

Eliminating the ones whose schools have been confirmed and forming the table once again,

Red Blue White Green Delhi Indore Lucknow Kolkata Ahm edabad

Qadir 2 0 0 1 × ×

Rishi 2 0 0 0 × ×

Tarun 0 1 1 2 × ×

Umesh 0 1 2 2 × × ×

Vipul 0 1 0 1 ×

Wasim 0 0 2 1 × ×

Yousuf 2 0 1 0 × ×

M em bers


Only Qadir or Rishi could be from Delhi school, because remaining blue and white medals for Delhi are 0. If Rishi is from Delhischool, one remaining green medal for this school would not be possible. Hence, Qadir is from Delhi school. This completesthe member list for Delhi school also.

The possible schools for the remaining members would further reduce:







SchoolsMedals

Red Blue White Green De lhi Indore Lucknow Kolkata Ahm edabad

Rishi 2 0 0 0 × × ×

Tarun 0 1 1 2 × ×

Umesh 0 1 2 2 × × ×

Vipul 0 1 0 1 × ×

Wasim 0 0 2 1 × × ×

Yousuf 2 0 1 0 × × ×

M em bers


If Tarun is from Indore school, remaining red and white medal tally for Indore school is 2,2 respectively. It could be seen thatnone of the remaining members have such a medal count. Even after combining the medal tally of two or more students,Red-White 2-2, combination cannot be formed. Hence, Tarun is not from Indore school. Tarun is not from AhmedabadSchool either, because in that case only single White medal remains for Ahmedabad School, which is not available to anyStudent. Instead, he is from Kolkata school. Other member from Kolkata school could be Rishi only. This makes Yousuf amember of Indore school, because he is the only one remaining with 2 red medals.The possible schools for the remaining members would further reduce as follows:







SchoolsMedals


Red Blue White Green Delhi Indore Lucknow Kolkata Ahmedabad

Umesh 0 1 2 2 × × ×

Vipul 0 1 0 1 × × ×

Wasim 0 0 2 1 × × ×

Members

Medals Possible School

The remaining medal tally of Indore school and Ahmedabad school are identical. Also, the total medals of Vipul and Wasimare identical to those of Umesh alone. Therefore, either Umesh is from Ahmedabad school and (Vipul + Wasim) are fromIndore school or vice-versa.In summary,

Delhi School Satyam, Zaheer, Qadir

Indore School Yousuf, (Vipul+Wasim)/Umesh

Lucknow School Xavier, Puneet

Kolkata School Tarun, Rishi

Ahmedabad School Umesh/(Vipul+Wasim)

1. d Option (a) is necessarily correct. Option (b) is alsonecessarily correct. Option (c) is necessarilyincorrect.Option (d) could be correct, if Vipul and Wasim arethe members of Indore school. Option (e) isnecessarily correct.

2. e Either Indore or Ahmedabad. Hence, cannot bedetermined

3. b Three representative members.

4. a Among the options, only Puneet is from Lucknowschool

5. e Option (a) is inconsistent because Satyam is notrepresenting Indore school.Option (b) is redundant because it is already knownthat Vipul and Wasim are fellows.Option (c) is consistent, but doesn’t give any datawhich could tell us the schools of Umesh, Vipul andWasim.Option (d) is also consistent, but not sufficient.Option (e) clearly tells that 3 members representedIndore school. This implies that Umesh is a memberof Ahmedabad school and (Vipul+Wasim) aremembers of Indore school.


Page: 65��

For Questions 6 to 10: Aggregate number of cars manufactured by all the companies across the twelve months of a year= 1200.Since aggregate number of cars manufactured by all the companies is same in each month, therefore number of cars

manufactured in each month = 1200

12 =100.

In January, Mitsubishi manufactured 7 cars and Renault manufactured 37 cars. The 6 remaining companies have tomanufacture at least 7 + 2 = 9 cars. But total cars in January = 100.Hence out of those 6 remaining companies, minimum cars any company could manufacture = 9 and maximum cars anycompany could manufacture = 11.Similarly, with the help of the given information, we can find the range of the number of cars manufactured by each of thecompanies in the 12 months.

Jan Feb Mar April May June July Aug Sep Oct Nov Dec Mercedes 9-11 7-26 32 8-19 27 7-20 8-23 10 -18 7-23 3 10-11 10

Mazda 9-11 7-26 9-16 8-19 6-22 7-20 8-23 10 -18 5 5-33 10-11 32

Nissan 9-11 7-26 9-16 8-19 4 40 8-23 10 -18 28 5-33 8 10

Mitsubishi 7 7-26 9-16 8-19 6-22 7-20 8-23 10 -18 7-23 38 10-11 10

Porsche 9-11 7-26 9-16 8-19 6-22 5 31 8 7-23 5-33 31 10

Ferrari 9-11 7-26 9-16 35 6-22 7-20 8-23 24 7-23 5-33 10-11 8

Renault 37 5 9-16 6 6-22 7-20 8-23 10 -18 7-23 5-33 10-11 10

Honda 9-11 34 7 8-19 6-22 7-20 6 10 -18 7-23 5-33 10-11 10

Total 100 100 100 100 100 100 100 100 100 100 100 100

The maximum and minimum number of cars manufactured by different brands alongwith their optimum index (O.I.) andaverage index (A. I.) is collated in the following table.

Mercedes Mazda Nissan Mitsubishi Porsche Ferrari Renault Honda Maximum 223 236 236 233 235 252 224 214 Minimum 138 116 146 127 136 135 120 119

Sum 361 352 382 360 371 387 344 333 O.I. 85 120 90 106 99 117 104 95 A.I. 180.5 176 191 180 185.5 193.5 172 166.5

Ratio (O.I./A.I.)

0.470 0.68 0.471 0.588 0.53 0.60 0.604 0.57

6. c Maximum possible number of cars manufactured by Mitsubishi across all the 12 months = (7 + 26 + 16 + 19 + 22+ 20 + 23 + 18 + 23 + 38 + 11 + 10) = 233.

7. e Mazda is the company for which the Optimum Index is highest.

8. c Statement B and D are incorrect.

9. d The ratio of O.I. to A.I. is least for Mercedes.

10. c The minimum number of cars that a company possibly manufactured throughout the year is second largest forMercedes.



11. c Sunday: 131 = 25 × 5 + 2 × 3.Minimum possible number of students who loggedinto the website = 27.

Monday: 49 = 9 × 5 + 1 × 4.Minimum possible number of students who loggedinto the website = 10.

Tuesday: 92 = 18 × 5 + 1 × 2.Minimum possible number of students who loggedinto the website = 19.

Wednesday: 157 = 31 × 5 + 1 × 2.Minimum possible number of students who loggedinto the website = 32.

Thursday: 105 = 21 × 5.Minimum possible number of students who loggedinto the website = 21.

Friday: 81 = 15 × 5 + 1 × 4 + 1 × 2.Minimum possible number of students who loggedinto the website = 17.

Saturday: 63 = 12 × 5 + 1 × 3.Minimum possible number of students who loggedinto the website = 13.

Therefore during this particular week the minimumpossible number of students who logged into thewebsite = 27 + 10 + 19 + 32 + 21 + 17 + 13 = 139.

12. b Maximum number of students who logged into theSIS on Sunday = (64 × 2 + 1 × 3) = 65Similarly, maximum number of students who loggedinto the SIS on Monday, Tuesday, Wednesday,Thursday, Friday and Saturday is 24, 46, 78, 52, 40and 31 respectively.Required difference between the maximum and theminimum possible number of students who loggedinto the SIS on Sunday, Monday, Tuesday,Wednesday, Thursday, Friday and Saturday is 38,13, 27, 46, 31, 23 and 18 respectively.

Therefore, on 4 days the required difference is notless than 27.

13. a Total number of hits on the SIS on Sunday is 131and total number of students who logged into theSIS on Sunday is 49.To maximize the number of New Users who loggedinto the SIS exactly two times we need to minimizethe aggregate number of New Users who loggedinto the SIS at least three times.Minimum number of New Users who logged into theSIS at least three times is when 11 New Userslogged into the SIS exactly five times.

Maximum number of New Users who logged intothe SIS exactly two times = 49 – 11 = 38.Total number of hits on the SIS on Thursday is 105and total number of students who logged into theSIS on Thursday is 52.So a maximum of 52 New Users can possibly loginto the SIS exactly two times on Thursday, this ispossible when only one new User logged into theSIS exactly thrice on Thursday.Required answer = 38 + 51 = 89.

14. d Since on Sunday and Thursday no Old User loggedinto the SIS, total number of students that loggedinto the SIS on the remaining five days of the weekis 19 + 33 + 61 + 37 + 25 = 175.So to minimize the number of New Users that loggedinto the SIS, we need to maximize the number of OldUsers who logged into the SIS on exactly threedays.175 = 57 × 3 + 2 × 2Therefore at least 59 Old Users.

15. b 175 = 86 × 2 + 1 × 3. Maximum possible number ofOld Users that logged into the SIS is 87.Therefore, number of Old Users that logged into theSIS lies between 59 and 87 (both inclusive).

For questions 16 to 20:Let a, b, c, d, e, f, g, h, i and j be the number of questions insections I, II, III, IV, V, VI, VII, VIII, IX and X respectively.From the data given in TABLE 1, we can draw the followingconclusions:

a b a c b c a d a e28, 21, 28, 26, 24

2 2 2 2 2

+ + + + += = = = =

a f a g a h a i a j35, 38, 30, 32, 28

2 2 2 2 2

+ + + + += = = = =

Solving the above we get the following values

a 21, b 35, c 21, d 31, e 27= = = = =

f 49, g 55, h 39, i 43, j 35= = = = =Similarly the number of questions solved correctly in eachof the given 10 sections can be easily calculated.The following table provides information about the numberof questions in each section, number of questions solvedcorrectly in each section and the difference of the numberof questions and the number of questions solved correctlyin each section.

Sections I II III IV V VI VII VIII IX X Total

Total 21 35 21 31 27 49 55 39 43 35 356

Correct 7 21 15 21 23 35 41 33 29 27 252

Total – Correct 14 14 6 10 4 14 14 6 14 8 104

Absolute Value of [Correct –

(Total – Correct)]7 7 9 11 19 11 27 27 15 19 —


Page: 67��

16. d In section V, the difference between the numberof questions and the number of questions solvedcorrectly is the least.

17. d Number of questions solved incorrectly and numberof questions left unattempted in section VI can be3 and 11 respectively. Hence option (1) is possible.Number of questions solved incorrectly and numberof questions left unattempted in section VI can be6 and 8 respectively. Hence option (2) is possible.Number of questions solved incorrectly and numberof questions left unattempted in section X can be 3and 5 respectively. Hence option (3) is possible.If option (a) is possible, then option (e) is also

possible 3 3 11 1411 11 11

+ = = �

Option (d) is not possible and hence the correctchoice.

18. b The number of questions in sections VI, VII, VIII, IXand X is not less than the number of questions insection II. Hence there are 5 such sections.

19. c In seven sections namely I, II, III, IV, V, VI and IX.

20. b Maximum possible number of questions solvedincorrectly in sections I, II, III, IV, V, VI, VII, VIII, IXand X is 1, 5, 3, 5, 5, 8, 10, 8, 7 and 6 respectively.So, total number of questions solved incorrectly= 58.Therefore, number of question left unattempted= 104 – 58 = 46.Minimum possible marks obtained by the student inthe entrance test = 252 × 4 – (58 × 2 + 46 × 1) = 846

21. c Since 909909 are divisible by 9 and 567 is alsodivisible by 9.∴ x is also divisible by 9.

22. e Number of books is not known. Thus, by usingboth statements also we do not get the answer.

23. c Solving the given data

x yy x

x y x y y x 1 x y 1x y x y y x+ −+ = + = + = +

Thus, both statements can give answersindependently.

24. d Statement I: (c + d)(c – d) = 5.This data is not sufficient independently.Combining with (II) we get c + d.

25. c Using statement I, the orthocentre lies outside thetriangle only in the case of an obtuse triangle.Thus, ABC is not an equilateral triangle.Using statement II, only in the case of anequilateral triangle does the circumcentrecoincide with the orthocentre. Thus, ∆ABC is anequilateral triangle.


For questions 1 to 5:From information B and E, it is clear that surname of Asha isSingh and she has ordered one San Marino Imbottiti and oneSicilian Fugitive pizza. From information A, at least one ofthem must be in Solo size. So, possible total bill amounts ofAsha Singh are Rs. 246, Rs. 372, Rs. 266 and Rs. 396 only.

From information E, it is also evident that Esha’s surname iseither Hazra or Malhotra.

Let us assume that Esha’s surname is Malhotra. Then, frominformation C, the pizza, which is common between Eshaand Isha, did not start with ‘M’. So the single Montreal Hammust have been ordered by Esha and from information F, thisis a Solo size pizza. From information E, Esha Malhotra’s billamount is Rs. 127 less than Asha. So this amount can be Rs.119, Rs. 245, Rs. 139 and Rs. 269. Deduct Rs. 87 from theseamounts, which Esha Malhotra must have spent on Solosize Montreal Ham. You will get Rs. 32, Rs. 158, Rs. 52 andRs. 182 respectively, which must be the price of the otherpizza Esha ordered. Out of these only Rs. 158 can be foundout in the menu for a double size Raffaele’s Bianco. But, thepizza, which is common pizza between Esha and Isha, musthave been ordered at least twice. So Raffaele’s Biancocannot be the common pizza between Esha and Isha. So,Raffaele’s Bianco must have been ordered by Usha and notby Esha.

Therefore, our assumption, that Esha’s surname is Malhotrais wrong. Esha’s surname is Hazra. She must have orderedHawaiian Honeymoon in Solo size and one other pizza inDouble or Family size. Again, the common pizza betweenEsha and Isha, must have been ordered at least twice. SoRaffaele’s Bianco cannot be the common pizza betweenEsha and Isha. So, Usha must have ordered Raffaele’sBianco. From information F, Usha must have orderedRaffaele’s Bianco in any size other than Double. Now, Usha’ssurname is either Gupta or Malhotra.

Let us assume that Usha’s surname is Gupta. Then she musthave ordered Godfather’s Choice in Double size. So, shemust have ordered Raffaele’s Bianco in Solo size. So,possible total bill amounts of Usha Gupta are Rs. 214 only.From information E, Usha Gupta’s bill amount is Rs. 81 morethan Esha. So Esha’s bill amount can be Rs. 133 only. DeductRs. 87 from these amounts, which Esha Hazra must havespent on Solo size Hawaiian Honeymoon. The remainingamount is Rs. 46, which must be the price of the other pizzaEsha ordered. There is no pizza with this price tag in themenu.

Therefore, our assumption, that Usha’s surname is Gupta iswrong. Usha’s surname is Malhotra and Isha’s surname isGupta. Usha must have ordered Raffaele’s Bianco in Solosize and Montreal Ham in Double. So her total bill is Rs. 245.That means Asha’s total bill is Rs. 372. So she must haveordered San Marino Imbottiti in Solo size and one SicilianFugitive pizza in Family size.


Now Isha Gupta must have ordered one Godfather’s Choice.Let us assume this was in Solo size. She has orderedanother pizza common with Esha. We know, that Esha Hazramust have ordered one Hawaiian Honeymoon in Solo sizeand one other pizza in Double or Family size. The price ofGodfather’s Choice is Rs. 12 less than Hawaiian Honeymoonin Solo size. But, from information E, Ms. Gupta’s total billwas Rs. 81 more than Esha. So, the other pizza must havecreated the difference of Rs. 93. But there is no such pizza,for which the difference of price for Double and Family sizeis Rs. 93. So our assumption was wrong.

Next, let us assume that Isha Gupta ordered one Godfather’sChoice in Double size. That means Isha must have orderedthe other pizza in Solo size. The price of Godfather’s Choicein Double size is Rs. 20 more than Hawaiian Honeymoon inSolo size. So, the other pizza must have created thedifference of Rs. 61. But there is no such pizza, for whichthe difference of price for Solo and Double/Family size isRs. 61. So our assumption was again wrong.

So, Isha must have ordered one Godfather’s Choice in Familysize. That means Isha must have ordered the other pizza inSolo size. The price of Godfather’s Choice in Family size isRs. 118 more than Hawaiian Honeymoon in Solo size. So,the other pizza must have created the difference of Rs. 37.Only for Verona’s Army, the difference of price for Solo andDouble size is Rs. 37.

Now, finally we can collate the derived information in the following table, which can help us to answer all the questions:

Asha Esha Isha Usha

Singh Hazra Gupta Malhotra

Type San Marino Imbottiti Hawaiian Honeymoon Verona's Army Raffaele's Bianco

Size Solo Solo Solo Solo

Type Sicilian Fugitive Verona's Army Godfather's Choice Montreal Ham

Size Family Double Family Double

372 262 343 245Total Bill (Rs.)

Name

Surname

Pizza 1

Pizza 2

1. c 2. a 3. b 4. c 5. e

6. b 30+25+40+5 for reminder mail, i.e. 100% space isused∴ free space is 0%

7. a The deleted mails will go to trash and would stilloccupy space. So all will bounce.

8. d On day 3, 20 MB of space will be created in trashfolder as mails deleted on day 1 will be automaticallycleared.So, none of the mail will bounce.

For questions 9 to 14:The maximum possible lecturers were short-listed.Since there are 5 age groups and 4 differentdisciplines; and there can be at most two lecturersof a particular age group representing a particulardiscipline, the number of lecturers short-listed= 5 × 4 × 2 = 40.Number of them selected = 5 + 20 = 25.Now,(i) Tells that out of these 25, the number of lecturersfrom various age groups is:

Young: 5

Middle-aged: 5

Senior: 4

Stalwarts: 4

Retired: 7

TOTAL 25

Now coming directly to (iii),

Physics Chem istry Mathem atics Biology

Young Young Young Young

Middle-aged Middle-aged Middle-aged Middle-aged

Senior Senior Senior Senior

Stalw art Stalw art Stalw art Stalw art

Retired Retired Retired Retired


Page: 69��

Each of them is having at least one lecturer.So assigning one to each of them, the table lookslike:

Physics Chemistry Mathematics Biology

Young (1) Young (1) Young (1) Young (1)

Middle-aged (1)

Middle-aged (1)

Middle-aged (1)

Middle-aged

Senior Senior (1) Senior (1) Senior (1)

Stalw art (1) Stalw art (1) Stalw art (1) Stalw art

Retired (1) Retired (1) Retired (1) Retired (1)

The balance lecturers are:

Young: 1

Middle-aged: 2

Senior: 1

Stalwarts: 1

Retired: 3

TOTAL 8

Going back to (ii), total number of lecturers forphysics, chemistry and biology can be counted tolie between 12 and 20. Moreover, the total is asquare.⇒ Total in these three is 16 and there are 9 lecturersfrom mathematics stream.Proceeding, (iv) tells us that Rocky and Platy areYoung physics lecturers.So the balance Young lecturer goes in Physics.And since none of the cells can have more than 2and total for Mathematics is 9, we get a betterversion of the table as:

Physics (5 or more)

Chemistry (5 or more)

Mathematics (9)

Biology (3 or more)


Middle Aged (1)

Middle Aged (1)

Middle Aged (2)

Middle Aged




And the balance lecturers now:

Middle Aged: 1

Retired: 2

TOTAL 3

Note that the column for Mathematics is complete.And in Biology, count for Retired can be 2 at max.And there is no provision for Middle Aged in Biology.

Thus, out of the two balance retired lecturers, atmost one can go in Biology and in that case, theremaining retired and middle aged ones must go intoPhysics or Chemistry or both.

9. d As of now, minimum lecturers for Physics as wellas that of Chemistry are 5. But since the number ofPhysics lecturers is greater than the number ofChemistry lecturers, at least one of the balancethree must go into Physics.

Case I: Middle-aged lecturer goes into Physics.Now, out of the remaining two retired lecturers onehas to go into Physics and the remaining one can goeither in Chemistry or Biology.

Case II: Retired lecturer goes into Physics.Here also, the count for retired in Physics is full (2).Middle Aged cannot go in Biology and if he goes intoChemistry, the initial condition that "the number ofPhysics lecturers is greater than the number ofChemistry lecturers" can never get satisfied.So the middle aged goes into Physics and we getthe same configuration as in case I

Physics (7) Chemistry (5 or 6)

Mathematics (9)

Biology (3 or 4)


Middle Aged (2)

Middle Aged (1)

Middle Aged (2)

Middle Aged




The one remaining retired lecturer goes into eitherChemistry or Biology.Hence the number of retired Chemistry lecturerscannot be determined.

10. b It is already explained that the middle aged lecturerand at least one retired lecturer goes into Physicsor Chemistry or both, the total is at least 12 and atmost 13. Only option (b) is correct.

11. d Since Kandy is the lone retired Chemistry lecturer,the balance 2 retired Chemistry lecturers go intoPhysics and Biology (one each). So there are tworetired Biology lecturers for sure.

12. c At least one of the retired lecturer must go into oneof Physics or Chemistry. Thus the count for both ofthem cannot be one. Option (c) is not possible.


13. d Laxman has scored 3,000 runs and has taken 60catches. He has not taken any wickets. Even if weassume that he has scored all runs in centuries,then also his points would be:Runs → 3,000, Catches → 180, Centuries → 30 ×50 = 1,500Total points 4,680, so grade E.

14. a Zaheer has scored 1,000 runs and taken 50 catches.He has taken 150 wickets (half of Kumble). 8 fivewicket hauls. If we assume that he has scored allruns in centuries, then his maximum points wouldbe:Runs → 1,000, Catches → 150, Centuries → 10 ×50 = 500, Wickets → 3,000, Five wicket haul → 400Total points, 5,050 , so at the most he could be ingrade D.

15. a Ganguly has scored 10,000 runs and taken 50wickets. If we assume that he has scored all runsin centuries, then his points would be:Runs → 10,000, Centuries → 100 × 50 = 5,000,Wickets → 1,000Total points 16,000, so grade A.

16. b Sehwag could score maximum 7,999 runs and taken50 catches. He has taken 40 wickets.And taken 4 five wicket hauls. He has scored 10centuries. His points would be:Runs → 7,999, Catches → 150, Centuries → 500,Wickets → 800, 5 wicket haul → 200Total points 9,649, so he cannot be in grade B.


UPA (30 + 5 + 7)% = 42%If Congress led ⇒ 42% + 2% (of CPI) = 44%NDA = (20 + 2 + 2)% + 10% (“Others”) = 34%

17. d (i) –(8% (of SP) + 4% (of BSP)) = –12%(ii) 34% (of NDA) + 5% (of RJD) = 39%(iii) 55 members ⇒ –11%If all (i), (ii) and (iii) happen, thenNDA = 39% which is more than 50% of(100 – 12 – 11) = 77%

18. d If it breaks away at least five members, the totalstrength gets reduced to 495, 60 abstain (40+20)so for simple majority you need half of 435, i.e. 218,Congress has backing of 220. CPM's support woulddefinitely be another way.

19. d If members of BJP, Others, CPM and RJD support aCongress led government then it would have a 75%majority. Hence four parties.

20. b Statement I is not sufficient as there are more thanone sequences satisfying the question condition.Statement II is sufficient to answer the question.Since the first term is 0, the 3rd term will be 6.

21. a Statement I alone gives the relationship the threeshapes in terms of size and is enough to answerthe question.Statement II gives the data to find the area of thecircle but not for the two others.Thus, the answer is (a).

22. c ‘C’ is a set of integers, and ‘x’ an element in C. Thenfrom statement I, (x + 3) can also be an element in‘C’ if we take x as negative integers and hence thecondition is fulfilled. Similarly, with the help ofstatement II alone we can get the answer.

23. e There can be many different relationships possible.So statement I does not provide an answer.Statement II gives us a very general piece ofinformation. The two statements combined togetherare not sufficient.

24. d From statement I, we get 15x + 20y = 130The only possibility for x and y are (6, 2) and (2, 5).From statement II, we get x = 2 and y = 5,i.e. we have 5 tickets of $20 and 2 tickets of $15.Thus, the answer is (d).

25. d From statement I, we get the lowest = 20, highest =10From statement II, we get the two series are in AP.20, 23, 26, 29, ..., 80

∴ Average = 80 20

5020+ =

Explanations:Fundamentals of Ratio & Proportion

Page: 71��

Solutions (Non MCQ)

Level – 1

1. (54 – x) : (71 – x) : : (75 – x) : (99 – x)Solving, x = 3

2. Ratio = 51

:31

:21

= 15 : 10 : 6

or first part = 2254653115 =×

Second part = 1504653110 =×

Third part = 90465316 =×

Hence, the three parts are 225, 150, 90.

3. Ratio of first, second and third is2x : 2x – 10 : 5xand 2x + 2x – 10 + 5x = 1709x = 180x = 20So the three parts are 40, 30, 100 respectively.

4. If numerator = x and Denominator = 3x

∴ Fraction x y 1

3x 3y 3+⇒ =+

5. Ratio of number of coins = 4 : 5 : 6Suppose we have 4, 5 and 6 coins of Re. 1,50-paisa and 25-paisa respectively. Then the totalvalue = Rs. 8.Since the aggregate sum is Rs. 32 (4 times the sumthat we have arrived at), the number of coins of thedenomination Re. 1, Re. 0.5 and Re. 0.25 must be16, 20 and 24 respectively.

6. Let the initial number of employees = n.

New number of employees = n98

Let initial wages = x. New wages = .x1415

∴ Ratio of total wages before and after the change

= (nx) :

x14

15n

9

8 = 21 : 20.

7. Let father’s present age be F.Son’s present age be S.∴ (F – 5) = 5[S – 5] and (F + 2) = 3(S + 2) or,

F – 5S = –20 and F – 3S = 4∴ 2S = 24; S = 12 and F = 40∴ Ratio of ages of father and son = 10 : 3.

8. Mohan’s expenditure = 4xMohan’s savings = x∴ Income = 5xNew income = (5x)(1.2) = 6x; new saving = 1.12xNew expenditure = 4.88xPercentage increase in expenditure

= − × =

4.88x 4x100 22%.

4x

9. Ratio of values of 1 rupee, 50 paisa and 25-paisa

coin is 3 4 12

: :1 2 4

, i.e. 3 : 2 : 3

∴ 3x + 2x + 3x = 18008x = 1800

⇒ x = 225Value of 50-paisa coin = 2 × 225 = 450and number of 50-paisa coins = 450 × 2 = 900

10.

A

B C

The angles of the triangle areIf ∠ A = 4x; ∠ B = 5x; ∠ C = 5x – 30∠ A + ∠ B + ∠ C = 180

⇒ 4x + 5x + 5x – 30 = 180

⇒ 14x = 210 ⇒ x = 15So ∠ A = 60°, ∠ B = 75°, ∠ C = 45°

11. The ratio of volumes of two cylinders is x : y andheights are p : q.

2r pV x1 12V y2 r q2

π= =

π

2r xq12 pyr2

= =2

12

2

d xqpyd

⇒ =

12. The given expression can be written as

Q – p Q – q Q – rk

2 5 7= = =

p = Q – 2k; q = Q – 5k; r = Q – 7k

So, Q – 2k Q – 5k – Q – 7k

Q2

+=

��


2Q = 3Q –14kQ = 14kp = 12k;q = 9 k;r = 7k;∴ p : q : r = 12 : 9 : 7

13. Ram Shyam

Incomes 8x 11x

Expenditures 7y 10y

Savings 500 500

∴ 8x – 7y = 500 ⇒ 88x – 77y = 5500

11x – 10y = 500 ⇒ 88x – 80y = 4000

∴ 3y = 1500, y = 500

⇒ x = 500

∴ Incomes of Ram and Shyam are Rs. 4,000Rs. 5,500 their expenditures are Rs. 3,500,Rs. 5,000 respectively.

14. If B joined for t months then profit ratio = (450 × 12): 300t = 2 : 1

∴ 2t30012450 =× ⇒ t = 9 months

∴ B joined after 3 months.

15. When A gets 60 points, B gets 40 pointsWhen A gets 60 points, C gets 45 points∴ When C gets 45 points, B gets 40 points

When C gets 60 points, B gets 31

53 points

∴ C gives B 32

6 points in 60

16. Time .days325

]hr8men15[]hr5days10men20[

t =×

××=

17. Time t = 25.1875.015

51020 ×××

××

= 13.88 days (approximately).Multiplication by 1.25 is because of 25% extra workload and 0.75 accounts for the efficiency.

18. Average weight of the chairs

= 10 8 12 6

49

+ + += kg

Now a 12 kg chair is replaced by an 8 kg chair

∴ Average = 10 8 8 6

48

+ + += kg

i.e. a decrease of 1 kg in average.

Alternative method:There is a loss of 4 kg on replacement of 12 kgchair by 8 kg. This loss will be equally compensatedbetween all the four chairs. So, there is a decreaseof 1 kg in average.

19. When the same number is added to all thequantities, the average increases by the numberadded.Therefore, increase = Rs.1,000

20. New average will also increase by 20% of thepresent average.Therefore, new average = 15,000 + 3,000

= Rs. 18,000

21. Turnover from January to March = 12 × 3 = 36Turnover from March to June = 14 × 4 = 56Turnover from January to June = 36 + 56 – Turnoverof March.Also average turnover from January to June is 12.∴ 36 + 56 – Turnover in March = 12 × 6So turnover in March = Rs. 20 crore.

22. If John leaves the group, number of students inthe group will be 3.So average height

168 4 170167.37 cm

3× −= = (approximately)

23. Let Tom’s score be x.

62x705

x472 =⇒=

+×∴

24. Cost of 20 kg apples = 20 × 40 = Rs. 800Cost of 10 kg apples = 10 × 50 = Rs. 500Total cost of 30 kg = 800 + 500 = Rs. 1,300

∴ Average cost = 30

1300= Rs. 43.33 per kilogram

25. Time taken to ride 20 km at 10 km/hr = 1020

= 2 hr

Time taken for another 20 km at 5 km/hr =520

= 4 hr

∴ Total time taken = 6 hrTotal distance covered = 40 km

Average speed = 640

= 6.67 km/hr

26. Average speed = timeTotal

cetandisTotal

= .hr/km10122010 =

++


Page: 73��

27. Let total number of units = 100Revenue when 50 units are sold at Rs. 20 per unit= Rs. 1,000Revenue when 30 units are sold at Rs. 40 per unit= Rs. 1,200Total price = Rs. 2,200

Average price = 2200

50 30+ = Rs. 27.50

28. If 2 friends leave the group, then 3 are left.

New average = ( ) ( )× − ×

=6.5 5 7 2

6.16 kg3

29. Average = Total score

Number of students

=× + ×

=30 70 20 60

5066%

30. Method 1:

Fraction of gold in first alloy = 107

.

Fraction of gold in second alloy = 207

.

Applying rule of alligation,

710

720

11

Average fractionalconcentration of the constituents

Resultant average

Quantities

∴ x – x207

107 −=

⇒ x = 4021

203

27

2107

207

=

=

+

∴ Ratio of gold to copper = 21 : 19.

Method 2:The first alloy is a 70% gold alloy and the second isa 35% gold alloy. If the two are mixed in equalquantities, the average concentration of gold in the

resulting alloy is %5.522

3570 =+.

Therefore, the ratio of gold to copper is52.5 :47.5 = 21 : 19.

31. To a mixture containing 10% water, pure water(100%) is mixed to get the resultant solutioncontaining 20% water.Applying the rule of alligation, average concentrationof constituents

10% 100%

80 10

20%

Average percentage

Resultant average

Quantities

Ratio of the volumes = 8 : 1 or 40 : 5.∴ To 40 L of solution, 5 L of water must be added.Alternatively:36 L of milk in the original mixture now becomes80% of the mixture. Hence, the total volume of thenew solution = 45 L. So the extra 5 L must be thewater that was added.

