Cat

Quantitative Ability

Faculty Manual 2010

Part I

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S. No. Title Page No.

1. QA Exercise 1-2: Percentages 1

2. Question Bank: Percentags 9

3. QA Exercise 3-6: Numbers 18

4. Question Bank: Numbers 30

5. QA Exercise 7-8: Ratio 49

6 Question Bank: Ratio 58

7. QA Exercise 9-11: TSD 66

8. Question Bank: TSD 77

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Page 1QA Faculty Manual 2010 ��

Topic: Percentages Lecture Number: 01 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Percentages” important in almost all the tests.� Help students to learn how to deal with different real-time applications of percentages.� Help students to understand how to go about the study material and what to do before coming to

the next lecture.� Help students to make use of LOGIC in place of conventional methods.

Step 1: Fundamentals - 11. Change of base2. Successive percentage change

Step 2: Fundamentals - 23. Simple and compound Interest4. Interest on interest5. Formulae for SI and CI6. Non-annual compounding

Step 3: Class ExerciseGive them sufficient time and discuss few problems from class exercise at the end of the lecture. Tell themto get the doubts solved with the faculty available at the center before the next lecture.

NOTE: Make necessary announcements.

All the things that are to be discussed in the lecture have been summarized in the

following few pages. Read them carefully and don’t miss anything.

QA Exercise 1 - Percentages: 1


1. A brief Introduction (5-10 mins)

2. Change of base (15 mins)Start with the simplest problem: If A is 20% more than B, by what percent is B less than A? Do thesame problem if A was 37.5% more than B. Explain why it is important to work with fraction equivalentsto save time & calculations.Explain that the above simple concept is used extensively even in other topics as follows:a. if speed increases by 10% over the same distance, by what percent does time decrease?b. If length of a rectangle decreases by 12.5%, by what percent should its breadth be increased to

maintain the same area?c. If the volume of a milk and water solution is increased by 25% by pouring just water, by what

percentage does the concentration of milk reduce?d. If prices decrease by 16.666%, how much percentage can a consumer consume more for the

same amount?[Q. 2, 6, 8 and 9]

Explain that the above can be used in any relation where M × N = constant.If prices of apples increase by 25%, I am able to purchase 4 apples less in Rs.80. What was theoriginal price of one apple? Explain this question orally in the following two methods: Since pricesincreased by 25%, I am able to consume 20% less. Thus 20% of original consumption = 4 apples.Alternately, to maintain consumption i.e. to purchase 4 more apples, I would need 25% of 80 = 16 Rsmore. Thus increased price of apple = Rs. 4 per apple. Now find original price.

3. Successive Percent changes (10 mins)Take a simple case of two successive increases say 10% and 20% and then explain why the totalincrease is 32%. State that the net increase of an (a%) and a (b%) change is a (a + b + ab/100%)change. Do mention that the formula works fine with any changes viz. increase or decreases. Askstudent to find the net percentage change of a x% increase and a x% decrease and also for a 10%increase and a 9.0909% decrease. Probably the last example can be more easily solved with multiplyingfactors. Thus one can use Multiplying factors or a + b + ab/100 interchangeably.Successive percentage change is also useful in any relation of the type C = A × B. If there is a (a%)change in A and a (b%) change in B, then C changes by (a + b + ab/100%). This has also applicationsin Data Interpretation. Thus if market share grows by 20% and even if the total market size declines by10%, the sales grows by 1.2 × 0.9 = 1.08 i.e. 8% as Sales = Market size × Market Share.The same relation appears many times in geometry. Thus if any quadrilateral has all its sides increasingby 10%, the area increases by 21% as area is proportional to square of linear dimensions. If sides ofa cuboid increase by 20%, volume increases by 72.8% and surface area increases by 44%.[Q. 1 and 3]

CAT 2003 (Leaked): Let A and B be two solid spheres such that the surface area of B is 300% higherthan the surface area of A. The volume of A is found to be k% lower than the volume of B. The value ofk must bea. 85.5 b. 92.5 c. 90.5 d. 87.5


4. Simple and Compound Interest (10 mins)Quickly ask students for the basic difference between SI and CI i.e. the amount at end of year acts asprincipal for next year in CI. Explain this difference using the following table :For P = 1000 and r = 10%,(draw this table such that you still have additional space on the board to write)

Year Simple Interest Compound Interest Principal Interest Amount Principal Interest Amount 1st 1000 100 1100 1000 100 1100 2nd 1000 100 1200 1100 110 1210 3rd 1000 100 1300 1210 121 1331

Points to be highlighted in this table:a. Simple Interest earned in any year is always the sameb. The amount with SI increases in a linear fashion i.e. by a fixed amount every year. Thus the graph of

amount and years will be a straight line.c. The SI and CI is same for the first yeard. CI earned in a year keeps increasing every year.

5. Interest on Interest (10 mins)With the above table still present on the board, explain point number d of above in more details.Explain that in any year the CI earned is higher because whatever was earned in the previous yearwould surely be earned and additionally the interest earned in previous year will also be added to theprincipal and start earning interest. Make sure this concept of Interest on Interest is understood byeveryone. Take a simple problem : If CI earned in nth year is 800 and in n+1th year is 864, what is therate of interest?Next, compare the first two years at SI and CI. For same principal and rate of interest, in the first twoyears, the total CI earned is higher than the total SI earned by an amount equal to the interest on firstyears interest. Take simple problems like :If total SI earned in first two years is Rs. 800 and total CI earned in first two years is Rs. 864, what isthe rate of interest?Explain the following:For same principal and rate of interest,

Year Simple Interest

Compound Interest

1st X X 2nd X X + r × X Total 2X 2X + r × X

Thus explain the difference is r% of X and also that the ratio of CI of first two years to SI of first twoyears is (2 + r) : 2Take another simple example : If the difference between CI and SI of first two years @ 10% is Rs. 500,what is the principal invested.If the ratio of SI to CI earned in first two years is 24 : 25, what is the rate of interest?


6. Formulae for SI and CI (10 mins)Explain the formulae :

nP r n r

SI At CI, A P 1100 100× × = = +

Explain that for CI, the formula is of Amount and for SI the formula is of Interest.In the formulae of SI, make sure to highlight the fact that SI is directly proportional to P, r and n. Thusif an amount becomes 3 times in 7 years, in how many years will it become 9 times? Please explainto all students who have answered 21 years that the Amount is not directly proportional to the numberof years but the Simple Interest is. In this case the SI earned is 2P in 7 years and hence to earn a SIof 8P it will take 28 years. However had it been the case of CI, the amount would have become 9 timesin 14 years itself because in every 7 years the amount triples, thus in next 7 years 3P will become 9P.Also spend some time explaining that for a r% increase the Multiplying factor is (1 + r/100). Thus for 2years at CI, the amount is nothing but P × MF × MF and thus CI is in essence a case of successivepercentage changes.[Q. 4, 5 and 10]

7. Non-Annual Compounding (10 mins)For non-annual compounding DO NOT introduce any new formula. Tell the students that the formula

remains n

rA P 1 .

100 = +

In this formula n does not refer to the number of years but to the number of

time periods and r refers to the rate per time period. Since all this time we were taking compoundingdone annually hence r became rate per annum and n was number of years. If compounding is donesemi-annually or monthly, n would become the number of half years or the number of months respectively.Correspondingly r will be the rate of interest offered per half year or per month respectively. A rate of12% p.a. corresponds to 6% per half year and to 1% per month.DO NOT explain any method of finding 1.26 or finding the 5th root of 1.1. There has never been anyquestion of CI in CAT in last 15 years which involves one to calculate. DO NOT spend any time inapproximation or in answering student queries about the calculation (specifically about with higherindices or higher roots). Believe me, it’s not worth the time.

8. Exercise (45 mins)Solve the exercise and clear any doubts in the problems of the exercise. If you usually fall short oftime, keep discussing the problems of the exercise while you are explaining the individual concepts.

Announcement: Ask all students to solve each problem in exercise of chapter on Profit Loss Discountand ratio Proportion of QA Fundamental Book and to get the book in the class in the next QA session.CAT questions based on percentage and applications that have appeared since 1999:I. CAT 1999: Forty percent of the employees of a certain company are men, and 75% of the men

earn more than Rs. 25,000 per year. If 45% of the company’s employees earn more than Rs.25,000 per year, what fraction of the women employed by the company earn Rs. 25,000 per yearor less?a. 2/11 b. ¼ c. 10/17 d. None of these


II. CAT 2000: A truck traveling at 70 kmph uses 30% more diesel to travel a certain distance than itdoes when it travels at speed of 50 kmph. If the truck can travel 19.5 km on a litre of diesel at 50kmph, how far can the truck travel on 10 litres of diesel at a speed of 70 kmpha. 130 b. 140 c. 150 d. 175

III CAT 2001: A college raised 75% of the amount it needs for a new building by receiving anaverage donation of Rs. 60 from the people already solicited. The people already solicitedrepresents 60% of the people the college will ask for donations. If the college is to raise exactlythe amount needed for the new building, what should be the average donation from theremaining people to be solicited?a. Rs. 300 b. Rs. 250 c. Rs. 400 d. Rs. 500

IV. CAT 2001: The owner of an art shop conducts his business in the following manner: Every once ina while he raises his prices by X%, then a while later he reduces all the new prices by X%. Afterone such up-down cycle, the price of painting decreased by Rs. 441. After the second up-downcycle the painting was sold for Rs. 1944.81. What was the original price of the painting?a. 2756.25 b. 2256.25 c. 2500 d. 2000

V. CAT 2003 (Leaked): At the end of year 1998, Shepard bought nine dozen goats. Henceforth everyyear he added p% of the goats at the beginning of the year and sold q% of the goats at the end ofthe year, where p>0 and q>0. If Shepard had nine dozen goats at the end of the year 2002, aftermaking the sales for that year, which of the following is true?a. p = q b. p < q c. p > q d. p = q/2

VI. CAT 2003 (Retest): A piece of paper is in the shape of right angles triangle and is cut along a linethat is parallel to the hypotenuse, leaving a smaller triangle. There was a 35% reduction in thelength of the hypotenuse of the triangle. If the area of the original triangle was 34 sq. inches beforethe cut, what is the area (in sq. inches) of the smaller triangle?a. 16.665 b. 16.565 c. 15.465 d. 14.365

Puzzle: An ant is at one end of a rubber strip of length 100 mts. The ant is continuously moving ata speed of 10 m/min and the rubber strip is stretched at the end of every minute by 20%. Wouldthe ant reach the other end?


Topic: Percentages Lecture Number: 02 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Percentages” important in almost all the tests.� Help students to identify and to be familiar with different terms used in questions as well as in the

real life.� Help students to learn how to deal with different real-time applications of percentages.� Help students to understand how to go about the study material and what to do before coming to


Step1: Revision� Make sure that the students are well aware of whatever they were taught in the last lecture of

numbers.� To check there progress as well as to be assured of the above stated fact you must throw some

questions and discuss them with the students.� “Revision” always helps in assessing the level of the class and to identify whether they (or some

of them) need extra sessions (Doubt solving).

Step 2: Fundamentals - 11. Terms in PLD2. Simple questions3. Marked price and discount

Step 3: Fundamentals - 21. Faulty balances2. Stocks and Shares

Step 4: Review Test and DiscussionConduct and discuss the test on Percentages. Conduct the test as an actual test i.e. do not discuss eachproblem one by one. Give the students the test and let them do it independently within stipulated time.


All the things that are to be discussed in the lecture have been summarized in thefollowing few pages. Read them carefully and don’t miss anything.

QA Exercise 2 - Percentages: 2


1. Terms in PLD (10 mins)The students would be knowing the terms and all you have to emphasize on is that profit % is apercentage of the CP and never of the SP (unless of course of the question mentions it so). When theprofit is expressed as a percentage of the SP, it is called as margin. One should be very flexible inchanging profit % to margin. Ask students to orally answer : What margin is equivalent to a profit % of12.5%? A margin of 16.66% is equivalent to a profit % of? Emphasize that this is nothing but changeof base and once should have just thought in their minds (8, 1, 9) and (5, 1, 6) for the above twoquestions.

Explain to students that the formula SP CP

P% 100CP− = ×

is same as SP

P% 1 100.CP

= − × Thus in many cases just by taking take the ratio of SP and CP and one can find the profit or losspercentage. Also this ratio is nothing but the Multiplying factor corresponding to the profit or loss. Thusif SP = 400 and profit percentage is 20%, one should mentally realise that 120% of CP = 400 and thusCP = 400/1.2 = 333.33

2. Simple questions (20 min)a. If SP of 8 = CP of 9, what is the profit or loss percentage? In this problem please do not leave it just

by saying that on every 8, profit of 1 is made and hence profit percentage is 12.5%. First explainit using CP and SP being k/9 and k/8 and ratio of SP to CP is 9/8 or 1.125. After this, explain theconcept on every 8, profit of 1 is made and detail this out as many students are genuinely confusedor not crystal clear about this.

b. If by selling 8 items, one makes a loss equivalent to CP of 2 items, what is the loss %?c. How many oranges should be sold for Re. 1 if they were purchased at a rate of 6 per Re. and a

profit % of 20% is desired?Lay emphasis on students solving each of the above orally.Do cover the following problem : If equal number were purchased of two lots of oranges, one at Rs.12 per dozen and other at Rs. 18 per dozen and all the oranges were sold at Rs. 15 per dozen,what is the profit or loss percentage?In the above case if equal amount of Rs. were spent on the two lots, what would have been theprofit%?Last do the problem : Because the CP increase by 20%, a shop-keeper increased the SP by Rs.340. Due to these changes the profit percentage decreased from 15% to 10%. What was theoriginal CP of the item?

3. Marked Price and Discount (10 mins)Explain the phenomena of how the list price is usually marked up over the CP and this is called asmark-up percentage. Make it clear that mark-up percentage is a percentage of CP. When a customerasks for discount the Discount is given as a percentage of Marked Price. Thus make it very clear thatMark-up and Discount are nothing but two successive percentage changes. Even though a discount isgiven, the trader does not necessarily make a loss because of the marking-up done earlier. The profitpercentage is the net percentage change of two successive percentage changes viz. m% and –d%and thus is equal to m-d-md/100.

Question: By what percentage should a shop-keeper mark up his goods if he desires a profit of 25%even after giving a discount of 9.0909%?Make sure that everyone is clear that discount is a percentage of Marked price. Then ask them whatis the discount percentage in the scheme “Buy 3, get 1 free”.[Q. 2 and 5]


4. Faulty Balances (20 mins)Unscrupulous traders may sell their wares at the rate at which they bought it and yet make a profit bycheating on volume. What is the profit % made by a trader who sells goods at the rate of CP but witha balance which reads 900 for 1000 gms? Many will answer 10% or 11.11%. Tell them that in suchcases pay a lot of importance to “reads 900 for 1000 gms”. The reading is that what is charged for andthe other value is the actual amount given. Thus in this case SP = 900x and CP = 1000x (CP will berelated to what is actually given). Thus in this case the shop-keeper makes a loss of 10%. Don’tassume that a shopkeeper will always make a profit, he might be an idiot.Take few more examples with mark-up% as well as faulty balances. Finally tell them the followingsituations are also related to faulty balances though may not be apparent by the wordings.a. A milkman mixes 100 lts of water with every 800 lts of milk and sells at a mark-up of 11.11%. What

is the profit%? (same as reading 900 for 800)b. The morning stores offers 2 items free with every 5 purchased. If it marks up items at 16.66%,

what is the profit/loss % when a customer purchases 5 items?c. A meter scale is rigged to measure 90 cms. However in summers it expands by 20% of its actual

length. The trader sells goods at a mark-up of 10% but he does not know that the wholesaler fromwere the trader buys goods uses a meter scale which measures 80 cms. What profit or losspercentage does the trader make?

This problem is simply taken to emphasize students to use multiplying factor and need not be takenwith any slow batch. In any problem of percentages, one just needs to identify the multiplying factorand then either multiply or divide with it based on whether you want an increase or decrease. It’s thatsimple, just keep your brains away and solve mechanically and you will never ever make a mistake.Multiplying factor corresponding to rigging the meter scale to read 90 cms = 0.9 or 9/10Multiplying factor corresponding to 20% expanding = 1.2 or 12/10Multiplying factor corresponding to 10% mark-up = 1.1 or 11/10Multiplying factor corresponding to meter scale reading 80 cms = 0.8 or 8/10Now use common sense to see what results in a profit and what results in a loss for the trader.

The trader makes a net profit of 10 10 11 8 229 12 10 10 27

× × × = i.e. a net loss 5/27 i.e. 5 × 3.7% = 18.5% loss

[Q. 1]

5. Stocks and Shares (20 mins)Explain basic terms like Face Value, Market Value, Selling Value, Purchase Value, Dividend, Brokerage,Income, Profit and Return.Explain in detail the difference between Income, Profit and Return.Make it very clear that though the topic is not very important as far as QA section is considered, it isrequired to deal with it in order to solve relevant questions from LRDI section.[Q 3 and 4]

6. Review Test and Discussion (35 mins)Conduct the test on percentages and applications. Ask students to evaluate their performance andrecord it. Discuss the doubts of the test.

Announcement: Ask students to revise all the concepts of percentages discussed so far. Ask themcome prepared in the next lecture as it is going to be a test. As after the test doubt solving session willbe there so also ask them to come with fundabook.


1. A trader sells fruits at a profit of 10%. Further, he uses a false weight and weights only 900 gminstead of 1000 gm. What is his overall profit percentage?a. 30% b. 25% c. 22.22% d. 42.22%

2. Arun has to pay off his debt by paying Rs. 2000 in the first month and subsequently, has to keeprepaying 80% of the amount he paid in previous month until he clears the total debt. The approximateamount of his debt is:a. Rs. 400000 b. Rs. 5000 c. Rs. 10000 d. Rs. 60000

3. Labor allocation is a very important process. A particular weaving section has 20 hours looms andwith five laborer’s loom efficiency is 75% and production of a loom at 100% efficiency is 10 meter perhour. Salary of a laborer is Rs. 11000 per month. I removed one laborer due to which efficiency camedown to 70%. How much did I gain or lose due to this action? Assume that profit on one meter clothis Rs. 4 and looms are working for 30 days in a month and 10 hours per day.a. Rs. 1000 profit b. Rs. 1500 lossc. Rs. 1500 profit d. Rs. 1000 loss

4. Fresh grapes contain 90% water by weight while dried grapes contain 20% water by weight. Whatis the weight of dry grapes available from 20 kg of fresh grapes?a. 2 kg b. 2.4 kg c. 2.5 kg d. None of these (CAT - 2001)

5. A man invested one third of an amount at an interest rate of 7% one fourth at a rate of S. I. 8% & theremaining at the rate of S.I. 10% . If the earns an overall 561 Rs. then the initial amount that he hadwas?a. 5500 b. 4400 c. 6600 d. None of these (FMS - MS 2006)

6. A father wanted to divide a sum of Rs. 12750 between his two sons, Jay and Ajay. Jay & Ajay whoare 23 and 24 years old respectively, divide it in such a way that if their shares are laid out atcompound interest at 4% per annum they will receive equal amount on attaining the age of 26 years.The amount which they will be receiving at the age of 26 years individually is equal amount onattaining the age of 26 years. The amount which they will be receiving at the age of 26 yearsindividually isa. Rs. 7030 b. Rs. 6250 c. Rs. 6500 d. Rs. 7780

7. A person invests Rs. 2000 in 3 months fixed deposit scheme of a bank at a rate of interest of 12.5%per annum. Due to some changes in government policies, the rate of interest changed in every threemonths, after the first period, to 12%, 11% and 10% respectively. If there was no changes in theinterest rate and the person withdraws the interest after every 3 months and continues deposit, thenhow much more interest would that person have earned in one year?a. Rs. 12.75 b. Rs. 17.5 c. Rs. 22.50 d. Rs. 20

8. In what time will Rs. 390625 amount to Rs. 456976 at 4% compound interest?a. 3 years b. 4 years c. 5 years d. None of these

Question Bank: Percentages


9. A sum of money at C.I. for 2 years at 20% would fetch Rs. 482 more if interest were half yearlyinstead of yearly. Find the principal.a. Rs. 10000 b. Rs. 20000 c. Rs. 40000 d. Rs. 50000

10. In an election 10% of the voters on the voters list did not cast their votes and 60 voters cast theirballot papers blank. There were only two candidates. The winner was supported by 47% of all votersin the list and he got 308 votes more than his rival. The number of voters on the list wasa. 3600 b. 6200 c. 4575 d. 6026

11. While traveling late in the night one has to pay 50% more than the basic auto rickshaw fare inIndore. Vivek traveled late night on the condition. However as the auto rickshaw meter was faulty,the meter showed the basic fare as what Vivek should have paid totally as night fare. How muchmore than the basic fare did Vivek end up paying?a. 100% b. 125% c. 150% d. 133.33%

12. A gardener has supply of fertilizer of type I which consists of 10% nitrogen and 6%phosphoric acidand type II fertilizer which consists of 5% nitrogen and 10% phosphoric acid. After testing he soilconditions, he finds that he needs at least 14 kg of nitrogen and 14 kg of phosphoric acid for hiscrop. If type I fertilizer costs 60 paisa per kg and type II fertilizer costs 40 paisa per kg. then what isthe minimum cost at which nutrient requirements are met?a. Rs. 82 b. Rs. 72 c. Rs. 60 d. Rs. 92

13. The price of sugar is raised by 10%.By how much percent must a man reduce his consumption ofsugar so as not to increase his expenditure?

a. 1

7 %11

b. 1

5 %3

c. 1

9 %11

d. 8.2 %

14. The cost of an apple is twice that of a banana and the cost of a banana is 25% less than that of aguava. If the cost of each type of fruit increaes by 10% find the percentage increase in the cost ofapples., 4 bananas and 3 guavas.a. 10% b. 12% c. 16% d. 18%

15. If the selling price of 10 oranges is equal to the cost price of 14 oranges, which in turn is equal to one- third of the total discount offered upon 70 oranges, then find the profit/loss percentage when themark-up percentage is shelved and the discount percentage is decreased by 5 percentage points.a. 12.5 % profit b. 20% profit c. 7.5 % loss d. 10% loss

16. Mahesh sells sugar at 20% more than the cost of purchase. He also cheats on weight and weighsonly 950 gm per kg of sugar. Suresh being a regular customer is offered a discount of 5% and paysonly Rs. 18 for a kg of sugar. The profit of Mahesh when Suresh purchases 1 kg of sugar from his is:a. Rs. 3.60 b. Rs. 3 c. Rs. 2.85 d. Rs. 2.70

17. A manufacturer sold a machine to salesman at a profit of 30% on the cost of manufacturer, thesalesman sold it to a merchant at a profit of 10%, an the merchant sold it to a customer at a profitof 40%. By what percentage did the cost to the customer exceed the cost of manufacturer?a. 100% b. 200% c. 50% d. 150%


18. An object is sold for Rs. 150. making a profit of 50% on the selling price. If the article is bought forRs. 25 less, what price must be marked so as to gain 40% by selling the object at market price?a. Rs. 75 b. Rs. 80 c. Rs. 50 d. Rs. 70

19. A dealer sells two articles at the same selling price, making 30% profit on one and incurring 30%loss on the other. If the sales prices of both the articles are Rs. 9100 each, what is the overall profitor loss for the dealer in both the transactions combined?a. Rs. 1638 loss b. Rs. 1638 profitc. Rs. 1800 loss d. Rs. 1800 profit

20. A shopkeeper’s gives 20% festival discount ion an article. Further he announces a discount of 5%on the reduced price if a customer makes a cash purchase. Fakir purchased an article in cash bypaying Rs. 2052. What was the “Marked price”{ of the article?a. Rs. 2500 b. Rs. 2700 c. Rs. 3000 d. Rs. 2736

21. An off season discount of X% is being offered at a store. An additional 12.5% discount is given if thevalue of purchase is more that Rs. 500. After the discount a person pays Rs. 525 for a pair of jeanswhose list is Rs. 750. What is the value of x?a. 30 b. 25 c. 20 d. 35

22. Krishan has 12 eggs with him. He sells x at a profit of 10% and remaining at a loss of 10%. Hegains 5% on the whole. What is the value of x?a. 7 b. 9 c. 8 d. 10

23. A man bought pens and sold them at a gain of x%. If he had sold it at a loss of x%, then he wouldhave lost Rs.x compared to the price for a gain of x%. Then which of the following statements istrue?a. The cost price is Rs. 50. b. The selling price is Rs. 50c. The profit is Rs. 50. d. The cost price depends on the value of x.

Directions for questions 24 and 25: Answer the following questions based on the information givenbelow:Shabnam is considering three alternatives to invest her surplus cash for a week. She wishes to guaranteemaximum returns on her investment. She has three options, each of which can be utilized fully or partiallyin conjunction with others.Option A : Invest in a public sector bank. It promises a return of +0.10%.Option B : Invest in mutual funds of ABC Ltd. A rise in the stock market will result in a return of + 5% whilea fall will entail a return of –3%.Option C: Invest in mutual funds of CBA Ltd. A rise in the stock market will result in a return of –2.5%,while a fall will entail a return of +2%.

24. The maximum guaranteed return to Shabnam is(a) 0.25% (b) 0.10% (c) 0.20% (d) 0.15% (e) 0.30%

25. What strategy will maximize the guaranteed return to Shabnam?(a) 100% in option A(b) 36% in option B and 64% in option C(c) 64% in option B and 36% in option C(d) 1/3 in each of the three options(e) 30% in option A, 32% in option B and 38% in option C


26. Mungeri Lal has two investment plans- A and B, to choose from. Plan A offers interest of 10%compounded annually while plan B offers interest of 12% per annum. Till how many years is plan Ba better investment?(a) 3 (b) 4 (c) 5 (d) 6 (e) 7

27. A salesman sells two kinds of trousers: cotton and woollen. A pair of cotton trousers is sold at 30%profit and a pair of woollen trousers is sold at 50% profit. The salesman has calculated that if hesells 100% more woolen trousers than cotton trousers, his overall profit will be 45%. However heends up selling 50% more cotton trousers than woollen trousers. What will be his overall profit?(a) 37.5% (b) 40% (c) 41% (d) 42.33% (e) None of the above.

Directions: Questions 28 to 30 relate to the EXCHANGE RATES problem given below:

The official ‘buy’ and ‘sell’ exchange rate for the US $, UK £ and EU ε with reference to the India INR arepresented in the table below:

You sell Bank Pays You

You Pay bank

You buy

$1 INR 40 INR 42 $1£ 1 INR 77 INR 79 £ 1ε 1 INR 60 INR 62 ε 1

Bank BUYs Bank SELLs

You lockl bank agrees to sell $ 0.023 or £ 0.012 or ε 0.015 for INR 1.

28. You wish to buy foreign currency with INR 1 lakh. Based on the values of each foreign currency youwill receive for the bank, arrange them in the descending order?

1 1 1Please note : 0.024; 0.013; 0.016

42 79 62 = = = (a) $> ε > £ (b) ε > £ > $ (c) £ > ε > $ (d) All are equal

29. When you buy foreign currency from your local bank, it will levy a transaction fee equivalent of INR500 and an additional INR 500 to deliver the exchanged money to the branch of your choice. Thistotal amount of INR 1000 will be deducted from the foreign currency payable to you. At the airport,the money changer is willing to offer $ 0.022 for INR 1.

What is the range of values of INR that can be exchanged for buying the $ which will get you a betterdeal at the airport than the bank?(a) INR 0 to 1000 (b) INR 1000 to 23000(c) INR 23000 (d) INR > 23000

30. The £ to ε ‘buy’ rate is £ 1 = e 1.222. Using the INR as the reference currency, determine by whatpercentage this ‘buy’ rate should change such that there is no arbitrage (or, differences among thethree pairwise exchange rates) across the three currencies?(a) Approx. 5% (b) Approx. 36%(c) Approx. 4.25% (d) Approx. 35.75%


31. The mean monthly salary paid to graduating MBA class of 2008 of a management institute is Rs.16,000. The mean monthly salary paid to students with work experience is Rs. 18,000. Thecorresponding figure for the students without any work experience is Rs. 12,000. Determine thepercentage of students with work experience and percentage of students without any work experiencein the class of 2008.(a) 66.67% with work experience, 33.33% without work experience(b) 33.33% with work experience , 66.67% without work experience(c) 75% with work experience, 25% without work experience(d) 25% with work experience, 75% without work experience

32. A large private airline increased the price of their air tickets by 20 percent to compensate for theincrease in airport charges. Due to increasing cost of Aviation Turbine fuel, the airline had to increasefurther the price of the ticket by 30 percent. By approximately what percentage the ticket priceshave gone up as a result of two price hikes by the airline.(a) 50% (b) 56% (c) 54% (d) 60%

33. Mr. Jeevan wanted to give some amount of money to his two children, so that although today theymay not be using it, in the future the money would be of use to them. He divides a sum of Rs.18,750/- between his two sons of age 10 years and 13 years respectively in such a way that eachwould receive the same amount at 3% p.a. Compound interest when he attains the age of 30 years.What would be the original share of the younger son?(a) 8959.80 (b) 8559.80 (c) 8969.80 (d) 8995.80

34. A space research company wants to sell its two products A and B. If the product A is sold at 20%loss and the product B at 30% gain, the company will not lose anything. If the product A is sold at15% loss and the product B at 15% gain, the company will lose Rs. 6 million in the deal. What is thecost of product B?(a) Rs. 140 million (b) Rs. 120 million(c) Rs. 100 million (d) Rs. 80 million

35. BSNL offers its share at a premium of Rs. 40, whereas its par value is Rs. 160. Parul Mehra investedRs. 50,000 in this stock. After one year BSNL declared a dividend of 19%. What rate of interest didMs. Mehra receive on her investment?(a) 15. 2% (b) 16.2% (c) 19% (d) 19.2%

36. In view of the present global financial crisis, the Finance Minister decided to slash the excise dutiesto boost demand and propel economic growth. The excise duty or cement was reduced by 30% ofits present amount to boost the spending in the infrastructure. What should be the percentageincrease in the consumption of cement so that the revenue of the government remains unchanged?

(a) 5

42 %7

(b) 6

42 %7

(c) 6

34 %7

(d) 5

34 %7

37. A merchant wants to make profit by selling food grains. Which of the following would maximize his profit?I. Sell product at 30% profitII. Increase the price by 15% over the cost price and reduce weight by 15%III. Use 700 gm of weight instead of 1 kg.IV. Mix 30% impurities in grains and sell it at cost price(a) III (b) II and I (c) II (d) All give the same profit


��

1. c Let the cost per kg be Rs. 1000∴ Selling price = Rs. 1100But the cost is only for 900 g = Rs. 900

Profit = 1100 900

100 22.22%900

− × =

2. c Debt = 2000 + 2000 × 0.8 + 2000 × 0.82 + ...

= 1

2000 100001 0.8

× =−

Hence c.

3. d Production by looms at 100% efficiency= 20 looms × 10 m/hr × 10 hrs X 30 days= 60000 m/month∴ Production at 75% efficiency= 60000 × 0.75 = 45000 m/month.Profit = 45000 × 4 = Rs. 180000But salary of one laborer is saved.∴ The cost price reduces by Rs. 11000∴ Change profit = (60000 × 0.7 × 4 × 11000)= Rs. (168000 + 11000) = Rs. 179000∴ Change in profit = Rs. 1000

Alternatively.The total production per month at 100% efficiency= Number of looms × production rate × Number of hours/day × number of days = 20 × 10 × 10 × 30 = 60000MetersNow, due to change in labor allocation, efficiencyreduces by 5% Hence, loss in production

= Loss in efficiency × total production 5

60000100

×

= 3000 mHence, total loss is = 4 × 3000 = Rs. 12000But I recover Rs. 11000 as the salary of one labor.Hence total loss is Rs. 1000.