32. Using alligation,

CP15 – 0 3II0 – (–10) CP 2I

= =

CP Cost price of second shirtIICP Cost price of first shirtI

= =

and CPI + CPII = 900

∴ CPI =2

900 3605

× =

3CP 900 540II 5

= × =

33. Fraction of water in the adulterated milk = th61

.

Fraction of water in pure water (added) = 1.

Fraction of water in resultant mixture = th83

.

∴ Apply rule of alligation.

16

1

58

524

38

Average fractional constitution of the constituents

Resultant average

Quantities


Ratio of the volumes of the solution to the wateradded = 3 : 1 = 66 : 22.∴ 22 kg of water should be added.

Alternative method:

55 kg of milk is now 85

th of the new mixture.

Hence, new weight = 5558 × = 88 kg.

Hence, the amount of water added = 88 – 66 = 22 kg.

34. The volume of the solution is 9

10 times the original

volume. Hence, if the original volume is 729 ml, thenew volume = 810 ml. The increase is due to theaddition of 81 ml of water.

(Hint: the factor 9

10 is the ratio of

97

and 107

,

which are the average milk content in the initial andthe final mixtures.)

35. Method 1:Let CP of 1 L of milk = Rs. 100.He mixed x litres of water.SP of 1 L of mixture = 100.∴ SP of (1 + x) L of mixture = 100(1 + x).

P% = 100CP

CPSP ×− = 100

100x100 × = 100x = 25

∴ Volume of mixture = 1 + 0.25 = 1.25.

Percentage of water = %2010025.125.0 =× .

Method 2:If the total cost of 100 L of milk is Rs. 100, themilkman is making a revenue of Rs. 125. Since he isselling at the same cost price, he must actually beselling 125 L. Hence, he is adding 25 L of water or20% of the mixture is water.

Level – 2

36. 2 men = 5 boys and 3 boys = 2 women

∴ 1 boy gets 52

of what man gets.

1 woman gets 2

31× or 1.5 times of what a boy

gets.

∴ A woman gets 53

of what a man gets

∴ 6m + 12w + 17b

= 6m + 12 × 53

m + 17

52

m = 20m

or m = Rs. 2.5. Similarly, b = Rs. 1, w = Rs. 1.5m, b, w are shares of a man, a boy and a woman,respectively.

37. Assume that the persons in 4 battalions are a, b, cand d.

Then d54

c43

b32

a21 ===

i.e. b = ,a43

c = a32

and d = a85

∴ a + 7300a85

a32

a43 =++

⇒ a = 2400, b = 1800, c = 1600 and d = 1500

38. Let the worth of the properties of A, B and C be2x, 3x and 5x respectively. Total = 10x

B sells to A = 31

(3x) = x and 31

(3x) = x to C.

∴ A has 3x, B has x, C has 6x

C sells x23

Ato)x6(41 =

∴ A has now x29

2x3

x3 =+ = Rs. 18,000

Hence, x = Rs. 4,000∴ Total worth of properties = 10x = Rs. 40,000.

39. Let A, B and C invested Rs. 5, Rs. 6 andRs. 8, respectively for a, b, and c months.Then 5a : 6b : 8c is the ratio of profits which is givento be 5 : 3 : 1.∴ 5a : 6b : 8c = 5 : 3 : 1.

∴ a : b : c = 1 : 81

:21

= 8 : 4 : 1.

40. Total amount deducted is 15 + 10 + 30 = 55.Amount left = 535 – 55 = 480.Now Rs. 480 is divided in the ratio 4 : 5 : 7.So after deduction

share of A = 4

480 12016

× = ,

share of B = 5

480 15016

× = ,


Page: 75��

share of C = 7

480 21016

× = .

∴ Initial share of A = 120 + 15 = Rs. 135.Initial share of B = 150 + 10 = Rs. 160.Initial share of C = 210 + 30 = Rs. 240.

41. Ratio of contributions expected from the three men= (110 × 6) : (50 × 9) : (440 × 3) = 22 : 1 5 : 44.

∴ Second man must pay 15 5

th or81 27

th of the total.

42. B should get =

5100 3

1248 Rs. 459(3200 4) (5100 3) (2700 5)

× = × + × + ×

43. 2 leaps of the dog = 5 leaps of the hareor 1 leap of the dog = 2.5 leaps of the hare∴ 6 leaps of the dog = 15 leaps of the hare

∴ Ratio of leaps of dog to hare = 15 : 4.

44. Let the original prices by 13x and 17x. So ratio for

new prices =13x 100% of 13x 3

17x 2500 5+ =

+

26x 317x 2500 5

=+

130x = 51x + 750079x = 7500x = 94.93So the original prices are 13 × 94.93= Rs. 1234.08and 17 × 94.93 = Rs. 1613.81.

45. In a race of 100 m, A travels 100 m while B travels95 m. This means in a race of 500 m, when A travels

500 m, B travels 95 × 100500

= 475 m.

Similarly, when B travels 200 m, C travels 190 m. So

if B travels 500 m C travels 190 × 200500

= 475 m.

Hence, when B travels 475 m, C travels 475 × 500475

=

451.25.Therefore, when A travels 500 m, C travels 451.25m.This means, A can give C a lead of 48.75 m.

46. Please recollect you would have solved thisproblem in the chapter dealing with SI and CI. It isrepeated here to explain solving the problem usingalligation. On an average Shiv pays a interest rateof 7.2%.

Using alligation, we have 8 7.2 0.8 27.2 6 1.2 3

− = =−

Thus amount lent at 6% = 2

10000 40005

× =

47. Total weight = 25 × 10 kg2 cases of weight 30 kg each are replaced by2 cases of weight 32 kg each.∴ Total weight = (25 × 10) – (2 × 30) + (2 × 32)

= 254 kgNumber of students = 10

∴ Average = 10254

= 25.4 kg

∴ The average will change by 0.4 kg.

48. Average = Total weight

Number of students

Average is decreased by 2 kg.∴ New average = 60 kg

So 60 = 62 5 70

560

× − +⇒ =

xx

Weight of new student x = 60 kg

49. If water weights x kg/unit volumeweight of gold = 19x/unit volumeweight of copper = 9x/unit volume∴ Weight of the mixture desired = 15x/unit volumeApply rule of alligation:

19x 9x

Q1 Q2

Average of gold and copper

Resultant average

Quantities

1Q

2Q

x9–x15

x15–x19=

3

2

6

4=⇒

50. Let the cost price of first variety be Rs. 2x perkilogram.So the cost price of second variety becomes Rs. xper kilogram.Selling price of mixture is Rs. 36 per kilogram aftermaking profit of 20%.

So the cost price of mixture = 36

30 per kilogram1.2

=


Applying the principle of alligation,

2x – 30 730 – x 3

=

6x – 90 = 210 – 7x13x = 300x = 23.07So the cost price of first variety = Rs. 46.14 andcost price of the second variety = Rs. 23.07.

51. Ratio of copper in first alloy = 39

.

Ratio of copper in second alloy = 59

.

Ratio of copper in new alloy = 12

.


5 1– Q9 2 I

1 3 QII–2 9

==

Quantity of first alloyQuantity of second alloy

Q 1IQ 3II

⇒ =

∴ Quantity of first alloy = 1

64 16 kg4

× =

and quantity of second alloy = 3

64 48 kg4

× = .

52. Milk proportion in the total solution of first variety

(VI ) = 5

13.

Milk proportion in the total solution of second variety

(VII ) = 9

13.

Milk proportion to total in the required mixture = 7

13.


9 7– V13 13 I

7 5 VII–13 13

=

V 2 1IV 2 1II

= =

53. Applying the rule of alligation,

12% –8%

19 1

11%

Average percentage

Resultant average

Quantities

The ratio of the items that he is selling at a profit of12% and at a loss of 8% is 19 : 1.

Hence, he sold 602019 × = 57 pens at a 12% profit

and 60201 × = 3 pens at a 8% loss.

54. Apply rule of alligation for fractional water content.

37

1

38

1156

58

Average fractional constitution of the constutuents

Resultant average

Quantities

The ratio of the volumes of the alcohol mixture towater = 21 : 11. If only 8 L of water was added,

then the original amount of the mixture = 11

821×.

Alcohol content in that = 1196

11821

74 =×× .

55. Let the volume of the mixture be x. Then

amount of A = ,x127

amount of B = x125

.

When 9 L of mixture is taken out,

amount of A withdrawn = 7 21

9 L;12 4

× =

amount of A left in the mixture = 421

x127 − .

∴ Final amount of B in the mixture = 421

x125 + .

(Since all the milk that was removed was substitutedby water.)


Page: 77��

Now ratio of A and B = ,97

421

x125

421

x127

=

+

−

x = 36.

∴ Amount of A = 7

36 21 L.12

× =

Alternative method:Use rule of alligation:After 9 L is taken out, the ratio of A and B is 7 : 5. Tothis is added pure B to obtain a solution containing Aand B in the ratio 7 : 9. Considering the rule ofalligation for fractional B content,

512

1

716

748

916

Average fractional constitution of the constituents

Resultant average

Quantities

∴ Ratio of the original solution left to that of thewater added = 3 : 1 = 27 : 9.Hence, the initial volume = 36 L.

∴ Amount of A initially = 127

× 36 = 21 L.

56. Let t be the time for which he travelled on foot.∴ 8t + 16(7 – t) = 80 or t = 4 hr.The distance travelled on foot = 32 km.

Alternatively:The problem could also be done using the principleof alligation. The average speeds on foot and on abicycle are given and the resulting average speed

80i.e.

7

is also given. We can find the ratio of

times and hence, the time taken for each part of thejourney.

57. If the cost of second liquid is Rs. x per litre, thencost of first liquid is Rs. (x + 2) per litre.Selling price of mixture is Rs. 120, after making aprofit of Rs. 20.

So the cost price of mixture =120

100 per litre1.2

= .

Use Alligation (x 2) – 100 5

100 – x 3+ =

x – 98 5100 – x 3

=

8x = 794x = Rs. 99.25 per litre∴ Cost of first solution = 99.25 + 2 = Rs. 101.25 perlitre and cost of second solution = Rs. 99.25 perlitre.

58. 9 L of orange juice contains = 312

104 calories3

= .

9 L of mango juice contains = 54 calories and 9 L ofmixture (orange and mango) contains

= 118

59 calories2

= .

Use alligation.

V104 – 59 Volume of mango juicem59 – 54 V Volume of orange juice0

= =

V45 m5 V0

=

or V 9mV 10

= =

So fraction of orange juice in 18 L mixture is 1

10.

59. Apply rule of alligation:

–6% 7.5%

7.5 6

0%

Average costs of horses

Resultant average

Costs of horses

∴ The ratio of the costs of the horses is7.5 : 6 or 5 : 4.∴ 5 : 4 is the ratio of cost of each horse.

∴ 750135095 =× is the cost of one which was sold

at a loss of 6% and 600135094 =× is the cost of the

other.


60. waterofamountfractionalTotalmilkofamountfractionalTotal

=

2 3 33 4 51 1 23 4 5

+ +

+ + =

40 45 36 12120 15 24 59

+ + =+ +

61. Ratio of quantities = 2 : 5.Ratio of prices = 3 : 1.∴ Ratio of the values of sugar to orange peels= 6 : 5.

∴ [ ] 4.2$8.02.5116 =− worth is sugar in marmalade

62. Fresh grapes contain 90% water. So 20 kg freshgrapes contain 18 kg water and 2 kg pulp. Now if 2kg pulp contributes 80% of dried grapes, then thetotal amount of dried grapes = 2.5 kg.

63. Final ratio of water to total = Initial ratio of water to

total xnV – X

V

As the process is repeated two more times.∴ The number of times we do the same operationwill be 3.

327 1 V – 31728 1 V

=

3 V – 312 V

= ⇒ V = 4 L

64. Final ratio of rum to total = Initial ratio of rum to total

×nV – X

V

= 41 240 – 80

1 240

4 4160 2 16240 3 81

= = =

∴ Quantity of rum left = 16

240 47.40 L81

× =

Level – 3

65. At the first stage, the concentration of milk is 60%

×= 100

53

. Since the volume increases by 50%,

the volume becomes 32

times of the original volume.

Hence, the concentration becomes ×

260%

3

= 40%. Now 30 L is withdrawn and replaced with

water. Final concentration = 30%

×= 100

103

.

Hence, 41

th of milk has been removed from the

mixture.

==−∴

41

4010

403040

Or 41

th of the mixture = 30 L

So the volume at the end of first state = 120 L.

This is 23

times of the initial volume.

Hence, the initial volume = ×2120

3 = 80 L.

Solutions Exercise 1 (Level – 1)

1. c A 3B 7

= . Let A = 3x, B = 7x

A + B = 45; 3x + 7x = 45, x = 45

4.510

=

B = 7x = 31.5

2. d Let the fraction be xy

x 17x 35y 27 or,

3 1 3y 277 35

= =

x 35 3 5y 27 7 9

= × =

3. a If a, b and c are in continued proportion, the meanproportional is b.

Therefore, b2 = ac, b2 = 8 × 72, b = 576 24=

4. b If a, b, c and d are in proportion, a cb d

= or 3 27

,15 d

=

or d = 135 is the fourth proportional.

5. e Third proportional, c = 2b

a =

30 3045

20× =

6. b Let the number added be x. Then,

4 x 29 x 3

+ =+

, 12 + 3x = 18 + 2x

x = 6


Page: 79��

7. d A’s share = x

Px y z

×+ +

= 3610 × 20004750

= Rs. 1,520Where P = Profit and x, y and z are respectiveshares of A, B and C.

8. cx 2y 7

= Let x’s share = 2a, y’s share = 7a

x 's share 2a 2x 's share y 's share 5a 5

= =−

9. d68 – x 349 – x 4

= , ⇒ 272 – 4x

= 147 – 3x ⇒ x = 125

10. aA B C2 3 6

= = or 3A = 2B = C

BB's part 3960

2B B 2B

3

= ×+ +

= 3

3960 3 360 108011

× = × =

11. d C’s share = 2 B’s share

A’s share = 23

B’ share

⇒ required ratio is

2B's share 23

(2 B's share – B's share) 3=

12. c Let the number be 5x and 3x, then 5x 9 233x 9 12

− =−

60x – 108 = 69x – 207; 9x = 99, x = 11The first number is 11 × 5 = 55

13. d 4x = 3y = 2z

x 3 y 2 4&

y 4 z 3 6⇒ = = =

x : y : z 3 : 4 : 6⇒ =

14. d In a ratio A:B, A is called the antecedent and B iscalled the consequent.Let antecedent = 4x & consequent be 9x.4x = 36 ⇒ 9x = 81.

15. aM 3W 1

= in 100 L mixture

Milk = 75 L, Water = 25 LAfter adding 200 L of water, water = 225 L andmilk = 75 L

Ratio = 75225

= 1 : 3

16. b Let a b ck

3 4 7= = =

a = 3k, b = 4k, c = 7k

a b c 3k 4k 7k 14k2

c 7k 7k+ + + += = =

17. b Option a:Numbers = 9x, 3x12x = 84x = 7; possible.

Option b:8x = 84 ⇒ No whole number x is possible.

Option c and d:4x = 84x = 4; Possible.

18. b 0.35 of x = 0.07 of y

x 0.07 1y 0.35 5

∴ = =

19. b B’s money = 5

800 10004

× =

C’s money = 3

1000 15002

× =

Therefore, total amount of money = Rs. 3,300

20. c Let the two parts be x, (68 – x)

( )x 1 68 x68 x

7 10 10 10= − = −

x x 68 17x 68or,

7 10 10 70 10+ = = , x = 28

21. b x’s share ⇒ 3x + 30y’s share ⇒ 4x + 20z’s share ⇒ 5x + 50Sum is 970012x + 100 = 9700, 12x = 9600, x = 800y’s share = 4x + 20 = 3200 + 20 = 3220

22. d Let B get’s Rs. x A get’s Rs. x + 70 C get’s Rs. x – 80

x + x + 70 + x – 80 = 5303x – 10 = 530, 3x = 540x = 180

A 's share 180 70 250 5C's share 180 80 100 2

+= = =−


23. b A = 12

B, B = 2C or A 1B 2

= , B 2C 1

=

A : B : C = 1 : 2 : 1

Shares of A, B, C are x, 2x, x

B 2x 2x 2A B x 2x 3x 3

= = =+ +

24. c Initially, Number of boys = 250Number of girls = 250

New Batch: 1

2505

× = 50 girls left & 25 boys joined

the class

Boys 275 11Girls 200 8

= =

25. a Let Father’s age be 7x, son’s age = 2xAfter 15 years,

7x 15 2,

2x 15 1+ =+

7x + 15 = 4x + 30, x = 5

Present age of father = 35Present age of son = 10Father’s age when son was born = 35 – 10= 25

26. d Let present age of son = x andpresent age of father = y(y – 4) = 6(x – 4) … (i)

(y 12) 2(x 12)+ = + … (ii)from (i) and (ii),y – 6x + 20 = 0y – 2x – 12 = 04x = 32, x = 8 years, y = 28 years

Ratio of their present ages = 28

7 : 28

=

27. d 1 2

1 2

C C5 5,

I 8 I 3= =

5 5Copper 40 65 10513 8

8 3Iron 64 39 10313 8

+ += = =++

28. c In 80 liters of mixture, milk = 5

80 50 liters8

× =

Water = (80 – 50) = 30 liters.When we take a 16 liter sample of this mixture.

Milk taken out 5

16 10 litres8

= × =

Water taken out = (16 – 20) = 6 litres.⇒ Milk available = 50 L – 10 L = 40 litres

Water available = 30 L – 6 L = 24 litres.Now, 16 litres of milk are added,⇒ Milk = 40l + 16 L = 56 LWater = 24 litres⇒ Milk : Water ratio is 56 : 24 = 7:3.

29. b Let copper = 13xZinc = 7xIn 500 kg, 13x + 7x = 500

x = 50020

= 25 kg

Copper = 13 × 25 = 325 kg

30. d Simplest form = 27

Numerator = 2xDenominator = 7x7x – 2x = 40, x = 8

Number is 1656

.

Solutions Exercise 2 (Level 1)

1. b B 7G 5

=

Let number of boys = 7xLet number of girls = 5x7x + 5x = 72, x = 6Number of boys = 42Number of girls = 3012 more girls should be admitted to make the ratioequal.

2. d Let A’s income = Rs. 3xB’s income = Rs. 2xA’s expenses = Rs. 5yB’s expenses = Rs. 3yA’s savings = 3x – 5y = 3000 … (i)B’s savings = 2x – 3y = 3000 … (ii)Solving (i) and (ii), x = 6000B’s income = 2x = Rs. 12,000

3. a Let the number to be added be, x.

Then, 7 x 1

19 x 2+ =+

, x = 5

4. a Take 7 L of the solution of each mixture,(LCM of 5 + 2, 6 + 1, 4 + 3)∴ net amount of water in the resulting mixture,

5 6 47 7 7

7 7 7× + × + × = 15 L

The net amount of alcohol is,

2 1 37 7 7

7 7 7× + × + × = 6 L

Ratio is, 15 : 6 = 5 : 2


Page: 81��

5. c If the ratio is a : b, then a + b should divide 12completely. Only 3 : 2 does not divide, the restare all possible ratios.

6. d If the ship weighs 5 kg, weight above water= 3 kg.Weight submerged = 2 kg.Therefore, desired ratio = 2 : 3.

7. c 5% of x = 30. Therefore, x = 600. Therefore,number of entrants who lost = 570.

8. c If the ratio is a : b, then a + b should divide 8completely. Only 36 does not fulfil this condition.

9. e Considering the ratio,B’s + C’s shares = 20x and A’s share = 5x.Hence, the second sentence gives no additionalinformation.Since the total amount is not given, we cannot sayanything about A’s share.

10. d LCM(15, 25) = 75 min is the time when both ofthem would meet at the starting point for thefirst time. In this time A takes a lead of two roundsover B. If they have to meet for the first time,then A has to take a lead of one round, whichhappens after 37.5 min.

11. a Let the incomes are 3x and 2x and theexpenditures are 5y and 3y.Therefore, 3x – 5y = 1000 and 2x – 3y = 1000.y = 1000, x = 2000.Therefore, incomes are Rs. 6000 and Rs. 4000.

Alternative method:In first step see which option gives a ratio 3 : 2 inincomes. Unfortunately, all three are in the ratio 3 :2.Now subtract Rs. 1,000 (savings) from each optiontoget the ratio of expenditures.We see that only option (a) gives a ratio 5 : 3.

12.b Plastic lenses are four times costlier than glasslenses.∴ If cost of glass lenses = x, cost of plasticlenses = 4x.Cost of age examination and frames = y.∴ y + x = 40 and y + 4x = 52⇒ 3x = 12 ⇒ x = $4

Alternative method:The price differential is due to plastic lenseswhich cost four times as much as the glasslenses.∴ Price differential = Cost of plastic lenses overand above glass lenses = 3 (Cost of glasslenses).

∴ Cost of glass lenses = ( ) 4$405231 =− .

(Compare with the above algebraic method.)

13.a Ratio of savings of A and B is 5 : 3. So it is

obvious that A’s savings is %32

66 more than that

of B.

14.b Here Sujith earns more but spends less sodefinitely his savings will be more than that ofPratima.

15.d 1 : 5 and 3 : 5 solutionsSince equal quantities of both are mixed, we cantake LCM of (1 + 5 = 6) and (3 + 5 = 8), i.e. 24.Assuming that 24 L of both solutions are mixedtogether in the resultant solution.Milk = 4 + 9 = 13; Water = 20 + 15 = 35Hence, the ratio of water and milk = 35 : 13.

Alternative method:

Amount 61

of milk in 1 : 5 solution = 61

.

Amount 83

of milk in 3 : 5 solution = 83

.

1x – 163 1– x8

=

2413

2494

83

61

x2 =+=+=⇒

4813

x =⇒

∴ Water : Milk = (48 – 13) : 13 = 35 : 13


16. c Mileage when petrol is adulterated

= 43

of (16) = 12 km/L

Hence, the cost increases to a factor

34

of the

original cost of maintenance. So it increases by

%33.3310031 =× .

Alternative method:We can safely assume that the car runs for thesamenumber of kilometres.

Now mileage is reduced to 34

of its original value,

which is, say, x.∴ Distance = Mileage × Number of litres

1 2

3x y x y

4⇒ × = ×

34

yy

1

2 =⇒ = 1.33

Therefore, we see that the consumption of fuel

increases by %3.33 .

17.a If the number is x, 125x185

x85 =− or x = 360.

18. e If A’s age is 5x, B’s age = 7x.Case I: 7x – (5x + 6) = 2So x = 4.Case II: (5x + 6) – 7x = 2So x = 2Hence, sum of their present ages = 12 × 4 = 48years or 12 × 2 = 24 years.

19. a This can be solved using alligation.

So the amounts invested are in the ratio 2.5 : 1.5= 5 : 3.Since the amounts invested at the two rates are inthe ratio 5 : 3, and Rs. 680 is invested at 6%,

amount invested at 10% =5

680 × 3 = Rs. 408.

Questions 20 and 21: Half of the volumes of the containersA and B are emptied into the third container. Hence, thevolumes emptied are 108 cm3 and 32 cm3 respectively.

20. c Hence, the ratio of water to milk is 108 : 32 or 216 :64.

21. e1401000

× 100 = 14%

22. a Father Elder son Younger son now

20x 5x 4x

30x 15x 14x

After 10x years,

So 30x = 2 × 14x + 3⇒ x = 1.5 ∴ 20x = 30 years.

23. a The final amount of milk after two operations

= L2.195053

54

54 =

×××

24. c The harmonic mean of 2 numbers x and y is ( )yxxy2+

and the geometric mean is xy .

1312

xy

yxxy2

=+∴ ,

2 xy 12x y 13

⇒ =+ or

xy 6x y 13

=+

x y 13 x y 13or or

6 y x 6xy xy+ = + =

From here, you can put options in the above equationand check. Only option (c) satisfies.

25. c Let X ml be added to the solution.

⇒ [ ]

[ ](15% of 400) X 32

400 X 100

+=

+

⇒ X = 100 ml

Alternative Method:15% of 400 ml = 60 mlRest is water = 400 – 60 = 340 mlSince, water is constant ⇒ 340 ml will be 68% (inthe new mixture)⇒ 340 is 68%

⇒ 100 is 340

100 500 ml68

× = ; Extra alcohol

required 500 – 400 = 100 ml


Page: 83��


1. d The question uses the same concept as in Q 17.Number of mugs = 12The sum of broken and unbroken mugs must beequal to 12.Sum of ratio = 3, 12, 5, 6, (5 does not divide 12)∴ 3 : 2 cannot be the ratio.

2. a A’s effective investment = 16000 × 6 = Rs. 96,000B’s effective investment = 12000 × 8 = Rs. 96,000C’s effective investment = 1000 × 12 = Rs. 12,000Profit sharing ratio = 96 : 96 : 12 or 8 : 8 : 1

3. e Ratio of their profits = Ratio of their investments= 4000 × 12 : 8000 × 9 : 12000 × 2

48 : 72 : 24 2 : 3 : 1

Bs’ share = 36

× 5200 = 2600

4. c For every 100 paise that A gets, B gets 65 paiseand C gets 35 paise.

35C's share 560 sum

200= = ×

∴ Sum = Rs. 3,200

5. d L : R : O = 5 : 7 : 3Let Labour cost be 5x, Raw Material Cost= 7xOverheads = 3x, Total cost = 15xProfits = 20% of 15x = 3x

Material cos t 7x 7Pr ofit 3x 3

= =

6. c Let A’s, B’s and C’s investments be 8x, 7x, 5xA’s effective investment = (8x) × 5 + (4x) × 7

= 40x + 28x = 68 xB’s effective investment = 7x × 12 = 84xC’s effective investment = 5x × 12 = 60x

B’s share = 84

26500212

× = 10,500

7. a The net rate of interest paid is,

2100r 100 10.5% per annum

20,000= × =

∴ using the rule of alligation,

9

10.5

11

0 .5 1 .5

Thus ratio of loan from first to second bank is0.5 : 1.5 = 1:3

∴ Loan from 1st bank 120,000

4= × = 5,000.

and Loan from 2nd bank = 15000.

8.b Let the required quantity be x kg.∴ 20 × 2 + 24x + 30 × 3 = 25 (3 + x + 2)⇒ x = 5

9. b Let the amount of second and third quality of sugarin the mixture be ‘x’ and ‘y’ kg respectively.

[ ]10015 4 18x 22y 22 4 x y

110∴ × + + = × + +

y x 10⇒ − =

10. a Mean rice of the mixture is 1008 Rs.16 /kg

50× =

Let the quantity of second and third variety be ‘a’and ‘b’ kg.∴ 15 × 6 + 18a + 20b = 16[6 + a + b]

a 2b 3⇒ + =a = 1 and b = 1 is the only integral solution.

11. e Let the quantity of second and third variety be xand y kg respectively.

[ ]10020 1 30x 24y 30 1 x y

120∴ × + + = × + +

5x y 5⇒ − = …(i)

All the four options (a), (b), (c) and (d) satisfyequation (i).

12. e Let’s take12 litre of 5 : 1 sample24 litre of 2 : 1 sample and36 litre of 3 : 1 sampleAmount of milk in the final mixture

5 2 312 24 36 53 L

6 3 4= × + × + × =

Amount of water in the final mixture = 72 – 53= 19 L⇒ milk : water = 53 : 19.

13. e Using alligation we find the amount of copper(average amount of copper in the final mixture isx)

x

2 1

25

710

7– x 2102 1x –5

⇒ =

1

x2

⇒ =

Since, copper is 12

of the new alloy the final ratio

of copper to nickel will be 1 : 1.


14. b The ratio between the distance and the speed isthe same. Since the ratios between the two are thesame, the time has to be constant and hence theratio needs to be 1 : 1 : 1 or mathematically,

SpeedcetanDis

Time =

Hence, Ratio of time = 1:1:1speedofRatio

cetandisofRatio = .

15. e Selling at Rs. 11 per kilogram, profit percentage= 25%.

Therefore, CP =25.1

11 = Rs. 8.8

Applying the rule of alligation,

The required ratio is 2 : 3.

16. c Let the maximum marks in the quizzes be 1 andthat in the examination is 3. Hence, the weightage

of the examination = 3 1 1

6 1 3 1 3× =

× + × .

17. b If he spends ‘t’ minutes on the second half, then

he spends 32

t on the first half.

Therefore, 32

t + t = 90 or t = 54.

18. d If there were 2, 5, 1 coins of the denominationsRs. 1.50, one-rupee and 25-paisa respectively,then value of the coins would be Rs. 8.25. Hence,if the total value has to be Rs. 330, then thenumber of coins would have to be multiplied bya factor of 40. There would be in all 80 of Rs.1.50 coins, 200 one-rupee coins and 40 25-paisacoins.

19. c Let the value of the turban be Rs. x.

Then 100 x 1265 x 9

+ = ⇒+

or x = Rs. 40.

20. c Suppose the person buys A apples and Mmangoes and cost price of an apple is Rs. x.Therefore, cost price of a mango will be 2x.Total cost price = Ax + 2Mx.Now selling price of an apple is 2x.∴ SP of a mango will be 6x.Total SP = 2Ax + 6Mx.Now we have

52Ax 6Mx (Ax 2Mx)

2+ = + or

M 1A 2

= .

21. a Let us assume that out of 100, he obtained 72marks. If he had attempted four more questions,he would have made one more mistake. Hence,three of them are correct. The three correctquestions secure him 12 marks more. This

means, each question carries 124

3= marks and

hence, number of questions is 25.

Alternative method:We can clearly see that three correct answersincrease the score by 12%. [Three correct questionsas four new questions and one mistake]100% score corresponds to

25%100%12

3 =× questions.

22. a 20 boys = 10 men. 15 girls = 7.5 men.Therefore, total number of men = 17.5Now 50 × 10 × 8 = (17.5 × 15)x, where x is thenumber of days.Therefore, x = 15.24 (approximately).

Alternative method:In these type of questions it is always easier toworkwith man-hours (or man-days, etc). 50 men working10 hr a day for 8 days is 50 × 10 × 8 = 4,000 man-hours.Now 35 (boys and girls) working 15 hr a day for

(say) x days is

××× x1535

21

man-hours

000,4x153521 =×××⇒

24.151535

8000x =

×=⇒ days.

23.c Suppose numbers of students who pass andfail are 3x and x respectively.∴ Total number of students appearing= 3x + x = 4x.


Page: 85��

In second case,total number of students who appear = 4x + 8;total number of students who pass = 3x – 6.∴ Total number of students who fail= (4x + 8) – (3x – 6) = x + 14.

Now we have 114x

6x3 =+− ⇒ x = 10.

∴ Total number of students is 40.

24.c Ratio of white to yellow balls = 6 : 5.Difference in the number of white and yellowballs= 6x – 5x = x = 45.Therefore, number of white balls now available= 45 × 6.Number of white balls ordered= (45 × 6) – 45 = 225.

25. c Ratio of white mice to total mice

1 12 8

1 1 1: 1 : 4 : 1

4= = = .

Ratio of gray mice to total mice

1 13 9

1 1 1: 3 : 9 1 : 3 : 1

3= = = = .

∴ Ratio of white mice to gray mice =

4:331

:41 = .