4 c Fresh grapes contain 10% pulp.∴ 20 kg fresh grapes contain 2 kg pulp.Dry grapes contain 80% pulp.∴ 2 kg pulp would contain

2 202.5

0.8 8= = kg dry grapes

5. c Let the total amount be, 12x rupees.Then

− − × + × + × = 12x 7 12x 8 12x 3x 4x 10

5613 100 4 100 1 100

⇒ x = 55 ⇒ 12x = Rs. 6600

6. a Suppose Jay gets Rs. x and Ajay gets Rs. (12750 – x)∴ Jay’s share when he will be of 26 years of age i.e.

after 3 years = 34

x 1100

+

Ajay’s share after 2 years = (12750 – x) 24

1100

+ According to given condition

( )3 24 4

x 1 12750 x 1100 100

+ = − +

⇒ 1 x

x 1 12750 x x 12750 x25 25

+ = − ⇒ + = −

⇒ x 51x

x x 12750 1275025 25

+ + = ⇒ =

∴ 25 12750

x Rs.625051

×= =

∴ Jay receives Rs. 6250 and Ajay receives Rs. 6500now.After 3 years Jay receives

= 3 34 4

x 1 6250 1100 100

+ = + = Rs. 7030.4 which

is same as Ajay receives after 2 years.

7. c The usual procedure adopted would be to computethe total interest earned for one year at 12.5% interest,the interest earned with the prevailing rates of interestand determine the difference. A quicker method wouldbe to compute only the increase or decrease in interestearned in every period over 12.5%.period interest rate inc or dec% interest differenceI 12.5% 0 0

II 12% 0.5% 2000 0.5

Rs.2.504 100

× =×

III 11% 1.5% Rs. 7.50IV 10% 2.5% Rs. 12.50Total difference Rs. 22.50


8. b ⇒ Pnr

1 A100

+ = ⇒ 390625

n41

100 +

= 456976.

⇒ n4 456976

1100 390625

+ =

⇒ n 426 26

25 25 =

⇒ n = 4 ⇒ the required time is 4

years.

9. b If the amount compounded at annual intervals the

220CI P 1 P

100 = + −

= 1.44P – P = 0.44P

If it is done semiannually then amount =

4202P 1

100

−

+

= 1.4641P and CI = 1.4641P P = 0.4641P.∴ 0.4641 – 0.44P = 0.0241P∴ 0.0241P = 482 (given)∴ P = Rs. 20000

10. b Let the total number of voters be n.10% of the voters did not vote, so only 90% of n voterscast their voters.60 voters left blank ballot papers, hence valid voteswere 0.9 n – 60 the winner got 47% of all votes = 0.47votes.So, the loser should have got 308 votesi.e. 0.47 n – (0.9 n – 60 – 47 n) = 308 ⇒ 0.9 n + 60– 0.9 n = 308.

248 1000.04n 248 n 6200

4×= = ⇒ = = .

11. b Let the basic fare = Rs. x∴ Night fare = 1.5 xIF basic fare = 1.5 x, then new night fare (i.e. faultyfare)= 1.5 (1.5 x) = 2.25 x∴ Percentage fare = 1.5 xc, then new night are (i.e.faulty fare)= 1.5 (1.5 x) = 2.25 x∴ Percentage increase = (2.25 – 1.00) × 100 = 25%Hence Vivek paid 125% extra fare instead of 50%more.

12. d Let, fertilizer of type I required = x kgs and fertilizer oftype II required = y kgs. Then by the conditions given inquestion0.1 x + 0.05 y = 14 ... (i)0.06 x 0.1 y = 14 ... (ii)

By solving not the equations we get 5

x y4

=

Putting this value in equation (i)y = 80 and x = 100.so minimum cost = Rs.92.

13. c Direct from formula : ( )10 1

100 9 %100 10 11

× =+ .

14. a The data is tabulated below.

Apple Banana Guava

Costs 6x 3x 4x

Number of fruits 1 2 3

Total cos ts 6x 6x 12x

As the cost of each type of fruit increases by 10%. thetotal cost of one apple, 4 bananas and 3 guavas i.e. 36x increase also by 10%.

15. a Given 10s = 14 c

Also ( )7014C M – S

3=

42C 70M – 98C⇒ =

140C 70M M 2C⇒ = ⇒ =

0.4CPr ofit% 40%

1C⇒ = =

2C – 1.4CDiscount% 30%

2C= =

2C – CMark up% 100%

C= =

now, mark up % is halved i.e. 50% and discount % is25 %.If cost price is Rs. 100 then marked up price is 150 andselling price is 150 × 75% = 112.5= Profit percentage is 12.5%.

16. b The 5% discount neutralises benefit of theunderweighting. Hence he makes 20% an the deal i.e.Rs. 3. Hence (b).

1.20xAletnately, if CP x,SP 0.95 18

0.95 = = × =

∴ x = 15 profit = Rs. 3.

17. a If manufacturer pays Re. 1, customer pays1.3 × 1.1 × 1. 4 × 1 = Rs. 2.002Thus C.P. of customer exceeds C.P. of manufacturerby

( )2.002 – 1100% 100.2%

1

× =

.


18. d SP = Rs. 150, Profit = 50% on SP∴ cost = Rs. 75.New CP = 50, required gain = 40%∴ marked price = 50 × 1.4 = Rs. 70.

19. c We have, 1.3 ×C. P1 = 9100 ⇒ C. P. = Rs. 7000 and, 0.7× C. P2 = 9100⇒ C. P2 = Rs. 13000Thus total cost price = Rs. 20000 and total selling price= Rs. 18200.Thus, total loss would be 20000 – 18200 = Rs. 1800.

20. b Let the marked price of the article be Rs. 100. With thefestival discount of 20% the reduced price would be(100 – 20% of 100) = Rs. 80.With a cash discount of 5%, the sales price would be(80 – 5% of 80) = Rs. 76.Thus, we get,

76 Rs. 2052 2052 100? Rs. 2700

76100 ?

= ×⇒ == .

21. c Final selling price = Rs. 525

= ( ) ( )100 – 12.5 100 – x

Rs.750100 100

× ×

( ) 525 100100 – x 100 80

750 87.5⇒ = × × = . Hence x = 20%.

22. b Let each egg cost Rs. 10. Then the overall profit=Rs. 6If x eggs are sold at a profit of 10% then (12 – x) aresold at a loss of 10% hence total revenue= 11 x + (12 – x) 9 = 2x + 108 = Rs. 126.So, 2x = 18 and x = 9.

23. a 100 xSP at x% profit CP. SP. at x% loss

100

100 – xCP.

100

+ = =

2x 100Difference CP. x CP. Rs.50

100 2 ⇒ = = ⇒ = =

.

For questions 24 and 25:To maximise Shabnam’s return we need to evaluate all thegiven options in the question number 7. Assume Shabnam hadone rupee to invest. Let the return be denoted by ‘r’.Consider the option (30% in option A, 32% in option B and 38%in option C): If the stock market rises thenr = 0.1 × 0.3 + 5 × 0.32 – 2.5 × 0.38 = 0.653If the stock market falls thenr = 0.1 × 0.3 – 3 × 0.32 + 2 × 0.38 = – 0.197Consider option (100% in option A): This will give a return of0.1%.

Consider option (36% in option B and 64% in option C): If thestock market rises thenr = 5 × 0.36 – 2.5 × 0.64 = 0.2If the stock market falls thenr = – 3 × 0.36 + 2 × 0.64 = 0.2Consider option (64% in option B and 36% in option C): If thestock market rises thenr = 5 × 0.64 – 2.5 × 0.36 = 2.1If the stock market falls thenr = – 3 × 0.64 + 2 × 0.36 = –1.2Consider option (1/3 in each of the 3 options): If the stockmarket rises thenr = 0.1 × 0.33 + 5 × 0.33 – 2.5 × 0.33 = 0.858If the stock market falls thenr = 0.1 × 0.33 – 3 × 0.33 + 2 × 0.33 = –0.297We can see that only in option (36% in option B and 64% inoption C), Shabnam gets an assured return of 0.2% irrespectiveof the behaviour of the stock market. So right option forquestions number 13 is (0.20%) and question number 14 is(36% in option B and 64% in option C).

24. c

25. b

26. b In case of compounded annually. Let P be the initialinvestment in plan A

nn

110

A P 1 P(1.1)100

= + = … (1)

In case of simple interest in plan B

2P 12 12

A P P P n 0.12100

× ×= + = + × ×

P(1 n 0.12)= + × … (2)

Checking at different values of n = 1, 2, 3, 4, 5 A1

becomes greater than A2 when n = 5.Hence upto 4 years, plan B is better than plan A.

27. b Let the cost price of 1 cotton trouser and 1 woollentrouser be ‘C’ and ‘W’ respectively.Case I: Number of woollen trousers sold is 100% morethan cotton trousers∴ 1.3C + 1.5 × 2 × W = 1.45 (C + 2W)⇒ 0.15C = 0.1W⇒ 3C = 2WCase II: Number of cotton trousers sold is 50% morethan woollen trousers

S.P. = 1.5 2W

1.3C3×+

or, S.P. = 1.3C + W = 2.8C

C.P. = 2

C W 2C3

+ =

2.8C 2CPr ofit 100 40%

2C− = × =


28. a (a) 1 1 1

42 62 79> >

$ > • > £Hence, option (a)

29. c Deal with exchanging money from Bank = 0.024 × (x– 1000).Deal with exchanges money from changer at airport =0.022 × x

Now, 0.023 (x – 1000) 0.22x<

⇒ 0.00 1 x 23<

⇒ x 23000<

30. c The reference currency is INR. In 1INR we can buy

179

£ or 1

62 �. From this relation the

£ to � rate come out to be £ 1 = 1.274

�. But the given rate is £ 1 = 1.222 �. Hence the rate

should change by 1.274 1.222

1001.222

− × = 4.25%.

Hence option ( c) is correct

31. a Let the total number of students in the institute be ‘x’Let the total number of students with work experiencebe ‘a’, therefore the total number of students withoutwork experience be ‘x – a’As per the information given in the question

( )16000a 18000x 12000 a x= + −

16a 18x 12a 12x⇒ = + −

4a 6x⇒ =

2ax

3⇒ =

Percentage of students with work experience will be

2ax 23100 100 66 %a a 3

× = × =

Hence, option (a) is the correct choice.

32. b Effective price increase

1.2 1.3 1100 56%

1× − = × =

Hence, option (b) is the correct choice.

33. a Let the amount given to younger son be Rs. x and theamount given to older son be Rs.(18750 – x). Theyounger son turns 30, after 20 years and the olderturns 30 after 17 years. As each of them will receive

the same amount, we must have:

( )20 173 3

x 1 18750 – x 1100 100

+ = +

Or 3x(1.03) = (18750 x)−

Or 1.092727x = 18750 x−⇒ 2.092727 x = 18750⇒ x = Rs. 8959.60 is the share of the younger son.Hence, option (a) is the correct answer.

34. d Let the selling price of A and B ‘a’ and ‘b’ respectively.Given that0.2a = 0.3b2a = 3b … (i)Also,0.15a – 0.15b = Rs. 6 milliona – b = Rs. 600 million … (ii)Solving equation (i) and (ii), we getb = Rs. 80 million.Hence, the correct answer is option (d).

35. a Cost at which Parul purchased the shares is Rs. 160+ Rs. 40 = Rs. 200

Total shares = Rs. 50000

200 = Rs. 2500.

But, she will get an interest only on the face value ofthe share, i.e. Rs. 160.∴ Total amount on which she will receive an interestis Rs. 40,000.

1940000 Rs. 7600

100∴ × =

But since, Parul invested Rs. 50,000.

So, %age gain will be 7600

100 or 15.2%50000

×

36. b Revenue = Price × ConsumptionNow excise duty decreases from 100% to 70% con-sumption will increase from 70 to 100 to keep therevenue uncharged which is change of

3 6100 42 %

7 7× =

Hence option (b) is the correct choice.

37. a Profit under option I = 30%

Profit under option II 115 85

100 35.29%85

−= × =

Profit under option III 1000 700

100 42.85%700

−= × =

Profit under option IV plus the cost of impurities.

1000 700100 42.85%

700−= × =

Hence, option III is the most profitable.


Topic: Numbers Lecture Number: 01 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Numbers” important in almost all the tests.� Help students to learn how to use the basic properties of the numbers in solving the questions.� Help students to understand how to go about the study material and what to do before coming to

the next lecture.

Step1: Introduction� Faculty is supposed to introduce him/herself if he has not done it ever before in the very batch.� Talk something over the importance of numbers (some very easy questions from last few years’

actual test papers can be used as examples)� Make them aware of the importance of answer options.� Tell them that their first aim will be to be thorough with the basic concepts and to develop a habit

of getting right answers in the first attempt only (Talk about the importance of a good accuracyrate)

� Tell them that speed is a thing that is achieved very slowly with constant efforts.

Step 2: Fundamentals - 1(Please check if they are already aware of identities, ways to operate with fractions etc)

1. Classification of the numbers2. Conversion of recurring number to p/q form3. Divisibility rules

Step 3: Fundamentals - 24. Cyclicity5. HCF, LCM and Applications

Step 4: Class ExerciseGive them sufficient time and if possible discuss few problems from class exercise at the end of thelecture. Tell them to get the doubts solved with the faculty available at the center before the nextlecture.


All the things that are to be discussed in the lecture have been summarized in the followingfew pages. Read them carefully and don’t miss anything.

QA Exercise 3 - Numbers: 1


1. Classification (20 mins)• Number Tree : (5 mins)

Start with counting numbers i.e. Natural numbers. Next, state that with the addition of zero thisbecomes the set of Whole Numbers. The positive and negatives of Whole numbers form the set ofIntegers. Zero is neither a positive nor a negative number. Thus set of Natural numbers is set of positiveIntegers and set of Whole Numbers is set of non-negative Integers. Integers and fractions togetherform rational numbers.(Make it clear that 10/2 is an integer and not a fraction) Rational numbers arethose that can be represented in p/q form where p, q are integers and q is not equal to zero. Just bymere observations any terminating and non-terminating but recurring numbers are Rational Numbers.The counterpart of Rational Numbers is Irrational Numbers and thus any non-terminating and non-recurring numbers are Irrational numbers. Spend not more than a minute stating that pi is an Irrationalnumber and not the 22/7 is, an approximation of it. Just this fact of pi is enough and do not go into thehistory of pi. Denominators containing only 3, 7, 11 or higher prime numbers as factor and not even asingle 2 or 5 give recurring decimal. Rational and Irrational together form Real Numbers. The counterpartof real numbers is Imaginary numbers i.e. which include i which is root of -1. DO NOT discuss anyfurther aspect of i i.e. how it was introduced, conjugate pairs, etc.Note : Tell them to refer to the fundabook (Numbers) for basic operations like rationalization,simplification etc.

• Odd and Even Numbers : (10 mins)Do not spend time in explaining what even and odd numbers are. Students know this. Just explainodd + odd = even, even + even = even, odd + even = odd,If we add even numbers any number of time, the result will be even.If we add odd numbers odd number of times, result will be odd, otherwise it will be even.Product of odd numbers is odd,If product of numbers is odd then all numbers are odd,product of an even number with any numbers will be even andIf product of numbers is even then atleast one of them has to be even.

CAT 2001: x, y, and z are distinct integers. x and y are odd and positive whereas z is even and positive.Which of the following statements cannot be true?a. (x-z)2y is even b. (x-z)y2 is odd c. (x-z)y is odd d. (x-y)2z is even

Exactly similar question also came in CAT 2000. Infact the wordings of the questions were exactlysimilar. Only the options were different.[Take any one of first three questions of the class exercise as example (preferably Q. 2)]

• Prime and Composite Numbers (5 mins)Again no need of explaining the terms. Just cover two things viz. 2 is the only even prime number andthat prime numbers greater than 3 are of the form 6k±1 (reverse is not necessarily true). DO NOT doany other aspect related to prime numbers except the following problems.

CAT 2000: Let S be the set of prime numbers greater than or equal to 2 and less than 100. Multiply allelements of S. With how many consecutive zeros will the product end?a. 1 b. 4 c. 5 d. 10


CAT 2003 (retest): If a, a+2 and a+4 are prime numbers, then the number of possible solutions for a isa. one b. two c. three d. more than three

CAT 2003 (retest): Let x and y be positive integers such that x is prime and y is composite. Then,a. y – x cannot be an even integer b. xy cannot be an even integerc. (x+y)/x cannot be an even integer d. None of the above statements is true

Question: There are seven prime numbers such that 1 2 3 4 5 6 7p p p p p p p< < < < < < and

1 2 3 4 5 6 7p p p p p p p 510510× × × × × × = , find the value of p1.

Solution : There is no need of factorizing. Since the product is even and we have only one even primenumber that too the smallest. So p1=2. Students can also be asked to find the value of p6.

Note : No need to discuss any other numbers.

2. Conversion of recurring number to p/q form: (15 mins)

• Start directly with a general example like 0.12454545… Explain the conversion to p/q form. Once thegeneral case is done any specific case is done automatically. Quickly ask students to find p/q form of0.4444…, 0.272727…, 0.0555…. Then explain the shortcut as given in the Quant Fundabook[Q. 4].

CAT 2000: With what value should 0.ababab…, where a and b are single digit whole numbers with botha and b not being zero simultaneously, be multiplied to get an integral value?a. 1/ab b. 1/99 c. 9 d. 198 (options in CAT were different)

3. Divisibility Rules: (10 mins)

• Do not spend any time in discussing rules of 2, 3, 4, 5, 8, 9, 10 and taking examples on them. Thestudent is expected to study these at home and come. Just discuss the rules for 6 and 11. Since 6 isthe LCM (and not because 6 is the product) of 2 & 3, any number divisible by 2 & 3 is also divisible by6. Thus to check divisibility by 24, we do not use 4 & 6 (product = 6 but LCM = 12). Rather we use3 and 8.[Q. 6]

4. HCF, LCM & applications : (35 mins)

• Understanding factors, multiples, HCF & LCM : (10 mins)HCF and LCM are not to be explained by the factorization method. Just explain the underlying meaningof HCF and LCM as it is necessary for applications of LCM and HCF.

Factors Number Multiples

1, 2, 3, 4, 6, 12 12 12, 24, 36, 48, 60, 72, 84, 96, 108…1, 2, 4, 8, 16 16 16, 32, 48, 64, 80, 96, 112…Common Factors Common Multiples1, 2, 4 48, 96, 144…HCF LCM4 48


Thus, make sure that everyone understands that HCF is the highest number that can divide each of thegiven numbers and that LCM is the least number that can be divided by each of the given numbers.DO NOT spend any time in discussing the factorization method to find HCF & LCM. At best just pickthree numbers (say 48, 60, 84) and explain how it is evident that 4 is surely common and that from thequotients (12, 15, 21) there is further 3 common and hence HCF is 12 and LCM is 12 × 4 × 5 ×7. Askthem to refer to fundabook for any other method including division process to find HCF.[Q. 5 and 9]

• Applications of HCF & LCM (30 mins)a. Of the types clocks striking together or people starting a circular race simultaneously meeting at

the starting point for the first time.[Q. 8]

b. The series of numbers which leaves a common remainder of 4 when divided by 6, 7, and 9c. The series of numbers which leaves a remainder of 4, 5 and 7 when divided by 6, 7, and 9

respectively.d. A general case viz. series of numbers which leaves remainder of 4, 6 when divided by 6, 7 respectively.e. The least number of equal pieces that the three cakes of weight 18 kg, 45 kgs and 36 kgs can be

cut so that no cake is wasted.[Q. 12]

• CAT 2001: A red light flashes 3 times per minute and a green light flashes 5 times in two minutes atregular intervals. If both lights start flashing at the same time, how many times do they flash togetherin each hour?

• There were two questions, straight based on fundas covered in point a and e above in CAT 2002. Socover these applications quite well.

5. Cyclicity (25 mins)Explain concept of cyclicity taking unit digit as 2 or 3. Explain procedure after making the process veryclear.[Q. 16, 20 and 25]

6. Exercise: (15 mins)Announcement: Ask students to solve each and every problem of exercise at end of chapter onNumber System of QA Fundabook 1. In the third session of Number systems, we would just solvethe doubts of this exercise.

Puzzle: 10 people rob certain number of coins. When they divide the coins among 2 thieves,1 coin is left over. When the coins are divided among 3 thieves, 2 coins are left over. When dividedamong 4 thieves, 3 coins are left. And so on. What is the least number of coins they must haverobbed.What is the least number of coins if the remainder in each case of division would have been 1 in theabove problem.



Objectives:� Help students to discover what makes the “Numbers” important in almost all the tests.� Help students to learn how to use the properties of the numbers in solving the questions.� Help students to understand how to go about the study material and what to do before coming to






Step 2: Fundamentals - 11. Factors and multiples2. Remainders

Step 3: Class ExerciseGive them sufficient time and if possible discuss few problems from class exercise at the end of thelecture. Tell them to get the doubts solved with the faculty available at the center before the next lecture.





Check the pre-class exercise of those students who had not solved it in the first class. For all those whohave not solved it, they have to sit back after the class and solve it and only then leave.

1. Number of Factors : (25 mins)

Take any example (23 × 52) and explain the process of finding the total numbers of factors. Take carethat everyone understands the process clearly and each student knows why we add 1 to the exponents.DO NOT derive or discuss the formula for the sum of factors. At best you can state the formula and askthe student to find the derivation at home. Lay emphasis on the fact that the number should be infactorised form.[Q. 12, 20 and 27]Take example of 24 × 33 × 52 and explain finding total number of factors, how many of these are evenand how many are odd, how many factors are perfect squares. While explaining perfect squares pointout that only perfect squares will have odd number of factors and also vice-versa.Lastly, explain the problem of identifying in how many ways can 72 be written as a product of twoNatural Numbers.

2. Remainders (60 mins)Start explaining with the simple problem that if x and y when divided by 7 leaves remainders 3 and 2respectively, what will the following expressions leave a remainder if divided by 7?a. x + yb. x – yc. x × yd. x × y × xe. y – x, explain the funda of negative remaindersWith the same setting also explain how to identify the remainders in the following cases:f. y36

g. y38

Find the remainder if 430 is divided by 13?Find the remainder if 433 is divided by 13?[Preferably don’t discuss binomial theorem here even if students ask for that as it takes a lot of timeand anyways we are going to deal with that in algebra lectures][Q. 7, 8, 9, 10, 11, 14, 17]

CAT 2003: Find the remainder when 496 is divided by 6? In this example, explain that when we cancelout a factor from the numerator and denominator, we need to multiply the remainder found with thisfactor at the end.CAT 2002: Let N = 1421 × 1423 × 1425. What is the remainder when N is divided by 12?a. 0 b. 9 c. 3 d. 6CAT 2001: What is the remainder when 784 is divided by 342. Change the question and ask studentsto also find the remainder when 786 is divided by 342 and the remainder when it is divided by 344.

CAT 2002: When 2256 is divided by 17, the remainder would bea. 1 b. 16 c. 14 d. None of these


3. Exercise and its Discussion: (35 mins)Announcement: Remind everyone that in the next class they are supposed to complete the exerciseof Fundabook.

Puzzle: Along a long corridor are 100 doors numbered 1 to 100, all of them initially closed. Person#1 changes the state of all doors. After this person #2 changes state of doors 2, 4, 6, … Nextperson #3 changes state of doors 3, 6, 9…And so on with each person changing state of doorswhich are multiple of his number. After person #100 has finished, how many doors are open andhow many are closed.



Objectives:� Help students to discover what makes the “Numbers” important in almost all the tests.� Help students to learn how to use the properties of the numbers in solving the questions.� Help students to understand how to go about the study material and what to do before coming to

the next lecture.� Help students to make use of LOGIC in place of conventional methods.� Help students to identify real-time applications of the numbers.





Step 2: Fundamentals - 11. Base System2. Calendar3. Highest power dividing a factorial

Step 3: Fundamentals - 24. Miscellaneous questions on numbers

Step 4: Class ExerciseGive them sufficient time and if possible discuss few problems from class exercise at the end of thelecture. Tell them to get the doubts solved with the faculty available at the center before the next lecture.

Step 4: Discussion and Doubt SolvingIn this lecture a faculty will get a good opportunity to discuss all the things taught so far in the numbers. Tryto solve as many doubts as possible from all the lectures done so far.


All the things that are to be discussed in the lecture have been summarized in the following

few pages. Read them carefully and don’t miss anything.



1. Highest power dividing a factorial (20 mins)Explain the entire process of identifying the highest power of 2 that can divide 12!. Explain the processfirst and then the procedure. Do not just mention the procedure and make sure everyone has understoodthe process. In the same problem then ask them if one needed to find the highest power of 4 thatdivided 12!, should one work on multiples of 4, then multiples of 16 and so on? Having explained this,ask student to find the highest power of 10 that can divide 50! (briefly remark that this will also be equalto the number of trailing zeroes in 50!). One could also ask students to find the highest power of 12 thatwill divide 40! And the highest power of 12 that will divide 33![Q. 5, 9 and 14]

2. Base System (20 mins)Explain the way base system works, using the analogy of an odometer (that which measures cumulativemileage in the speedometer). Just mention the most common bases used 2, 8 and 16. In hexadecimalsystem explain A, B, C, D, E and F as 10, 11, 12, 13, 14 and 15.Convert from other base to base 10 : Start with decimal system and explain the position value. Useposition value to convert a number in any base to base 10. Explain the position value of decimals also.Also while explaining this do point out that in base b, 10 (in base b) will represent b (in decimalsystem), 100 (in base b)will represent b2 (in decimal system), and so on.Next, explain the conversion from base 10 to any other base. If time permits, also do addition, subtractionin other bases. No need to do any multiplication as it is just an extension of addition.[Q. 10 and 11]

CAT 2001: In a number system the product of 44 and 11 is 1034. The number 3111 of this system,when converted to the decimal number system, becomesa. 406 b. 1086 c. 213 d. 691

CAT 2000: Convert the number 1982 from base 10 to base 12. The result isa. 1182 b. 1912 c. 1192 d. 1292

3. Calendar (20 mins)

Explain the importance of reference day (1st January of the first year was a Monday is taken asuniversal reference, if reference is not given in the question).Explain how the day changes after 7 complete days, 70 complete days, 100 complete days, 1 year,2 years, 20 years, 100 years and so on. Also explain if 1st January of a year is Monday then 31st

January of the same year will be Wednesday and 31st December of the same year will be eitherTuesday (non-leap year) or Wednesday (leap year).[Q 1 to 4]

4. Miscellaneous (30 mins)Given below are a few miscellaneous questions, most of them from CAT. Discuss this in this class iftime permits or else discuss it at start of next class. It is given here because if you follow the timingsstrictly, you will have time left.a. Explain the number of times any particular digit appears in units or tens places in any block of 100

numbers (obviously starting from greater than 9)


b. Explain successive division and remaindersCAT 2002: After the division of a number successively by 3, 4 and 7, the remainders obtained are2, 1, and 4 respectively. What will be the remainder if 84 divides the same number?a. 80 b. 76 c. 41 d. 53

c. CAT 2003 (leaked): How many even integers n, where 100≤n≤200, are divisible neither by sevennor by nine?a. 40 b. 37 c. 39 d. 38

d. Properties of numbers (use of identities, observation and equations while solving the questionsbased on numbers)

e. CAT 2001: What is the value of the following expression?(1/(22 – 1)) + (1/(42 – 1)) + (1/(62 – 1)) +……+ (1/(202 – 1))a. 9/19 b. 10/19 c. 10/21 d. 11/21If the students cannot get this problem, they have not done their MCQs. There are two questionsexactly similar to this in the MCQs

f. CAT 2002: The integers 34041 and 32506 when divided by a three-digit integer n leave the sameremainder. What is n?a. 289 b. 367 c. 453 d. 307

g. CAT 2003 (leak): Let T be the set of integers {3, 11, 19, 27,……, 451, 459, 467} and S be a subsetof T such that the sum of no two elements of S is 470. The maximum possible number of elementsin S isa. 32 b. 28 c. 29 d. 30

Again if this problem cannot be solved, tell the students that they have not done the fundabookthoroughly. There is an exactly similar question in the exercise at end of number system chapter.

5. Exercise (30 mins)Solve & discuss the exercise.

Announcement: Tell the students in the next class they will have a test on Number Systems andshould come prepared. Also they can get all the doubts on Number Systems.

Tell them to see the set of online tests available in their SIS and start taking the tests as per thesession plane.

Puzzle: If AYE + AYE + AYE + AYE = YES + YES + YES, which digit does each alphabet standfor. Each alphabet stands for different digit.



Objectives:� Help students to identify the effective way solving a test paper as well as what they have learnt in

the first three lectures of numbers.� Help students to identify the requirement of periodic Self-assessment.� Help students to boost their confidence up and to identify which areas they need to focus more

upon.

NOTE: Testing process is not to compare the students with each other rather it is to judge one’s ownweaknesses and strengths.

Step1: Review Test� Give them stipulated time but just before that make it very clear that they have to try to solve

“RIGHT” questions with good accuracy. (To be selective is to be effective)� Tell them to focus on the accuracy first and then on speed.

Step 2: Discussion� Give the students some time to calculate their own scores and to be aware of their own perfor-

mance and mistakes.� Don’t discuss about what should have been their score or what one was expected to do.� Tell them how to select the questions, how to proceed, how to scan the question paper and how

to decide which questions are to be attempted in the first go, second go etc and which questionsare not to be attempted at all.

� Solve as many questions as possible with “shortcut/alternative” methods and teach them tomake use of answer options.

Step 3: Doubt Solving� Solve the doubts.� Always solve the questions with fastest method first, tell them alternative ways later on.




1. Tests (25 mins)

Conduct and discuss the test on Number System. Conduct the test as an actual test i.e. do notdiscuss each problem one by one. Give the students the test and let them do it independently withinstipulated time.

2. Discussion (30 mins)

After the time is over ask them to evaluate the performance. After this discuss all problems and alsotalk about how to choose questions.

3. Doubt Solving (60 mins)Solve doubts from Fundabook.

Puzzle: A traveler has a chain with 63 links (links means rings such that each ring is looped toanother ring, except the last ring). He takes up a room in an inn with the condition that he has topay the inn-keeper one link at the end of each day. What is the least number of links that he hasto cut such that he can honour the condition.


Question Bank: Numbers

1. In a collage sports meet, there were 841 participants numbered from 101 to 941 and there wereeight events - X1, X2, X3, X4, X5, X6, X7, and X8. The following is the list of the participants who took partin each events.

1X 101,102,103,104,105,106.........⇒

2X 101,103,105,107,109,111.........⇒

3X 101,104,107,110,113,116.........⇒

4X 101,105,109,113,117,121.........⇒

5X 101,106,111,116,121,126.........⇒

6X 101,108,115,122,129,136.........⇒

7X 101,109,117,125,133,141.........⇒How many students participated only in X

1?

a. 288 b. 240 c. 192 d. 144

2. A set S consists of 143 natural numbers, each of which is a perfect cube. The maximum number ofelements of S that one can always find such that each of them leaves the same remainder whendivided by 13 isa. 26 b. 27 c. 29 d. 28

3. A box contains 100 tickets, numbered from 1 to 100. A person picks out three tickets from the box,such that the product of the numbers on two of the tickets yields the number on the third ticket. Howmany tickets can never be picked up?a. 10 b. 11 c. 25 d. 26

4. All integers from 1 to 2000 are written on a blackboard. A single operation consists of erasing anytwo of the integers on the board and then writing their product on the board. After 1998 suchoperationsa. precisely two numbers will be left on the board but their product will depend on the order in which the operations are performed.b. precisely two numbers will be left on the board and their product is unique.c. only one number will be left on the board and it is unique.d. only one number will be left on the board but it will depend upon the order in which the operations are performed.