26. e When 5 L brandy is transferred, volume ofbrandy left in vessel I = 15 L.Volume of brandy in vessel II = 5 L.Volume of water in vessel II = 20 L.Now, 4 L of mixture containing brandy and waterin the ratio 1 : 4 is poured into pure brandy ofvolume 15 L.4 L of mixture contains 3.2 L water and0.8 L brandy.∴ Volume of brandy left in vessel II= 5 – 0.8 = 4.2 L.Ratio of water in vessel I to brandy in vessel II= 3.2 : 4.2 = 16 : 21.

27. a In 100 L pure milk, let him add ‘a’ litres water.Let CP of 100 L pure milk = Rs. 100.CP of (100 + x) L adulterated milk = 100.

CP of 100 L adulterated milk =

+ x10010000

.

SP of 100 L adulterated milk = 100.

Profit percentage = ( )( )

100100 x

10000100 x

100 1100+

+

−×

or x = 30.

Therefore, 30 L is added or %3010010030 =×

of water is added to pure milk.

28. e No price values (either of profit amount or ofcost price) is given. Obviously, SP cannot bedetermined.

29.c v41

xv74 =− ⇒ 9v = 28x

Since by adding 35 L the level rises from aquarter to a half, the volume of the vessel = 35 ×4 = 140 L.

Therefore, 4 1

x 140 140 80 35 457 4

= × − × = − = L.

30. c Six people contribute a total of Rs. 180.Let the seventh person contributes Rs. x.Eighth person contributes Rs. 55.Total contributions of these eight persons= 235 + x.

Now 235 xx 10

8+= +

( )7 1x 235 10 x 45

8 8⇒ = + ⇒ =

∴ Total collection = 235 + 45 = Rs. 280.


1. c Percentage of whisky in mixture = 75%.Percentage of whisky in water = 0%.

So mixture and water should be mixed in the

ratio 2 : 1 or we need to replace 1

3 of the mixture

with water.


2. a Let x be the distance travelled by bus.

∴ 1636

x100080x =−+

⇒ x = 770 km (approximately)

3. d Applying the rule of alligation:

∴ Ratio of volumes of two solutions = 1 : 2.

4. e Tin content in the alloy = 100 – 77.78% = 22.22%.Using the principle of alligation,

Ratio of volumes = 18:4.3218 : 1878.27

50 =

× .

∴ Total tin content in the new alloy= (22.22% of 32.4) + 18 = 25.2 kg.Total copper = (77.78% of 32.4) = 25.2 kg.Hence, none of these is the correct option.By the way, is there any need to do anycalculations? In the question it is given that thenew alloy has 50%, i.e. equal amount of copperand tin.

5.b Quantity of milk after first draw

= 90

80 72 kg100

× = .

Quantity of milk after second draw

= 90 90

80 64 kg100 100

× = .

Quantity of milk after third draw

= 32.5880109

109

109 =×××

∴ Ratio = 100 : 81.

Alternative method:We know that

New amount of liquid XOriginal amount of liquid X =

nAmount of liquid taken out and replaced

1Total amount of liquid

−

where n = Number of times the process is repeated

= 1 –

2

808

2

109

=

∴ Ratio = 22 9:10 = 100 : 81

6. b Initial respective amount of milk and water is 28L and 7 L.7 L of water added means now we have 14 L ofwater. To keep the ratio M : W same, amount ofmilk must be doubled, i.e. 28 L of milk must beadded.

Alternative method:In order to keep the solution of the same ratio, weneed to add milk in the same ratio as already there inthe solution.

∴ We need to add 28714 =× L of milk.

7. a Rule of alligation for fractional content of water:

or 11 : 1 or 33 : 3∴ 3 L of water must be added to 33 L of solutionof milk.


Page: 87��

8.b In fact, she takes out 5% of the existing volumeof wine in every operation.After the first operation concentration of wineis 95%.After the second operation concentration ofwine is 95 × 0.95 = 90.25%.After the third operation concentration of wineis90.25 × (0.95) = 85.7375%.Obviously, after the fourth operat ion theconcentration of wine will be less than 85%.

9.b The ratio of incomes of Rashmi and Divya is3 : 5. Ratio of their expenditures is 2 : 3, i.e.3 : 4.5. Had the ratio of expenditures been 3 : 5,ratio of savings also would have been 3 : 5; butsince ratio of their expenditures is 3 : 4.5 only,obviously savings of Divya will be something

more than 53

of savings of Rashmi and thus

Divya will save more.

10.c Rohit Sunil

Income 5x 7x

Expenditure 5y 7y

Saving 5x – 5y 7x – 7y

Ratio of savings is 7:5)yx(7)yx(5 =

−−

.

11.b Sum of squares of first n natural number is

6)1n2)(1n(n ++ .

Square of sum of first n natural numbers is

325:174

)1n(n 22=+

Now 325:17

4

2)1n(2n

6)1n2()1n(n

=+

++

.

On solving, we get n = 25.

12.c If IBM initially quoted Rs. 7x lakh, SGI quoted 4x lakh.IBM’s final quote = (4x – 1) lakh

43

x41x4 =−

or x = 1

IBM’s bid winning price = Rs. 3 lakh.So IBM wins the bid at 4x – 1 = Rs. 3 lakh.Note: Whoever quotes the minimum price, wins thebid.

13. b SP = 4.5 and profit = 20%.

Hence, CP =120450

= 3.75

Using the alligation method,

0 .25 0 .25

i.e. 1 : 1Hence, in a 40 kg mixture, there are 20 kg of the firstvariety and 20 kg of the second.

14. a Here we calculate the number of birds in terms ofnumber of trees. We know that there are 630 birds.Let the number of trees be x.

∴ Number of sparrows = 22x × .

Also number of pigeons = 141

2x ×

× .

Other birds = 443

2x ×× .

Thus, x + 240x630x821

2x3

8x =⇒==+

Hence, other birds are there in 908x3 = trees.

15.a 6a2 = 2(lb + bh + lh)l : b : h = 1: 2 : 4. Therefore, b = 2l; h = 4lHence, 6a2 = 2[2l2 + 8l2 + 4l2]6a2 = 28l2

a =

1214

l3

323 14

a3

∴ = l

3

Ratio of the volumes of the cube and cuboid

= a3 : lbh =23

314

: 8


16. a If three kinds of birds are taken to be 3x, 7x and 5xrespectively, then 7x – 3x = 63y (where y is anypositive integer). As the number is a multiple of both9 and 7, it has to be a multiple of 63.

4y63

x =⇒

Minimum value of y for which x is a natural numberis 4.⇒ x = 63Hence, the number of birds = 15x = 945.

17. e Price = k × Weight, where k is any constant.Original weight of the stone = 28x.Original price = 28kx.New price = k(15x + 13x) = 28kx.Hence, there is no profit, no loss.Concept: Please note that since we have a linearfunction, there is neither profit nor loss.

18. d In a mixture of 1,000 ml, milk : water = 3 : 1.Hence, milk = 750 ml, water 250 ml.A 250 ml of 3 : 2 solution contains 150 ml milk and100 ml water.Total milk = 900 ml, total water = 350 ml.After using 250 ml to make curd milk used

= 250

9001250

× = 180 ml.

Pure milk left = 900 – 180 = 720 ml.

19. d Let income and expenditure of the first person be5x and 9y respectively.Then income and expenditure of the second personare 4x and 7y respectively.5x – 9y = 4x – 7y = 500or x = 2y.Therefore, y = 500, and x = 1000.Incomes are Rs. 5,000 and Rs. 4,000 respectively.ORBy checking the choices you will find that only forchoice (d) is the ratio of incomes 5 : 4.

Alternative method:In the given choices, the choices (a) and (c) are notinthe ratio 5 : 4. Now we can subtract Rs. 500 fromincomes and check the ratio of expenditures inchoice (b). Ratio of expenditures= (3750 – 500) : (3000 – 500)= 3250 : 2500 = 13 : 10Hence, the solution is (d).Please note that we need not calculate the ratio ofexpenditures in (d).

20. d A, B and C are to share profit in the ratio 4 : 3 : 7.But, since A is to receive 10% of profit, only 90% isto be shared.

A’s total share = 10% of profit + 144

× 90% of profit

= 6000

or (10 + 7

180)% of profit = 6000

or 7

250% P = 6000 ⇒ P = Rs. 16,800

B and C’s share combined

= 1410

of 90% profit 1680010090

1410 ××=

= 10,800

21. a In the original mixture,

spirit = 43

× 40 = 30 L,

water = 41

× 40 = 10 L.

To get the ratio 5 : 2, let’s assume that we add xlitres of water.

25

)x10(30 =+ or x = 2

Hence, 2 L of water has to be added.

22.d

i.e. 2 : 1Hence x litres must be 12 L, i.e. (6 × 2).

23. c This is a classic case for approximations. We seethat the ratio 1075 : 1000, when converted topercentage terms, is approximately 52% : 48%. Thus,we can calculate the literacy percentage as

24 852 48 16.3%

100 100× + × ≈

Now as 100 – 16.3 = 83.7%, only choices (c) and(e) are left. We calculate the number of literatepeople as

16.3311250 50,700

100× = .


Page: 89��

We could have approximated the ratio of men towomen as 50% : 50% also.

24. b Let x be the number of children employed.Then numbers of men and women are 8x and 5xrespectively.Let y be the common factor for wages.Then wages of men, women and children are Rs.5y, 2y and 3y respectively.Hence, total wages of different categories will be40xy, 10xy and 3xy respectively.Total wages = (40 + 10 + 3)xy = 53xy = 318.So xy = 6.Thus, they will be getting 40 × 6, 10 × 6 and 3 × 6respectively.

Alternative method:

Sum of all three wages have to be equal to Rs. 318.

Only (b) displays this property.

25. a Speed in upstream= Speed of the boat in still water – Speed of the river= b – r.Speed in downstream= Speed of the boat in still water + Speed of the river= b + r.Therefore, b + r = 3(b – r)

or 2b = 4r or b = 2r

Since r = 4 km/hr, b = 8 km/hr.

26. a Milk in final/Milk in original = [(a – b)/a]n,where a = Quantity of mixture,b = Amount taken out during each operation,n = Number of times the operation is repeated.

Milk in final/Milk in original =

33

43

401040

=

−

.

Milk in final solution = 40 ×6427

= 16.875 L.

Water = 23.125 L.Required ratio = 16.875 : 23.125 = 27 : 37.

27. b This question can either be done by substitutingchoices, or by using weighted averages.If x be the number of members in club initially, thensum of ages of (x + 3) members now = 26x + 87

= 27(x + 3)Hence, x = 6.So number of members now = 9.

Alternative method:This method is called sum of deviations, as follows.“The sum of all weighted deviations from the averageis zero.”

Thus, assuming we have x students initially.x(26 – 27) + 3(29 – 27) = 0

– x 6 0⇒ + =x = 6Please note that always subtract the average fromthe number and not vice versa.

28. a Assuming that there is 1 L of spirit and 3 L of water,the total volume of the mixture is 4 L.After increasing its volume by 25%, volume of themixture is 4 + 25% of 4 = 5 L.So the ratio of volumes of spirit to water = 2 : 3.

29.c A ratio 21

: 31

:41

is equivalent to 6 : 4 : 3.

So in this case, A, B and C would have got Rs. 54,Rs. 36 and Rs. 27 respectively.But actually the money was divided in the ratio2 : 3 : 4 and shares of A, B and C in this case was26, 39 and 52 respectively.Hence, the answer is (c).

Alternative method:

25.0:33.0:5.041

:31

:21 =

Thus, C, who was getting the least, finally got themaximum amount.This eliminates (a) and (b).Thus, gain of

C = Rs.117

++++ 5.03/125.025.0

–234

4

= Rs. 25Hence, the answer is (c).

30. d

i.e. 7 : 1Hence, milk and water are in the ratio 7 : 1 sincethere is 28 L milk. Hence, water should be 4 L.



1. a If A, B and C represent the number of threedenominations in increasing order of their values,then A = 7B, C = 4B.Since C = 12, B = 3 and A = 21, the total money in thepurse = 21 × 1 + 3 × 10 + 12 × 20 = Rs. 291.

2.c If there are x number of one-rupee coins in the bag,then there are 8x and 16x, 50-paisa and 25-paisecoins respectively.Therefore, money in the bag = x + 4x + 4x = 9x =495.So x = 55.Hence, there are 8 × 55 = 440 50-paisa coins in thebag.

3. a There are 880 25-paisa coins.Value of the money = 0.25 × 880 = Rs. 220.

4.c

V 1 V2 V3

Initial volume: 0 L 40 L milk 80 L W water

After operation 1: 0 20 L M + 20 L W 60 L w + 20 L milk

After operation 2: 0 15 L M + 25 L W 55 L w + 25 L milk

5.a Check from the above data.

6.a Suppose he invested Rs. x and Rs. y in thematches o f South Af r ica and Ind iarespectively. From the first match he will get 3x, but from the second he will get nothing.

Now 3x = (x + y) + (x + y)100150

or yx

= 15

.

7.a Suppose he invested 3X on the first matchand 4X on the second match.He will get back 3X × 3 = 9X .So gain = 9X – 7X = 2X = 200000or X = 100000:. Total investment = 7X = 7 × 100000 = 700000

8.c Investment = x + 2x = 3xHe gets back 3x + 8x = 11x. So gain = 8x.

Gain percentage = 8x

1003x

× = 266.66%.

9. c C = 3

CBA ++ ⇒ Originally B had an amount

= 2C – A

Now A′ = 2A

, B′ = 1B

3, C′ = 0

∴ 2A

+ 3B

= 21

(2C – A) = 2B

∴ B = 3A, C = 2ASo amount spent

= (A + 3A + 2A) – 3A2

= 4.5A

Percentage of amount spent = 75%.

10.e We cannot determine the final ratio unless we knowthe volume in 3 jars.

11. b Mixing two lumps, we have 18 g from each metal.So price of the second metal

= 18

)187.660.7887( ×−+ = Rs. 2.50 per gram

∴ 6.7x + 2.5(18 – x) = 87∴ x = 10 g and 18 – x = 8 g

12.c2

x54 1 24

54 − =

⇒ 2

x1

54 −

=5424

= 94

⇒x

154

− =32

⇒ x = 18

13. a When 25 L is withdrawn, half the solution isremoved. Hence, 9.6 L of milk is left.

14. c Let Tina’s share be T, Issan’s be I, Abhishek’s be Aand Fatima’s be F.

Given that T + 3 = I + I 80A

F 43 100

= = −

You getT = F – 7 ... (i)

I =43

(F – 4) ... (ii)

A =45

(F – 4) ... (iii)

Also given that T + I + F + A = 80 ... (iv)Substituting the values from (i), (ii) and (iii) in (iv),we get

+++ F

4F5

4F3

F – 7 – 3 – 5 = 80 or F = 23.75

15. a Let the price of one tea = x

xthe price of other tea=

2⇒

Price of 1 kg 2x 3 x 7x5 5 2 10

= + × =

But CP = 17 50 100

12514

. ×=


Page: 91��

(Only the first option satisfies the first condition.)

⇒ = ⇒ =710

14 20x

x

So the price of the tea’s are 20 and 10.

16. c If the person who gets 41

of the whole gets thrice

of what the others get on an average, each one willget

= 121

41

31 =× of the whole.

Therefore, if there are x persons other than theperson who gets one-fourth of the whole, then

1 x1

4 12+ =

So x = 9Hence, total number of people = 10

17. c 6p + 8q = 7.5(p + q); 6q + 8p = n(p + q)

5.6nqp)n5.7()qp(14 =⇒++=+

18. b Amount of 16% iodine in the mixture

= 16

735 gm100

×

Now the amount of iodine becomes 20% of themixture.Amount

evaporated16 100

735 735100 20

= − × × = 147 g

19. b Whether you add them in the ratio 2 : 3 or 1 : 1, weget the same concentration.Hence, both of them must have a 35% concentration.So (b) is the answer.

20. c Only 5 women were college graduates while 20women were employed. Therefore, maximumnumber of ‘employed graduate women’ can be 5.Since number of ‘unemployed men’ was (55 – 15) =

40. Answer cannot be greater than 81

.

21. b Only possibility is if 650 is divided by a multiple of13, because the other jars have a capacity that isnot a multiple of 13. 403 is the only choice possible.

Alternative method:As the beakers filled by same fractions,

200650250cba

200c

650b

250a

++++===

682cba =++� 1100682

650b =∴

b = 403 ml

22. d 10 kg copper is from alloy II.Hence, 28 kg copper is from alloy I.In alloy I, tin : copper = 3 : 4Since, 4 units = 28 kgAlloy = 7 units = 49 kg

23 c Use alligations.

Resultant average

Average percentageof acid in so lutions 90% 75%

X YRatio of quantitiesof acid in so lutions

Solution 1 Solution 2

XY

75787890 =

−−∴

41

YX

XY

312 =⇒=⇒

So ratio of quantities of solutions = 1 : 4Total quantity = 30 L∴ Quantity of 90% solution of concentrated acid

= L651

30 =×

24. d k4

cs7

bs1

as =−=−=−

Simplifying this, we geta : b : c = 11 : 5 : 8


1.c After first transfer,amount of wine in first vessel = 9 L,amount of water in first vessel = 3 L,amount of wine in second vessel = 3 L,amount of water in second vessel = 1 L.

3 L drawn from first vessel contains 49

343 =× L

of wine and 43

341 =× L of water.

3 L drawn from second vessel contains 49

L of

wine and 43

L of water.


∴ Amount of wine in first vessel after second

transfer = 949

49

9 =+− .

Amount of water in first vessel after second

transfer = 343

43

3 =+− .

Similar ly, for second vessel, wine =

349

49

3 =+−

Water = 143

43

1 =+−

∴ Ratio is same in both the vessels.

Alternative method:Ratio of wine to water in first vessel after firsttransferis (12 – 3) : 3 = 3 : 1.Ratio of wine to water in second vessel after firsttransfer is 3 : (4 – 3) = 3 : 1.Since the ratio of wine to water in both the vesselsis equal. Any amount of such exchange any numberof times will not alter the ratio of wine to water.

2.e Consider two cases:

i. Princy Kunjumol

Income 100 200

Expenditure 1 5

Saving 100 – 1 = 99 200 – 5 = 195

Thus, Kunjumol is saving more.

ii. Princy Kunjumol

Income 100 200

Expenditure 39 195

Saving 100 – 39 = 61 200 – 195 =5

Thus, Princy is saving more.Hence, their savings depends upon theirrespective expenditures.

3. c 5a + b > 51 … (i)3a – b = 21 … (ii)8a > 72 or a > 9.

From (ii), 3

b21a

+=

6b93

b21 >⇒>+⇒

Combining a > 9, b > 6.

4. b Average age = 13. Number of students = 50W ∝ H and H ∝ A ⇒ W ∝ A

∴ W = 1K H and H = K2A ⇒ W = K1K2A

For A = 11, H = 165 and W = 33

K1K2 = 1133

= 3 ∴ W = 3A

Since W ∝ A, the averages should also be in directproportion.∴ Wavg = K1K2 A avg = 3 × 13 = 39 kg

Explanations: Fundamentals ofTime, Speed & Distance

Page: 93��

Solutions (Non MCQ)

Level – I

1. Distance to be covered = 100 m.Speed of the person = 36 km/hr = 10 m/s.

Therefore, time taken = 100 m

10 s.10 m/ s

=

2. Average speed = [ ] hr/km605075

50752 =+

××

3. One revolution means circumference of the wheel= 250 cm = 2.5 m.In 10 revolutions 25 m is covered by the wheel.

Speed of the wheel = 5

184

25s/m

425 ×= km/hr =

22.5 km/hr

4. Method 1:The man takes 2 hr to travel 14 km, where he stopsfor 10 min. Therefore, time between the start ofjourney and start after first stoppage= 2 hr 10 min.This pattern of his travel repeats. Now, 98 km= 14 × 7. But when he completes 98 km, the case ofhis stopping for 10 min does not arise. Therefore,total time taken= 6 (2 hrs 10 min.) + 2 hr = 15 hr

Method 2:If the man had travelled non-stop he would havecovered the distance in 14 hr. But since he stops 6times on the way he would take an extra 6 × 10 = 60min to cover the same distance. Hence, he wouldreach the destination after 15 hr.

5. Total relative distance travelled by the car = [40 + 60] m = 100 m.

The relative speed = 100 m

5 m/ s.20 sec

=

The relative speed is also equal to(30 – v) km/hr

= (30 – v)×

185

m/s, where ‘v’ is the speed of the

bus.

Therefore, (30 – v) × 185

= 5 or, v = 12 km/hr.

6. Speed of first train = s/m10200

= 20 m/s.

Speed of second train = s/m15200

= 340

m/s.

Relative speed of the two trains (when theymove in opposite directions)

= 20 + 3

100340 =

Time taken by the two trains to completely pass

each other = 3/100

200200 += 12 s

7. As distance covered by faster one is 4 timesthe distance covered by slower one while theirspeed ratio is 3 : 2. So faster one is movingdownstream and slower one is movingupstream.

So + =

6 u 44 – u 1

They will meet first time in (6 + u) t + (4 – u)t = 100;t = 10 s.For second time meeting, faster is coming backto its starting point after reaching starting pointof slower one.

20 80

Slower Faster

B A

20 m will cover by A to reach slower’s starting

point. So he will take 820

= 2.5 s to reach B. In

this time slower one has cover 2 × 2.5 = 5 m.Now, A has to cover (20+ 5 = 25 m) withrespect to B to catch B. Now A is also movingupstream so his speed is now 6 – 2 = 4 m/s. Sorelative speed of A with respect to B is4 – 2 = 2 m/s.

So A will take 25

12.5 s2

= .

So in all = 10 + 2.5 + 12.5 = 25 s.They will meet second time after 25 s.

��


8. Speed downstream = v + u (where, v = Speed ofthe boat in still water, and u = Speed of the stream).Speed upstream = v – u.Distance downstream = Distance upstream(v + u) 45 = (v – u)75 or, u : v = 1 : 4.

9. a. Ratio of distances = Ratio of speeds= 20 : 23 :27.

b. Let us calculate the time of meeting of A and Bfor the first time. A will meet B after

2023200

− = 3

200s.

This means that A and B meet every 3

200s.

Similarly, B meets C every 4

2002327

200 =−

= 50 s.Therefore, LCM of these time periods will giveus the time when all the three meet the first time

which is LCM

50,3

200

= 200 s or (3 min. 20 s).

10. A’s speed is 2 m/s with respect to B. (i.e. relativespeed)So to catch B he has to overcome 250 m because B

is 250 m ahead of A. So A will take = s1252

250 = .

11.

P

A

B

Q

500 – x

x

At the first time they will meet at point P. After thatthey meet for the second time at point Q. Up to pointP distance travelled by A and B is add up to 250 m.After that when they meet at Q together they havetravelled one circle. So total distance travelled bythem is 750 m [250 + 500]. So time taken by them tocover this much distance is (5 + 3)t = 750

t = 75.938

750 =

12. Let the train increase its speed after ‘t’ hours.Total time taken by the train = 1 hr 15 min = 1.25 hr.Therefore, 40t + 50(1.25 – t) = 55 km.Or t = 45 min.

13. At 10 a.m. a train from station B is just coming tostation A. That train must have started at5 a.m. from station B. The train that starts at 10a.m.. will reach at 3 p.m. at station B.So from 5 am to 3 p.m. every train started fromstation B will cross the train. So total number oftrain = 11.

14. From A a train is starting at 10 a.m. So from B every

train will start at an (integer + 12

) hr. interval i.e. 5 :

30, 6 : 30, 7: 30 .... and so on.

So the 10 a.m train from A will meet first train whichis suppose to reach at 10 : 30 starting from B at5 : 30.Also, 10 a.m. train from A will reach station B at3 p.m. So the last train it will cross is the one startedat 2 : 30 p.m. from station B. So from5 : 30 to 2 : 30, we have a total of 10 trains.

15. So at 10 a.m. a train is just reaching station A thatstarted from B at 7 a.m. So, the train that left at10 a.m. train will reach at 3 p.m. Thus, from 7 a.m. to3 p.m. each train will cross the train from B thatstarted at 10 a.m.So total number of trains = 8 + 1 = 9 trains.

16. In this case we cannot find the exact number oftrains crossing because we do not know thestarting time of any train started from station B.May be possible it will reach station A at 10 a.m. orat 11 a.m. If it reaches at 10 a.m. to station A, thenit starts at 5 a.m. and 10 a.m. train will reach at 3p.m. So total number of trains crossing.= 5 a.m., 7 a.m., 9 a.m., 11 a.m., 1 p.m., 3 p.m. = 6Or if train started from B reach A at 11 am shouldstart from Bat 6 a.m. So number of crossing trains:= 6 a.m., 8 a.m., 10 a.m., 12 a.m., 2 a.m. = 5 trains.

17. They will finish 100% of the work in 30203020

+×

= 12 days.Therefore, 75% of the work will be done in 75% of12 = 9 days.

18. Method 1:Let a, b and c be the times taken by A, B and C to dothe work alone.

Also let x = a1

, y = b1

, z = c1

.

Therefore, x + y = 201

, y + z = 151

, z + x = 181

.

From the last two equations, y – x

= 151

– 181

= 901

.


Page: 95��

Therefore, 2x = 201

– 901

= 180

7, which means x

= 3607

or a = 7

360days.

Method 2:Assume that the number of units of work is 180units.A and B put in 9 units per day B and C put in12 units per day; and A and C put in 10 units per day.

So A, B and C working together put in 2

)12109( ++

units per day, i.e. 15.5 units per day. So A alone puts

in 15.5 – 12 = 27

units per day. So he

takes7

36072

180 =× days to complete the job

working alone.

Level – II

19. Let the distance travelled by the man be x km.Therefore, time taken at a speed of 10 km/hr

= 10x

hr.

Time taken at a speed of 20 km/hr = 20x

hr.

Hence, 6045

20x

10x =−

Solving, we get x = 15 km.Therefore,

x x 15 15– – 0.225 hr 13.5 min

25 40 25 40= = =

20. Let us say ‘t’ hr is the usual time. When Sanju

moves at 10 km/hr, time of travel = 606

t +

When he moves at 12 km/hr, time of travel =609

t −

Difference in time taken in the two cases

= 41

203

101

609

t606

t =+=

−−

+ hr

If x km is the distance between school and house,difference in time taken

= 60x

12x

10x =−

Therefore, .hr41

60x = or x = 15 km.

21. Ratio of distance = Ratio of speed = 32 : 41.Therefore, distance travelled by both the trains= 32x and 41x.According to given condition, 41x – 32x = 45 kmTherefore, x = 5. Hence, distance betweenA and B = 41x + 32x = 365 km

22. Let the trains meet at some place M betweenA and B. Time taken from A to M = time taken from Bto M.Therefore, ratio of speed = ratio of distance= 4 : 3.

23. A takes a lead of (3.5 km/hr) (2.5 hr) = 8.75 km overB.Hence, when B starts he has to travel a relativedistance of 8.75 km and at a relative speed of(4.5 – 3.5) km/hr = 1 km/hr, which means B will take8.75 hr to catch up and then overtake A.

24. In one day, the time gap between the two faultywatches is 6 + 4 = 10 min.The two watches will show the same time nextwhen the faster watch takes a lead of exactly12 hr over the slower. This would happen after

1060121 ××

= 72 days. In 72 days the faster watch

would gain

72 44.8 hr 4 hr and 48 min.

60× = =

So the time according to the faster watch wouldbe 7:48 p.m. and the slower would be 12 hr behindbut would also show the same time.

25. Normal day it travels a distance = 25 (50 + u)

On that particular day it completes 12

the journey in

same time (12.5 hr).

But in half time due to engine problem it takes 1

22

hr

extra.So it take 15 hr to complete the remaining halfjourney.

So, 15(40 + u) = 1

25(50 u)2

× +

30(40+ u) = 25 × 50 + 25 u30u – 25u = 1250 – 12005 u = 50; u = 10 km/hr


26. Ratio of speed of Bhim and Arjun = 7 : 4.(Since the ratio of the times = 4 : 7)a. If the length of circular track = 28m, the speeds

of Bhim and Arjun are 7 and 4 m/min.The time when they are together the first timewill be when Bhim (the faster one) has takenone round more than Arjun (the slower one).Therefore, if time when they meet is ‘t’, then 7t

– 4t = 28, which means t = 328

min.

b. They will meet at the starting place the first timeat a time which is the LCM of the times eachone of them takes to reach the starting place.Therefore, LCM of 4 and 7 is 28 min.

c. Diametrically opposite point is at a circulardistance of 14 m. Bhim reaches this point in

714

= 2 min and Arjun reaches this point in 4

14=

3.5 min. Bhim reaches this point in the 2ndmin, 2 + 4 = 6th min, 6 + 4 = 10th min ... so on.Arjun reaches after 3.5 min, 10.5 min, 17.5min. so on.

The time after the start when Bhim reaches thepoint is a natural number, whereas the timewhen Arjun reaches this point will always bea non-natural number.So they will never meet.

Alternatively for (a) and (b):If the time when they would meet for the first time atthe starting point = LCM(4, 7) = 28 min, in this timeBhim does 7 rounds and Arjun completes 4 rounds.So Bhim takes a lead of 3 rounds. Hence, he would

take

328

min to take a lead of one round. This is

the time when they would meet for the first time.

27. When Bhim gives Arjun a lead of 4 min, in 4 minArjun travels 16 m. Bhim now starts running towardsArjun at a relative speed of 3 m/min. To catch up

with a distance of 16 m, Bhim takes 3

16 min.

28. Let x be the distance between A and B.The train which leaves A travels a total distance ofx + (x – 200) = 2x – 200.The train which leaves B travels a total distance ofx + 200.The ratio of distances travelled by the trains= Ratio of the speeds.Therefore, (2x – 200) : (x + 200) : : 40 : 60Or x = 250 km.

29. Method 1:Let the total distance between A and B be x.Since B had to travel 145 km to reach A, this meansthey met at a distance of 145 km from A.Hence, distance between the meeting place and B= (x – 145) km.Ratio of speed = Ratios of distance25 : 49 : : 145 : (x – 145).Therefore, x = 429.2 km.

Method 2:

(49 25)x 145 429.2 km

25+ = × =

30. As A starts puffing, the length of burnt cigarette in

one cycle is gap gapA B → →

= 2 × 3 + 3 + 3 × 3 + 3= 6 + 3 + 9 + 3 = 21 mm.

As there are three such cycles 63

321

= � ,

time taken in one cycle = 2 + 3 + 3 + 3 = 11s.So total time = 3 × 11 = 33 s.

31. 1 1

2 2

3 3

4 4

5 5

G oing out Com ing in

So the hour hand and minute hand move 360° in thistime (together).

As the speed of the hour hand = °1

2per minute and

speed of minute hand = 6° per minute. Let t is therequired time in minutes.

So 1

6t t 3602

+ = °

13t 360

2= °

720t min.

13=

32. Let us say total work is of 20 × 20 units.Each woman does 1 unit per day.Each man does 2 units per day while each boy

does 12

unit per day.

So on first day 20 units are completed.On second day 21 units, on third day 20.5 units andon fourth day 21.5 units are completed.


Page: 97��

Combining two terms, we get 41 + 42 + 43 + ... + 48+ 49 = 405 > 400.So there are 9 such pairs. So on the 18th day workwill be complete.

33. Let the total work be of 4,00 units.So a woman does 1 unit per day,while a man does 2 units per day.So work done on first day = 20 unitsand work done on second day = 21 units.Thus, the pattern goes like 20 + 21 + 22 + ... + 30= 275275 + 31 + 32 + 33 = 371 up to 13th day, 371 unitshas been completed. On 14th day job will becomplete.