5. A number N has 6 factors, one of which is 81. In which range does the minimum possible value ofthe number N lies?a. 100 N 200≤ ≤ b. 200 N 300≤ ≤ c. 300 N 400≤ ≤ d. 400 N≤

6. In the number 1604320, after how many digits from the right will you subsets of A, such that P is asubset of Question. Find the number of ways of choosing the subsets P and Q.a. 4n b. 3n c. 2n d. n2


7. The L. C. M. of 12th, 2569 and X is 2424. How many possible integral values does X have?a. 73 b. 25 c. 1225 d. 1825

8. If in a certain number system the difference of 5333 and 555 is 4445, then the sum of the numbers3555 and 333 in that system isa. 5444 b. 5554 c. 4221 d. 2441

9. The number of natural numbers n such that ( )2n 1

n 7

++

is an integer is

a. 4 b. 5 c. 6 d. None of these

Directions for questions 10 and 11 : Answer the questions on the basis of the information given below.

Set A is formed by selecting some of the numbers from the first 100 natural numbers such that the HCF ofany two numbers in the set is the same.

10. If every pair of numbers for set A has to be relatively prime and set A has the maximum numbers ofelements possible, then in how ways can the set A be selected?a. 64 b. 96 c. 72 d. 108

11. IF the HCF of any two numbers in set A is 3, what is the maximum number elements that set A canhave?a. 10 b. 12 c. 11 d. 14

12. If m is a positive integer and 100m perfectly divides 1000! what is the largest possible value of m?a. 249 b. 124 c. 125 d. 250

13. Let N = 121212 .... upto 300 digits. What is the reminder when N is divided by 999?a. 12 b. 121 c. 216 d. 666

14. Let f(n) be the product of all the composite numbers less than n. What is the number of consecutivezeroes at the end of f(102)?a. 25 b. 24 c. 23 d. 22

15. What is the when 13400 is divided by 187?a. 137 b. 1 c. 50 d. 186

16. Last two digits of the number 3400 area. 39 b. 29 c. 41 d. 01


Directions for questions 17 and 18: refer to the data and below and answer the question that below.

A, B, C, D, E and F single digit numbers such that A C E

, , andB D F

are proper fractions. LCM of

A C E 10A B, ,

B D F A+ =

I.10A B

A+

cannot be simplified further.

II. B is multiple of A.

17. B can be __________.a. 3 b. 5 c. 7 d. 9

18. The highest possible value of A + B is:a. 9 b. 10 c. 13 d. 12

19. (212 – 46), is divisible by:a. 3 b. 5 c. Both (a) and (b) d. None of these

20. The sum of the digits of an integer is called its digital root. x, y and z are three integers with thesame number if digits such that x + y = z. If the digital root of x is p and that of y is q, then what isthe digital root of z given that there were exactly n ‘carries’ when that addition is performed?a. p + q – 7n b. p – q + n c. p + q – 9n d. p + q + 2n

21. 113 + 213 + 313 + ....... + 6013 is divisible by:a. 61 b. 63 c. 65 d. 60

22. What is the remainder when 433 + 323 – 173 + 73 is divisible bya. 91 b. 35 c. 65 d. 55

23. If x1n += where x is the product of four consecutive positive integers, then which of the followingis/are true? (CAT 1999)A. n is oddB. n is primeC. n is a perfect squarea. A and C only b. A and B onlyc. A only d. None of these


Directions for questions 24 to 26: Answer the questions based on the following information.A young girl Roopa leaves home with x flowers, goes to the bank of a nearby river. On the bank of the river,there are four places of worship, standing in a row. She dips all the x flowers into the river. The number offlowers doubles. Then she enters the first place of worship, offers y flowers to the deity. She dips theremaining flowers into the river, and again the number of flowers doubles. She goes to the second place ofworship, offers y flowers to the deity. She dips the remaining flowers into the river, and again the number offlowers doubles. She goes to the third place of worship, offers y flowers to the deity. She dips the remainingflowers into the river, and again the number of flowers doubles. She goes to the fourth place of worship,offers y flowers to the deity. Now she is left with no flowers in hand. (CAT 1999)

24. If Roopa leaves home with 30 flowers, the number of flowers she offers to each deity isa. 30 b. 31 c. 32 d. 33

25. The minimum number of flowers that could be offered to each deity isa. 0 b. 15 c. 16 d. Cannot be determined

26. The minimum number of flowers with which Roopa leaves home isa. 16 b. 15 c. 0 d. Cannot be determined

27. Each of the numbers 1 2 nx , x , , x , n 4,≥� is equal to 1 or –1. Suppose

1 2 3 4 2 3 4 5 3 4 5 6 n 3 n 2 n 1 n n 2 n 1 n 1 n 1 n 1 2 n 1 2 3x 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 0,− − − − − −+ + + + + + + =�

then (CAT 2000)a. n is even b. n is oddc. n is an odd multiple of 3 d. n is prime

28. Let N = 553 + 173 – 723. N is divisible by (CAT 2000)a. both 7 and 13 b. both 3 and 13c. both 17 and 7 d. both 3 and 17

29. Convert the number 1982 from base 10 to base 12. The result is (CAT 2000)a. 1182 b. 1912 c. 1192 d. 1292

30. If 09/12/2001(DD/MM/YYYY) happens to be Sunday, then 09/12/1971 would have been a (CAT 2001)

a. Wednesday b. Tuesday c. Saturday d. Thursday

31. At a bookstore, ‘MODERN BOOK STORE’ is flashed using neon lights. The words are individually

flashed at the intervals of 1 1 1

2 s, 4 s and 5 s2 4 8

respectively, and each word is put off after a second.

The least time after which the full name of the bookstore can be read again is (CAT 2002)a. 49.5 s b. 73.5 s c. 1744.5 s d. 855 s


32. Three pieces of cakes of weights 1 3 1

4 lb, 6 lb and 7 lb2 4 5

respectively are to be divided into parts of

equal weight. Further, each part must be as heavy as possible. If one such part is served to eachguest, then what is the maximum number of guests that could be entertained?a. 54 b. 72 c. 20 d. None of these

33. 6n 6n7 – 6 , where n is an integer > 0, is divisible by (CAT 2002)

a. 13 b. 127 c. 559 d. All of these

34. F(x) is a fourth order polynomial with integer coefficients and with no common factor. The roots ofF(x) are –2, –1, 1, 2. If p is a prime number greater than 97, then the largest integer that divides F(p)for all values of p is:(a) 72 (b) 120 (c) 240 (d) 360 (e) None of the above.

35. If a, a + 2 and a + 4 are prime numbers, then the number of possible solutions for a is (CAT 2003 (Re-Test))

a. one b. two c. three d. more than three

36. If ( )3 3 3 3x 16 17 18 19= + + + , then x divided by 70 leaves a remainder of (CAT 2005)a. 0 b. 1 c. 69 d. 35

37. The rightmost non-zero digits of the number 302720 is (CAT 2005)a. 1 b. 3 c. 7 d. 9

38. Let X be a four-digit number with exactly three consecutive digits being same and is a multiple of 9.How many such X’s are possible?(a) 12. (b) 16 (c) 19 (d) 21 (e) None of the above.

39. How many natural numbers are there for which the remainder is 41 when the dividend is 1997 ? (XAT 2006)

a. 5 b. 12 c. 10 d. 6

40. When 4101 + 6101 is divided by 25, the remainder is (JMET 2006)a. 20 b. 10 c. 5 d. 0

41. If x = – 0.5, then which of the following has the smallest value? (CAT 2006)

a. 1x2

b. 1x

c. 2

1

xd. 2X e.

−1

x


42. Which among 122 ,

133 ,

144

, 166 and

11212

is the largest? (CAT 2006)

a. 122 b.

133 c.

144 d.

166 e.

11212

43. When 1012 – 1 is divided by 111, the quotient isa. 9009009 b. 9000009 c. 9009009009 d. 9000000009

44. Deepali and Priya have some marbles with each of them, such that the number of marbles with thePriya is twice that with Deepali. If Deepali distributes the marbles with her equally among certainnumber of bags, then she is left with 41 extra marbles. If both Deepali and Priya were to pool themarbles with them and then distribute the total marbles equally among the same number of bags asDeepali did, they will be left with only 10 extra marbles. What is the least possible number ofmarbles with Priya and Deepali put together, if Deepali says that she has at least 1000 marbles withher?a. 3051 b. 3081 c. 3174 d. 3297

45. If ‘q’ is a prime number such that ‘q + 4’ and ‘q + 14’ are also prime, how many such q’s are there?a. 0 b. 1 c. 3 d. Infinite

46. There are three bags with 100 balls in each bag. Each bag contains balls of two different colours.Three boys, X

1, X

2 and X

3 are given these bags say B

1, B

2 and B

3 respectively. Let B

1 contain white

and black balls. B2 contains white and blue balls, B3 contain white and red balls. X1 picks a whiteball after 3 black balls, X2 picks a white ball after 2 blue balls and X3 picks a white ball after 7 redballs. How many balls would each of them have to pick before all of them pick a white ballsimultaneously?a. 23 b. 11 c. 12 d. 24

47. The digit in the unit’s place in 15 + 25 + 35 + .... 995 isa. 0 b. 1 c. 2 d. 3

48. The last digit of (13 + 23 + 33 + ... 103)64 isa. 2 b. 5 c. 9 d. 0

49. Three persons A, B & C go hunting with a monkey. They had a pile of oranges. One of them tries todivide the oranges into 3 parts and finds one extra which he gives to the monkey and takes one thirdof the remaining. Now another person tries to divide the oranges into three parts and finds one extrawhich he gives to the monkey and takes the one third share of the remaining. Now the third persontries to divide the remaining oranges into three equal parts and finds one extra which he gives to themonkey and takes one third of the remaining. Later all of them sat together, divided the pile intothree and gave one extra orange to the monkey. Thus, no orange is left. The minimum of oranges inthe pile could bea. 85 b. 70 c. 88 d. 79

50. The remainder (in decimal system) when (47648)9 is divided by 5 isa. 2 b. 3 c. 5 d. 4


51. For which of the following values of n in the remainder of 491n when divided by 7 equal to theremainder of 492n when divided by 7?a. 585 b. 592 c. 611 d. 623

52. How many numbers less than or equal to 500 are there, each of which is the product of more thanthree distinct prime numbers?a. 3 b. 5 c. 4 d. None of these

53. The numbers (9)n, (11)n, (P)n and (20)n are four numbers, in the number system to the base n,forming an arithmetic progression. Find P.a. 15 b. 16 c. 17 d. 18

54. Rahul writes the number 458 on the blackboard, after which each one of his friends walks up to theboard and is allowed to perform exactly one operation. The operation can be either to double thenumber on the board (after the erasing the earlier number)or to erase the last digit of the number onthe board. If after sometime the number 14 was on the board, then what is the minimum possiblenumber of friends that Raghu has? Assume that no friend walks up to the board twice.a. 4 b. 6 c. 8 d. None of these

55. If ( )3

10729 is expressed as (1001)x, then which of the following will be expressed as (11011)x?

a. 15 b. 27 c. 13 d. 18

56. In how many zeroes does the number ( )2

2002!

1001!end?

a. 0 b. 1 c. 2 d. 200

57. In a number system, the product of 122 and 41 is 5442. The number 4434 of this system whenconverted to decimal system becomesa. 1030 b. 1010 c. 1020 d. 1040 e. None of these

58. The number 10001000100010001 isa. a multiple of 11 b. a prime numberc. a composite number d. a perfect square

59. Which of the following is true?a. 9950 + 10050 = 10150 b. 9950 + 10050 > 10150

c. 9950 + 10050 < 10150 d. None of these

60. The number of ordered pairs (a, b) of positive integers such that a + b = 90 and their greatestcommon divisor is 6 equalsa. 15 b. 14 c. 8 d. 10


61. The prime factor of 237 – 1 lying between 200 and 300 isa. 227 b. 223 c. 251 d. 271

62. If x, y and z are positive numbers with no common factors and such that 1 1 1x y z

+ + , then which of

the following is true?a. (x + y) is a square b. (y + z) is a squarec. (z + x + y) is a square d. None of these

63. The H.C.F of 1.08, 0.36 and 0.9 isa. 0.03 b. 0.9 c. 0.18 d. 0.108

64. If n is a natural number, find the remainder when 999 ... 9, upto (2n + 1) digits, is divided by 550.a. 549 b. 449 c. 499 d. 399

65. In a party, all the guests were distributed tags, labelled with the consecutive numbers starting from1. At the end of the party, chocolates were distributed such that guests with a tag labelled 1 gets 1chocolate, the one with a tag labelled 2 gets two chocolates and so on. One of them gets thechocolates twice. The total number of chocolates that were distributed was 1200. What was the tagnumber of the guest who got the chocolates twice?a. 35 b. 28 c. 24 d. 21

66. MTNL has a waiting list of 5005 applicants for its recently launched mobile phone scheme. The listshows that there are at least 5 males between any two females. The largest possible number offemales in the waiting list is:a. 920 b. 835 c. 721 d. 1005

(JMET - 2005)67. The smallest positive value of x for which the fractions

x 2 x 13 x 26 x 41 x 1913 x 2002, , , , , ,

10 11 12 13 49 50+ + + + + +− − −− are in their simplest form is

a. 47 b. 49 c. 51 d. 53 (JMET - 2006)

68. The age of Mr. Chetan in 2002 was 1

90 of his birth year. What is his age in 2006?

a. 30 b. 28 c. 26 d. 22 (JMET - 2006)

69. If the sum of 7 consecutive = 13524, Find the first year?a. 1919 b. 1939 c. 1929 d. 1918

70. A three digit natural number abc is written side by side to form a six digit natural numberN = abcabc. Then which of the following is necessarily true?a. N is perfectly divisible by each of 7, 11 and 13b. N is perfectly divisible by 7 and 11 but not 13c. N is perfectly divisible by 7 and 13 but not 11d. N is perfectly divisible by 11 and 13 but not 7 (ATM - 2000)


71. How many numbers between 5 and 95 (both inclusive) have odd number of distinct factors?a. 7 b. 9 c. 16 d. None of these (ATM - 2003)

72. A teachers writes 101 numbers on the blackboard, of which 50 are zeros, and 51 are ones. Astudent is asked to perform the following operation 100 times on the board: strike out any twonumbers. If they are equal, write another zero. If they are unequal, write a one. What are thenumbers left on the board?a. Single One b. Single Zero c. Two Ones d. None of the above (ATM - 2005)

73. Let S be a set of any five distinct numbers chosen from the set {1, 2, 3, ..., 7, 8}. Then S containsa. two numbers which add to 9. b. at least three prime numbers.c. at least two non-prime numbers. d. at least two odd numbers. (ATM - 1999)

74. Some children stand in a queue and share a box of chocolates in the following manner:

First, Child 1 takes 100 chocolates plus 1

10th of whatever remains in the box. Then Child 2 takes

200 chocolates plus 1

10th of whatever remains, then Child 3 takes 300 chocolates plus

110

th of

whatever remains, and so on for each child in the queue. It turns out that each child gets the samenumber of chocolates. Thena. there must be exactly 7 children in the queue.b. each child must have received 900 chocolates.c. the total number of chocolates initially in the box must have been 6300.d. none of the above is necessarily true. (ATM - 1999)

75. The sum of four consecutive two-digit odd numbers, when divided by 10, becomes a perfect square.Which of the following can possibly be one of these four numbers?(a) 21 (b) 25 (c) 41 (d) 67 (e) 73

76. The number of employees in Obelix Menhir Co. is a prime number and is less than 300. The ratio ofthe number of employees who are graduates and above, to that of employees who are not, canpossibly be:(a) 101 : 88 (b) 87 : 100 (c) 110 : 111 (d) 85 : 98 (e) 97 : 84

77. A confused bank teller transposed the rupees and paise when he cashed a cheque for Shailaja.giving her rupees instead of paise and paise instead of rupees. After buying a toffee for 50 paise,Shailaja noticed that she was left with exactly three times as much as the amount on the cheque.Which of the following is a valid statement about the cheque amount?(a) Over Rupees 13 but less than Rupees 14(b) Over Rupees 7 but less than Rupees 8(c) Over Rupees 22 but less than Rupees 23(d) Over Rupees 18 but less than Rupees 19(e) Over Rupees 4 but less than Rupees 5


78. The integers 1, 2, …, 40 are written on a blackboard. The following operation is then repeated 39times: In each repetition, any two numbers, say a and b, currently on the blackboard are erasedand a new number a + b – 1 is written. What will be the number left on the board at the end?(a) 820 (b) 821 (c) 781 (d) 819 (e) 780

79. What are the last two digits of 72008?(a) 21 (b) 61 (c) 01 (d) 41 (e) 81

80. Suppose, the seed of any positive integer n is defined as follows: seed(n) = n, if n < 10

= seed(s(n)), otherwise,where s(n) indicates the sum of digits of n. For example,seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed (1 + 4) = seed (5) = 5 etc.How many positive integers n, such that n < 500, will have seed (n) = 9?(a) 39 (b) 72 (c) 81 (d) 108 (e) 55

Directions for Questions 81 and 82:Mark (a) if Q can be answered from A alone but not from B alone.Mark (b) if Q can be answered from B alone but not from A alone.Mark (c) if Q can be answered from A alone as well as from B alone.Mark (d) if Q can be answered from A and B together but not from any of them alone.Mark (e) if Q cannot be answered even from A and B together.