34. C can start the work on first day but there isnothing to demolish then. So this day is wasted.The work effectively starts from second dayonwards.Let the total work be of LCM(10, 15, 20) = 60 units.Now A can do 6 units in a day.B can do 4 units in a day.C can demolish 3 units in a day.So in 3 days (starting from A) 6 + 4 – 3 = 7 units willbe complete. So in 8 × 3 days, 8 × 7 = 56 units ofwork will be done by A, B and C. The rest 4 units

will be done by A in 2

day3

. So total number of days

= + × + =2 21 8 3 25 days.

3 3

35. Let the total job be of LCM(10,15) = 30 units.A + B can do 3 units per day.B + C can do 2 units per day.A + B + C can do 5 units a day.A + 2B + C can do 5 units.Here it can be clearly observed that B’s share in thetotal work is null. Hence, he cannot complete the jobby himself.

36. As A alone can do the job in 10 days and A and Btogether also taking 10 days to complete the job. So100% of A = 60% of A + B’s efficiency. So B will take

1025 days

0.4= .

37. Let the leakage take ‘t’ minutes to empty the tankcompletely. When all the three pipes (two inlets and

one leakage) are open, then 251

t1

–201

101 =+

or, t = 100/11 = 9.09 minutes (approximately).

38. Method 1:The outlet pipe will discharge the water in

3200

302000 = minutes alone.

Let the inlet pipe take ‘t’ minutes to fill the tank alone.

Then 401

2003

–t1 = . Therefore, t = 25 min.

Method 2:The net inflow is 50 l per minute. Hence, the actualinflow = 80 l per min (50 + 30). So the inlet pipe will

take

802000

= 25 min to fill the tank independently.

39. A → First pipe

B → Second pipe

C → Third pipeA can fill in t hr.B can fill in 2t hr

C can fill in t 2t 3

t hr2 2

+ = .

So,

1 1 2 1t 2t 3t 3

+ + =

1 1 2 11

t 2 3 3 × + + =

1 13 1t 6 3

× = t = 6.5 hr.

Level – III

40. 36 km/hr ⇔ 10 m/s

18 km/hr ⇔ 5 m/sIt is mentioned that the reduction in speed of thefaster truck = 1 m/s per second. Note that the brakesare applied successively.Thus, the faster truck travels 9 m in the 1st second,8 m in the 2nd second, 7 m in the 3rd second and soon.Final speed of the faster truck = 0 m/s⇒ Truck stops at the 10th second.Since both the trucks stop simultaneously, speed ofslower at 10th second = 0 m/sIf the reduction in speed of the slower truck= ‘a’ m/s per second, (5 – 10a) = 0

⇒ a = 12

m/s per second.

Thus, reduction in the speed of the slower truck =

12

m/s per second.


Alternative Method:Ratio of speeds = 18 : 36 = 1 : 2Since both the trucks are equally efficient, thereduction in the speed of the slower truck must be

12

m/s per second

Now, distance travelled by the first truck in 10seconds = (9 + 8 + 7 + 6 + … + 1 + 0) = 45 mDistance travelled by the second truck in 10 seconds= (4.5 + 4 + 3.5 + 3 + … + 0.5 + 0)= 22.5 mThus, net separation between the trucks when theynotice each other = 45 + 22.5 = 67.5 m

41. The thief travels for 2 hr (7.00 p.m. to 9.00 p.m.)and takes a lead of (4.5 km/hr)(2hr) = 9 km.The policeman is required to cover up a distance of9 km at a relative speed of1.5 km/hr. Therefore, the policeman will take6 hr after 9 p.m. to catch up with the thief, i.e at3 a.m.

42. The rate of burning of first candle = 1 cm/hr.The rate of burning of second candle = 1.5 cm/hr.They will be of the same length when the longercandle melts by a relative length of (10 – 8)= 2 cm.Therefore, time after which they will be of the same

length = ]hr/cm1–hr/cm5.1[cm2

= 4 hr.

After 4 hr, the candle of length 8 cm is now8 – 4 = 4 cm long which would be of the samelength as the length of the other candle.

43. Rate of flow for the first pipe = λ(1)2 = λ.Rate of flow for the second pipe = λ(2)2 = 4λ.Rate of flow for the third pipe = λ(3)2 = 9λ.Since rates are inversely proportional to time takenby each to fill the tank, time taken by the 3 pipes are

9 min, 49

min and 1 min respectively. Therefore,

when all the three pipes are opened, the fraction ofthe tank filled in 1 min

= 9

1411

94

91 =++

The time taken by all the pipes to fill the tank is

149

min. (λ is the constant of proportionality)

44. Tap 1 can fill the tank in 10 hr.

So tap 1 gives 40

4 lit / hr10

= .

Similarly, tap 2 leaks = 40

2 l /hr20

= .

Evaporation which is similar to a leakage

2.540 1 l /hr.

100= × =

So total leakage is 3 l/hr.So added water in an hour is 4 – 3 = 1 lSo it will take 40 hr.

45. 3t = 6Hence, t = 2Distance travelled by bus is equal to area under

the curve, i.e. 1 1 11 1

Vm 2 Vm 2 Vm 22 2

× × + × + × ×

= 4Vm1 [Vm1 = Bus (maximum speed)]Distance travelled by jeep.

m2 2 m21 1

V 2 Vm 2 V 22 2

× + × + × ×

m24V= [ m2V = jeep’s maximum speed]

So 4Vm1 + 4Vm2 = 280Vm1 + Vm2 = 70Vm1 – Vm2 = 10Vm1 = 40 km/hrVm2 = 30 km/hr

46. Average speed by bus

Dis tance travelled by busTime taken by bus

=

14Vm6

=

2 8040 km/hr

3 3= × =

Average speed by jeep

= Dis tance travelled by Jeep

Time taken by jeep

24Vm6

=

230 20 km/hr

3= × =


Page: 99��


1. a A takes 7 × 9 = 63 hr.

∴ in 1 hour A does = 1

63 of work

B takes 6 × 7 = 42 hr.

∴ in 1 hr B does = 1

42 of work

A + B in 1 hr = 1

63+

142

A + B in 425

hr = 1 1 42

+ ×63 42 5

= 0.33

∴ Number of days = 1

0.33 = 3

2. e In 1 day, A does = 1

80 part

∴ 10 days’ work = 1080

= 18

part

Work to be done by B = (1 – 18

) = 78

Let ‘B’ do ‘x’ part of the work in 1 day

x × 42 = 78

, x = 7

42×8 =

7336

Number of days required by B = 336

7= 48 days

A + B = 1 1 128

+ =80 48 80×48

∴ Number of days = 80×48

= 30128

days

3. b Let us assume that work done by A and B in 1 dayare 2 and 1 unit respectively.Work done in 1 day by (A + B) = 3 units∴ Work done in 14 days by (A + B)= 14 × 3 = 42

∴ A finishes the work in = 422

= 21 days

4. b Assume total wages = 21 × 28A’s wages for 1 day = 21 × 28/21 = 28B’s wages for 1 day = 28 × 21/28 = 21A + B’s wages of 1 day = 49.A + B’s wages will last for 21 × 28/49 = 12 days

5. e Assume that units to fill up= LCM of (20, 15, 12) = 60.Units filled up by A, B, C in 1 min= 3, 4, 5 respectively.

Time taken by (A + B + C) to fill up = 6012

= 5 min.

6. c Assume that capacity of the tank= LCM (10, 12, 20) = 60 units.

Rate of A, B, C per hr = 60 60 -60

, ,10 12 20

= −6, 5 and ( 3) units respectively.

∴ A + B = 11 units, C = –3 units∴ Per hr intake = 11 – 3 = 8 unitsTime to fill up = 60/8 = 7.5 hr.

7. d Capacity in units = LCM (9 ,10) = 90.

Inflow, when there is no leak 909

= 10 units per hr.

Inflow, when there is a leak = 9010

= 9 units per hr.

∴ Resultant outflow = 10 – 9 = 1 unit per hr.Total outflow = 90 units, Min. = 90/1 = 90 hr.

8. a Say, capacity of tank = LCM (12, 15)= 60 units

1 min total intake in units = 60 60

+12 15

= 9

3 min total intake in units = 9 × 3 = 27Units to be filled = 60 – 27 = 33Time taken by B = 33/4 = 8.25 (8 min 15 sec)

9. c Pipes A + B both are there,

1 1 1+ =

A B 12∴ , say A takes x hr to fi l l then

B will take x – 10 hr to fill the tank.

( )1 1 1 x –10 + x

+ =x x – 10 12 x x –10

⇒

( )1 2x 10 1

12 x x 10 12−= ⇒ =−

24x – 120 = x2 – 10xx2 – 34x + 120 = 0x2 – 30x – 4x + 120 = 0x(x – 30) – 4 (x – 30) = 0(x – 4) (x – 30) = 0 x = 4, 30∴ 30 hr (� x cannot be 4)


10. d (a) Distance = Speed × Time = 48 × 10 = 480 km

(b) To cover the same distance in 8 hr.

Speed = d 480

=t 8

= 60 km/hr

∴ Speed must be increased by 60 – 48= 12 km/hr

11. a Let his usual time be t hr and his usual speed be s km/hr.

Distance, d = st = 34

s × (t + 2.5)

Or, 4t = 3t + 7.5t = 7.5 hr

12. c By the time the trains crossed each other, one ofthem had covered 110 km and the other hadcovered 90 km.Ratio of speeds = Ratios of the distances covered

= 110

=11 : 990

13. a Thief starts at 2.30 p.m.Policeman starts at 3.00 p.m.In 30 min (= 1/2 hour), the thief covers

160 × = 30 km

2At this instance, the distance between them = 30kmdifference of their speeds = 75 – 60= 15 km/hr

∴ Policeman will overtake the thief in 3015

hr

= 2 hr, i.e. at 5.00 p.m.

14. d Since ratio of speeds of A : B = 3 : 4. So, ratio oftime taken would be 4 : 3. If A takes 30 min more,then 4x – 3x = 30 min⇒ x = 30 min.A takes 4x min, or 4 × 30 = 120 min = 2 hr.

15. b Average speed = 2xy 2 24 36x y 24 36

× ×=+ +

= 144

5 = 28.8 km/hr

16. d Average speed = Total distance

Total time

2500 1200 500 4200420 kmph

2500 1200 500 10500 400 250

+ += = =+ +

17. d A = 2B, B = 2C. C takes 54 min.So, B would cover the same distance in half of

the time = 54

= 272

min.

So, time taken be A = 27

min2

= 13.5 min.

18. bA

B

S 3=

S 4

Speed of A = 3x km/minSpeed of B = 4x km/mindistance is same.d = SA × tA = SB × tB3x × tA = 4x × 36tA = 48 min

19. b To cross the platform the train has to cover165 + 110 = 275 m.

132 km/hr = 132 × 5 110

= m/sec18 3

t = 275

×3 = 7.5 sec110

20. a t = 10 sec,

Speed = 90 km/hr = 90 × 5

18 = 25 m/s

Length = s × t = 25 × 10 = 250 m

21. d Total length to be covered = Length of train +Length of platform = 900 + 300 = 1200 mTotal time = 60 + 12 = 72 sec

Speed = 1200 50

m/ s72 3

= = 50 18

×3 5

= 60 km/hr

22. b A + B = 72 days, B + C = 120 days, A + C = 90 days.

Let total units of work = 360

… ( )( )LCM 72,120, 90=

∴ (A + B) do units of work in 1 day = 36072

= 5.

(B + C) do units of work in 1 day = 360120

= 3.

(A + C) do units of work in 1 day = 36090

= 4.

∴ 2(A + B + C) units in 1 day = 5 + 3 + 4 = 12.

A + B + C units in 1 day = 122

= 6

∴ A units in 1 day = 6 – 3 = 3

A will finish 360 units in 360

3 = 120 days.


Page: 101��


1. d Let total units of work = LCM of (12, 16) = 48.Work done by:(A + B) in 1 day = 48/12 = 4 units.(B + C) in 1 day = 48/16 = 3 units.(A + B) worked for 5 days ⇒ 4 × 5 = 20 unitsdone.(B + C) worked for 2 days ⇒ 3 × 2 = 6 units done.Work remains = 48 – 26 = 22C finishes in (13 – 2) days = 11 days.

∴ Units of work in 1 day by C = 2211

= 2

∴ Number of days required for C = 482

= 24

2. c Let total work = 12 × 8 = 96.In 6 days, 12 men complete work = 12 × 6= 72∴ Work left after 6 days = 96 – 72 = 24Now, number of men = 12 + 4 = 16

∴ Time = 24/16 = 112

= 1.5 days

3. b 10 men in 20 days or 20 women in 15 days.∴ 5 men in 40 days and 10 women in 30 days.

⇒ 5 men + 10 women in 1

1 140 30

+

days

= 1200

70 days = 17

17

days.

4. b Let units of work = 18 units.In one hour, man completes work = 9 units(= 6 units + 3 units)∴ Time taken = 2 hr.

5. c Total time = 60 + 24 = 84/60 hr

At 4 km/hr 1

2 / 3d 1t d

4 6= = hr

At 5 km/hr 2

1/ 3d 1t d hrs

5 15= =

Total time = d d 846 15 60

+ =

Or, d = 6 km

6. d Let his normal speed be s km/hr.Let his normal time be t hr

d = st ⇒ 41

t6

− = 3 1

t6

+

4t – 46

= 3t + 36

, t = 76

hr

Distance = 4 × 7 16 6

− = 4 km

7. d Let normal speed be s km/hr.Let normal time be t hr.

d = st ⇒ 40 11t

60 +

= 50 5t

60 +

4t + 4460

= 5t +2560

t = 19/60 hr = 19 min

8. c d = s × t

( ) ( )= =1 3journey 40 km in of time 6 hrs

2 5 Total distance = 80 km

Total time = 10 hr80 = 40 + S2 × 4, S2 = 10 km/hr

9. d Length of train= 110 m,

Speed of train = 58 × 5

18 m/s = 16.11 m/s

Speed of man = 4 km/hr = 4 × 5

18= 1.11 m/s

Time = 110 m 110

16.11 1.11 15=

− = 7.33 sec

10. a Length of first train = 108 m. Its speed= 50 km/hrLength of second train = 112 m.Let its speed = xtime to cross = 6 sec

( ) ( )108 112 220 18

65 5 50 x50 x

18

+ ×⇒ = =++ ×

250 + 5x = 660, 5x = 410x = 82 km/hr

11. b Length of the faster train

= (36 + 45) × 5

18× 8 m = 81 ×

518

× 8 = 180 m

12. d Let x be the rowing speed of the man in still watersSpeed of the river = 2 km/hr = ‘y’Speed upstream = x – y km/hrSpeed downstream = x + y km/hr(x – y) 2t = (x + y) t2 (x – 2) = x + 2x = 6 km/hr


13. b Let length of train be � m and its speed be s m/s.It crosses a pole in 15 sec

∴ � = 15 s … (i)It crosses a platform 100-m long in 25 sec

10025

s+∴ =� … (ii)

Solving (i) and (ii)100 + 15s = 25 s

∴ s = 10010

=10 m/s, ⇒ � = 150 m.

14. a The time taken to cover 240 km without any stops =6 hr. Since he stops every 80 km, he would stoptwice before reaching the destination.Hence, total time taken = 6 hr 40 min.

15. c Let ‘l’ = Length of train in metres. Then speed of the

train = s/m15l

= s/m30100l +

.

∴ l = 100 m

16. c1 1 1 1a b c 15

+ + = … (i)

1c25

b25

a5 =++

⇒ 251

c1

b1

a51 =++ … (ii)

Subtracting (ii) from (i),

=4 25a 75

⇒ a = 30 days

Alternative method:Let the total work be of 15 units. They complete thiswork in 15 days. It means one day work is 1 unit. In5 days, they will complete 5 units of work.Remaining 10 unit of work, B and C complete in 20days. It means one day work of (B + C)

= 21

unit

⇒ A’s one day work is 21

21

1 =− unit.

So total days taken by A alone to complete15 units of work = 30 days.

17. a1 1 1 1

22 33 44 x+ + =

when x = Number of days in which they togethercan finish the work.

or 1 1 1 1

22 33 44 x+ + =

or x = 11 12 132 2

10 days6 4 3 13 13

× = =+ +

18. b Here the total distance = 180 km.

Total time = 30 60 90 2 2 6 38

hr45 90 75 3 3 5 15

+ + = + + = .

Average speed = 38 1

180 71 km/hr.15 19

÷ =

19. d When Sujit runs 100 m, Rishi runs 95 m.When Rishi runs 100 m, Praveen runs 95 m.∴ When Rishi runs 95 m, Praveen runs 90.25 m.When Sujit runs 100 m, Praveen runs 90.25 m and isbeaten by 9.75 m.

Alternative method:The ratio of speeds of Sujit and Rishi = 100 : 95= 20 : 19.Similarly, the ratio of speeds of Rishi and Parveen =20 : 19.∴ The ratio of speeds of Sujit and Parveen= 202 : 102.⇒ When Sujit goes 100 m, Parveen goes

361

100 90.25 m400

× = .

∴ The lead that can be given is 100 – 90.25= 9.75 m.

20. c Suppose Pallavi takes t seconds to finish the race. Itmeans Anuva and Richa would be taking (t + 50)sand (t + 90)s respectively to finish the race. WhenPallavi runs 1,000 m, Richa will run 550 m⇒ Richa runs 1,000 m in (t + 90) s.

So we have 550

1000t90t =+

of t = 110 s

= 110 + 90 = 200 s.

21. a Take (+) for the person walking down and(–) for walking up. Therefore, his relative positionfrom the starting point = +4 – 3 + 6 – 2 – 9 + 2 = –2.Hence, the answer is 2 steps above the startingstep.


Page: 103��

22. a Since they move at the same time on different days,the situation is equivalent to the that of two personswhere one moves uphill and the other downhill,both starting at the same time, viz. 7 a.m.Ratio of speeds of uphill to downhill = Ratio ofdistances = 10 : 15 = 2 : 3.

Therefore, 3

of 25 15 km5

= from the top of the hill

they will meet or 52

of 25 = 10 km from the bottom of

the hill is the point where they will meet.


1. a Let speed of boat be x km/hr.Let speed of stream be y km/hr.

Rowing upstream = 135

= x – y … (i)

Rowing downstream = 285

= x + y … (ii)

Solving for y, y = 1.5 km/hr

2. d Let speed of boat in still waters be ’x’ = 6 km/hrLet speed of stream be ‘y’.x – y = 4.5 km/hr⇒ y = 1.5 km/hr∴ Rate along the stream = 1.5 + 6 = 7.5 km/hr

3. c Let the total work be (LCM of 45,40) = 360 units.In one day,

A does360

= 8 units45

and

B does 360

= 9 units40

Both A and B together do 8 + 9 = 17 units of work.In last 23 days, B does 9 × 23 = 207 units of work.So work done by both A and B together= 360 – 207 units = 153 units.

Number of days = 153

= 917

days.

4. d A takes 12 days,B is 60% more efficient

⇒ B will take time12

= = 7.5 days1.60

5. d In 10 day’s A completes work = 1 2

×10 =25 5

of the

total work.

Remaining work = 35

of the total work.

1 1+

25 20

work is done by (A + B) in 1 day

⇒35

work is done by them in 100 3

6.669 5

= × =

daysNumber of days = 10 + 6.66 = 16.66

6. b Due to stoppage, it covers 9 km less.

Time taken to cover 9 km = 9

6054

× min

= 10 min

7. d A in 1 hr = 2000B in 1 hr = 1333.33⇒ (A + B) in 2 hr = 3333.33

⇒ (A + B) in 4 hr = 6666.66In 5th hr work to be done = 8000 – 6666.66 = 1333.33.

Number of hr required by A = 1333.33

2000 = 0.66

⇒ Total number of hr = 4.66

8. d A : B = 1 : 2C : (A + B) = 1.5 : 3If C does 1.5 units of work per day, A does 1 and B does 2.Total units done by C = 40 × 1.5 = 60 unitsIn a day work done by (A + B + C) = 4.5⇒ Number of days taken by (A + B + C) to complete

work = 604.5

= 13.33

9. e Let r lt/hr be the rate at which the hole empties thetank and let the tank has a capacity of V liters.The inlet pipe fills the tank at 4 liters/min or at 240liters/hour, we have two equations,r × 6 = V …(i)& V + 8 × 240 = r × 8 …(ii)V = 5760 liters

10. b Length of bridge = 1 kmLength of train = 0.5 km

Time to clear the bridge = 2 min = 2

60 hr

Speed = 1 0.5 1.5

60 45 kmph2/ 60 2+ = × =

11. a Speed of stream = 1 km/hrLet speed of boat in still waters = x km/hrTotal time = 12 hr

12 = 35 35

x 1 x 1+

− +Now put the value of x and check the options, only(a) satisfies.


12. a Ratio of distances travelled by the trains when theymeet = 9 : 10.Suppose first train travels 9x. Then the secondwould travel 10x km.∴ Total distance = 10x + 9x = 19x km.We have 10x – 9x = 120 or x = 120 km.∴ 19x = 19 × 120 = 2,280 km.

Alternative method:It is given that speed is 45 and 50 km/hr respectively.So in 1 hr, the faster train moves 5 km extra than theslower one. It is given that the faster one moves120 km extra than the slower one.

So they travel for =12024 hr.

5There relative speed is 95 km/hr.So distance travelled by them = 95 × 24= 2,280 km.

13. e In half an hour, the thief travels 60 × 21

= 30 km,

which is the relative distance to be covered by theowner moving at relative speed of(70 – 60) km/hr = 10 km/hr.∴ Time taken by the owner to catch up with the

thief = 30 km10 km/hr = 3 hr which is at 5 p.m.

14. c Suppose the distance between the college andresidence is d. Then

− =d d 210 15 3

=

240 min hr

3 ⇒ d = 20 km

15. d Average speed

+ += =(50)(1) (48)(2) (52)(3) 150 km/hr.

6 3

16. e If the person had not moved towards the source ofthe gunfire, he would have heard the second shot12 min after the first shot. Since the person isactually moving towards the source, the shot isnow heard after 10 min. It means the sound wouldhave taken 2 min more to reach the initial position ofthe person; but this very distance was travelled bythe train in 10 min. It means it is the speed of the

train in 10 min. It means speed of the train is 51

of

the speed of the sound, i.e. s/m5

330

= 66 m/s 66 3600

237.6 km/hr1000×= =

Alternative method:Solve using the relative velocity concept, V is thespeed of train.10(V + 330) = 330 × 12

330V 66 m/ s

5= =

330v

17. c When he travels the entire journey at 5 km/hr, totaldistance travelled = 35 + 2 = 37 km.

∴ Time of travel =537

= 7.4 hr, which is same as

the time taken at the original rate.

18. e Let ‘w’ is the time of walking and ‘r’ is the time ofriding∴ w + r = 7 hr.Riding both ways he gains 2 hr. Therefore, he takes5 hr to ride both ways or 2.5 hr to ride one way.∴ r = 2.5 hr ⇒ w = 4.5 hr.Time taken to walk both ways = 2 × 4.5 hr = 9 hr.

19. b From 7 a.m. to 1 p.m., i.e. in 6 hr, the first traintravels a distance of 60 × 6 = 360 km.∴ Distance between the trains at 1 p.m.= 1200 – 360 = 840 km.∴ Time taken to cover this relative distance

= 140840

= 6 hr, i.e. 7 p.m.

20. c In 6 hr, the second train travels 6 × 80 = 480 km.∴ They meet at 480 km from Q.

21. d Speed in upstream =

34 601

114

× km/hr.

Speed in downstream =

34 601

72

× km/hr.

Speed of man in still water

= 21

(Speed in downstream + Speed in upstream)

= 1 3 1 1

60 51 12 4 11 74 2

× + =

km/hr.


Page: 105��

22. b Let x be the distance per litre of petrol and v= Speed at which it is driven.

Then 21

xv

∝ or xv2 = k = Constant. 25(36)2

= 36 ( v′ )2 ( v′ is the required speed)

⇒ v′ = 30 km/hr

23. b Let a = Speed in upstream and b = Speed in downstream

5b90

a40 =+ and 5

b60

a60 =+

∴ a = 20 km/hr and b = 30 km/hr.∴ Speed of the water (current)

= 1

(b a) 52

− = km/hr

= + = −� W B B W{b V V and a V V }

24. a Let the job be completed in x days. Then the youngman worked for 2 days.His father worked for x – 1 days and the grandfatherworked for x days.

∴ 120x

151x

102 =+−+

or 160

x34x412 =+−+ or x =

73

7 days

25. c To complete the given work, 10 × 18= 180 man-days will be required.In 6 days, amount of work done= 60 man-days.Number of men now available = 10 + 5 = 15.∴ Number of days in which the job can be done =

15120

= 8 days more.

Alternative method 1:After 6 days, 12 days of the job is left. If the number

of people becomes

23

times the earlier number,

the job would be done in

32

times the earlier

number of days, i.e. 8 days.

Alternative method 2:Let the total work be 180 units. 10 men completethis in 18 days. So one day work of 10 men= 10 units.(All men are equally efficient, so one man’s work inone day is 1 unit.)In 6 days, 10 men will complete 60 units work.Remaining 120 unit is to be done by (10 + 5) men in

15120

days = 8 days.


1. c 3 men = 4 women. 4 women complete the job in 12days. Hence, 5 women and 3 men, i.e.

9 women can do the job in days31

59

412 =×.

2. c Earning of 1 girl = Rs. 50.Earning of 10 girls = Rs. 500.Earning of 8 boys = Rs. 500.

Earning of 6 boys = Rs. 6

500 Rs. 3758

× = .

Earning of 4 women = Rs. 375.Earning of 3 men = Rs. 375.Earning of 1 man = Rs. 125.

Alternative method:

The man earns 5 3 4

50 Rs.1254 2 3

× = .

3. b Amount of work to be done = 10n,where n = Number of workers originally available.Now 10n = 12(n – 5) ⇒ 2n = 60. Therefore, n = 30.

4. a31

of a man =21

of a woman = 1 child.

⇒ 2 men = 3 women = 6 children.20m + 30w + 36c = 60c + 60c + 36c = 156c

If 156 children get Rs. 78, 1 child gets 15678

.

Now 15m + 21w + 30c = 45c + 42c + 30c= 117c.

∴ 117 children should get Rs.

×11715678

per day.

∴ For (18 × 7) days, they should get

= Rs. (18 × 7)

×11715678

= Rs. 7371.


Alternative method:Let a man, a woman, a child put in 3, 2, 1 units perday.Then they are paid Rs. 78 for putting in(20 × 3 + 30 × 2 + 36 × 1) units, i.e. 156 units.Hence, if (15 × 3 + 21 × 2 + 30 × 1), i.e. 117 units is

put in, then the payment = Rs. 2

117 per day or Rs.

7,371 in 18 weeks.

5. b181

c1

b1

a1 =++ … (i)

c2

b1

a1 =+ … (ii)

b3

c1

a1 =+ … (iii)

(i) and (ii) ⇒ c2

c1

181 =− or c = 54 days

(i) and (iii) ⇒ b3

b1

181 =− or b = 72 days

∴ a = 43.2 days

Alternative method 1:Let the number of units of work = 18.

C puts in

31

of that in 18 days = 6 units.

B puts in

41

of that in 18 days = 4.5 units.

Hence, A does 7.5 units in 18 days.

In order to finish it he takes 2.43185.7

18 =×

days.

Alternative method 2:Let the total work be 18 × 12 = 216 units.Ratio of (A and B) : C = 2 : 1.

∴ C does 1

12 4 units3

× = .

Ratio of (A and C) : B = 3 : 1.

1D does 12 3 units

4∴ = × = .

∴ A does = 12 – (4 + 3) = 5 units.

Total work will be done in 2.435

216 = days.

6. a 12n = 18(n – 5) ⇒ n = 15

7. c45

of work is done by them in 7 days.

Hence, they take

547

= 8.75 days to complete the

job.

∴ 1 1 1a b 8.75

+ = and 8 10

1a b

+ =

⇒71

75.825.1

175.8

10a2 ==−=

a = 14 days

Alternative method 1:A and B work together for 8 days.

WA B+ in 7 days is 4W

5 [W is total work]

⇒ WA B+ in 8 days is 4 8 32

W W5 7 35

× =

Now A leaves and B does 3

W35

in 2 days.

⇒ B will do W in 35 70

23 3

× = days.

W W – WA A B B= +

= 1 3 5 1

–10 70 70 14

= =

So A will do the work in 14 days.

Alternative method 2:Let the total work be 35 units (LCM of 7 and 5).

64

of work = 28 units, is done by A and B in

7 days.∴ 1 day’s work of A + B = 4 units.B completes (7 – 4) units in 2 days.(Out of a total of 10 days, 7 + 1 = 8 days’ work = 8× 4 = 32 units)∴ 1 day’s work of B = 1.5 units.∴ A’s 1 day’s work = 2.5 units.Total time taken by A alone to complete 35 units of

work = days145.2

35 = .

8. d In 30 days, only half the work could be done.Therefore, number of men must be doubled to getthe work done in half the time (i.e. half of 30 = 15days, because only 15 days are remaining).


Page: 107��

9. c 15m = 24w = 36bx men must be associated.

∴ (x)m + 12w + 6b =

++ 618x

1536

boys.

Therefore,

×=×

+

1812

)36(

41

2

63024

15x36

⇒ x = 8 men

10. a Suppose the work is of 150 units. So in a day A andB together do 15 units; B and C together do 10 units;and A and C together do 6 units.We have A + B = 15, B + C = 10 and C + A = 6.⇒ A = 5.5, B = 9.5 and C = 0.5In first 4 days, A, B and C together will do4(5.5 + 9.5 + 0.5) = 62 units.In next 5 days B and C will do 5(9.5 + 0.5) = 50 units.So C alone has to do 150 – (62 + 50) = 38 units, that

he will do in days765.0

38 = .

11. e Let us assume that the distance between the twoplaces be 60 km. This means that to go 120 km at

12 km/hr, she can use 1012

120 = hr.

Also, as she goes at 6 km/hr, the time taken for this

journey is 106

60 = hr.

Thus, there is no time to come back.

Alternative method:If distance from Kashipur to Bareily is D, Nupur takes

D6

hr to travel distance D.

Since average speed of journey is 12, total time

taken is 2D D12 6

= .

Hence, there is no time to return.

12. c Suppose they meet after (100 + x) m. Shivku startsrunning when Pawan has already run 100 m. Nowif Pawan runs x km, Shivku will run 2x km.So we have 100 + x = 2x or x = 100.Thus, they will meet at 100 + 100 = 200 m from thestarting point.If the length of the track is 150 m, they will meet at adistance of 50 m from the starting point.

13. a Suppose the total time was x hr.Since distances travelled by both would be same,we have (8 × 4) + (x – 4)16 = (13 × 4) + (x – 4)12or x = 9

Alternative method:In 4 hr, lead taken by Shivku is (13 – 8) × 4 = 20 km.Now speed of Sujeet = 16 km/hr and of Shivku = 12km/hr.Sujeet has to cover 20 km with relative speed 4 km/hr in 5 hr. So total time = 4 + 5 = 9 hr.

14. dA BX

G ollu M ayank

Let X be the point where they meet on the way.

AX 6XB 5

⇒ = [As their speeds are in this ratio]

Now m

G

G

m

GG

mm

G

m

S

S

dd

S/dS/d

tt ×==

⇒ Since d AX 6md XB 5G

= = 2536

56

56

tt

G

m =×=⇒

2.5 36t 3.6m 25

×= = hr = 3 hr 36 min.