In a single elimination tournament, any a player is eliminated with a single loss. The tournament is playedin multiple rounds subject to the following rules :

(A) If the number of players, say n, in any round is even, then the players are grouped into n/2 pairs. Theplayers in each pair play a match against each other and the winner moves on to the next round.

(B) If the number of players, say n, in any round is odd, then one of them is given a bye, that is heautomatically moves on to the next round. The remaining (n–1) players are grouped into (n–1)/2pairs. The players in each pair play a match against each other and the winner moves on to the nextround. No player gets more than one bye in the entire tournament.

Thus, if n is even, then n/2 players move on to the next round while if n is odd, then (n+1)/2 playersmove on to the next round. The process is continued till the final round, which obviously is playedbetween two players. The winner in the final round is the champion of the tournament.

81. What is the number of Matches played by the champion?A. The entry list for the tournament consists of 83 players?B. The champion received one bye.

82. If the number of players, say n, in the first round was between 65 and 128, then what is the exactvalue of n?A. Exactly one player received a bye in the entire tournament.B. One player received a bye while moving on to the fourth round from the third round.

83. Four digits of the number 29138576 are omitted so that the result is as large as possible. Thelargest omitted digit is(a) 9 (b) 8 (c) 7 (d) 6 (e) 5


For question 84 and 85, a statement is followed by three conclusions. Select the answer from thefollowing options.(a) Using the given statement, only conclusion I can be derived.(b) Using the given statement, only conclusion II can be derived.(c) Using the given statement, only conclusion III can be derived.(d) Using the given statement, all conclusion I, II and III can be derived.(e) Using the given statement, none of the three conclusion I, II and III can be derived.

84. An operation “#” is defined by

a # b = b

1–a

Conclusion I. (2 # 1) # (4 # 3) = –1Conclusion II. (3 # 1) # (4 # 2) = –2Conclusion III. (2 # 3) # (1 # 3) = 0

85. A, B, C and D are whole numbers such thatA + B +C = 118B + C + D = 156C + D + A = 166D + A + B = 178Conclusion I. A is the smallest number and A = 21.Conclusion II. D is the smallest number and D = 88.Conclusion III. B is the smallest number and B= 56.

86. If March 1, 2006 was Wednesday, which day was it on March 1,2002?(a) Wednesday (b) Thursday (c) Friday (d) Saturday

87. Each page of a book is printed and numbered serially on both sides in the usual manner, that is , thefront side of the first page is numbered 1, the back side of the first page is numbered 2, and so on.Of course, as usual , there is a possibility that the book ends at the front side of the last page inwhich case only the front side of this page is numbered and the back side of this page is left blank.The book has one of its pages missing. The total of the remaining page numbers is 15000. What isthe last page number in the complete book?a. 172 b. 173 c. 174 d.Cannot be determined (ATM)

88. If p is the product of four consecutive positive integers, then which of the following statements inNOT true?a. p is perfectly divisible by 24 b. p is not a perfect squarec. (p + 1) is a perfect square d. (p –2) is a perfect square (ATM)

89. In a certain year, the month of January had exactly four Wednesdays and four Sundays. ThenJanuary 1 of that year was aa. Saturday b. Monday c. Thursday d. Friday (ATM)


1. c There are 841 participants identified with numbersfrom 101 to 941. The following tabulation gives thedetails of the events and the identify numbers of theparticipants taking part.

1

2

3

4

5

Identification number of the participantsEvent

taking part in The event

E All the participants

X Every sec ond participant starting from 101

X Every third participant starting from 101

X Every fourth participant starting from 101

X

6

7

8

Every fifth participant starting from 101

X Every sixth participant starting from 101

X Every seventh participant starting from 101

X Every eightparticipant starting from 101

While dealing with these number, the relative positionwith respect to the first numbers is of greatersignificance than the number itself. so, let us subtract101 from the difference as their ‘New’ identify. So, theparticipants with the identification numbers of, 101,102, 103, ..... 940, 941 will have their ‘New’ identify as0, 1, 2, 3 ..... 839, 840 respectively.

1

2

3

4

5

6

7

The 'New ' identify of the participants whoEvent

took part in the event

X All the numbers from 0 to 840.

X All the mutiples of 2.

X All the multiples of 3.




X All the mul

8

tiples of 7.


The participant with the ‘New’ identify 0 participated inall the events. There are 840 participants remaining840 = 23 × 3 × 5 × 4 ( 2, 3, 5, 7 are the prime factors of840).So, all the participants numbered with the multiples of2. 3, 5 and 7 took part in the event X1, and at least oneother event.∴ The number of participants who took part only in E1

= The number of natural numbers les than 840 and arerelative prime to 840.

1 1 1 1840 1– 1– 1– 1–

2 3 5 7 =

1 2 4 6840 192

2 3 5 7= × × × × =

2. c Any number when divided by 13 may leave 13 distinctreminder of 0, 1, 2, ..... 12.But cubes of natural number will leave only five distinctreminder which can be arrived at by finding theremainder of the cubes.03, 13, 23, ...... 123, when divided by 13 leave 5 distinctremainders.the five distinct reminders are 0, 1, 5, 8, 12.Hence, if 143 cubes of natural numbers are present .

At least 143

15

+ Elements will have the same reminder 28 1 29⇒ + =Hence at least 29 elements which leave the samereminder are there.

3. b The tickets are numbered from 1 to 100.Let the numbers on the 3 selected tickets be a, b, N 1,2, 3, ..... 99, 100(N) = (a), (b)Where a, b, N are three distinct numbers. Clearly noneof a, b, N can be 1.Again, if a takes the 2. then b can take any value from3 to 50.All the composite numbers from 51 to 100 can beobtained by taking suitable value of a and b. But noneof a, b, N can be a prime number, greater than 50 andless than 100. Required number of numbers.= (Total prime numbers between 50 and 100) + 1= 10 + 1 = 11

4. d After 1 operation the number of integers decreasesby 1. after 1998 operation there are exactly twointegers whose product is always 2000.

5. b A number N has 6 factors one of which is 81.Lwt N = P1

x P2y where P1 and P2 are two of its distinct

prime factors. If there are three distinct prime factorsthe total number of factors is at least 8.]The number of factors of N is (x + 1) (y + 1)= 6 = 1 (6) = 2 (3)∴ (x + 1) = 1 and (y + 1) = 6 ....... (A)OR (x + 1) = 2 and (y + 1) = 3 ...... (B)If B is true, 81 can’t be a factor of N.

∴ (A) is true x 0 and y 5⇒ = = . As one of the factors

is 81 (i.e. 34) the value of P2 should be 3.∴ N = P1° (3)° = 243.N is uniquely specified and it satisfies choice.

6. b Any power of any number ending with 6 (With indexgreater than 1) always ends with a 6 and the tensdigit of that power is always odd.Hence 1604320 has 4320 zeroes, a six before that andbefore that an odd number before that. Hence weencounter the for 1st odd number after. 4321 digits ifwe start from the right.

��


7. d The factors of the 2 s and 3’s in X. There could anynumber of 2’s from 0 to 72 and for each of thesepossibilities. There could be 0 to 24 3’s. Thus X canhave 73(25) or 1825 values.

8. c Given 5333 – 555 4445 or 555 + 4445 = 5333Consider the sum 5 + 5 = 10 But we have 3 in the rightmost place. So, 10 – 3 = 7 is the base of the numbersystemIn base 7.

3 5 5 5

3 3 3

4 2 2 1

+

9. a2 2(n 1) (n 7) (12n 48)

n 7 n 7+ + − +=+ +

n 4 36(n 7) 12 (n 7) 12

n 7 n 7+ = + − = + − + + +

∴For n = 2, 5 and 29; 36

n 7+ would be an integer.

For Questions 10 and 11:In set a as the HCF of any two numbers is the same, all thenumbers in set A are of the from ha, hb, hc, .... where a, b, c,.... are co-primes. Also as we require maximum number ofelements in set a, we can take a, b, c, d, .... as 1 an primes.

10. b As every pair of HCF of any two numbers is a productof distinct combination of the 25 prime numbers lesthan 100. As we want the largest set A, we shouldtake each prime number on its, own rather than incombination with other primes. For the lower primes2, 3, 5, 7, we have choices. For 11, 13, ..... , 97 (theother 21 primes) we have no choices.

The target 1st a has 26 elements, which are a powerof 1, 2, 3, 5, 7 less that 100 can be selected in 6, 4, 2,2 ways respectively. Thus there are (6) (4) (2) (2) =96 ways of selecting the largest set A.

11. b Given HCF of the numbers in set a as 3.Hence the numbers are 3, 6, 9, 15, 21, ... 93.These are the numbers of the form 3p where p is a

prime number less than100

3 (i.e. primes less than 33).

Further p can be 1. There are 12 numbers in set A.

12. b The number of zeroes at the end of 1000!

1000 1000 1000 10005 25 125 625

= + + + = 200 + 40 + 81 + 1 = 249∴ The largest power of 100 that divided 1000!

249124

2 = =

.

13. d When 1000 is divided by 999, reminder is 1.

∴ We can be use reminder theorem to answer thisquestion./ But first we need to write need to write N inthe f(x). Where n – 1 = 999. i.e. n = 1000For e.g. 121212 can be written as121 × (1000)99 + 212 × (1000)98 + .... 121 (1) + 212 = 121 × 50 + 212 × 50 = (121 + 212) 50 = 333 × 50= (333 × 48) + (333 × 2) = (999 × 16) + 666.∴ When the above value is divided by 999. Thereminder will be 666.

14. c Consider 101.

It has 101 1015 25

+ = 20 + 4 = 24 zeroes at the end.f(102) is the product of all the composite numbersbetween 1 and 101 (both inclusive)Since the product of primes between 1 an 101 hasonly on e zero (2 × 5) in the end. So, the remaining 23zeroes at the end of f(102).

15. b We see that Ram

400 40013 2Rem

11 11

=

8032Rem

11

=

( ) ( )80 32–1 1 Rem –1

11 = = = �

Further, Rem 400 20013 169

rem17 17

=

( )200 169–1 Rem –1

17 = = �

= 1

40013 11M 1 17K 1∴ = + = +

40013Rem 1

187

∴ =


16. d 3400 = (34)100 = (1 + 80)100 = 1 + 100C2 (80)2 + .... + 100C100

(80)100.= 1 + 800 + (last two digit in each term is 00). Therefore,last two digits = 01.

For questions 17 and 18:

Statement I A 1⇒ = or A does not divide B.But statement II rules out of the second possibility.

A 1⇒ =

10A B⇒ + is the number between 11 to 19.

Now LCM (1, C, E) = 10A + B ⇒ 10 + B ⇒ is a composite

number i.e. an y one of 12, 14, 15 16 and 18 ⇒ B canbe 2, 4, 5, 6 or 8.

17. b.

18. a

19. c 212 – 46 = (212 – 1) – 45If ‘n’ is an even integer then 2n – 1 is divisible by 3 and22n – 1 is divisible by 5 (212 – 1) is divisible by both 3 and5 an d45 is also Hence, (c).Alternatively.212 – 46 = 4096 – 46 = 4050Which is divisible by both 3 and 5.

20. c During the addition of two integers when a digit iscarried over, we are actually subtracting 10 from thesum of the 2 digits being added and adding 1 to over,9 is subtracted from the sum of the digital roots of the2 integers.⇒ the sum of digits of z = p + q – 9n

21. c Let N = 113 + 213 + 313 + ..... + 6013

Now 113 + 6013 is divisible by 61(... an + bn is divisible by a + b when ‘n’ is odd)Similarly, 213 + 5913 is divisible by 61 and so on.Therefore, the whole expression, i.e. 113 + 213 + 313 + ...+ 6013 is divisible by 61.

22. c We can write it is433 + 73 + 323 – 173

= (43 + 7) (432 – 43 × 7 + 72) + (32 – 17)= (322 + 32 × 17 + 172)= 50 (x) + 15(y)= 5 (10 x + 3y) which is divisible by 13.∴ Combining the given term is divisible by 5 × 13 = 65.

23. a Use the method of simulation, viz. take any samplevalues of x and verify that n is both odd as well as aperfect square.

For questions 24 to 26:

foecalPpihsrow

forebmuNsrewolferofebgnireffo

forebmuNsrewolfdereffo

forebmuNsrewolf

tfel

1 y)8/51( y y)8/7(

2 y)4/7( y y)4/3(

3 y)2/3( y 2/y

4 y y 0

Starting from the fourth place of worship and movingbackwards, we find that number of flowers before

entering the first place of worship is 15

y8

.

Hence, number of flowers before doubling = 15

y16

(but this is equal to 30)Hence, y = 32Answer for 24 is (c)

The minimum value of y so that 15

y16

is a whole number

is 16.Therefore, 16 is the minimum number of flowers thatcan be offered.Answer for 25 is (c).

For y = 16, the value of 15

y 1516

= .

Hence, the minimum number of flowers with whichRoopa leaves home is 15.Answer for 26 is (b).

27. a Each term has to be either 1 or –1.Hence, if the sum of n such terms is 0, then n is even.

28. d N can be written either (54 + 1)3 + (18 – 1)3 – 723 or (51+ 4 )3 + 173 – (68 + 4)3 .The first form is divisible by 3, and the second by 17.

29. c

12 198212 165 212 13 9 1 – 1

––

The answer is 1192.

30. d In 30 years from 1971 to 2001, number of odd days= 30 + (8 from leap years) = 38 and 38 ≡ 3 mod 7So December 9, 1971 is Sunday – 3 days= Thursday


31. b Because each word is lit for a second,

5 17 41 7 21 49LCM 1, 1, 1 LCM , ,

2 4 8 2 4 8 + + + =

LCM(7, 21, 49) 49 373.5 s

HCF ( 2, 4, 8) 2×= =

32. d9 27 36 HCF(9, 27, 36)

HCF , ,2 4 5 LCM (2, 4, 5)

=

920

= lb

= Weight of each pieceTotal weight = 18.45 lb

Maximum number of guests = 18.45 20

419× =

33. d 6n 6n7 – 6Put n = 1

6 6 3 3 3 37 – 6 (7 – 6 )(7 6 )= +

This is a multiple of 3 37 – 6 127= and 73 + 63 = 559and 7 + 6 = 13

34. d Given that ( )( )( )( )F(x) x 2 x 1 x 1 x 2= + + − −Putting x = P, we have

( ) ( )( )( )( )F P P 2 P 1 P 1 P 2= + + − −

Now P is in the form 6K 1± where K is positive integer

( ) ( )( )( )( )F 6K 1 6K 3 6K 2 6K 6K 1+ = + + −

( )( )( )( )( )36 2K 1 3K 1 K 6K 1= + + − …(1)

( ) ( )( )( )( )F 6K 1 6K 1 6K 2 6K 6K 3− = + + −

( )( )( )( )36 6K 1 3K 1 K 3K 1= + + −…(2)Please note that the value of K 17≥ and expression F(6K + 1) and F (6K – 1) always bear the factor 10.Hence 360 is the correct choice. Therefore option(D) is the correct choice.

35. a As any prime number greater than 3 can be expressed

in the form 6n 1± , minimum difference between threeconsecutive prime numbers will be 2 and 4. The valuesthat satisfy the given conditions are only 3, 5 and 7,i.e. only one set is possible.

36. a 3 3 3 3x 16 17 18 19= + + + is even number

Therefore 2 divides x.

3 3 2 2a b (a b)(a – ab b )+ = + +⇒ a + b always divides a3 + b3

Therefore 163 + 193 is divisible by 35183 + 173 is divisible by 35Hence x is divisible by 70.Hence option (a)

37. a ((30)4)680 = (8100)680. Hence the right most non-zerodigit is 1.

38. e Let the four digit number be ‘aaab’ or ‘baaa’Since, the number has to be a multiple of 9, therefore3a + b should be either 9, 18 or 27.

Case I: 3a + b = 9Possible cases are:(1116, 6111, 2223, 3222, 3330, 9000)

Case II: 3a + b = 18Possible cases are:(3339, 9333, 4446, 6444, 5553, 3555, 6660)

Case III: 3a + b = 27Possible cases are:(6669, 9666, 8883, 3888, 7776, 6777, 9990)

Hence total number of cases 20.

39. ddy

r

a

c

We have a relation:- d a c r= × +

So, we should have 2 numbers a & b such that:-1997 = a × b + 41or ab = (1997 – 41) = 1956

ab 1956=

ab = 1956 = 23 × 31 × 1631

The no. of different (a & b); satisfying above is givenby:-

( )( )( )2 1 1 1 1 1

2

− + + = 6

40. b101 1014 6

25+

101 101 20 4 254 6 4(1024) 6.(6 )25 25 25 25

= + = +

254 6 (1296)25 25

×= +

256( 4)4

25−= +

=4 + (– 6) (–1) = 10.]

41. b Go by option, put −= 1

x2

(1) − =2 1

24

(2) ⇒ = −−

1 12

x 1/ 2


(3) ( )⇒ =

−2 21 1

4x 1/ 2

(4) − =1/ 2 1

22

42. b LCM of 2, 3, 4, 6, 12 = 12

12 12 1212 126 4 3 2 12 3 4 6 12

∴ 34 is greatestNote: n1/n is maximum when n = e (2.718). Among theoptions n = 3 is closest to the value of e.

43. c 1012

– 1 = (106 – 1) (10

6 + 1) = (10

3 – 1) (10

3 + 1) (10

6 +

1) = 999 × 1001 × 1000001

So, (1012 – 1 ÷ 111) = 999 1001 1000001

111× ×

= 9 ×

1001 × 1000001 = 9009009009.

44. c Let the number of marbles with Deepali = N (N ≤ 1000)⇒ The total number of marble with Priya and Deepalitogether = N + 2N = 3NLet the number of bags = NN = nQ + 41 where Q is the number of marbles per bagin the first case

⇒ 3N = 3nQ + 41 × 3 = 113

n 3Qn

+ + 10

(... If total marbles are distributed in n bags, 10 marble

are extra)i.e. in the second case each bag must have

1133Q

n +

marbles and this must be a natural number.

Hence n must be a factor of 113 for 113

n to be an

integer.⇒ n = 113, since 113 is prime⇒ Minimum four-digit value of N = 1130 – 113 + 41 =1057⇒ Minimum value of 3N = 3174

45. b Put q = 3 ⇒ 7 and 17 are prime numbers. Only q = 3satisfies this condition.

46. a The 4th, 8th, 12th, 16th ... balls that X, picks are white.The 3rd, 6th, 9th, 12th ... balls that X

2 picks are white.

The 8th, 16th, 24th .... balls that X3 picks, are white.

Hence L.C.M. of (4, 3, 8) = 24∴ At the 24th ball, all of them will pick a white ball atthe same time.Hence, before that each of them had picked 23 balls.

47. a The sum 15 + 11

5 + 21

5 + ... 99

5 ends in 0 (10 one’s).

The sum 25 + 12

5 + 22

5 + ... 92

5 ends in 0 (10 two’s)

......

.... ⇒ The sum 15 + 2

5 + .... 99

5 ends in zero

48. b Let S = 13 + 2

3 + ... 10

3 =

210 112×

= (55)

2

⇒ S is a number ending with 5S, for all integral (+ve) powers ends with 5.∴ (1

3 + 2

3 + 3

3 + .... + 10

3)

64 ends in 5.

49. d Let the number of oranges in the pile be x.

( )

Distribution A B C Monkey Remaining

x 1 2I – – 1 x 1

3 32x 5 4x 10

II – – 19 9

4x 19 8x 38III – – 1

27 278x 65 8x 65 8x 65

IV 1 081 81 81

− −

− −

− −

− − −

For minimum number of oranges

8x 65 1461 x

81 8− = ⇒ = non-integer

For integer value of oranges we get

8x 65 567 657 x 79

81 8− += ⇒ = =

Short-cut: Work backward from the options to getthe answer quickly.

50. a 4 × 94 + 7 × 9

3 + 6 × 9

2 + 4 × 9 + 8 × 9

0 = 26244 + 5103

+ 486 + 36 + 8 = 31877 = 2mod(5)

51. a The remainder if 491 i.e. 490 + 1 divided by 7 is 1 as490 is divisible by 7. The remainder of 491

2 i.e. (490 +

1)(490 + 1) = (490)

2 + 490 + 490 + 1 divided by 7 is 1.

Thus the remainder 491 to any power of divided by 7is 1. Using a similar procedure 492

n when divided by 7

will have a remainder of 2n. These two remainders

can be equal only if n is divisible by 3, 585 is the onlychoice divisible by 3.

52. c Numbers which are products of more than 3 distinctprime factors will have at least 4 distinct prime factors.Starting from the least possible values, the four distinctprime numbers are 2, 3, 5, 7.a) 2 × 3 × 5 × 7 = 210, which is the least of the

numbers of the required type.b) If 7 is replaced by 11, the number is 330.c) Similarly 2 × 3 × 5 × 13 = 390.d) 2 × 3 × 5 × 17 = 510 which is more than 500,

hence not admissible.e) If 2, 3, 7, 11 are considered the number is 42 × 11

= 462, admissible.f ) 2, 3, 7, 13 given a number 42 × 13, which is

greater than 500, not admissible


g) 2, 5, 7, 11 gives the number 770, not admissibleh) 3, 5, 7, 11 gives the number greater than 500, not

admissible.i) When the first 5 prime numbers are considered,

the product is 2 × 3 × 5 × 7 × 11, and this is greaterthan 500.

Hence, the only possible numbers are as given in (a),(b), (c) and (e) i.e. a total of 4 numbers.

53. d (9)n = 9 = say a

(11)n = n + 1 = (a + d) and

(P)n = a + 2d

(20)n = 2n + 0 = a + 3d

From above equations.[(11)

n – (9)] × 2 = (20)

n – (11)

n⇒ (n – 8) × 2 = n – 1 ⇒ n = 15 ⇒ (9)n = (9)

10(11)n = (16)

10 and

P = (16)10

+ (16 – 9)10

= 23 ⇒ k = (18)15

54. c The required result 14 ca be achieved in exactly 8steps as shown below 1 2 3 4 5 6 7 8458 → 45 → 90 → 180 → 360 → 720 → 72 → 7 → 14

55. b Given that ( )13729 = (1001)

x

9 = (1001)x. This could be written as 9 = x

3 + 1.

x3 = 8. Hence x = 2.

56. b The number of zeros at the end of n! is the largestpower of 10 that is a factor of n!. It is also the largestpower of 5 that divides n! The largest power of 5 that

is a factor of ( )22002!

1001!

2002 2002 2002 20025 25 125 625

1000 1000 1000 10002

5 25 125 625

+ + + − + +

= (400 + 80 + 16 + 3) – 2 (200 + 40 + 8 + 1) = 1

57. a Suppose base is x,(122)

x × (42)

x = (5442)

xi.e (1 × x2 + 2 × x

1 + 2 × x

0) (4 × x

1 + 1 × x

0)

= [5 × x3 + 4 × x

2 + 4 × x

1 + 2 × x

0]

i.e. 4x3 + x

2 + 8x

2 + 2x + 8x + 2 = 5x

3 + 4x

2 + 4x + 2

i.e. x = 6 now (4434)6 = 4 × 6

3 + 4 × 6

2 + 3 × 6

1 + 4 × 6

0

= 1030

58. c The required number can be written asN =1 + 10

4 + 10

8 + 10

12 + 10

18

= ( )( )

( ) ( )( )54 20 10 10

4 44

10 1 10 1 10 1 10 1

10 1 10 110 1

− − + −= =

− −−

= ( )( )( )( )

5 5

2 2

100 1 100 1

10 1 10 1

+ −

− +

(1005 + 1) is a multiple of (100 + 1) and (100

5 – 1) is a

multiple of (100 – 1)So, the original number can be written as a product oftwo numbers. The number can’t be a perfect squarebecause it leaves a remainder of 5 when divided by 9,while perfect squares can leave only 0, 1, 4 or 7when divided by 9.

59. c (101)50

= (100 + 1)50

= 10050

+ 50.10049

1 + ( )

( )4850.49 .100

1.2 + ... (1)

(99)50

= (100 – 1)50

= 10050

– 50.10049

+ ( )

( )4850.49 .100

1.2

.... (2)Subtracting (2) from (1), we have

(101)50

– (99)50

= 2[50.10049

+ ( )

( )4750.49.48 .100

1.2.3 ... ]

⇒ (101)50

– (99)50

= 10050

+ some positive number⇒ (101)

50 > (100)

50 + (99)

50.

60. c Given a, b are positive integers such that a + b = 90and their HCF is 6.Let a = 6k

1 and b = 6k

2 then 6k

1 + 6k

2 = 90 ⇒ k

1 + k

2 = 15

So, k1 + k

2 = 15 for which we have (1, 14) (2, 13), (4,

11), (7, 8)But the value of k

1 and k

2 can be interchanged.

61. b As 37 is a odd number is in power of 2.∴ each prime divisor of 2

37 – 1 is of the form 1 (mod 2

× 37).i.e., 1 (mod 74)Or ⇒ 1 (mod 74 × 3) (to get between 200 and 300).1 (mod 222) or 223 is the number that divides 2

37 – 1

and is between 200 and 300.(The theorem that follows is ⇒ If P is an odd primenumber then each prime divisor of 2

P – 1 is of the form

1 (mod 2P).

62. a1 1 1x y z+ +

⇒ z(x + y) = xy ⇒ xy – z(x + y) = 0⇒ xy – zx – zy + z

2 = z

2 (adding z

2 both sides)

⇒ (x – z) (y – z) = z2

[if (x – z) is a factor of z2 and (y – z) is a factor of z

2

then they are known as complementary factor]let x – z and y – z share a common factor P⇒ P

2 is a factor of z

2 or P is a factor of z.

⇒ that is x and y are also factors of z which is nottrue.


∴ x – z and y – z are co-prime numbersIf product of two co-prime numbers is a square thisshow that x – z and y – z are also squares.

Let x = u2; y – z =

2

2u

z (where u is a factor of z)

⇒ x + y = u2 +

2

2u

z + 2z ⇒

2zu

u +

∴ x + y is a square.

63. c The given numbers are 1.08, 0.36 and 0.90.The HCF of 108, 36 and 90 is 18.∴ The HCF of the given numbers = 0.18

64. b Since there are no “cannot be determined” among thechoices, the answer must be independent of n.Hence, put n =1 ⇒ 2n+ 1 = 3

⇒ 999550

, remainder = 449

Alternatively: i.e. if’s ‘cannot be determined’ were tobe one of the choices, then the following approach isuseful.when 999 .... 9 (99 digits) is divided by 11 the remainderis 9.N = 9999 .... 9 (99 digits) = 11k + 9 ... (1)When the given number is divided by 50, we get 49 asremainder as shown below.N 9999...9 9999...900 99= = +��

N 999....900 50 49= + +��

N 999....900 50 4950 50 50

+= +

N 491

50 50= +

∴ N = 50I + 49 ... (2)k and I are positive integers,From equations (1) and (2), we get11k + 9 = 50I + 49∴ 11k = 50I + 40 ...(3)The minimum value of I for which k is an integer is 8,also when I = 8, we get k = 40.From equation (1) & (2), the minimum value of N is 449.Therefore N can be written asN = 11 × 40 × m + 449 (where m is +ve integer)This the remainder is 449

65. c We have to find x such that

( )n n 1x 1200

2

++ = where 1 ≤ x ≤ n.

( )n n 11200

2

+<

n2 + n – 2400 < 0

for n2 + n – 2400 = 0,

1 1 9600n

2− ± +=

∴ 1 9604 1 9601 1 9601

n2 2 2

− − − − − +< < <

< 1 9604

2− +

99 97n

2 2− < <

– 50 < n < 48∴ The maximum value can n take is 48.

∴ If n = 48, ( )n n 1

2

+ =

48 472×

= 1176

∴ x = 1200 – 1176 = 24. Hence c,Alternatively,

( )n n 1x 1200

2

++ = where x ≤ n

So, ( )n n 1

2

+ should be a little less than 1200.

∴ n(n + 1) should be a little less than 2400.Now, 50 × 51 = 255049 × 50 = 245048 × 49 = 235247 × 48 = 2256∴ n = 48 and x =24

66. b FMMMMMFMMMMM...��

Hence, largest possible number of females is 5005

6= 834 + 1 = 835.

67. c Checking by options and putting x = 51, one can easilysee that at x = 51, the ratios are already in their simplestforms.

68. c Let age of Chetan in 2002 = x

2002 xSo, x

90− =

⇒ x = 22So, Chetan’s age in 2006 = 22 + 4 = 26 yrs.

69. c Let, the first year is ‘a’then, a + a + 1 + a + 2 + a + 3 + a + 4 + a + 5 + a + 6= 135247a + 21 = 13524

a = 13503

7

a 1929⇒ =


70. a 71. a 72. a 73. a 74. b

75. c By options checking option (c), four consecutive oddnumbers are 37, 39, 41 and 43. The sum of these 4numbers is 160.When divided by 10, we get 16 which is a perfectsquare.∴ 41 is one of the odd numbers.