Alternative method 1:Conventional method of solving

xA

G M

BC

D

Let Golu and Mayank met at point C which isx kilometres from A, and A and B are D kilometresapart.

The ratio of speeds = V 6GV 5M

= .

They take same time to reach point C.

x D – x6 5

⇒ =

6x D

11⇒ =

Now Golu covers5

D11

in52

hr.

So 5D 2D

VG 5 11112

= =×

km/hr

5 5 2 5DV V DM G6 6 11 33

⇒ = × = × =

Mayank covers 5D in 33 hr.

So he will cover 6

D11

in 33 65 11

× = 3.6 hr

= 3 hr 36 min


Alternative method 2:G M

G

M

M

G

tt

V

V=

G

M2

tt

56 =

⇒

min36hr325

2536

tM =×=∴

15. a Suppose the winning point is x metres far if thegirls finish the race at the same time. Then Pallaviwill have to cover x metres when Richa covers (x– 350) m.Ratio of their speeds is 20 : 13.

So we have 350xx

1320

−=

or x = 1,000 m = 1 km.

Alternative method:Pallavi has to cover 350 m with a relative speed of7x.

Required time will be x7

350.

In this time, the actual distance covered by Pallavi

is x20x7

350 × = 1,000 m, i.e. length of the track.

16. a It happens when both of them meet for first time atthe starting point. Ratios of speeds= 4 : 12. Hence, ratios of distances covered= 1 : 3. Hence, when A makes one round, B makes3 rounds, so the answer is 1.

17. b Let length of the journey = x km andspeed of the train = v km/hr.

Then

vx1 3 x 12 12v2 4 v 23

−+ + − =

[Converting minutes to hours]

Also

v v60 x 603 x2 2 1

2vv 4 v3

+ − −+ + − =

Solving for x, we get x = 120 km and v = 60 km/hr.

Alternative method:Overall, the train was late by 1 hr 30 min.Out of this for 45 min, the train was stopped. So

due to 23

of the speed, the train delayed by 45 min

only.

Normal time min452t

tt23 ==−

or t = 90 min.Had the accident occurred 60 km away, the trainwould have been delayed by 1 hr only.

Due to 23

of the speed, the delay will be 15 min only

(1 hr – 45 min stoppage)∴ Normal time to cover the remaining distance⇒ t = 2 × 15 = 30 min.It means the train covers 60 km in 60 min(90 – 30 min).Speed of the train = 60 km/hr.Total journey = 30 km + 90 km = 120 km.

18. a Vibhor had lost 4 gallons.Since he travelled at 50 mph for 4 hr = 200 miles, he

used 8 gallons

25200

for the journey.

19. c We know that in 60 min, the minute hand gains 55min over the hour hand.At 3 o’ clock, the minute hand and the hour hand are15 min apart.To cross the latter, the minute hand must gain 15min.Since 55 min is gained by the minute hand in 60 min,15 min is gained by the minute hand in

5560

× 15 = 16 114

min.

114

16x =∴ min

Alternative method:At 3 o’ clock, the angle between the hour hand andthe minute hand is 90°. The minute hand will coverthe distance with a relative speed of (6 – 0.5) = 5.5°per minute, i.e.

90 416 min

5.5 11= .

20. b Let’s assume the distance to be 24 km.(LCM of 6, 4, 3)Then time taken to cover that distance will be in the

ratio 324

:4

24:

624

= 4 : 6 : 8 = 2 : 3 : 4.

Hence, the ratio of their speed will also be 2 : 3 : 4.


Page: 109��

Alternative method:Since the distance travelled by the three is same,the ratio of speeds will be inverse of the ratio oftime taken.

Hence, the ratio of speeds = 31

:41

:61

= 4 : 6 : 8 = 2 : 3 : 4.

21. c Ratio of speeds = 3 : 4.Distance remaining constant, the ratio of time taken= 4 : 3.A takes 0.5 hr more than B.Hence, time taken by A = 4 × 0.5 = 2 hr

22. a Work done by A in 2 days 31

62 == .

Work done by B and C in 2 days

= 1 1 5 4 9

2 28 10 40 20

+ + = = .

Remaining work

6013

60272060

209

31

1 =−−=−−= .

Hence, 6013

part of the job will be done by C in

613

101

6013

= .days61

2=

Total time taken to complete the job

= + + =1 12 2 2 6 days.

6 6

Alternative method:Let the total work be LCM (6, 8, 10)= 120 units.A’s 1 day’s work = 20 units.B’s 1 day’s work = 15 units.C’s 1 day’s work = 12 units.In 2 days A will work = 40 units.(B + C) will work in 2 days = 54 units.Work remaining after 4 days = 26 units.

∴ Time taken by C = 6

131226 = .

Total time taken = days6

134 + .

Answer is days61

6 .

23. a Fraction of the tank filled in 2 hr = 209

51

41 =+ .

So in 4 hr, fraction of the tank filled = 2018

.

The remaining 2 1

or20 10

of the tank will be filled by

pipe A.

Pipe A fills 41

of the tank in 1 hr.

So it will fill 101

of the tank in 4.0101

4 =×

= 0.4 hr = 24 min.Hence, the total time taken to fill the tank= 4 hr 24 min.

24. c A can give B a lead of 200 m for every 1,000 m, i.e.every one kilometre. This means that when A covers1,000 m, B covers 800 m.Suppose the speed of A is 25 m/s and of B is 20 m/s.A will meet B for the first time after

1000200 s

25 20=

− .

In 200 s, A will have made

100025200

251000

200×=÷ = 5 rounds.

Alternative method:From the given data, in 1,000 m race A can give B astart of 200 m. It means when A has covered 1,000m, B can cover 800 m. The ratio of their speeds is 5: 4. From this ratio of speed, it can be easily foundthat they will meet only at the starting point. So Ahas covered 5 rounds.

25. a When B covers 800 m A has covered 1,000 m.So when B has covered 3 rounds = 3,000 m, A willcover 3,750 m.But to meet at some point, the overall lead should beof one round = 1,000 m.So there must be an initial lead of 250 m.



1. d The hands of the clock are 35 min (= 35 × 6= 210º) apart at 7 o’ clock.For the hands to be together, the minute hand hasto gain 210° min over the hour hand.Speed of the minute hand relative to the hourshand= 6º – 1/2º = 5.5º per minute.Hence, the hands will meet after

11382

11420

5.5210 == min.

OR

Between x and (x + 1) o’ clock, the two hands will

be together at 5x ×

1112

min past x.

i.e. 5 × 7 ×

1112

min past 7 = 38112

min past 7.

Alternative method:Since the hour hand is between 7 and 8, the minutehand has to show the time between 35 min and 40min. Hence, there is only one choice in betweenthem.

2. a If original speed and time are s and t, and newspeed and time are t1 and t1 respectively, then

s1 = s107

⇒ t1 = 1t7

t10 =−

or 7t hr

3= = 2 hr 20 min.

3. c Let X minutes be the time required for going toschool from home. Usually the car leaves home by(4 p.m. – X) min and the children are back home by(4 p.m. + X) min.On Saturday, the children reached home 20 minearlier, i.e. by (4 p.m. + X – 20) min.⇒ The total time for which the car was in use= 4 p.m. + X – 20 – [4 p.m. – X] = (2X – 20) min.Hence, the car was used for (X – 10) min to goand come back from school. If the children werewalking for Y minutes, then⇒ 3 p.m. + Y + (X – 10) = 4 p.m. + (X – 20) min⇒ Y – 10 = 4 p.m. – 3 p.m. – 20.Hence, Y = 60 – 20 + 10 = 50 min.

Alternative method:We see that the car reduces its travelling time by 20min. So it reduces its one-way travelling time by

220

= 10 min.

Thus, it would have met the children10 min earlier, i.e. 3.50 p.m. Now this means that thechildren walked from 3 p.m. to 3.50 p.m., i.e. 50 min.

4. d In 24 hr, i.e. 1 day, the snail goes up by only11 inch. height of the pole = 12 × 25 = 300 inches.In 25 days, i.e. 600 hr it will go up (effectively) by25 × 11 = 275 inches.Now it can creep up 27 inches in 12 hr.So to climb up 25 inches it will take

25 × 2712

= 11.11 hr.

Hence, the total time = 600 + 11.11 = 611.11 hr.

5. d Two of the dog’s leaps cover 4 m,three of the fox’s leaps cover 3 m,i.e. each time the dog runs 4 m.The distance between them is reduced by4 m – 3 m = 1 m.The initial distance between them is 30 m.Hence, the dog will catch up with the fox when itcovers 4 × 30 = 120 m.

6. b The first train takes 5.18

148 = 8 hr, i.e. it reaches 8 hr

after 7 a.m., i.e. 3 p.m.So the second train reaches at 3.15 p.m.Ratio of speeds of two trains is 8 : 5.Hence, the ratio of time taken (distance beingconstant) is 5 : 8.Therefore, time taken by the second train

=

85

× 8 = 5 hr, i.e. it starts 5 hr before 3.15 p.m.

⇒ 10.15 a.m.

7. c Since Ajay is faster than Mallu and they starttogether, to meet at 10 m from B, Ajay would havecovered a distance from A to B and would meetMallu on his way back to A.Mallu would be on his way from A to B.So Ajay covers 200 + 10 = 210 m in 10 s.Hence, Ajay’s speed = 21 m/s.

So he will take 19021

s to cover the remaining 190

m.The time required for Ajay to reach A will be

190 400

10 s21 21

+ = .


Page: 111��

8. c In the first 15 min, the thief will cover

156041 =× km.

Hence, to cover this 15 km, a policeman will take =

Dis tance 15 153

Relative speed 65 60 5= = =

−hr.

9. a When Zigma-S leaves Jaipur, the thief has covereda distance of 15 km.Distance between them = 300 – 15 = 285 km.Relative speed of Zigma-S with respect to thethief= 60 + 60 = 120 km.So time taken to catch the thief

= 375.2120285 = hr = 2 hr 22.5 min.

According to question 8, Sigma-Z takes 3 hr tocatch the thief. Hence, Sigma-Z takes 37.5 minmore than Zigma-S in catching the thief.

10. d Let the total distance covered by them when theyfirst meet is d kilometres.When they meet for the second time, the totaldistance covered by them is 3d.Hence, they will meet after 12 × 3 = 36 min.

11. a In 1 hr the first photocopier will complete 81

of the

work.

Similarly, the second photocopier will complete 121

of the work in 1 hr.If both work together, then fraction of the workcompleted in 1 hr

=121

81 +

245

2423 =+= .

Hence, both the photocopiers working

simultaneously will take524

= 4.8 hr.

12. d Duration of the journey = 24

384 = 16 days.

Thus, the person starting from Delhi (at 3 p.m.) isbound to meet the 17 persons already enrouted,and 16 more (who will have started after he starts)during his journey.Therefore, the total number of people he meets =17 + 16 = 33.

Alternative method:

It takes 24

384= 16 days for New York-Delhi travel.

Since the speed gets added up, when two peoplecome from opposite sides, anybody going from NewYork to Delhi will meet two people going from Delhi

to New York every 2

24= 12 hr.

∴ A person will take 12384

= 32 intervals to complete

his journey.⇒ He will meet (32 + 1) = 33 people during thejourney (As he will meet the first person as soon ashe leaves New York and last person as soon as hereaches Delhi.)

13. b Let the speeds of two persons are 3x and 7x.

Y takes x7

16 hr to complete the race.

Since they move in the same direction, relative speed= 4x.

Time when they first meet =x4

4 = .

x1

Therefore, number of times they meet in the entire

race = 7

16

x1

x7

16

= = 2.3.

Therefore, they meet twice before the race finishes.

14. d When they move in opposite directions, the time

when they meet the first time = x5

2x10

4 = .

Therefore, they meet 1480

x52

x7

16

= = 5.7 times,

i.e. 5 times.

15. c In 1 hr, pipe A fills 60 m3, and pipe B fills 61

of the

tank. Hence, if the capacity of the tank = V m3, 60

+4v

6v = (Portion of the tank filled in 1 hr when both

A and B are open)720 + 2V = 3V ⇒ V = 720 m3


16. d Pipe A can fill the tank in 60

720 = 12 hr.

Therefore, 1 hr net task of all three pipes, whenopened simultaneously, is

81

24342

81

61

121 =−+=−+ .

Hence, the tank will be filled in 8 hr.

17. e Let the speed of the boat in still water be u km/hrand that of the river be v km/hr.If the distance from A to B is d, then we have

2vu

d =+ and 3

vud =− or

32

vuvu =

+−

or 15

vu = .

Thus, we see that we can get the ratio of the speedsand not the exact speeds.

18. c Let the job consists of 120 units [i.e. LCM(20, 24, 15)]∴ A + B do 6 units per day.B + C do 5 units per day.A + C do 8 units per day.A + B + C do 9.5 units per day.So B + C in 8 days do 40 units. A in 4 daysdoes 18 units.

Hence,58 29

120 60= part of the job is completed.

19. e Average speed of hr/km19

18019

9102A =××=

Ratio of times of B and A = 19:1512:19180 ⇒

Hence, if A takes 10 min more than Bibek

= min4

19019

410 =×

Hence, total distance covered

= km5.7601

4190

19180 ⇒××

PQ = 3.75 km

20. a Let the speed of the boat in still water is uLet the speed of the flowing river is vwhen boat goes upstream (against the flow of river)speed of the boat = u – vfor downstream (with the flow of the river)= u + v4 : 1 = (u + v) : (u – v) ⇒ u + v = 4u – 4v⇒ 3u = 5v

21. a Time taken = 1x5

4.2x5

4.2 =+

+−

⇒ x = 1 km/hr

22. c Assume speed of the river and the boat and makean equation by equating their time.Let M be the speed of the motor boat and R be thespeed of the raft/river. Then

R4

)RM(328

)RM(340 =

−+

+ 3

17RM =⇒

23. c Each maid takes 7 hr to clean a floor. Hence,3 maids would also take 7 hr.Here the concept of chain rule can also be used,but the mop and the person should be taken as oneentity.Since no work can be done if there are only mopsor only persons without mops.

24. c Area per revolution = (2πR) × L = 1.32 sq. m⇒ Total area = 1.32 × 400 = 528 sq. m⇒ Total cost = Rs. 5280

25. d In a mile race Akshay can be given a start of 128mby of Bhairav. Or in 100m race Bhairav can give astart off 8 m to Akshay.So, for 100 m race:

Akshay Bhairav Chinmay92 100 96

So Chinmay is faster than Akshay.Now Chinmay can give a start off 4 m in 96 m raceto Akshay.So in 2,400 m race start given to Akshay

m1002400964 =×= mile

161=


1. a Let the speed of the goods train be X m/s, and thatof the passenger train be Y m/s. In 28 s the goodstrain covered 28X (m), and the passenger train,28Y (m). Therefore, 28X + 28Y = 700.

The goods train passes the signal lights in X

490 s

and the passenger train in Y

210 s.

Therefore, 490 210

– 35X Y

= .

Solving the two equations, we get X = 36 km/hr andY = 54 km/hr.


Page: 113��

Short cut:Going from the given answer choices, we see thatoption (a): 36 km/hr = 10 m/s and 54 km/hr = 15 m/s.

So the goods train takes 49 s

10490

and the

passenger train takes 14 s

15210

.

Hence, the goods train takes 35 s longer than thepassenger train.

2. c Let’s assume that A takes x days to finish the job.Then B will take 3x days.3x – x = 60 ⇒ x = 30 days

Work done by A in one day =301

.

Work done by B in one day = 901

.

Both combined would do 452

901

301 =

+ of the

job in one day. Hence, they together will do the job

in 2

45 days = 22.5 days.

3. a Since the two trains have the same speed, theywould take same time to cover the same distance(between A and B). Hence, they would meet exactly6 hr after they start, i.e. at 3 p.m.

4. b If the distance between A and B is assumed to be12 km, then relative speed of the two trains = 1 km/hr.Therefore, by 1 p.m., they would have both covered4 km each. Distance left between them = 4 km.Speed of the first train = 0.5 km/hr; speed of thesecond train = 1 km/hr.(Relative speed = 1 + 0.5 = 1.5 kmhr)Hence, time taken to meet, after the accident

= 5.1

4=

32

2 hr = 2 hr 40 min.

So they meet at 3.40 p.m.

5. c The train is late by 10 hr due to reduction in speedafter the accident as the speed is half. So timetaken now will be twice that of the original. Hence,the accident must have occurred 10 hr before theactual arrival time at the destination, i.e. at 11 o’clock. Now till 11 o’ clock, if we assume the speedof each train to be 1 km/hr and total distance = 12km, in 2 hr they together will cover 4 km. Theremaining distance of 8 km will be covered with

relative speed of 1 km/hr and 21

km/hr.

8 16Time hrs

3 32

∴ = =

The two trains would meet after = 16

11 hr3

+ or

at 4 : 20 p.m.

6. d As in question 5,speeds of the first train before and after theaccident would be 1 km/hr and 0.5 km/hrrespectively.Since they meet 7 hr after they leave, the distancecovered by the first train till 4 p.m.= 12 – 7 = 5 km.

Average speed of the first train till 4 p.m. = 75

km/hr.

Using alligations,

57

12

Hence, the time for which the train travels at twospeeds is in the ratio 3 : 4, i.e. after 3 hr of departure,the accident takes place.Hence, the accident takes place at 12 noon.

Alternative method:Assume that the total distance d = 12 km.Speed of the train from A to B and B to A is 1 km/hr.

A d BNow if the accident occurred after t hours fromstart, the distance covered by the first train in thours = t km.

And distance covered in (7 – t) hr 21

)t7( ×−=

(Because after the accident the speed is half)(9 to 4 → 7 hr)Distance covered by the second train in 7 hr= 7 km.To meet at 4 a.m., the total distance = 12 km.

1221

)t7(t7 =×−++∴

⇒ t = 3 hr after start, i.e. at 12 o’ clock.


7. b Let the length of the wall be 30 m. Bamdev candemolish 1 unit per day and his son can demolish0.5 unit per day. Manas can construct 3 units perday. If they work simultaneously with Manas startingthey would do effectively 1.5 units per day. So in 36days, 18 × 1.5 units is completed, i.e. 27 units iscompleted. Hence, on the 37th day, Manas wouldcomplete the job. Manas would have worked for 19days and constructed 19 × 3 m.Hence, the fraction of the job completed

= 2038

1019

3057 == .

Alternative method:Manas can construct a wall in 10 days.

∴ He does 101

work in 1 day.

Also Bamdev and his son can demolish

201

601

301 =+ of the wall in next day.

Thus, in 2 days Manas effectively constructs

201

201

–101 = of the wall.

∴ He constructs 109

of the wall in 210

920 ××

= 36 days and on 37th day, he finishes theconstruction. Since he has worked for 19 days, he

constructed 2038

1019 = portion of the wall.

8. b If Manas started when there were 24 units to becompleted, then working at 1.5 units per 2 days, 21

units would be completed in

5.121

= 14 pairs of days or 28 days.Hence, on 29th day Manas would complete the job.

Alternative method 1:

This is identical to question 22, only that 102

of the

wall, i.e. 8 days of work is already done.∴ So 37 – 8 = 29 days are required.

Alternative method 2:From choices it is evident that there is only one oddchoice. We know that the answer has to be oddsinceon the last day Manas has to complete the job.

9. d In his first shift, he uses the container six times.(45 glasses in 5 rounds and the remaining 6 glassesin the 6th round)Similarly, he prepared 73 glasses in 9 rounds, and112 glasses in 13 rounds.Hence, the required average

Total timeTotal number of rounds

=

4 min 18s 7 min 13 s 12 min 24 s6 9 13

+ +=+ +

= 23 min 55 s28

+ ~ 51 s.

10. a Let us calculate the tonne-hours done.3 tonnes × 30 trucks × 8 hr = 720 tonne-hours.5 tonnes × 9 trucks × 6 hr = 270 tonne-hours.∴ Total tonne-hours = 720 + 270 = 990 tonne-hours∴ Hours required for a single three-tonne truck

= 3303

990 = hr

Hours required for five-tonne truck

= 1985

990 = hr.

11. b Let the rate of growth of the grass be x straws perday and the initial amount of grass be y straws.Assuming that one cow grazes on one straw ofgrass in a day, (40 cows) × (40 days)= 1600 = y + 40x.(30 cows) × (60 days) = 1800 = y + 60x.⇒ y = 1200 and x = 10.So for 20 cows,(20 cows) × (n days) = y + nx = 1200 + 10n

⇒ n = 120 days.

Alternative method:Let G be the amount of grass in the field and g bethe growth rate of grass per day.

G 40g 40 40⇒ + = ×and G + 60g = 60 × 30

g 10, G 1200⇒ = =

1200 N 10 20 N⇒ + × = ⇒ N = 120 days

12. d t1 t2

Let 1t be the time at which B switches the speed

and 21 tt + be the total time between start and finish.

Let x be the speed of B initially. So A’s speed= 1.2x and B’s final speed = 1.44x.


Page: 115��

Now lag of B in time 1t

( ) 11 t2.0tx–x2.1 ×== … (i)

Also gain of B in time 2t

( ) 22 t24.0tx2.1–x44.1 ×== … (ii)

Since both reach at the same time,∴ Lag = lead

1

2

t 0.24 6t 0.20 5

⇒ = =

∴ A covers ( )1760 t1

t t1 2

×+ of the distance

×= =1760 6960 m.

11

Alternative method 1:Let the speed of B → u.Speed of A → 1.2uNew speed of

B = 20 2020 20 44%

100×+ + = more

= 1.44 u

x1

1.44 uuSince race ended in dry heat, equating time of A andB

u2.11760

u44.1x1760

ux 11 =−+ 1x 797.819 m⇒ = .

Alternative method 2:Assume that A runs 20% faster than B for t1 timeand for t2 time B runs 20% faster than A.Speed of B initially is VB for t1 time.So VA = 1.2 VB for (t1 + t2) timeVB’ = 1.44 VB for t2 time

1760)tt(V2.1 21B =+∴

1760tV44.1tV 2B1B =++

21 t56

t =⇒

Now distance covered by A in t1 time = 1.2 VBt1

1760t65

tV2.1 11B =

+

B 11760 6

1.2 V t 960 m11

×= = .

13. d The time taken by 5 children to complete the job =

520

= 4 days.

So time taken by 4 men to complete the job= 4 × 2 = 8 days.And time taken by 4 women = 8 × 2 = 16 days.Therefore, 12 men + 10 children + 8 women

complete = 83

+21

+81

=88

8)143( =++

= 1 of the

work.Hence, time taken = 1 day.

14. a x = Length of train, y = Speed of train (m/s)Relative speed with respect to first one

= 9

5y995

y−=− (As 2 km/hr =

5m/ s

9)

9

9

5y9

x=

− ⇒ x = 9y – 5 ... (i)

Similarly, for second 10

910y9

x =−

⇒ 90y – 9x = 100 ... (ii)Solving (i) and (ii) x = 50 m

y = 955

m/s

15. b When Ram completes second round, they pat eachother once. In the 20th round, Ram will finish therace and the total number of pats = 20 – 1 = 19This solution is only valid when greater ones speedis less than twice the first one speed.

16. c They reverse directions on reaching A.Ram with 12 km/hr speed will reach A first andimmediately reverse directions.Suraj will reverse direction only after he reaches A.Time taken by Ram to complete the circle= 1.2 km/12 km/hr = 0.1 hr = 6 minTime taken by Suraj to complete the circle= 1.2 km/8 km/hr = 0.15 hr = 9 minIn those extra (9 – 6) = 3 min, Ram would havealready reversed direction and done another halfcircle in the same direction as Suraj.

The effective distance between them = 22.1

= 0.6 kmTime taken to meet after Suraj reverses direction

= hr03.0)812(

6.0 =+

= 1.8 min

Therefore, time taken to meet second time= 9 + 1.8 min = 10.8 min


17. b Time taken will be double the time taken to meet thefirst time = 3.6 × 2 = 7.2 minSince each time the effective distance betweenthem will be 1,200 m.

18. c The time taken by P and Q to meet after their start is

100400

6040400 =

+ = 4 hr

Hence, they must have started at 11 a.m.

Alternative method:

Q covers 35

of the distance, i.e. 240 km.

It takes 4 hr to cover this distance.

19. e R would take 60 480 60

×−

⇒( )

12 hr to meet Q.

By that time, Q would have already reached Guntur.

20. a Normally P and Q meet after 4 hr of journey.Now they are meeting after 5 hr.Hence, the distance travelled byQ = 400 – (40 × 5) = 200 kmIf the accident occurred after x hours, then60(x) + 20(5 – x) = 200

or 40x = 100 or x = 21

2 hr


1. a Let ship sends the radiowave from point A andreceives it at point B

500m

O cean bed

BMA

O

0.5 1time taken 1min

30 60= = =

Distance travelled by radiowave in 1 min.= 200 × 60 = 12 km

OA OB 6km∴ = =

In AOM∆OM2 = OA2 – AM2

22 1

62

= − 143

OM2

⇒ = km

2. e For every day's work, he can afford to miss3 days. Hence, to break even it has to be 7 days outof 28 days.

3. e In 1 day A and B together produce

1 1 315 12 20

+ = units

∴ In 20 days,

A and B together will produce 3

20 3 units20

× = ,

which will fetch themRs. 270.

Now, Efficiency of B 15 5Efficiency of A 12 4

= =

⇒ Share of B is 59

× 270 = Rs. 150

4. a 12 bottles have to be shared by has 10 people, i.e.a man consumes 1.2 bottle and 50% has beenconsumed⇒ 6 bottles or (0.6 bottle by each).2 bottles out of 6 have to be scarified to give themtheir original share.

31

62 =∴ of the remaining is to be scarified by

each.

5. b V = Speed, N = Wagons∴ The quantity by which speed is diminished = 40– V

⇒ 40 – V = NK ; V = 28 and N = 16

⇒ K = 3

∴ 40 – V = 3 NWhen V ≥ 10, the least value of V = 10 km/hr

⇒ 40 – 10 = 3 N ⇒ N = 100

6. d192

x28161

121

x =

+

Left capacity = 1 – 192

x28

This is filled by P in 5 min and fills 121

in 1 min

⇒125

192x28192 =− ⇒ x = 4 min


Page: 117��

7. eS 2 S 1

10.5m

L

Deep endShallow end18.5mOO

First they meet at point O,

for which 2

1

S L 18.5S 18.5

−= ...(i)

as both stay at the ends for equal time, so timetaken from the starting up to the second meeting(atO’) will be same.

2

1

S L (L 10.5)S L 10.5

+ −⇒ =+ ...(ii)

from (i) and (ii) L 18.5 2L 10.5

18.5 L 10.5− −=

+apply componendo & dividendo-

L 3LL 37 L 21

=− − L 45m⇒ =

8. b The two trains start at the same time.If they meet after x hr, then 21x – 16x = 60⇒ x = 12

⇒ Distance = (16 × 12 + 21 × 12) = 444 km

9. c Ratio of time taken is 3 : 2.If difference is 1 min, A takes 3 min.If difference is 10 min, A takes 30 min.⇒ At double speed A takes 15 min.

10. c 3 men = 4 boysHence, 27 boys can reap a field in 15 days.

So 20 boys + 16 boys = 36 boys will take 43

of the

time, i.e. 11.25 days.

11. c If A takes x seconds and B takes y seconds torun 1 km, then

x + 19 = y301000

x960andy

1000960 =+

⇒ y = 150 s and x = 125 s

⇒ answer = 150

5000 750s1000

× =

12. b Trained labourer Days Hours Work

52

× 33 15 12 1

y 11 9 1.5

Thus, y = 52

× 33 × 1115

× 9

12 ×

23

= 36

13. d 2v

36u8 =+ and

47

v33

u6 =+

Solving these, we get u = 16 and v = 24.Hence, rate of stream is 4 km/hr.

14. b LCM of 224, i.e. (25 × 7) and 364, i.e. (22 × 7 × 13) =25 × 7 × 1325 × 7 divides this 13 times.So A does 13 rounds before meeting again.22 × 7 × 13 divides this 8 times.So B does 8 rounds before meeting again.Thus, the difference in the number of laps is(13 – 8) = 5

15. b

X's rate Time taken by Y to reach Jodhpur

=Y's rate Time taken by X to reach Jaipur

or 45 200 5

Y 's rate 288 6= =

⇒ Y’s rate = 54 km/hr

16. d If the cross-sectional circumference of pipe A is 1unit, that of B would be 2 units and that of C wouldbe 3 units.So areas would be in the ratio 1 : 4 : 9 or the areasof B and C together is 13 times that of A.

So it takes 1

16 min13

× or min133

1 to fill the

original tank.But now we have to fill the second tank. Thus, time

required will be min1332

13216 =×

.


17. c Let him walk at 4 km/hr for t hr and at 3 km for h hrNow 4 × t + 3 × h = 36Also 4 × h + 3 × t = 34Solving, we get h = 4, t = 6Therefore, h + t = 6 + 4 = 10 hr

18. c 1 man = 2.5 boys; 1 woman = 1.5 boys; convert allwork in terms of boys.Thus, group 1 = 43 boys, group 2= 107.5 boys.

Hence, group 2 would take 35.107

4325 ××

= 30 days

19. d It is given that 87

of the distance = 420 km or total

distance = 78

420× = 480 km.

It is given that he stopped for 9 min and when heincreased his speed by 20 km/hr, he reached hisdestination by 6 min earlier. Hence, there is adifference of 9 + 6 = 15 min. Now take options intowork.Taking option (d), i.e. 60 km/hr,

time taken =2060

6060420

++ = 7 hr 45 min.

If he travelled the entire 480 km at the rate 60 km/hr, then time taken = 8 hr.Since there is a difference of 15 min, (d) is theanswer.

Alternative method:Let x be the initial speed. Equating time

420 60 9 480 6–

x x 20 60 x 60+ + =

+

420 60 1 480x x 20 4 x

+ + =+

60 1 60x 20 4 x

⇒ + =+

Check from options, x = 60.

20. e Average speed = Total dis tance

Total time taken

= 480

7 hr 9 min 45 min+ + = 480 480

7 hr 54 min 7.9=

= 60.75 km/hr.

Explanations:Fundamentals of Grammar

Page: 119��

Unit – 1

Chapter 1

Exercise 1

1. time - subject,may be - verb

2. mail - subject,was - verb

3. letter- subject,has been - verb

4. men - subject,were - verb

5. tracks - subject, were – verb

Exercise 2

1. compound verb

2. compound sentence


4. compound subject

5. compound verb


7. compound object of the preposition


9. compound verb


Exercise 3

1. who listens to his men modifies leader

2. which I loved dearly modifies dog

3. who takes responsibility well modifies person

4. who purchased tickets modifies individuals

5. that you bought for me modifies shirt

6. who baked the winning pie modifies woman

7. when I was unable to answer modifies time

8. for whom you are looking modifies one

9. who are willing to serve others modifies those

10. to whom much is given modifies one

Exercise 4

1. How the prisoner escaped = subject

2. that the robbery was an inside job = predicatenominative

3. how he could just disappear = direct object

4. that he had escaped = appositive

5. whoever finds the stolen diamonds = indirect object

6. where he might be = object of the preposition

7. That we were ready to go = subject

8. whoever wants to go = indirect object

9. That you are losing ground = subject

10. Whoever injured the handicapped woman = subject

Exercise 5

1. that we have had = adjective clause modifying thepredicate nominative year

2. until we received word of his rescue = adverb clausemodifying the verb waited

3. whom I saw on Mount Kilimanjaroo = adjective clausemodifying the subject hiker

4. that he will win the lottery = noun clause used as thedirect object

5. Who lost this wallet = noun clause used as the subject

Exercise 6

1. If the manager is unable to help = adverb clausemodifying the verb try

2. whom you should write the letter = noun clause usedas the object of the preposition

��


3. whose neck was broken = adjective clause modifyingthe subject man

4. that the ozone levels were dangerous = noun clauseused as the direct object

5. when the governor changed his mind = adverb clausemodifying the verb objected

6. that Sarika will not return = adverb clause modifyingthe predicate adjective unfortunate

7. Why you don’t do your work = noun clause used asthe subject.

8. where your Aunt is buried = noun clause used as thepredicate nominative.

9. that the island is under water = noun clause used asthe appositive.

10. whoever told the truth = noun clause used as theindirect object.

Chapter 2

Exercise 1

1. Mistakes are made by us.

2. Extensive research is done to determine by whichgene autism is caused in the body.

3. People all over the world speak English.

4. The dog bit Latika.

5. Characters are revealed by manners.

6. Why did Sheena say such a thing?

7. He was laughed at by everyone.

8. I will be obliged to go by the circumstances.

Chapter 3

Exercise 1

1. Vandana says that her husband loves her a lot.

2. Julia Roberts asked him about his plans for thatday.

3. Richa said that all of us would go there in themorning tomorrow.

4. She said, “How long have you worked here?”

5. She said that she did not know where her shoeswere.

6. She asked, “Why are you studying English?”

7. She said, “May I open a new browser?”

8. She said she had to have a computer to teachEnglish online.

Unit – 2

Chapter 5

Test 1

1. She wants an orange from that tree. (there can bemany oranges on the tree and the noun is startingwith a vowel sound).

2. The edifice on the corner is huge. (some specificbuilding, which is huge).

3. The Chinese that Mrs. Sarla speaks is very easy tolearn. (that specific Chinese, which Mrs Sarla speaks).

4. I borrowed a pencil from your sister’s pile of pensand pencils. (She borrowed one of the many pencilsthat her sister has).

5. One of the officers said, “ The chairman is late today.”(The officers are talking about someone specific, theirchairman)

6 Ankur likes to play volleyball. (No article is requiredhere.)

7. I bought an umbrella recently. (the modified noun isbeginning with a vowel; u)

8. Kirti is learning to play the violin at her school. (Definitearticle comes before a)

9. Please give me the book that is on the counter. (Thespeaker is asking for a particular book, which is lyingon the table.)

10. We lived on Pali Street when I first came to your town.(No article is required before names of cities, states, countries or streets)

11. Delhi is the capital of India. (No article is requiredbefore names of cities, states , countries or streets)

12. My colleague’s family speaks Polish. (No article isrequired before names of languages.)


Page: 121��

13. The apples in my basket are red. (Those particularapples, which are in my basket)

14. Our friends have a cat and a dog. (our friend hasboth, one dog and one cat)

15. Go to a university near your house. (The noun isstarting with a vowel but is pronounced with a ‘u’sound, therefore, a instead of an is used.)

16. a lot

17. several of

18. a lot of

19. some

20. only, a few

21. a majority of, enough

22. many

23. much of

Test 2

1. a In 'A' it should be 'a tedious loser' and in 'C' it shouldbe 'a similar level…'.

2. c In 'E' it should be 'in …the sunshine".

3. c In 'B' 'enough' is not required. The correct expressionis 'large enough' not 'larger enough'. In 'D' 'various' isinappropriate. It should be replaced by 'several'.

4. c 'A' should have 'a friend'. 'C' should have 'gardeningis a joy....'and 'D' should have 'pick a better.."

5. a In 'B' it should be 'ever more....' and not 'even..'. In 'C'it should be 'many' not 'more'.

6. e 'E' should have '..the fleece'.

7. a In A it should be 'a prudent board..' because it can beany board.

8. e In E it should b 'a mistake'.

9. c Money is quantified in terms of ‘little’.

10. d ‘People’ are numbered. ‘Few’ is used for numbers.

Test 3

1. d A is incorrect use an 'a' before store, B is incorrectuse 'the' before commodities, E is incorrect, use 'an'before enormous.

2. c Options A, C & D are incorrect, all need the article 'a'before 'far' in A, 'spirit' in C and 'watch tower' in D.

3. d B is incorrect, needs 'the' before hero, E is incorrect,needs 'A' before crush.

4. b Option A is incorrect because it should be 'an' epigramnot 'the' as it embodies the generic, C is incorrectsince it uses an before earliest which should be 'the'instead.

5. d A is incorrect, use 'the' before 'West', C is incorrect,use 'the' before the word 'roots'.

6. a Option B is incorrect, use 'a ' instead of 'the', E isincorrect, use 'The' instead of 'A' to describe thedefinite.

7. b Ans. b: A is incorrect as the article 'a' should be usedbefore certain, C is incorrect 'the' should be used toemphasize 'nutritional'.

8. d A is incorrect; use 'the' before Olympics, B is incorrect,use the article 'the' before 'big contract'.

9. b A is incorrect, use definite article 'the' before'dangers', D is ambiguous, needs a determiner before'people' use 'some' to make the sentence correct.

10. e C is incorrect the definite article (the) should precede'two' and E is incorrect as 'the' should precede theword 'pleasure'.

Unit – 3

Chapter 6

Exercise 1

1. petrol - mass; class - collective

2. group - collective; bus - count

3. orchestra - collective; arena and evening - count

4. water and oil - mass; beach - count


Exercise 2

1. Objective case

2. Objective case

3. Objective case

4. Subjective case

5. Subjective case

6. Objective case

7. Possessive case

8. Possessive case

9. Objective case

10. Possessive case

Exercise 3

1. People and wall are concrete nouns. Home run is acompound noun.

2. Son is a concrete noun while post office and UdaipurLake city are compound nouns.

3. Mumbai Island is compound; success is abstract;Singapore is concrete.

4. Respect, honesty and policy are abstract nouns.Compound nouns can also be concrete or abstract.

Test 1

1. e All the parts of the sentence are grammaticallycorrect.

2. d In D it should be ‘another’ and not other since it istalking about one more.

3. c In C ‘barren century’ should be preceded by ‘some’ or‘a’.

4. b In B the right way of writing will be ‘each called ascissor’.

5. a In A the correct way of writing is ‘information doesnot..’ since ‘information’ is always treated as asingular.

6. c In A the correct word is ‘damage’ and not damagesand in E the right word is ‘illness’ since it is ‘a chronic’.

7. a It should be ‘Sumeet, Rekha and I’ because in suchcases the sequence should be third person, secondperson and finally first person.

8. e In A the right word is ‘baggage’; in C the right word is‘equipment’ and in E the right word is ‘data’.

9. b In B the correct word is ‘strength’ and not ‘strengths’,and in E it will not be ‘shore’ but ‘shores’.

10. e ‘Suspicion’ will not take ‘s’.

Test 2

1. c C is the correct option. In A it should be ‘Soviet dictator’(singular). In B the correct word should be ‘they’ sinceit refers to ‘people’ and ‘people’ is plural. In D ‘often’should come after ‘Stalin’ and in E it should not be‘Soviet’s’; rather it should be ‘Soviet’ because it istelling about the nationality of the concerned personand thus qualifying the noun.

2. c C is the correct option because ‘our house’ is beingcompared to ‘that of’ our neighbours. B & D suggestthat ‘our houses or house’ is bigger than ‘ourneighbours’.

3. a B is incorrect because the ‘apostrophe’ should comewith the ‘novelist’ and not ‘Charles Dickens’. In caseof two nouns coming together the apostrophe isalways placed with the second noun. D is incorrectbecause ‘respect’ is an abstract noun and it can neverbe in plural. In E ‘emotions’ will be followed by ‘has’and not ‘have’ because it is referring to the ‘portrayal’and not to ‘emotions’.

4. e In A ‘boomers’ should have an apostrophe. In B ‘civilright’ is a wrong expression because it’s always‘rights’ in the given context. In C ‘credits’ is incorrectbecause here ‘credit’ means ‘trustworthiness;credibility’ and it cannot be used as plural. In D ‘toomuch’ is not appropriate in the given context becauseit changes the positive tone of the sentence into anegative one.

5. b In ‘A’ ‘coverages’ is incorrect. The right word is‘coverage’. Certain words remain same in both singularand plural states. For example, ‘aircraft’, information’,‘sheep’ etc. ‘C’ has improper structure. In D it shouldbe ‘coverage makes..’ because ‘coverage’ is acollective noun here. In E it should be ‘days’ since thecontext is referring to a general kind of atmospherefor which the author yearns therefore ‘days’ will bemore appropriate.


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6. c A is incorrect because ‘hairs’ is incorrect. B is incorrectbecause it says ‘not the…’. A process can be anyprocess therefore it will be ‘an ..extremely…’. D isincorrect because ‘growing’ will not take ‘of’ here. Eis incorrect because ‘are’ and ‘processes’ areincorrect out here, since the noun (gerund) referredto, in this is ‘growing’.

7. a ‘Tourists’ and not tourism should be used.

8. c The climate of Delhi can be compared to the climate ofJaipur and not to the city of Jaipur.

9. a ‘alike’ means resembling each other. Similar wouldmean not exactly the same.

10. c When the sentence is positive, the tag has to benegative.

Test 3

1. c In A ‘politician’ should be in plural because the sentencetalks about ‘other’ which is an undefined number. Incase it were a singular then it should have been ‘..theanother’. In B ‘a time’ is incorrect; it should have been‘some..’

2. c In C ‘colonial’ is an incorrect word to be used becauseit is an adjective and not a noun.

3. d In A instead of ‘enough’ it should be ‘some’. In D anarticle ‘the’ should precede ‘Atlantic’.

4. a In A it should be ‘a degree of showmanship’ becausedegree has to be somewhat quantified andshowmanship does not require a ‘the’ before it.

5. d In A it should be ‘a rebuke’. In D it should be ‘indiscretion’and not ‘indiscrete’, because the previous word is anoun and the latter is an adjective.

6. a In A it should be ‘surge’ and not ‘surging’ since thecontext requires a noun. In E logging will not take an‘s’.

7. c In C prediction should be in plural form and in D it willbe ‘effects’ and not ‘effect’.

8. b In B ‘experience’ is the right word and not ‘experiences’.

9. e In E, it will be ‘imminence’ and not ‘imminent..’ since inthe context the emphasis is on the very fact of‘imminence’ which is a noun. If we put ‘imminent’ themeaning will be entirely out of context.

10. e In D it will be ‘tank bombardment’; since ‘bombarding’is a verb.

Chapter 7

Exercise 1

1. She, him, his

2. I, you, your

3. He, himself, our

4. It, me, you

5. They, her, me

Exercise 2

1. ‘He’ is the antecedent for ‘his’.

2. ‘Madonna’ is the antecedent for ‘her’.

3. ‘Rabbit’ is the antecedent for ‘its’.

4. ‘They’ is the antecedent for themselves, and you isthe antecedent for ‘your’.

5. ‘Teacher’ is the antecedent for she, and us is theantecedent for ‘our’.

Exercise 3

1. his

2. yours

3. theirs

4. his, hers

Exercise 4

1. myself

2. herself

3. themselves

4. himself

Exercise 5

1. ourselves

2. himself

3. herself

4. themselves


Exercise 6

1. ‘Which’ is the relative pronoun.

2. ‘That’ is the relative pronoun.

3. ‘Who’ is the relative pronoun.

4. ‘Whom’ is the relative pronoun.

5. ‘Whose’ is the relative pronoun.

Exercise 7

1. that

2. those

3. this

4. these

Exercise 8

1. everybody, someone

2. both, everything

3. anyone

4. no one, others

5. somebody, one, neither

Exercise 9

1. what

2. who

3. which

4. whose

5. whom

Exercise 10

1. whom - interrogative,you - personal,that - demonstrative

2. neither - indefinite, my - personal,me - personal

3. you - personal, someone - indefinite,who - relative, others - indefinite

4. what - interrogative,this - demonstrative,me - personal

5. I - personal, that - relative, you - personal

Exercise 11

1. he - personal, himself - personal, intensive, my -personal, possessive. He is the antecedent forhimself. (something is a noun)

2. Which - interrogative, this - demonstrative

3. These - demonstrative, mine - personal, possessive,Whose - interrogative, these - demonstrative

4. this - demonstrative, that - relative, I - personal, you -personal

5. everyone, some, many, no one, none - all are indefinite

6. he - personal, himself - personal, reflexive, his -personal. He is the antecedent for himself and his.

7. I - personal, myself - personal, intensive, him -personal, himself - personal, reflexive, everybody -indefinite. I is the antecedent for myself, and him isthe antecedent for himself.

8. neither - indefinite, them - personal, anyone - indefinite,who - relative, us - personal

9. who - interrogative, that - relative, that - demonstrative

Test 1

1. Here the subjective case of pronoun will be used.

2. ‘She’ is also a subject in this case along with ‘Payal’.

3. In this sentence ‘They’ is the subject and the action isdirected towards ‘him’ and ‘me’ therefore these twowords are in objective cases.

4. The ‘call’ is directed towards ‘him and her’ thereforethese two are the objects of the call in the sentencethus in objective case.

5. Same explanation as in ‘4’.

6. Here both ‘I’ and ‘he’ are the subjective cases ofpronouns.

7. ‘Rohit’ and ‘she’ both are subjects in this sentence.

8. In this sentence ‘we’ will be used because the actionof ‘decide’ is being done by them in the sentence;thus the subjective case.


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9. Here ‘I’ will be used because ‘you’ is the subjectand the condition referred to is ‘if you were ..’.

10. ‘One’ is the pronoun antecedent therefore it shouldbe followed by ‘oneself’.

Test 2

1. ‘whom‘ is the correct pronoun.In this sentence a relative pronoun that is in theobjective case because the woman is the object ofthe action and ‘we’ is the subject.

2. Similarly in the second sentence the subject is ‘you’and the object has to be ‘whom’ and not ‘who’ becausethe latter is a subjective case of pronoun.

3. Same explanation as ‘2’.

4. ‘Whomever’ is the objective case.

5. Same explanation as ‘4’



8. Politicians are the subjects out here, therefore therelative pronoun should be in the objective case.

9. In this sentence the action is ‘write’, therefore thedoer has to be in the subjective case. And in this case‘whoever’ is the doer.

10. Similar explanation as ‘9’.

11. In this sentence a particular and definite set of book isbeing referred to, therefore ‘that’ has been used.

12. The sentence is referring to one of the books,therefore ‘which’ is being used.

13. Similar explanation.

14. Here the reference is to a definite law therefore‘that’.

15. The law referred to is one of the laws thus ‘which’.

Test 3

1. e C is incorrect, uses the pronoun ‘myself’ , ‘me’ is thecorrect pronoun to be used here, E is incorrect use ‘I’instead of ‘me’.

2. e B is incorrect; use singular verb with ‘each one’ shouldbe works, E is incorrect there should be a pronounbefore ‘young ones’, use ‘their’ to make it correct.

3. b A is incorrect, use Everybody / Everyone instead of‘Everyman’, B is incorrect use ‘who’ instead of ‘whom’– whom is used as the object of a verb or preposition.D is incorrectuse ‘they’ instead of them.

4. e B is incorrect, use ‘what’ instead of which, ‘what’here is used relatively to indicate that which; D isincorrect, use ‘This’ instead of the plural ‘these’, ‘this’is used to indicate a person, thing, idea, state, event.

5. b B is incorrect, should be the possessive pronoun ‘its’,C is incorrect , the pronoun should agree in numberwith the noun that is the object of the preposition, use‘is’ instead of ‘are’.

6. e D is incorrect, use singular pronoun instead of ‘their’,E is incorrect use ‘your’ instead of the possessiveyours.

7. c E is incorrect, should be ‘her speech’- with compoundantecedents joined by or, nor, either...or, neither...nor,use pronouns that agree with the nearest antecedent.

8. a A is incorrect- When a pronoun is used along with anoun, choose the pronoun case that matches thenoun’s function, should be ‘us’, B ‘They’- When apronoun is part of a compound element, choose thepronoun case that would be correct if the pronounwere not part of a compound element, C- Should be‘he’ instead of himself - when a personal pronoun isused in a comparison, choose the correct pronouncase by carrying the sentence out to its logicalconclusion.

9. e B is incorrect uses the contraction of it is should bethe possessive ‘its’, D is incorrect should use ‘which’instead of ‘who’.

10. e B is incorrect use the reflexive form ‘herself’ insteadof her, C is incorrect use ‘they’ instead of which.

Unit – 4

Chapter 8

Exercise 1

1. are going,

2. have been resting

3. must be

4. will be finished.


Exercise 2

1. has

2. do

3. w a s

4. are being

‘Do’ is action verb, and was, has and are being arestate of being verbs.

Exercise 3

1. plays - action

2. will return - action

3. is - linking or state of being

4. have been - state of being

5. should have been playing - action

6. go - action

Exercise 4

1. Can understand

2. must have told

3. shall go

4. was howling

The use of helping verbs causes certain changes inverb phrases. One change is the use of contractions.

Exercise 5

1. ‘ve (have) done

2. are going,

3. ‘s (is) staying.

Have and is are in contracted form. Are is connectedto the contracted form of not. All three-verb phrasesare action verbs.

Exercise 6

1. have been driving

2. was parked

3. can be.

Exercise 7

1. have helped

2. will be

3. had choked.

The first and third sentences are action verbs, andthe second sentence a state of being verb.

Exercise 8

1. is reading

2. can be altered

3. is being planned

4. should be

5. ‘ve (have) run.

There are twenty-three helping verbs viz. is, am, are,was, were, be, being, been, have, has, had, do, does,did, shall, will, should, would, may, might, must, can,and could.

Exercise 9

1. will be finished

2. should have been worked

3. would give. Not and n’t are not part of the verb phrase.

Verb phrases can have one, two or three helpingverbs in themVerb phrases with two or more helping verbs alwayskeep a definite order. Most helping verbs can combinewith other helping verbs but will not combine with allof them. e.g., is being said, has been said, will be said,could have been said, may have said, had been said.

We can change the form of a verb: A verb can havean s added to it as in eat, eats or run, runs. Otherchanges could be eating, ate, or eaten for the verbeat. Run could be changed to running, or ran. SomeIrregular verbs have several confusing changes.


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Exercise 10

1. am coming

2. came

3. comes

4. send

5. was sent

6. am sending

Exercise 11

1. can understand - action

2. is going - action

3. can be held - action

4. have seen - action

Exercise 12

1. roared - intransitive complete (no receiver of theaction)

2. was - intransitive linking (captain is a predicatenominative)

3. will be needed - transitive passive (be is the helpingverb and dress receives the action)

4. did forget - transitive active (title receives the actionand is the direct object)

5. has - transitive active (camera receives the actionand is the direct object)

Exercise 13

1. failing

2. bordering

3. to consult

4. searching

5. to match

Exercise 14

1. farming = predicate nominative

2. traveling = appositive

3. writing = subject

4. saving = object of the preposition / traveling = directobject

5. gossiping = indirect object

Exercise 15

1. to skate = subject

2. to enjoy = direct object

3. to win = predicate nominative

4. to marry = direct object

5. to kill = appositive

Exercise 16

1. to sit in judgment = subject

2. to waste time in class = subject

3. to party/to sleep in = subjects

4. to build a home/(to) raise a family = direct objects

5. to be home soon = direct object

6. to play cricket/to visit friends = predicate nominatives

7. to save money for a rainy day = subject

8. to go to college/to study management = predicatenominatives

9. to be rewarded for one’s efforts = direct object

10. to live happily = object of the preposition

Exercise 17

1. broken modifying spoke

2. smiling modifying face

3. frightened modifying child

4. growling modifying dog

5. squeaking modifying wheel


Exercise 18

1. are = verb, you = subject, important = predicateadjective modifying you, too = adverb modifyingimportant, to help the poor = adverb infinitive phrasemodifying important, poor = direct object to the verbalto help, the = adjective modifying poor

2. had upset = verb, child = subject, the = adjectivemodifying child, crying = participle modifying child,everyone = direct object, in the room = prepositionalphrase modifying everyone, in = preposition, room =object of the preposition, the = adjective modifyingroom

3. jumped = verb, he = subject, across the gap =prepositional phrase modifying jumped, across =preposition, gap = object of the preposition, the =adjective modifying gap, without knowing the distance= prepositional phrase modifying jumped, without =preposition, knowing = gerund used as the object ofthe preposition, distance = direct object of the verbalknowing, the = adjective modifying distance

4. is = verb, exercising = gerund used as the subject,good = predicate adjective modifying exercising, foreveryone = prepositional phrase modifying good, for= preposition, everyone = object of the preposition

5. loves = verb, Jasmine = subject, to dance constantly= noun infinitive phrase used as a direct object,constantly = adverb modifying to dance

6. is = verb, teasing by your friends = gerund phraseused as the subject, by your friends = prepositionalphrase modifying teasing, by = preposition, friends =object of the preposition, your = adjective modifyingfriends, hard = predicate adjective modifying teasing,to take = adverb infinitive modifying hard

7. fled = verb, people = subject, the = adjective modifyingpeople, fearing reprisal = participial phrase modifyingpeople, reprisal = direct object to the verbal fearing,from the city = prepositional phrase modifying fled,from = preposition, city = object of the preposition, the= adjective modifying city

8. is = verb, eating out = gerund phrase used as a subject,out = adverb modifying the verbal eating, thing =predicate nominative, the = adjective modifying thing,to do tonight = adjective infinitive phrase modifyingthing, tonight = adverb modifying to do

9. do know = verb, I = subject, n’t = adverb modifying doknow, whether/or = correlative conjunction, to tellhim/to keep quiet = noun infinitive phrases used asdirect objects, him = direct object to the verbal to tell,quiet = adverb modifying to keep

10. should be done = verb, job = subject, our/next =adjectives modifying job, to run to the shop = nouninfinitive phrase used as an appositive, to the shop =prepositional phrase modifying to run, to = preposition,shop = object of the preposition, the = adjectivemodifying shop, quickly = adverb modifying should bedone.

Test 1

1. Both “to spend” and “spending” could be correct

2. to have gone

3. Both “to call” and “calling” are correct.

4. Answer E. (the participial phrases in A and B correctlymodify Rahul.)

5. The overloaded car gathered speed slowly.

6. The opening participial phrase is misplaced becauseit is intended to modify him, not the proposal. A possiblerevision would be: Espousing a conservative point ofview, he was bothered by the proposal for morespending on federal social programs.

7. When I return to Mumbai next year, I will be veryhappy.

8. Rahul goes to school every day.

9. Sapna is visiting her family right now.

10. I studied/was studying Economics in 1994.

Test 2

1. He has spoken/has been speaking French since hewas a child.

2. Raj had visited many places before he came here.

3. We saw terrible things back then.

4. Sometimes I still have dreams like I did twenty yearsago

5. Japan had never had democracy until 1945

6. The father will call the family together if he thinksthere is disharmony.

7. When I was young, I never cooked because myparents had two servants.

8. d ‘hearing’ is not used in present continuous. It is ‘hear’even in continuous tense — ‘I hear her’.


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9. b The verb with amounts of money and time periods issingular. Therefore, ‘twenty thousand’ is required.

10. e ‘ordered for three cups of tea’ is correct.

Test 3

1. a ‘a’ is the right option. In ‘A’ it should be ‘off’ not ‘of’. In‘B’ the correct word will be ‘seemed’ not ‘seemingly’.And in ‘D’ it should be ‘at’ not ‘in’.

2. e ‘e’ is the correct option. B and C are correct becausein both cases ‘its’ does not refer to Alan Davey, but tothe company that is not mentioned.

3. b ‘b’ is the right option. In ‘A’ it should ‘was’ and not‘were’; in ‘B’ it should ‘than’ and not ‘then’; and in ‘D’ itshould be ‘conservatives’’ and not ‘conservative’s’.

4. a In ‘B’ the correct verb should be began because ‘..asI nosed..’ happens simultaneously with the rest,therefore past perfect is not appropriate in this case.

5. d It should be ‘changing ..’

6. e In ‘C’ it should be ‘attracts us..’ and in ‘E’ it should be‘combine’ because it talks about a habitual action.

7. c In C it should be ‘you’ll bump..’; ‘have bump’ is incorrect.

8. a In ‘A’ it should be ‘helped’ because the sentence isreferring to past action.

9. a In ‘A’ it should be ‘of making’ and in ‘C’ ‘being’ is notrequired.

10. e In ‘A’ it should be ‘did..’ and not ‘do’ because thestatement is in past tense. In ‘C’ it should be ‘like..’ andnot ‘alike’.

Chapter 9

Test 1

1. a In (a), instead of ‘using’ it should be ‘use’.

2. d Rather than ‘precipitate’ it should be ‘precipitating’because the action talked about is in presentcontinuous.

3. a Instead of ‘giving’, it should be ‘given’ because ‘theefforts’ have already been made therefore past tenseshould be used.

4 . d ‘Being’ is not required here.

5. a I go. Simple present is correct here rather thancontinuous tense to indicate habit.

6. e ‘ate … left’ the parallelism in the sentence is correct.

7. a Present perfect continuous is correct here.

8. e The sentence in indirect speech is correct.

9. a ‘with aplomb’ is correct.

10. c (c) is the correct way of asking a question.

11. a ‘Reach’ doesn’t take a preposition.

Test 2

1. a In A the correct helping verb is ‘is’ and not ‘was’,because the whole sentence is in present tense. In Einstead of ‘gives’ it should be ‘given’ because the actis in past tense.

2. b In B it should be ‘...will be’ and not ‘...have to be…’because the sentence is indicating towards the futureaction, besides ‘have’ is used for a plural subject.

3. d In A it should be ‘rears’ since the sentence is talkingabout a routine action.

4. e In A, it should be only ‘happened’ and not ‘will havehappened’; since the statement is referring to actionthat has already taken place.

5. c In A it should be ‘I have’ because ‘will be followed’ isan incorrect expression. In D it should be ‘gathered’and not ‘gathers’, since the statement talks about pastaction.

6. a In A it should be ‘to encompass’. In E it should be‘dampen’ and not ‘dampens’.

7. e In E ‘does’ is the incorrect word. It should be ‘do’.

8. d In ‘B’ it should be ‘..have been’, since the sentence isin present tense.

9. b In B the correct tense will be present, so it should be‘They work…’.

10. a In A, since ‘may’ is already there, ‘will’ is not required;and since the idea is related to probability, ‘will’ is toostrong a word to be used here.


Test 3

1. b Option B is the correct answer because ‘by’ shouldprecede ‘involving’ to give a proper meaning to thesentence.

2. d 'Has been' is the correct expression here since thecontext suggests that the work has got over firstnow. Thus present perfect should be used here.

3. e Option E is correct in all respects of the verb usage,‘becomes’ is the correct usage.

4. c correct usage of the verb is ‘is’.

5. b Correct usage is ‘ motivated’ which means to providewith a motive. In ‘C’ the meaning of the sentence isentirely changed.

6. c Option C is correct with respect to the form of theverb as well the subject, since the sentence is in pasttense.

7. d Option D is correct because once the results comeout then only analyzing discoveries can the madethus the use of past tense.

8. e Only option is E is correct in its usage of the verbs‘are’ and ‘explore’.

9. a Correct usage is ‘have’.

10. d Option D is the correct answer becaue have shouldprecede ‘held and hold’ which is a phrase here.

Chapter 10

Test 1

1. In the newspaper, an interesting article appeared.

2. Across the road lived her boyfriend.

3. Around every cloud is a silver lining.

4. Neither he nor his brother are (change to is) capableof such a crime.

5. The teacher or student is going to appear on stagefirst. (No change required)

6. The mother duck, along with all her ducklings, swim(change to swims) so gracefully.

7. Each of those games is exciting. (No change required).

8. The file, not the documents, were (change to was)misplaced.

9. Here is (change to are)the three books you wanted.

10. Five hundred rupees is/are all I am asking.

Test 2

1 d.

2. The majority of the Parliament is/are Congressmen.(The majority of the Parliament… what follows the ofis singular. Hence singular verb)

3. c (‘who’ refers to what is immediately before it‘students’, hence ‘are’)

4. d

5. The original document, as well as subsequent copies,was/were lost.

6. c (what follows the ‘of’ is plural) ‘... are going to thepolls.”

7. Almost all of the magazine is/are devoted toadvertisements.

8. Here is/are Manish and Mandar.

9. Taxes on interest is/are still deferrable.

10. b (five rupees is singular, hence is)

Test 3

1. d In D it should be ‘have’ since it is referring to bothgraffiti and its street art cousins.

2. b The correct verb is ‘come’ because the subject is‘irreverence’ which is singular.

3. b In B ‘carries’ is incorrect. It will be ‘carry’ since thesubject i.e. is in plural.

4. d It says ‘Mr. Lott, along with Speaker …’, therefore in Dit should be ‘was ..’ because the subject is ‘Mr. Lott’.

5. e In E it should be ‘that are ...’, since the subject referredto is ‘techniques’.

6. b It should be ‘there is ...’ since it is referring to ‘good aswell’.

7. c In C it should be ‘a hacker could get into…’.

8. c In C the correct helping verb is ‘has’ and not ‘have’since the subject referred to is ‘No one’.


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9. e In E it should be ‘is’ since the referred subject is ‘each’.

10. d In D it should be ‘nail scissors were..’ and in E itshould be ‘channel was..’ because here the subjectis ‘breach’; ‘some’ here has a different contextualmeaning.

Unit – 5

Chapter 11

Exercise 1

1. Heavy, red, fifty.

2. My, two, my, graduation

3. That, small, Indian, the, next, fresh

4. Little, black, the, well-dressed

5. Old, wood, several, discarded, packets

Exercise 2

1. red

2. Those, brown

3. Two, what, many, their

4. Third, city, another

5. That, good, this

Exercise 3

1. New

2. Stormy, spring, flash

3. New, UK

4. December, dangerous Himachal

Exercise 4

1. Soaking

2. Broken, crying

3. Great, giving

4. Laughing

5. Eager, torn

Exercise 5

1. Either and the first any are pronouns, this and thesecond any are adjectives.

2. The first each and their are adjectives, and thesecond each is a pronoun.

3. This and no one are pronouns.

4. Both and one are pronouns, and many and neitherare adjectives.

5. What and neither are pronouns, and your is anadjective.

Exercise 6

1. Our, first, many, strong

2. Happy, three, frolicking, their, his

3. Exciting, most

4. Flooded, terrible

5. Kaushik, hot, exhausting

Exercise 7

1. jolly, jollier, jolliest

2. honest, more honest, most honest

3. dim, dimmer, dimmest

4. friendly, friendlier, friendliest

5. little, less or lesser or littler, least or littlest (Little whenreferring to amount uses less, lesser and least; whenreferring to size uses littler and littlest.)

Exercise 8

1. interesting, more interesting, most interesting

2. critical, more critical, most critical

3. splendid, more splendid, most splendid

4. delicious, more delicious, most delicious

5. outstanding, more outstanding, most outstanding


Exercise 9

1. many, more, most

2. ill, worse, worst

3. much, more, most

4. perfect - cannot be compared since there is nomore perfect or most perfect.

5. bad, worse, worst

Exercise 10

1. worst

2. hungrier

3. shortest

4. best

5. happiest

Exercise 11

1. a wife and doctor

2. a girl and a boy

3. a green and red

4. a rock star or a lawyer

5. a bat and a ball

Exercise 12

1. this or that

2. this or that

3. these or those

4. these or those

5. these or those

Test 1

1. a In B ‘dried’ is not appropriate, since it cannot be usedas a predicative adjective here and it does not match‘clean’ and ‘safe’. Similarly C is incorrect because‘cleaned’ is not appropriate. In D ‘and’ is not requiredafter ‘dry’ because when using adjectives in a stringwe can separate them by commas. Therefore correctway of writing will be ‘dry, clean and safe’. In E‘effected’ is not the right adjective to be used.

2. e In A ‘relenting’ is not the right adjective because itchanges the meaning of the sentence. In B‘relentlessly’ is incorrect because it is an adverb andnot an adjective and contextually we require anadjective here to qualify the noun ‘increase’. In C‘sophisticated’ should come before ‘systems’ since itis qualifying ‘systems’ and is used as an attributiveadjective. In D ‘sophistication’ is a noun and not anadjective therefore incorrect.

3. c In A instead of ‘Kings eye’ it should be King’s eye’because here possessive case is required. In B theapostrophe is missing from ‘family’s’. In D instead of‘family’s’, ‘families’ is given. In E ‘king eye’ is incorrect.

4. a In B ‘pushy’ should be preceded by ‘and’ becauseonly two adjectives have been used here. In C ‘whichseems’ should be followed by ‘surprising’ which is anadjective and it qualifies the unwritten noun in thesentence that is ‘love’; whereas ‘surprisingly’ is anadverb which cannot qualify a noun. In D ‘velvety’ isnot an appropriate word because ‘brocade’ is a nounhere thus requires a noun ‘velvet’ as the other wordin the pair, whereas ‘velvety’ is an adjective; though‘Velvet and Brocade’ has been used here as a ‘Nounadjunct’ (refer to Practice Exercise 2). In E ‘velvet-and-brocade’ should precede ‘backdrop’ rather thanfollowing it since the former is qualifying the latter.

5. c In A, ‘original’ is an appropriate word because thecontext requires an adjective. In B ‘in probing hisopponent’s defences’ should follow ‘duke’s skill lay..’.In D it should be ‘persuasively’ and not ‘persuasive’because the former qualifies the verb ‘argues’ thus itis an adverb whereas the latter is an adjective andcannot qualify a verb.. In E, ‘when a weaknessopened up..’ should follow ‘quickly and committingtroops’.

6. a In B ‘enjoyed fervently the support ..’ does not makeany sense; rather it should be ‘the fervent support’because ‘fervently’ is an adverb and ‘fervent’ is anadjective. In C ‘of Americans of all colours’ shouldcome after ‘’fervent support’ or else it will not be clearas to preposition ‘of’ is referring to what. In D ‘billed asa contest’ is referring to ‘a fight’ therefore it shouldfollow ‘a fight’. In E, ‘similar’ is not an appropriate


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adjective. Instead ‘same’ should be used. If at all ‘similar’has to be used it should be preceded by ‘a’ becauseit can be any.

7. d In A rather than ‘irrelevantly’ it should be irrelevant,since the former is an adverb and an adverb cannotqualify a noun. In B ‘irrelevance’ is used incorrectlybecause it is a noun and the context requires anadjective. In C ‘publicly’ is incorrect. In E ‘nuisance’should follow ‘irrelevant’.

8. a In B a comparative adjective is not required, therefore‘crisper’ is incorrect. In C ‘admiring’ is incorrectbecause an adverb (admirably) is required here whichis qualifying the adjective ‘self- deprecating’. In D ‘bothemotional and professional’ is placed incorrectly. Itshould come after ‘..life’. In E ‘self-deprecatingly’ is anadverb which is inappropriate since ‘vignettes’ is anoun which requires an adjective.

9. d In A ‘populist’ is an incorrect adjective since thestatement talks about ‘sovereignty’. In B ‘with the idealof popular sovereignty’ is placed incorrectly; it shouldcome in the end. In C ‘habitual’ is incorrect since it isan adjective and the sentence requires a noun that is‘habit’. Again in E ‘popularly’ is an adverb and thecontext requires an adjective not an adverb.

10. c A doesn’t require an adverb ‘extraordinarily’; rather itrequires an adjective ‘extraordinary’. In B‘demographic’ is placed incorrectly before‘extraordinary’ but the ‘growth’ in this case pertainsto ‘demography’. In D ‘in which settler pressure’ shouldprecede ‘boosted by extraordinary demographicgrowth’, because the second phrase qualifies the‘..settler pressure..’. In E ‘demography’s’ is incorrectbecause it is a noun and an adjective is required outhere since ‘growth’ is a noun here.

Test 2

1. e In ‘A’ it should be ‘rural poor have..’ and not ‘has’.Similarly in ‘C’ ‘the rich..are..’; since collectiveadjectives are taken as plural.

2. d ‘A’ is incorrect because it’s probably not a good ideato use this construction with an adjective that is alreadya negative: “He is less unlucky than his brother,”although that is not the same thing as saying he isluckier than his brother. In ‘C’ when we are alreadyusing ‘harder’, ‘more’ is not required.

3. a In ‘B’ ‘more ardent’ is incorrect because ‘ardent’ cannothave any comparatives or superlatives.

4. a In C, ‘party’s preference’ is incorrect, rather it shouldbe ‘party preference’ otherwise the meaning will beentirely changed. In d, ‘running-down’ is an incorrectphrase. In should be ‘run- down’.

5. c In ‘C’ ‘unsightly’ should precede ‘beggars’ since it ismodifying ‘beggar’.

6. e In ‘A’, ‘exuberance’ is a noun whereas the contextdemands an adjective, therefore it should ‘exuberant’.In ‘E’ rather than ‘huge’ it should be ‘hugely’ because‘welcome’ is an adjective out here qualifying ‘start’.‘Huge’ is also an adjective therefore it cannot qualifyanother adjective whereas ‘hugely’ can since it is anadverb.

7. e In ‘A’ ‘particularly’ is an adverb and thereforeinappropriate. It should be ‘particular’ which is anadjective. In ‘E’ ‘extensive network’ is meaningless.Here ‘network’ will be treated as a verb rather than anoun and ‘extensive’ will become an adjective byadding ‘-ly’ to it. Thus it will be ‘network extensively’.

8. b In ‘B’ ‘most absolute’ is incorrect because ‘absolute’ initself is an absolute term and thus cannot have anysuperlative form. In ‘D’ ‘red facedly’ is an adverbtherefore inappropriate out here. It should be ‘redfaced’.

9. d In ‘D’ the appropriate adjective will be ‘irrational’. Sincein ‘E’the statement is talking about ‘all the new sitcoms’it should be ‘least favorite’. If it were a comparisonbetween ‘two’ then it would have been ‘less’.

10. d In D it should be impetuousness since the contextdemands a noun and not an adjective. In E, it shouldbe ‘an Elvis Presley wig’. An apostrophe with Presleywill change the meaning. It will mean that the wigactually belongs to Elvis Presley whereas the contextimplies that the wig should be of Elvis Presley style.

Test 3

1. a In ‘A’ it should be ‘cleverer’ since here a comparison ismade between what the Robots were and what theyare now.

2. c ‘no more longer’ is incorrect because longer will nottake ‘more’.

3. e In ‘E’ ‘tempting’ should precede ‘special’ since ‘special’describes the offer, and ‘tempting’ is the nature of the‘special offer’.

4. d Rather than the adjective ‘apparent’ an adverbapparently should by used.

5. a In ‘A’ ‘harder’ is the correct form of adjective since thesentence refers to ‘this time’, which implies a‘comparison’ is being made.

6. d In ‘D’ rather ‘solid linking’, it should be ‘solidly linked’.


7. d In ‘D’ using more with ‘humility’ and ‘grace’ is incorrect,since these two words do not require to be usedwith comparatives.

8. c Rather than professionalised it should be professional.

9. a It should be ‘most reliable’ and not ‘reliable most’.

10. b In ‘B’ ‘freer’ is an incorrect; ‘free’ is a more appropriateadjective.

Chapter 12

Exercise 1

1. quickly (how)

2. too (how much), softly (how)

3. soon (when), yesterday (when)

4. then (when), there (where)

5. why (why), so (how much), often (when)

Exercise 2

1. Daily, very

2. totally

3. hourly, rapidly

4. suddenly

5. unusually, outside

Exercise 3

1. completely modifies the adjective exhausted tellinghow much, quickly modifies the verb was pulled tellinghow, aboard modifies the verb was pulled tellingwhere

2. once/twice modify the verb has called telling when

3. usually/slowly modify the verb work telling how,rather modifies the adverb slowly telling how much

4. surely/n’t modify the verb were telling how, verymodifies the predicate adjective expensive tellinghow much

5. greedily modifies the verb had telling how, toomodifies the adjective much telling how much

Exercise 4

1. not, correctly (both words modify the verb did do)

2. never, so, deeply (never and deeply modify theverb was pleased; so modifies deeply telling howmuch)

3. actually, almost (actually modifies the verb likes;almost modifies every telling how much)

4. recently, n’t (recently modifies the verb found; n’tmodifies the verb would help)

5. not, tomorrow, very (not and tomorrow modify theverb will go; very modifies scary telling how much)

Exercise 5

1. n’t, very, quickly, foolishly

2. suddenly, quietly

3. laughingly, happily, together

4. rapidly, accurately

5. today, tomorrow

Exercise 6

1. wisely

2. seldom

3. often

4. now, again

5. now

Exercise 7

1. n’t (when/how), often (when), here, (where),before, (when). They all modify the verb havestopped.

2. faithfully (how), carefully (how). They both modifythe verb does.

3. sometimes (when), always (when), highly (howmuch). Sometimes modifies the verb say. Alwaysmodifies the verb have been. Highly modifies theadjective critical.

4. yesterday (when), by, (where), once (when), twice(when). They all modify the verb came.


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5. there (where), very (how much), safely (how). Thereand safely modify the verb lay. Very modifies theadverb safely.

6. today (when), rather (how much). Today modifiesthe verb seemed. Rather modifies the adjectiverestless.

Exercise 8

1. not, now

2. yesterday, very

3. never, there, before

4. too

5. always

Exercise 9

1. today, tomorrow, technically

2. n’t, alone

3. where, yesterday

4. too, again

5. finally, away

Exercise 10

1. any

2. can

3. ever

4. have

5. any

Exercise 11

1. extremely modifies the adjective tired, and verymodifies the adjective sore

2. yesterday and hardly modify the verb had completed,very modifies the adjective hard, and rudely modifiesthe verb was interrupted

3. gradually modifies the verb reached, and beforemodifies the verb had climbed

4. just modifies the adverb now, now modifies the verbremembered, and rather modifies the adjectiveimportant

5. often modifies the verb go, too modifies the adverbfar, and far modifies the verb go

Exercise 12

1. sure

2. surely

3. surely

4. sure

5. surely

Exercise 13

1. well

2. surely

3. very

4. good

5. really

6. badly

7. really

Test 1

1. e Since none of the sentence/s or part/s thereof have/has any error.

2. b In 'A' the adverb to be used is 'doubly'. It doesn'trequire an adjective since 'irrational' is an adjectiveand 'doubly' as an adverb will qualify it. On the contraryoption C will have an adjective i.e. 'huge' and not'hugely', which qualifies the noun 'businessopportunity'.

3. e 'B' is correct since 'downright' can be used as bothadjective and adverb.

4. c In 'D' 'evenly' is incorrect because it changes themeaning of the sentence. The correct adverb will be'evenly'.

5. d In B 'far' is an adverb that is describing the adverb'worse' and structurally it should come before 'worse'and not after it. In c since the context requires anadjective and both 'obsessive' and 'obsessed' areadjectives. But the right choice here will be 'obsessed'that means 'having an obsession', whereas'obsessive' means 'excessive in degree'. Here withachievement 'obsessed' parents will be appropriate


6. e In D instead of 'humanly' it should be 'human'. There isno word such as 'humanly'.In E 'nonetheless' shouldcome after 'robots', because 'nonetheless' is an adverband not an adjective. Usually an adjective precedesthe noun.

7. d In D 'anyways' is not appropriate since it is not usedin standard English usage. The correct adverb will be'anyway'.

8. a In 'A' 'deep' is an adjective and changes the meaningof the sentence therefore it should be replaced by anadverb 'deeply' which qualifies the verb 'held' here.

9. b In a 'very unique' is incorrect because 'unique' in itselfis a superlative adjective and 'very' will not make itmore so. In c the context demands an adverb since itdescribes the verb 'explaining' therefore 'originally'should replace 'original'.

10. b In C 'fully' is incorrect because the context requiresan adjective since it has to describe a noun (ac-count). Therefore 'full' is the right word and it shouldcome before account.

Test 2

1. c Correct adverb will be 'parallel'.

2. a It should be 'bump along..' because using 'alongside'here will imply that the surface is also bumping along.

3. e The use of an adverb 'deeply' is incorrect out here.The context requires the comparative form of an ad-jective.

4. d The correct adjective in this part will be 'worse' whichis the comparative form of 'bad' and in this sentencethe comparison is made with 'useless'.

5. d It should be 'moving quickly' rather than 'moving quick'.'Quickly' an adverb will modify 'moving' which is averb.

6. e Rather than 'relative' which is an adjective, 'relatively'(an adverb) will be more appropriate

7. b In part (b), 'helpful' (an adjective) should be precededby 'politically'(an adverb).

8. c 'Rather' itself means 'somewhat', therefore 'some-what' is not required here.

9. a It should be 'rapidly' because 'rapid' is an adjectiveand cannot qualify a verb i.e. 'developing'.

10. e Rather than an adverb the context requires anadjective; therefore it should be 'official' and not'officially'; since it is qualifying the noun 'neglect'.

11. b 'Discounting' is a verb and the context requires anoun adjunct. Therefore 'discount' is a better option.

12. c 'although' is not a required because 'yet' hasalready been used in the sentence.

Test 3

1. b 'a' is incorrect because 'yet' is incorrect in thissentence because it means 'for the present'; whereas'still' means 'nevertheless'. Option c is also incorrectbecause the placing of 'nevertheless' is incorrect.Option d inappropriate because 'albeit' means'notwithstanding' and though is not required, it doesn'tmake any sense in the sentence. In 'e' the use of 'yet'is incorrect.

2. a In 'a' 'such' which is an adjective,, means 'of the kind'.In 'b' 'as' with 'such' changes the meaning. In 'c' 'still' isincorrect because it doesn't make any sense here.Again in 'd' 'yet' is inappropriately placed after 'but'. In'e' 'as such' again changes the meaning of thesentence.

3. e In 'a' instead of 'more old' it should be 'older'. In 'b' acomparative form is required therefore 'older' isrequired. In c 'reassured' is incorrect because thecontext requires an adverb. In d rather than'reassuring' an adverb is required thus 'reassuringly'is appropriate.

4. c In 'a' 'enormous' is used which implies that the meaningof the sentence will convey that markets have becomeenormous, though the stress in the sentence is onthe 'growth'. 'Enormously' is apt because it is anadverb, describing the growth. In b 'possibly' is anadverb and inappropriate in the given context. In d'vigorously' an adverb is incorrectly used. Since theword is talking about the 'capital markets'- a noun, itcan be qualified only by an adjective. In e 'enormously'should follow 'grown'.

5. b In 'a' instead of 'poor' it should be 'poorly' because it isdescribing the verb 'enforced'. In 'c' 'wretched' willbe correct because wretchedly is an adverb andthus inappropriate since the context is dealing with'China disclosures' which is a noun.

In d 'poor' is incorrect because it changes the meaningof the sentence and it is also placed incorrectly. In e'enforcing' is incorrect because the 'laws' talked abouthave already been in past tense.

6. a In b 'awful' is an adjective and changes the meaningof the statement. In c 'awfully' should be preceded by'an', because the former is an adverb which isqualifying an adjective 'long'. In d 'worried' changesthe meaning. It appears that the 'signs' are 'worried'.In e 'awfully' is not placed properly.


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7. e In 'a' 'urbane' is inappropriate because though anadjective its meaning is 'having the polish and suavity'whereas the correct word should be 'urban'. Forsimilar reasons 'b' is incorrect. In 'c' 'costing' isinappropriate. In 'd' 'costlier' should precede 'fuel'.

8. d In 'a' 'fast' is inappropriate because here Yuan iscompeting against dollar. The requirement is that ofthe comparative form of adjective. In 'b' 'instead of' isinappropriate. Because instead (an adverb) means'as a substitute' whereas 'instead of' means 'in lieuof'. The context here demands 'instead'. In c 'annually'an adverb is incorrect since it cannot qualify a nouni.e. rate. Therefore the correct word is 'annual'. In e'despite' is an incorrect adverb to be used.

9. a 'Smaller' doesn't make any sense in option b becausethere is no comparison being made. In c 'opposing'does not make any sense. 'Opposite' will give ameaning to the sentence. In d 'complete' is an adjective,whereas the context requires an adverb. Similarly ine 'utter' is incorrect. It should be 'utterly'.

10. c In 'a' 'seldom' is incorrectly placed. It should be before'dissuade us'. Similarly in 'b' 'seldom' is placed between'dissuade' and 'us'. In d also seldom is incorrectlyplaced. And in e 'seldomly' i s an incorrect word.

Chapter 13

Test 1

1. b Only option (b) is correct. Option (b) clearly definesthe reason for going on trial and develops theconsequences in the correct order, a misplacedmodifier.

2. a Option (a) conveys the meaning of the sentencecompletely, defines the efficacy of houses 'with'central air conditioning.

3. c Option (c) is correct. The sentence describes the'mirror with the brass inlaid figures', only option (c)conveys the meaning correctly.

4. b Only (b) is the correct option. All other options make itseem that Federer has blue stripes, a case ofmisplaced modifier.

5. d Only (d) conveys the complete meaning of thesentence. Here the subject is the rider who wasthrown while jumping the obstacle, and example ofdangling modifier.

6. b Only option (b) is meaningful and conveys thecomplete meaning, an example of misplaced modifier.

7. e Only option (e) is correct, place the adjectives afterthe words they modify, the air is the subject which isbeing described here.

8. b Only option (b) is correct in all respects, the subject isthe type of Indian architecture in central India, anexample of misplaced modifier.

9. a Only option (a) correctly modifies the subject inquestion, an example of misplaced modifier.

10. c The sentence is correct in all respects, the subject isthe new experimental device, a case of misplacedmodifier.

Test 2

1. Looking through the telescope, we could see Venusclearly in the night sky.

2. Flying out the window, the papers were grabbed byhim.

ORHe grabbed the papers as they flew out the window.

3. Dhas arrived with the keys as I was waiting outside.OR

While I was waiting outside with the keys, Dhasarrived.

ORWhile I was waiting outside, Dhas arrived with thekeys.(The ambiguity in the original is removed in all thesesentences. First and third sentences are preferred.)

4. While walking on the grass he was bitten by a snake.

5. I tried calling half a dozen times to tell you about theCareer Launcher Seminar.

OR I called half a dozen times to tell you about the CareerLauncher Seminar.(The ambiguity in ‘tried calling’ is eliminated in the secondsentence. Choose this over the first sentence if bothare given as options)

6. Dhas manged to finish the soup although it wasextremely spicy.

7. While walking across the street, she was surroundedby them and was robbed of her purse.

ORShe was surrounded by the and was robbed ofher purse while walking across the street.

8. In her lunch box, she has some cake (that) she baked.

9. I really/very glad to be of help to you.

10. The baby smells very sweet.


Test 3

1. No change required.


3. Life in the city is exciting, but life in the countryside isbetter.

4. Drive more slowly as work is in progress.


6. Speak a little more slowly or you will not beunderstood.

7. When he spoke at a press conference on Saturdaynight, the Home Minister acknowledged the role playedby the men who subdued the gunman.

8. To improve company morale, the consultantrecommended three things.

9. In reviewing the company’s policy, the board identifiedthree areas of improvement.

10. Baked, boiled, or fried, potatoes make a welcomeaddition to almost any meal.

Unit – 6

Chapter 14

Exercise 1

1. on the wall, of the house

2. in the shade, of the apple tree, of the jobs, for theday

3. over the mound, behind the farm, into the street.

4. but you, from home, with parental permission

5. around the yard, for miles, except junk

Exercise 2

1. in, with

2. up, on, through

3. by, at

4. with, near

5. from

Exercise 3

1. of the new book modifies “title”/ about morals modifies“book”

2. on the planning commission modifies “work”/ of ideasand concepts modifies “kinds”

3. on the west side modifies “houses”/ of town modifies“side”

4. in the next room modifies “man”

5. of the citizens modifies “few”

Exercise 4

1. with sharp thorns modifies “tree”/ beside the wallmodifies “grew”

2. above the people modifies “soared”/ on the fieldmodifies “people”

3. of the village modifies “owner”/ past the housemodifies “rode”

4. by its tracks modifies “followed”/ in the snow modifieseither “tracks”(telling which tracks) or “followed”(telling where we followed it)

5. over the fence / into some bushes modify “tumbled”

6. of wreckage modifies “tons”/ after the tsunamimodifies “were left”

7. over a hill / through a beautiful valley modify“wound”

Exercise 5

1. from her desk modifies “took” telling where / of onemodifies “edition” telling which / of the classicsmodifies “one” telling what kind

2. in the display case modifies “was placed” tellingwhere / in the corner modifies “case” telling which /of the library modifies “corner” telling which

3. of mysteries and detective stories modifies “books”telling what kind / in the library modifies “are found”telling where

4. about magic modifies “story” telling what kind / inour literature book modifies “appears” telling where

5. to the solution modifies “clues” telling which / of themystery modifies “solution” telling which


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6. by my uncle modifies “stories” telling which / aboutSherlock Holmes modifies “stories” telling what kind

7. of ancient Pompeii modifies “wall” telling which / byan ordinary peasant modifies “was discovered” tellinghow

Test 1

1. This is the sort of English that I cannot put up with.

OR

I cannot put up with this sort of English

2. I don’t know where she will end up.

3. It’s the most curious book I have ever run across.


5. India became free on 15th Aug. 1947.

6. India is independent for more than 50 years.

7. India is free since 1947.

8. Where did you get this?

9. If we split it evenly among the three of us, no one willbe unhappy.

10. You can’t just walk into the class without permission.

Test 2

1. e The correct idiom is ‘light at the end of the tunnel’,which means seeing some hope after period ofdespair.

2. a ‘Step down’ means to resign from a high post or toreduce, especially in stages. Therefore it is correctcontextually. Other options are incorrect. ‘Step aside’means “to resign from a post, especially when beingreplaced.”

3. c ‘Chipped away at’ is the right way of writing. It meansto reduce or make progress on somethingincrementally. ‘Chipped in’ means too contribute moneyor labor and doing that to ‘that legislation’ is somewhatillogical.

4. d ‘With a straight face’ is the right way of writing, whichmeans ‘to do or say something confidently without atrace of guilt’.

5. a ‘Naval gazing’ is the correct expression. It meansindulge in self-absorbed pursuits.

6. b ‘Crystal-ball gazing’ is the correct expression. Itinvolves ‘the use of a crystal ball, a seer stone orother crystal as a divine tool through which one seeksto receive visions or information about the future.

7. d ‘Sell himself’ is the right expression. Other phrasalverbs used in the options have different meanings.‘Sell off’ means to get rid of by selling, often at reducedprices. ‘Sell out’ means to betray one’s cause orcolleagues. ‘Sell down’ means to betray the true trustor faith of.

8. e ‘Cast a shadow’ is the correct idiom.

9. a ‘Lo and behold’ is the correct idiom which means lookand behold; an expression of surprise andamazement.

10. a ‘Faced down’ is correct because it means to confrontboldly or intimidate.

Test 3

1. Cut the pizza into six pieces.


3. Where did he go?

4. Where did you get this?

5. I will go later.

6. Cut it into small pieces.

7. We will arrive on the fourth of next month.




11. Tanya entered the room.

12. She dived into the pool.

Test 4

1. e It means ‘Criticism’.

2. a ‘keeping an eye’ is the correct idiom.

3. b ‘Let loose’ means without any restriction (free).

4. e ‘Shell-shocked’ is the correct idiom which means‘stunned’.


5. c ‘At a low ebb’ is the correct expression which meansa bad state.

6. e ‘Wrapping up’ means to complete or stop doingsomething.

7. d ‘Middle of the road’ means having a balanced approach,between the two extremes.

8. a ‘Horse-trading’ takes place in parliament whenmembers change sides.

9. c ‘Lame duck’ means a person or company that is introuble and needs help.

10. d ‘Out in the cold’ means exposed and vulnerable. Inoption A, ‘cord’ is the correct word and not ‘cords’.

Test 5

1. a In A-Correct usage is 'in' business and in D it shouldbe 'in' the share market values.

2. e A is incorrect, should be 'In' January, B is incorrectshould be was found 'in' a street, C is incorrect shouldbe 'in' his family's care.

3. d B is incorrect should be 'think of…' C is incorrectshould be 'in a thing…'

4. c C is incorrect, should be 'outcome of…' , D is incorrect,should be 'us into'.

5. e A is incorrect should be 'in your neighbourhood…'and B is incorrect should be 'across the globe…'.

6. c A is incorrect, use the preposition 'by' instead of from,D is incorrect, use the preposition 'for' instead ofthrough.

7. b B is incorrect, use 'of the health' instead of 'with…', Cis incorrect use 'in time…' instead of 'from time…'

8. a A is incorrect, should be 'ultimate form of….', D isincorrect should be ' appeal as strongly to …'

9. c A is incorrect should be 'moved by', D is incorrectshould be 'through this…"

10. d A is incorrect, should be 'of' instead of 'from', C isincorrect, should be 'in earthquakes…'.

Test 6

1. e It should be ‘blink of ..’.

2. e It should be winning streak.

3. a It should be ‘pigeonholed..’.

4. b It means ‘to make an effort to understand and dealwith a problem or situation’. come/get off your highhorse- to stop talking as if you were better or moreclever than other people.come/go along for the ride- to join in an activity with-out playing an important part in it. come/go cap inhand- to ask someone for money or help in a waywhich makes you feel ashamed.come/go down in the world- to have less money anda worse social position than you had before.

5. d Cut the Gordian knot means to deal with a difficultproblem in a strong, simple and effective way.cut some slack- to allow someone to do somethingthat is not usually allowed, or to treat someone lessseverely than is usual.cut the (umbilical) cord- to stop needing someoneelse to look after you and start acting independentlycut the ground from under- to make someone or theirideas seem less good, especially by doing somethingbefore them or better than them.Cut the crap!- an impolite way of telling someone tostop saying things that are not true or not important.

6. b It implies something that you say which means thatfailing to do something when you almost succeededis no better than failing very badly.miss a trick- to not fail to notice and use a goodopportunity. miss out- to fail to use or enjoy anopportunity miss the boat- to be too late to getsomething that you wantmiss the point- to fail tounderstand what is important about something

7. a still and all- despite that. still waters run deep-something that you say which means people whosay very little often have very interesting andcomplicated personalities. sting in the tail- anunpleasant end to something that began pleasantly,especially a story or suggestion stink up- to make aplace smell unpleasant; to do something very badly.stitch in time- something that you say which means itis better to deal with a problem early before it gets toobad

8. e Sticks and stones may break my bones (but wordswill never hurt me)- something that you say whichmeans that people cannot hurt you with bad thingsthey say or write about you. sink like a stone- to failcompletelyleave no stone unturned- to do everything that youcan in order to achieve something or to find someoneor something.set in stone- firmly established and very difficult tochange get blood from a stone- to do something verydifficult

9. a hang the cost/expense - if you say that you will do orhave something and hang the cost, you mean thatyou will spend whatever is necessary.at all costs -if something must be done or avoided at


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all costs, it must be done or avoided whateverhappens.cost (someone) a pretty penny -to be very expensive.cost (someone) an arm and a leg (informal) to be veryexpensive.count the cost - to start to understand how badlysomething has affected you

10. b copper-bottomed- a copper-bottomed plan,agreement, or financial arrangement is completelysafe. cordon bleu- cordon bleu cooking is food whichis prepared to the highest standard and a cordonbleu cook is someone who cooks to a very highstandard corner the market- to become so successfulat selling or making a particular product that almost noone else sells or makes it. cotton-picking- somethingthat you say before a noun to express anger corridorsof power- the highest level of government where themost important decisions are made.

Chapter 15

Exercise 1

1. and - joining Akbar/I

2. but - joining slow/strong

3. or - joining Jatin/Lalit

4. nor - joining like/appreciate

5. or - joining You/I

Exercise 2

1. neither-nor

2. not only-but also

3. either-or

4. whether-or

5. both-and

Exercise 3

1. Cox and Kings is open today so we’re going to buyour tickets to Australia. (so is an adverb)

2. As he read the letter he laughed. (There is noconjunction As is an adverb: He laughed as he readthe letter)

3. So he told me but I didn’t believe him.

4. She did not reply, nor did she make any gesture.

5. We ran from the building when we noticed the time.(when is an adverb)

6. Either accept our conditions or leave.

7. We rested until the storm was over and we felt better.(until is an adverb)

Test 1

1. c Rather than 'in lieu of', the appropriate conjunctionwill be 'because', since the sentence is talking aboutthe reason that why 'pipeline for public offerings hasdried up.'

2. a In A the correct conjunction should be 'however' ratherthan 'wherever' because the latter doesn't make anysense in the given context.

3. b 'Moreover' is an incorrect conjunction out herebecause the sentence talks about a contradiction.Therefore 'but ' should replace 'moreover'.

4. c 'Due to' should be replaced by 'since' because thelatter gives the meaning that 'due to the tiny town',which has no meaning.

5. a The word 'for' is most often used as a preposition, ofcourse, but it does serve, on rare occasions, as acoordinating conjunction. Other options change themeaning of their respective sentence.

6. e The correlative conjunction 'not only' is alwayscoupled with 'but also'.

7. d The appropriate conjunction will be 'as long as..'

8. c In A 'as' is following 'though' which is incorrect. In B'like' has been incorrectly used. 'Like' as a conjunctionmeans 'as if', therefore doesn't make sense here.Similarly in D, 'though' and 'yet' are used together thusrendering the sentence incorrect. Same problem iswith E.

9. a Other options change the meanings of their respectivesentences.

10. e Other options change the meanings of their respectivesentences.

Test 2

1. c Instead of 'neither' it should be 'nor'.

2. d The correct conjunction will be 'otherwise'.

3. a The appropriate conjunction will be 'despite'. With 'yet','its size' doesn't make any sense.

4. d In this part the conjunction 'nor' is incorrect. In thiscase with 'whether' 'or' should be coupled.


5. d Here 'such' should be followed by 'as' or else it doesnot make any sense.

6. a Rather than alongside the conjunction 'along with' willbe more appropriate.

7. a Rather than 'even', the appropriate conjunction herewill be 'although'.

8. e 'So' is an incorrect adverb here. It should be replacedby 'like'.

9. b 'Wherever' doesn't make any sense here. It should bereplaced by 'however'.

10. c The appropriate conjunction here will be 'as muchas'.

Test 3

1. b In A 'like' is an inappropriate adverb. It should be re-placed by 'as'. In D rather than 'therefore', 'and' will bemore appropriate.

2. b In B, 'henceforth' will be replaced by 'while', sincethe sentence is talking about the ongoing mood ofpessimism.

3. b In C, rather than 'then', 'than' should be used becausecomparison is being made here. In E 'unless' is inap-propriate because it means 'except', whereas thesentence talks about 'till'. Therefore 'until' will be ap-propriate.

4. c In E, 'even' should be followed by 'before' or 'when',rather than 'if'.

5. d In C 'though even' is incorrect. Either it should be'though' or 'even though'. In D 'while' is inappropriate.It should be replaced by 'and'.Usage notes: Among some conservatives there is atraditional objection to the use of though in place ofalthough as a conjunction. However, the latter (ear-lier all though) was originally an emphatic form of theformer, and there is nothing in contemporary Englishusage to justify such a distinction.

6. d In E, rather than 'nor' it should be 'or'.

7. e In B, 'despite' is incorrect because it is always fol-lowed by 'of'. It should be replaced by 'instead'. In E,rather than 'or' should be 'nor' because the previoussentence has already discussed something in nega-tive therefore 'nor' can appropriately follow it; sincethe very first sentence says 'But this is not it'.

8. b In D 'whether' is inappropriate and should be replacedby 'if'.

Unit – 7

Chapter 16

Test – 1

1 I like hiking, skiing, and snowboarding. (A verb hasbeen mentioned with gerunds. So, ‘to snowboard’should be changed to ‘snowboarding’.)

2. A low-fat diet, most experts agree, is the best way toreduce artery blockage and achieve weight loss. (Thewriter describes the second benefit of a low-fat dietas “the achievement of weight loss”—a noun phrase.Clearly, following the first infinitival phrase withanother to create “to reduce artery blockage and toachieve weight loss” improves clarity.)

3. When one takes the CAT, it’s perfectly natural to be alittle nervous, irritable and sweaty. (“Nervous” and“irritable” are adjectives. “Sweaty” is also an adjective,but, because it is followed by “palms,” the third elementis a noun phrase modified by an adjective. Parallelismcould be achieved simply by removing “palms.”)

4. Eating huge meals, snacking between meals, andexercising too little can lead to obesity. A noun herehas been used with gerunds. So, exercise should bechanged to exercising.)

5. Our coach is a former champion batsman, but nowhe is overpaid, overweight, and over forty. (The factof being a champion batsman cannot be mentioned inthe same breath with negative attributes like overpaid,obese and over fifty.)

6. Mustaine likes people who have integrity andcharacter.(The verb ‘have’ should modify both nouns.)

7. I like editing books more than just reading them.(Here, an infinitive is paired with a gerund. A gerundshould be paired with a gerund.)

8. Career Launcher needs teachers who are ambitious,self-motivated, and dedicated. (Here, a verb form hasbeen given with one adjective and one noun. Theparallelism can be achieved by converting all threewords into adjectives)

9. As an artist, he drew, painted, and sculpted. (For theparallelism to be there, all three activities should bementioned as verbs.)

10. Aggression and melancholy are behaviours that manysteroid-users exhibit.(In this sentence the parallel elements should both benouns that function as the subject of the sentence.So they have to be in the same grammatical form.)


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Test 2

1. e There is no error in the statement. The first part of thesentence before hyphen (-) is the main clause andrest are the ‘elements’ (subordinate clauses) that themain clause includes. All of them are parallel- ‘balance’,‘age’ and ‘wearing’. ‘Wearing’ is correct because it isa noun (gerund) here and not a verb.

2. b Following the principles of parallelism, ‘to keep’ shouldbe match by ‘to supply’.

3. c Again according to the rules of parallelism, in a stringof nouns should have all singulars or all plurals. Sincewe have ‘whispers’, ‘tingles’ and ‘shocks’; ‘shout’should also be in plural.

4. d Instead of using an infinitive ‘to herd’; a gerund shouldbe used that is ‘herding’; because there is a string ofgerunds used in the sentence, which are ‘singing’,‘irritating’ and ‘barking’. Secondly ‘at’ cannot be fol-lowed by ‘to’.

5. d It should be ‘selling’ and not ‘to sell’.

6. d To make the structure parallel ‘who will benefit’ shouldbe matched with ‘to what extent’ because ‘and theextent’ does not have the component of questioningthough the context requires that.

7. d. In D rather than ‘consolidating’ it should be ‘consoli-date’ keeping the parallel structure in view becausethe verbs should match when in a string.

8. e Both should be followed by ‘manufacturers and deal-ers’ to keep the parallel structure at its best.

9. b All the verbs in the sentence are in simple past; thusit doesn’t make any sense to use past continuous inthe middle of the sentence. (Opossums- a prehen-sile-tailed marsupial, of the eastern U.S., the femalehaving an abdominal pouch in which its young arecarried: noted for the habit of feigning death when indanger.)

10. e ‘Running’ is a gerund here whereas the requirementis that of an infinitive to match ‘to have’ and ‘to buy’.

Test 3

1. a In C, rather than ‘running’ it should be ‘runs’ since thecontext talks about universal truth thus has to besimple present tense following the rules of parallelism.Similarly in D, instead of ‘cleared’ it should be ‘to clear’.

2. d In B ‘will have’ is incorrect because it is not parallelwith ‘has launched’ and ‘have fought’. Since the frameof reference pertains to present perfect, the insertionof ‘will have’ is totally out of context here.

3. b In C ‘to accept’ is incorrect. It should be replaced by‘accepting’

4. a In E ‘the crucible …’ should be preceded by ‘as’, sothat the parallel structure of the statement is intact.The correlative ideas here should be joined by ‘eitheras the …or as the’.

5. b This tradition began in ancient Rome and continuesinto modern times. (‘Begun’ is a participial adjectivewhile ‘continues’ is an active verb.)

6. b I acquired my wealth by investing carefully, workinghard and finding a rich father-in-law. (Here, a nounhas been mixed with a pair of gerunds.)

7. e We can either drive to Shimla or fly to Chennai.(Correlative conjunctions, here ‘either . . . or’, introduceclauses that must be parallel.)

8. c I do not like hot water or cold milk. (Here, unequalstructures are being used to explain equal ideas. Con-verting ‘milk that is cold’ to ‘cold milk’ rectifies the prob-lem.)

9. e Please write a long story, a short novel and an epicpoem. (Again, unequal structures are being used toexplain equal ideas. Converting ‘a poem that is epic’ to‘an epic poem’ rectifies the problem.).

Chapter 17

Exercise 1

1. Most graciously,

2. Dear Madam: (a business letter)

3. B-52, Okhla Industrial Estate, New Delhi-110020?

4. 1 March, 2006, at

5. (no comma needed - only one part)

6. (no comma needed - only one part)

7. Charu Gera, Sr., South Extension, New Delhi

8. New Delhi, India

Exercise 2

1. Football, basketball, track, and tennis require running.

2. The numbers 8, 16, 32, and 48 are called evennumbers.

3. Eat, drink, and make merry, for you will soon die.

4. I like shopping, my friend likes dining, and the familylikes activities.


5. Working hard, saving some money, and providing fora family should be important for a father.

6. I saw him run up the mountain, jump off the cliff, andland in a pine tree.

7. He was from Mahableswar and she was fromKanyakumari.

8. She likes to sing, to play the piano, and to read novels.

9. The search party looked along the road, up the hill,and down the alleys for clues.

Exercise 3

1. Ila, indeed, is a good mother.

2. I hope, Vishal, that you don’t go to jail.

3. My son-in-law Salim will be able to vote in the comingelection. (a closely related appositive or use commasaround Salim if you thought it was a noun of address)My son-in-law, Salim, will be able to vote in the comingelection.

4. Oh, Girish, I hope that you, on the other hand, will behappy with your decision, your move to Europe.

5. We sat in the shade beneath a broad green tree,Lara.

6. It was a lovely, happy, memorable time.

7. I know, after all, you will be successful.

8. Mr. Dinesh Bobby, the boy next door, has been fightingwith your brother Harish. (Harish is a closely relatedappositive)

9. Of course, we could hear immediately that you, afterall, will be going to Santos, a great city in Brazil.

10. Well, Vishal, I hope to see you, by the way, in Goa onour return from our vacation, a trip to Mumbai.

Exercise 4

1. “I wish the election was over,” said Farah.

2. “Will they finish this week?” asked Farheen.

3. Avinash added, “It is becoming a joke!”

4. “We can now see that every vote counts,” concludedNidhi.

5. “Yes, we know that we should vote every time,”commented Rajiv.

Exercise 5

1. Chandan is tall; his brother is short.

2. He knocked several times; no one came to the door.

3. The siren blew loudly; I rushed to the window; thepolice raced past as I looked out.

4. I waited several hours for you; you did not return; Ibecame concerned.

5. My sister loves mysteries; my brother likes technicalmanuals.

Exercise 6

1. Since you asked my opinion, I will tell you; and I hopeyou will listen well.

2. Although he is highly qualified, he is not dependable;and I am afraid to hire him.

3. Because Preeti is absent a great deal, she has a hardtime keeping up; but she is willing to work overtime.

4. Although I prefer English, I know that math is important;and I will work hard in both classes.

5. When you arrive on the train, take a taxi to the metrostation; or I can meet you at the platform.

Exercise 7

1. I am looking for the poem “The Path Not Taken”; I needit tomorrow.

2. Ram sings bass; Hari, tenor.

3. I have visited U.S.A., Australia, Canada and Bhutan.

4. I will steal, cheat, and lie for you; but I will not kill foryou.

5. There was a sudden noise; everything stoppedimmediately.

6. Although we may need more time, I believe we will bevictorious; and I believe you feel that way, too.

7. We can trust him implicitly; nevertheless, we shouldnot be careless.

8. The house looked like what we wanted; on the otherhand, we had not been inside.

9. I had food, clothing, and furniture; but I didn’t have myfamily.

10. He was such a “bore”; I couldn’t stand him.


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Exercise 8

1. The boy’s bike is in the back yard.

2. Gaurav’s car was in the accident yesterday.

3. Mr. Gupta’s talk was the best yet.

4. What happened to that horse’s leg?

5. That woman’s umbrella is blowing away in the wind.

Exercise 9

1. These women’s hats are sold in this store.

2. The children’s party was a great success.

3. The mice’s tracks were everywhere in the dust.

4. We followed the two deer’s tracks in the snow.

5. The geese’s flight was smooth and graceful.

Exercise 10

1. The men’s and boys’ boots were all mixed together.(separate ownership)

2. Shashi’s mother lives next door to us.

3. The dog’s growl scared the baby in the neighbor’syard.

4. Both Vinod’s and Shyam’s hair is red. (separateownership)

5. Meera’s and Seeta’s mother came to the performance.(joint ownership)

6. The babies and the children’s fun ended with theparents’ return. (joint ownership)

7. The men’s hoods covered their faces.

8. The coop was covered with several chickens’feathers.

9. I could hardly hear the puppy’s bark.

10. The wolves’ howls came sharply to the deer’s ears.

Exercise 11

1. Could I buy fifty rupees’ worth of sweets for thekids?

2. Somebody’s shoes have been left in the living room.

3. His shoes are here, but where are yours?

4. His aunt’s nephew will be on television with Ahuja’sgroup.

5. The cow’s udder was cut from jumping theneighbour’s fence.

6. Rahul and Sanjeev’s store will be open on Diwali

7. Everybody else’s help will be appreciated by mymother’s family.

8. Just two days’ work will finish this room.

9. Anika’s and Rani’s costumes were the prettiest ofeveryone’s.

10. The women’s and girls’ ages were revealed toeveryone. (could be girl’s)

Exercise 12

1. we’re it’s you’ve who’s hasn’t

2. I’ll I’m she’ll she’ll I’ll

3. I’ve we’ll they’re aren’t didn’t

4. he’s you’ll you’re isn’t hadn’t

5. wasn’t haven’t couldn’t we’d they’ll

6. shouldn’t doesn’t there’s they’ve you’d

7. weren’t wouldn’t that’s I’d won’t

Do not confuse the contractions (it’s, who’s, they’re,you’re) with the possessive pronouns (its, whose,their, your).

Exercise 13

1. It’s about time you started looking for your shoes.

2. They’re coming at about nine for their children.

3. Its mouth was sore because it’s chewing all the time.

4. Whose briefcase will you be using for your papers?

5. You’re going to be late, but who’s going to be on time?

Exercise 14

1. It used to be that one had to be twenty-one to vote.

2. When adding thirty-four and forty-two, you getseventy-six.

3. One hundred thirty-seven people were killed in thatcrash.

4. The sixty-fourth running of that race was cancelleddue to weather.

5. Many more privileges come to people who are sixty-five or older


Exercise 15

1. We will invite Sushma—she is the new girl next door—to our party.

2. The dog slid on the vinyl—his nails acting like skates—and crashed into the trash can.

3. When our stockpile was sold—indeed, dumped forsurplus—all our sales were compromised.

4. Today has been—but I will not bore you with mytroubles.

5. Let me tell you about—watch where you are going!

Exercise 16

1. We fished (or should I say mashed worms) in themurky pond.

2. They listened to the teacher’s stories (they were verydull) which gave some background for the book.

3. Gauri and Prial (you remember them) moved to a newhouse last week.

4. Even though he was not qualified (according to histranscripts), he knew more than most of the others.

5. Another possibility (the possibilities seem endless)was suggested by a person at the back of the room.

Test 1

1. The girl’s (girls’) vitality and humor were infectious.

2. New clients’ accounts showed an 11 percentincrease in sales.

3. These M.D.s’ credentials are excellent. (These M.D.s– plural)

4. Several M.D.s (Or M.D.’s) agreed that one bacterialstrain caused many of the symptoms.

5. You asked for forgiveness; he granted it to you.

6. We ask, therefore, that you keep this matterconfidential.

7. The order was requested six weeks ago; therefore Iexpected the shipment to arrive by now.

8. I need a few items at the store: tissues, a bottle opener,and some milk.

9. I needed only three cards to win, namely, the ten ofhearts, the jack of diamonds, and the king of hearts.(or win; namely,)

10. Correct. or colours;

Test 2

1. “How,” I asked, “can you always be so forgetful?”

2. The girl who is standing there is his fiancée.

3. Correct.

4. Finish your job; it is imperative that you do.

5. You may, of course, call us anytime you wish.

6. You signed the contract; consequently you mustprovide us with the raw materials.

7. “Stop it!” I said. “Don’t ever do that again.”

8. Because of his embezzling, the company wentbankrupt.

9. Correct.

10. Nature lovers will appreciate seeing whales, sea lions,and pelicans.

Test 3

1. c Uses the correct punctuation marks for distinction.

2. a Put a colon to introduce a list.

3. d Only commas are required to separate the item list,the colon is incorrect here; do not use apostrophesfor sweets, marshmallows and toffees.

4. e Semicolon is used to separate sentences which areclosely connected in thought.

5. b No interrogation mark is used after an indirect ques-tion, commas are not required.

6. e Use comma to separate co-ordinate clause in a com-pound sentence.

7. b Use comma after adverbial phrases of absolute con-struction.

8. c Use comma to separate words/ phrases/clauseinserted into the body of a sentence.

9. d Use inverted commas for reported speech and putcommas for short pauses.

10. b Use commas to mark off phrases in apposition.


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Unit – 8

Chapter 18

Test 1

1. d Options B, D & E are correct. A is incorrect as thecorrect phrase is 'as much as' , C is incorrect as thecorrect verb to be used is 'there are..'.

2. a Only A& B are correct, C is incorrect because it shouldbe 'recent months', D is incorrect because it shouldbe 'at the moment', E is incorrect because it should be'pants'.

3. d C is incorrect because the correct usage is 'their', E isincorrect an article should precede the word leader,should be 'a leader'.

4. d C uses incorrect parallelism, should be 'transporting',E use the article 'the' before China which is incorrect.

5. b A is incorrect should be 'Greeks 'since it's addressingthe entire race, D is incorrect, the verb should be'seek', E is incorrect because it should be 'than' notthen.

6. a A is incorrect, should be 'built', error of tense, C isincorrect should be 'on Monday' as the deed is alreadydone.

7. e A is incorrect should be 'in business', C is incorrectdue to incorrect subject verb agreement, and use'has 'here.

8. e B has incorrect usage of the pronoun should be 'in'instead of 'at', C is incorrect should be 'candidatesinstead of the singular 'candidate'- should agree withthe verb, E is incorrect because the collective usageof 'skills; is being discussed here and should agreewith the verb 'are'.

9.c C is incorrect, should be 'a good ...' the defining articleis missing, D is incorrect should be 'friends' in theplural.

10. a B is incorrect, should be 'setting' instead of set: errorof parallelism, D is incorrect should be 'said' insteadof 'says' incorrect tense usage, E is incorrect becausethe article 'a' is missing before the word negative.

Test 2

1. b A is incorrect, use the adverb form 'increasingly' here,in sentence B the correct usage is 'eight times' and inE the correct form is 'it's' instead of the possessiveform 'its'.

2. a A is incorrect, use 'between' instead of 'among', D isincorrect, use the article 'the' before the word 'plant'.

3. c B is incorrect, the correct phrase is 'akin to' , akins asword does not exist, C is incorrect, correct usage is'of' after the word silhouette.

4. b A is incorrect should be 'stop' instead of stops, errorof subject verb agreement. B is incorrect; the correctphrase is day-today activities.

5. c A is incorrect should be 'about' instead of 'over', C isincorrect should be 'the' public not 'a' public. E is in-correct uses a comparative 'greater', correct usageis 'great'.

6. c B is incorrect, use 'dreamt ' instead of 'dream' in keepingwith the tense of the sentence, C is incorrect uses'which' instead of 'that', which is mostly used forinanimate objects.

7. d A is incorrect should be 'resonant' in the adjectiveform rather than 'resonance' in the noun form, C isincorrect, use 'have' after Patriarchs as the correctform of subject verb agreement.

8. c A is incorrect, incorrect usage of the preposition 'upon'which means on something, something elevated,correct usage is 'on'. B should be 'scores' instead of'score', should agree with plural verb.

9. a D is incorrect, requires an article before 'user's' , E isincorrect due to the incorrect usage of the phrase'real time'.

10. e C is incorrect, should be the contraction 'it's' insteadof 'its', D is incorrect, uses the verb 'is' instead of'are'.

Chapter 19

1. Eating insects does far less damage. For one thing,the habit could help to protect crops. Some 30 yearsago the Thai government, struggling to contain a plagueof locusts with pesticides, began encouraging itscitizens to collect and eat the insects. Officials evendistributed recipes for cooking them. Locusts werenot commonly eaten at the time, but they have sincebecome popular. Today some farmers plant corn justto attract them. Stir-frying other menaces could helpreduce the use of pesticides.

But insect populations vary with the seasons, and itis hard to control the amount on offer at a given time.“There is very little knowledge or appreciation of thepotential for managing and harvesting insectssustainably,” notes Patrick Durst, a Bangkok-basedsenior forestry officer at the FAO. Those looking for areliable source of protein might prefer to farm them.Protein makes up a high proportion of most insects’weight. That makes them much more efficient at


converting feed to protein than livestock. For example,a cow yields only 10lb (4.5kg) of beef for every 100lbof feed it eats, whereas the same amount of feedwould produce tens times as much cricket.

2. While IT managers are understandably leery aboutrecommending the iPhone for company employees,individual owners are going to be sneaking their prizedpossession into the office to access e-mail andcorporate information. Bosses, too, who becomeenamoured of the iPhone will be insisting the firmsupport it. Against their better judgment, many ITmanagers will doubtless comply.

No question, Apple’s innovative design has sent awake-up call to smartphone makers everywhere.From tiny Palm, the struggling smartphone pioneer, toResearch In Motion (RIM), the BlackBerrymanufacturer, and the dozen or so handset makerswho build smartphones based on Microsoft’sWindows Mobile, as well as the 800-pound gorilla ofthe business, Nokia—all have been forced to react tothe iPhone’s touch screen, intuitive software andiconic appeal, not to mention its rapid penetration ofthe market.

Apple has sold more than 5m iPhones over the pastyear. Practically all have been bought by individualsat their own expense, rather than their employers’.The iPhone now accounts for 19% of the smartphonemarket in the United States. That’s ahead of Palm’s13%, but still well behind RIM’s 45%.

Over the past few months, competing 3G smartphoneswith touch screens and a host of features have beencoming thick and fast. Sprint has started to offerSamsung’s Instinct, which seeks to trump the iPhonewith a higher download speed, better video, picturemessaging, navigation and applications, plus a batterythat can be removed.

3. In truth, EU firms have been investing heavily in centraland eastern Europe since soon after the Berlin Wallcame down, and Italy was home to about 350,000Romanian migrants before Romania joined the union.Yet public fears about Polish plumbers and otherbogeymen are real enough. Even though Germanexporters have flourished by selling to the newmember states, 63% of Germans, according toEurobarometer, think that enlargement is making Europeas a whole less prosperous.

Some (0r most) of the newcomers have not helpedtheir cause since joining. Nasty populists have donewell in elections in several countries, and Romania,Bulgaria, Slovakia and the Czech Republic haveshown prejudice against the Roma too. But thenprejudice, bad government, corruption and organisedcrime are not the exclusive preserve of the new

members. Some (or many) existing members havebeen setting a bad example for them.

Nor was the fifth enlargement a simple matter ofcountries governed by former dissidents acceptingthe democratic embrace of the West. Plenty (or mostor many) of ex-communists smoothly relabelledthemselves and hung on to power across the block.Brussels is full of talk about “backsliding” to describethe way that politicians in the new member countriesforgot, or actively undermined, reforms that the EUdemanded during accession negotiations. Corruptionand organised crime blight many of the newcomers.Parliaments and ministerial suites shelter too manybad men.

All this has led some to suggest that enlargementhappened too soon, and that many of these problemscould have been avoided by waiting until the accessioncountries were better prepared. This report will arguethe opposite: that enlargement came in the nick oftime. Inside the candidate countries the first victims offurther delay would have been reformers who foryears had been pushing painful changes as vital forachieving EU membership. Had the public started todoubt that entry was fairly imminent, the drive forreforms would have been undermined.

For the existing member countries, three big reasonswould have made enlargement far more difficult if ithad come any later than it did. These can besummarised as migration, money and Moscow.

4. The banks’ problems feed back into the credit markets.For if the banks cannot raise as much capital as theyneed, they may feel they have to sell assets. And thatmay mean off-loading corporate and asset-backedbonds, even if the banks must accept fire-sale prices.The ABX index of asset-backed bonds is below thelevel that it sank to in March.

Analysts tend to agree that spreads are much higherthan are needed to compensate investors for thelikely default rate. According to Moody’s, a ratingsagency, the default rate over the past 12 months onglobal speculative, or junk, bonds was just 2%; theonly company outside America to walk away from itsdebt was a Bulgarian steelmaker. Jim Reid, a creditstrategist at Deutsche Bank, quips that, if the spreadson investment-grade debt accurately reflect defaultrates, everyone will end up living in caves.

But spreads often fail to reflect the likelihood ofdefaults. This was true during the credit boom, whenthey fell to historically low levels, as investors chasedanything with a yield higher than government bonds.And it is true now because, although an investorcould make money by buying corporate bonds and


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holding them for five years, few are able to followsuch a strategy. The big risk investors face is buyingtoo soon. Traditional bond managers fear losing theirclients if they have another poor quarter. Hedge-fundmanagers may find that the prime brokers who lendthem money will cut off their funding if their positionsdeteriorate.

Just like last summer, the risk is of a downward spiral,in which nervous investors sell bonds, driving pricesdown, only increasing the general nervousness. Thesystem is still deleveraging, at least according to thelatest figures from America—which show bank loanscontracting at an annualised rate of 8% over the 13weeks to June 25th.

To make matters worse, the fundamentals for thecorporate-bond market have worsened since lastAugust. Expectations for corporate profits are beingrevised down, as margins come under pressure fromslower growth and higher commodity prices. Headlineinflation is well above target, so central banks areunlikely to ride to the rescue with interest-rate cuts;indeed, many prefer to tighten monetary policy. Thisis one sequel where the scriptwriters are going tohave to work extra hard to come up with a happyending.

5. IT came to Seneca by his own hand in a warm bathafter falling foul of the emperor Nero, to Boethius byorder of the Ostrogothic King Theodoric and toGiordano Bruno for affronting the Inquisition.Pneumonia saw off both Francis Bacon when hetried to deep-freeze a chicken with snow and RenéDescartes on having to rise before dawn in a Swedishwinter to instruct Queen Christina. In modern times,Nietzsche’s friend Paul Rée fell off an Alp and MoritzSchlick, a logical positivist, was shot to death by aderanged student.

An eye-catching list perhaps. But quiteunrepresentative. Save when at work makingdistinctions and arguing relentlessly, philosophers aremuch like everyone else. They lead on the wholeworthwhile but humdrum lives. Most die of old age intheir beds. Philosophers’ deaths make an unpromisingtheme for a book.

A handful of philosophers, it is true, have diedphilosophically, by which people usually mean theyhave figured in memorable accounts of exemplaryends. Socrates was one. Unjustly condemned, hecould have left Athens for exile. But, as Plato recounts,he chose city over self, virtue over expediency, anddrank the hemlock, surrounded by friends.

Another philosopher who had a philosophical end inthis sense was David Hume. James Boswell, a famousdiarist, visited him as he was dying in Edinburgh in

1776. He found Hume in good cheer, pressed him onlife after death but could not shake the great man ofhis disbelief. The thought of annihilation at death, Humetold Boswell, caused him no more uneasiness thanthe thought of non-existence before birth.

Making a success out of such material requires styleand wit. Simon Critchley’s “The Book of DeadPhilosophers” shows leaden playfulness. His tonecan be portentous or giggly, as if he is unsure whohe is writing for or why. The reader will “die laughing”,he says. But most of the deaths he describes on hispadded-out list of 190 philosophers are banal, andvirtually none is funny. The real trouble is that MrCritchley, a professor at the New School of SocialResearch in New York, cannot decide whether he iswriting about philosophers’ own deaths, exemplarydeaths or philosophers’ thoughts on death in general.

6. All of these aspects of Noguchi’s career will beexplored in an exhibition opening Friday at theYorkshire Sculpture Park, near Wakefield in England.The stars of the show are a hundred or so of thepaper and bamboo Akari light sculptures that he beganmaking in the early 1950s, and that became his best-known work. Lovely to look at and surprisingly robust,the Akari lights not only fuse Noguchi’s Japanese andAmerican influences, but art and design,craftsmanship and industry. They were also thecatalyst for economic regeneration of a decliningJapanese industry and, last but not least, theirdramatically shaped mulberry-bark paper shades emita very beautiful light.

The Akari project came about by chance, after Noguchiwent back to Japan in 1950. By then his father wasdead, and the Japanese welcomed him as a famousAmerican artist. He visited the city of Gifu, where thetraditional candlelit paper lantern industry wasdeclining dramatically as more and more Japanesehomes introduced electricity. The mayor askedNoguchi how to revive it.

Noguchi’s solution was to modernize the old paperlanterns. Settling in an ancient teahouse with his then-wife, the Japanese movie star Yoshiko Yamaguchi,he designed a series of lamps powered by electricity,rather than candles. For the shades, he used thesilky Mino paper that had been made in a nearbyvillage from locally grown mulberry bark since theeighth century, but replaced the recently adopted wireframes with traditional bamboo. The design processwas traditional, too. Noguchi began by making awooden mold in the shape of the finished shade andwound fine strands of bamboo around it. Strips ofMino paper were glued to the bamboo, and the moldremoved once the glue had dried. A slender metalstructure was designed to hold the bulb and supportthe shade, both at the top and the bottom, where itseemed to float above the floor on spindly legs.


7. Like Tempelhof, the Beijing air terminal boasts asweeping concourse that evokes the glamour of airtravel while enclosing a surprisingly intimate interior.But Foster pushes the ideal of mobility to a newextreme. Guided by twinkling lights embedded in theterminal’s ceiling, arriving visitors glide up rampedfloors and across broad pedestrian bridges beforespilling out onto the elevated concourse. From therethey can disperse along a fluid network of roads,trains, subways, canals and parks whose tentaclesextend out through the region.

This sprawling web has completely reshaped Beijingsince the city was awarded the Olympic Games sevenyears ago. It is impossible not to think of the enormouspublic works projects built in the United States atmidcentury, when faith in technology’s promiseseemed boundless. Who would have guessed thenthat this faith would crumble for Americans, pavingthe way for a post-Katrina New Orleans just as thedream was being reborn in 21st-century China at 10times the scale?

Yet your sense of marvel at China’s transformation iseasily deflated on the drive from the airport. A banallandscape of ugly new towers flanks both sides.Many of those towers are sealed off in gatedcompounds, a reflection of the widening disparitybetween affluent and poor. Although most of themwere built in the run-up to the Olympics, the poorquality of construction makes them look decrepit anddecades old.

8. There may be recession closing in around the world,but nobody has told the major soccer clubs. This maybe the “closed season” of big European leagues, butthe players, the teams, the organizers are trading atfull pelt.

Any moment now, Ronaldinho, the wonderfully giftedBrazilian who has obliged Barcelona to sell himbecause he has frequented parties more than he hasplayed games over the past season, will get thetransfer his agent brother has negotiated. When thebartering stops, AC Milan will announce its acquisitionof Ronaldinho for a “cut price” •15 million, or$23.9 million.

It will grant his wish to play in the Olympics, and thenwill start paying his $10 million a season salary. Milanwill build its attacks through the much coveted Braziliantrio of Kaká, Ronaldinho and the teenager Pato.

That’s show business. That’s where soccer hastaken over from Hollywood as the industry that tradesthe talents of men for the length of contract barteredbetween them.

To paraphrase Sam.uel Goldwyn, the ownerssometimes pay too much but consider the playersworth it.

9. Some might say, of course, that when history losesits focus on the representation of straight facts, itceases to be history and becomes historical fiction.Empathy, after all, is hardly a neutral subject in historycircles, and has long been held responsible in somequarters for the dumbing-down of school historyteaching. Is Figes not wary of inviting such chargesto be brought to his own work? Does not the activeengagement of his researchers and himself in thegeneration of the sources compromise their findings?

“This is the big issue we had to deal with all the time,”he says. “The danger is that by collaborating with thesubjects on the painful retrieval of these memories,you somehow make them say things. So, yes, there’salways a risk of something.

“But you’ve got to remember that, without some kindof empathy, the retrieval process wouldn’t take placeat all. The important thing is to prepare the ground sothat the subjects can tell their stories in their ownwords. Those interviews, in several cases, lasteddays, with many pauses where the discussionbecame too painful for the subject. But it’s only whenyou start discussing the content of the interviewafterwards that you come to put your own intentionson the material.

10. I wasn’t expecting such a tender kiss on our firstdate, but April was determined. She had alreadysignalled her intentions by swimming past me in thepool, sliding her smooth skin next to mine, flippingover to give me a tantalising glimpse of her whitetummy. But I knew better than to be flattered. It wasn’tme she was interested in but the dead sardine in myhand.

She nuzzled my cheek, her chin as slippery as a wetmushroom, then reared effortlessly out of the water.I slipped the fish past her rows of peg-like teeth andwatched her plunge back beneath the surface... April,if you were in any doubt, is not a mermaid but a four-year-old dolphin. She lives in a marine park calledXcaret on Mexico’s Caribbean coast and swimmingwith her and her friends in a large sea-pen is just oneof the many grin-provoking magical experiences onoffer within the wild haven of the park.

Xcaret, pronounced Escaray, is a sustainable touristdevelopment set up in 1990 on a picturesque inletabout 45 miles south of Cancún. Its far-reaching aimwas to conserve both the flora and fauna of the area- some 4,000 species can be found on the 80-hectaresite - and to promote the area’s colourful culture andMayan heritage. All of which makes for a bizarrehybrid of serious conservation zone and thrill-seekers’paradise. It’s what might happen if Mickey Mousesuddenly became an eco-warrior.

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1. Explanations: Fundamentals of Logical Reasoning …media.careerlauncher.com.s3.amazonaws.com/mba/mba_e-lib...Hence, margin = 20%. Profit amount = 20% of sales = 0.2 × 22.5 = Rs