76. e Using options, the sum of the numerator anddenominator of the ratio should be a prime number.Only option (e) satisfies [97 + 84 = 181]

77. d Suppose the cheque for Shailaja is of Rs. X and YpaiseAs per the question: 3 × (100X + Y) = (100Y + X) – 50⇒ 299X = 97Y – 50

299X 50Y

97+⇒ =

Now the value of Y should be a integer.Checking by options only for X = 18, Y is a integer andthe value of Y = 56

78. c Total sum of the numbers written on the blackboard

40 41820

2× =

When two numbers ‘a’ and ‘b’ are erased and replacedby a new number a + b – 1, the total sum of thenumbers written on the blackboard is reduced by 1.Since, this operation is repeated 39 times, therefore,the total sum of the numbers will be reduced by1 × 39 = 39.Therefore, after 39 operations there will be only 1number that will be left on the blackboard and that willbe 820 – 39 = 781.Hence, option (c) is the correct choice.

79. c The last two digits of any number in the form of 74n willalways be equal to 01.For example 74 = 2401 and 78 = 5764801.Hence, option (c) is the correct choice.

80. e seed(n) function will eventually give the digit-sum ofany given number, n.All the numbers ‘n’ for which seed(n) = 9 will givethe remainder 0 when divided by 9.For all positive integers n, n < 500, there are 55 multiplesof 9.Hence, option (e) is the correct choice.

81. d Using statement A:The question cannot be answered because we donot know the number of byes got by the champion.Hence, statement A alone is not sufficient to answerthe question.Using statement B:The question cannot be answered because we do

not know the exact number of players in thetournament.Hence, statement B alone is not sufficient to answerthe question.

Combining both the statements together:If there are 83 players, then there will be 6 rounds inthe tournament and we know that the championreceived only one bye, therefore the total number ofmatches played by the tournament will be 6 – 1 = 5.Hence, option (d) is the correct choice.

82. d Using statement A:When n = 127, exactly one bye is given in round 1.When = 96, exactly one bye is given in round 6.As no unique value of n can be determined hence,statement A alone is not sufficient.

Using statement B:As we do not exactly many bye’s are given, in total,we cannot determine the value of n, uniquely.

Combining statement A and B:There is a unique value of n = 120, for which exactly1 bye is given from the third round to the fourth round.Hence, option (d) is the correct choice.

83. e Removing 2, 1, 3, 5 makes the result largest as 9876.Therefore the largest omitted digit is 5.option (e) is the correct choice.

84. e None of the conclusion can be derived. Hence option(e) is the correct choice.

85. e Given thatA + B + C = 118 … (i)B + C + D = 156 … (ii)C + D + A = 166 … (iii)D + A + B = 178 … (iv)Adding all the four equations, we get3 (A + B + C + D) = 618or A + B + C + D = 206… (v)Subtracting (i) from (v) we getD = 88Similarly A = 50, B = 40None of the conclusions can be derived . Hence option(e) is the correct choice.

86. c If → March 1 2002 is a MondayMarch 1 2003 will be TuesdayMarch 1 2004 will be ThursdayMarch 1 2005 will be FridayMarch 1 2006 will be SaturdaySo when March 1 2006 is a Saturday, March 1 2002will be Monday.So when March 1 2006 is a Wednesday, March 12002 will be a Friday.Option (c) is the correct choice.

87. b 88. d 89. c


Topic: Ratios Lecture Number: 01 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Ratios” important in almost all the tests.� Help students to learn how to use ratios and proportions in solving the questions.� Help students to discover the importance of assuming convenient values and avoiding fractions.� Help students to understand how to go about the study material and what to do before coming

to the next lecture.

Step1: Introduction� Faculty is supposed to introduce him/herself if he/she has not done it ever before in the very

batch.� Talk something over the importance of ratios (some very easy questions from last few years’

actual test papers can be used as examples)� Make them aware of the importance of answer options.

� Tell them that their first aim will be to be thorough with the basic concepts and to develop a habitof getting right answers in the first attempt only (Talk about the importance of a good accuracyrate)


Step 2: Fundamentals - 11. Basics of ratios2. Partnership3. Proportions

Step 3: Fundamentals - 24. Chain Rule

Step 4: Class ExerciseGive them sufficient time and if possible discuss few problems from class exercise at the end of thelecture. Tell them to get the doubts solved with the faculty available at the center before the nextlecture.NOTE: Make necessary announcements.


QA Exercise 7 - Ratio: 1


1. Basics of ratios (15 mins)Explain how a ratio does not give us actual values but gives us the relative sizes. Take a simpleproblem on ages to explain the usage of ratio : If the ratio of ages of father and son is 4 : 1, what will bethe ratio of their ages after 5 years if the difference between their ages after 5 years will be 30 years?Do take the example of savings and expenditure: If the ratio of incomes of A and B is in the ratio of3 : 4 and the ratio of the expenditures is in ratio of 2 : 3, what is the ratio of their savings. What will bethe answer if the ratio of the expenditures was also 3 : 4 in this example?Quickly ascertain that everyone knows dividendo and componendo and proceed to explain that

1 2 3

1 2 3

k a k c k ea c eb d f k b k d k

+ += = = =

+ +�

��

�

Ask the students that if a c

k,b d

= = what will the ratio 2 2

2 2

a cb d

−−

be?

Explain if a : b, b : c, c : d is given how we can find a : d and also a : b : c : d.[Q 2 and 3]

2. Partnership (10 mins)Explain how profit is divided in the ratio of product of capital invested and time invested for. Do somesimple problems on this. Finish it off fast and do not spend too much time on this simple topic[Q. 4 and 27]

3. Proportions (35 mins)Take the following three cases with an example of each :a. Direct Proportion (10 mins)

Make sure that in addition to ‘increase in one causing an increase in other’ you also mention thatthe increases are proportional. Thus if one variable becomes 7/5 times or increases by 15%, theother variable will also become 7/5 times or increase by 15%. E.g. Cost of picnic is directlyproportional to number of people going on the picnic. If 10 people go, the cost per head is Rs. 250.What will be the cost per head if 15 people go? Remind them that SI was directly proportional toP, r, n, i.e. if principal doubled, so would SI, if rate of interest became 1/3rd so would SI, etc.

Solve the question of diamond falling and breaking into pieces with the value of a diamond beingdirectly proportional to the square of its weight. The problem is there in your exercise. Solve theproblem now itself.[Q. 5]

b. Direct Relation (20 mins)In this case also an increase in one causes the other to increase but the increase is not proportional.E.g. Amount at SI is directly related to P, r, n. If Rs. 1000 is kept at 10% SI, the interest in 3 yearsis Rs. 300 and the interest in 6 years is Rs. 600 i.e. years have doubled and SI has also doubled– SI has a direct proportion. But the Amount after 3 years would be 1300 and after 6 years wouldbe 1600, i.e. amount has increased but not doubled. This is a relation of corresponding change but


not proportionality is called Direct Relation. Do° mention that in this case the equation becomes y

= k1x + k2 E.g. Cost of a picnic is directly related to number of people going on the picnic. If 10people go, the cost per head is Rs. 250 and if 15 people go the cost per head is Rs. 200. What isthe cost per head if 20 people go? Emphasise that since there are two unknowns (k1 and k2) directrelation needs two equations to be solved. For a good batch it would be worthwhile even to mentionthat in this same example it could also have been stated that the decrease in cost per head isdirectly proportional to number of people. This example is same as that of fixed price and variableprice per unit. Thus Total Cost = Variable cost × # of units + Fixed Cost is a example of DirectRelation.

CAT 1999: Total expenses of a boarding house are partly fixed and partly varying linearly with thenumber of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders andRs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100boarders?a. 550 b. 560 c. 540 d. 580Another difference between Direct proportion and Direct relation is that in Direct proportion whenone variable is zero the other variable is also zero (y = kx) but in Direct Relation even when onevariable is zero, the other need not be (y = k1x + k2)Solve the problem of the exercise – maximum number of bogies that can be attached so that trainmoves given that reduction in speed is directly proportional to square of number of bogies attached(observe the similarity – reduction in speed is directly proportional with reduction in cost per headis directly proportional. Since total cost was directly related, in this case speed is going to bedirectly related). Solve the problem now itself rather than at the end of class.

CAT 1999: The speed of a railway engine is 42 kmph when no compartment is attached, and thereduction in speed is directly proportional to the square root of the number of compartmentsattached. If the speed of the train carried by this engine is 24 kmph when 9 compartments areattached, the maximum number of compartments that can be carried by the engine isa. 49 b. 48 c. 46 d. 47No need to solve this problem. Just state it and as a similar problem just done, ask students to doit at home.

c. Inverse Proportion (5 mins)Quickly explain that Inverse proportion boils down to xy = constant. Apart from examples ofspeed do cover the following two examples as well. The number of days taken to complete a jobis inversely proportional to the number of workers. And so also when the amount of a milk andwater solution is increased by pouring just water, the concentration of milk is inverselyproportional to the amount of solution i.e. C1 × V1 = C2 ×V2. Thus if in 30 lts of solutionconcentration of milk is 40%, and if the solution is increased to 40 lts by adding water,

concentration of milk will be 30

40% 30%.40

× =


4. Chain Rule (15 mins)These are problems of the types “If 10 hens lay 10 eggs in 10 days, how many eggs will 1 hen lay in1 day”. These types of problems are a direct application of direct and inverse proportion. All one needsto identify is the relation (direct or inverse) between the variables. But there is a common-sense way totackle these problems without explicitly using any formula. Explain the working of this type of problemswith an example : If 15 carpenters can make 10 chairs in 5 days working 8 hours per day, how manychairs can 12 carpenters make in 15 days working 6 hours per day. It is ok if you explain the man-

hours-days approach but a better way would be to find the answer directly as 12 15 6

10 18.15 5 8

× × × = In

this you have to find number of chairs made and hence start with number of chairs that could be madei.e. 10. Now multiply this 10 with ratios of corresponding variables. Mentally just get the idea if morechairs or less chairs can be made because of the changes. If more chairs can be made, the ratio hasto be taken such that it is greater than 1 and if less chairs can be made the ratio has to be less than1. e.g. since carpenters have decreased from 15 to 12, less number of chairs can be made and hencemultiply with 12/15. Since more days are available now (15 compared to earlier 5), more chairs can bemade and hence multiply with 15/5 and so on.Give one more example to students: If it requires 5 pumps of 250 watts to raise 1000 lts of water toa height of 15 mts, how many pumps of 500 watts will be needed to raise 750 lts of water to a heightof 8 mts?[Q. 1]

CAT 2002: It takes 6 technicians a total of 10 hours to build a new server from a Direct Computer, witheach working at the same rate. If six technicians start to build the server at 11 AM, and one technicianper hour is added beginning at 5 PM, at what time will the server be complete?a. 6:40 PM b. 7 PM c. 7:20 PM d. 8 PM

CAT 2002: 3 small pumps are filling a tank. Each of the three small pumps works at 2/3rd the rate of thelarge pump. If all the four pumps work at the same time, they should fill the tank in what fraction of thetime that it would have taken the large pump alone?a. 4/7 b. 1/3 c. 2/3 d. 3/4

5. Solving the Class Exercise (25 mins)Tell them to solve all the undiscussed questions in the stipualetd time.

6. Discussion (20 mins)Explain questions from class exercise.

Announcement : Ask all students to solve all problems of the exercise on Mixtures and Solutions inthe QA Fundamental Book and get the book in the next class.


Topic: Ratios Lecture Number: 02 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Mixtures and Solutions” important in almost all the

tests.� Help students to learn how to deal with questions based on Mixtures and Solutions.� Help students to understand how to go about the study material and what to do before coming

to the next lecture.� Help students to make use of LOGIC in place of conventional methods.


ratios.� To check there progress as well as to be assured of the above stated fact you must throw some

questions and discuss with them.� “Revision” always helps in assessing the level of the class and to identify whether they (or some


Step 2: Fundamentals - 11. Averages2. Weighted average3. Alligation (Simple Examples)

Step 3: Fundamentals - 23. Alligation (Tougher Examples)4. Mixing a pure component to a solution5. Removal and replacement6. Average Speed and other unusual mixtures



QA Exercise 8 - Ratio: 2


1. Averages (10 mins)Directly start with an example as all students are well aware of what average is. If a person with age 65joins a group of 5 persons with an average age of 62, what will the new average age of the group be?While all students will be able to solve this, tell the students that (62 × 5 + 65)/6 is not the mostefficient way. The problem should be solved as follows : The new person is 3 more than the average andhence this extra 3 has to be divided now among the six members (so that all are equal) and hence theaverage will be 62 + 3/6 = 62.5What is the person joining was 60 years old? If the group also had an average age of 60, the newperson joining would not make any difference to the average. Now the extra with the group i.e. 2 × 5 =10 has to be divided among the six person and hence new average will be 60 + 10/6 = 61.666Two persons with weights 50 and 54 leave a group and hence the average of the group falls from 48 to46. How many persons were there in the group originally? The two persons are taking away with them2 + 6 = 8 and each of the remaining persons is giving 2. Thus number of remaining persons is 4 andoriginally the group had 6 members.Tell the students that questions on averages should be done orally.[Q. 3, 26]The average of x successive natural numbers is n. If the next natural number is included in the group,the average increases bya. depends on xb. depends on starting number of seriesc. both a and bd. ½

2. Weighted Average (10 mins)Start the class by explaining what weighted average is. If one group has an average age of 16 yearsand another group has an average age of 20 years, would the average age of both the groups combinedbe 18 years? If the first group had 6 members and the second group had 2 members, the average age

of both the groups combined could be found out by 16 6 20 2

178

× + × = This is nothing but the weighted

average formula. The group of 16 pulls the average towards itself as it exerts (influences) a greaterweight because it has more members.Take another example: If a trader mixes 42 kgs of wheat costing Rs. 12 per kg with 18 kgs of wheatcosting Rs. 17 per kg, what is the average cost price of the mixture. In this example explain thatinstead of using 42 kgs and 18 kgs, we could just use 7 and 3 to reduce our calculations as 42 : 18 is

same as 7 : 3. Thus average cost per kg will be 12 7 17 3

13.5.7 3

× + × =+

Ask students to solve all examples of the exercise based on weighted averages right now. Give themthe problem numbers they have to solve and clear any doubts if they have.


3. Alligation (20 mins)Take the following problem: If 10 lts of 40% milk solution is mixed with 15 lts of 55 % milk solution,what is the concentration of milk in the mixture? It’s a straight forward example of weighted averageand the answer can be found directly as (80 + 165)/5 i.e. 16 + 33 = 49% without introducing anyvariable. But if the data in the problem is slightly changed as : If 10 lts of 40% milk solution is mixedwith x lts of 75% milk solution and the resultant mixture is of 50% milk concentration, what is x? If wefollow the above approach of weighted average we will have to write the equation on paper, transpose

and then calculate : 40% 10 75% x

50%.10 x

× + × =+

Alternately, to avoid calculation or use of pencil, we can use Alligation. Explain the process of Alligation.PLEASE MAKE SURE that students don’t go back with the idea that Alligation is a cure-all solution formixtures. Alligation would not be of any help in finding the weighted average of 10 and 15 with weightsin ratio of 2 : 3. In this case simple formula of weighted average can give the answer much faster thanAlligation. Thus ENSURE that the students understand that Weighted Average and Alligation is thesame. Any problem that can be done by weighted average can be done by Alligation and vice-versa.Alligation is helpful when the resultant average is given and Weighted Average is helpful when resultantaverage is asked.Ask students to solve all examples of the exercise based on Alligation right now. Give them theproblem numbers they have to solve and clear any doubts if they have.[Q. 1, 2]

4. Mixing a pure component to a solution (20 mins)How many litres of water needs to be added to 30 lts of 3: 1 milk and water solution such that themixture has milk and water ratio of 2 : 3?Write this problem on the top of the board and keep the board clean, you have to solve the aboveproblem in 5 different methods and all of them have to be on the board simultaneously, so use theboard effectively.

Method 1: Equation Method : 22.5 2

7.5 x 3=

+

Method 2: Alligation with concentration of milk as 0% in water. Thus 75% 40% x40% 0% 30

− =−

Method 3: No Change component : Amount of milk is not changing. Thus milk i.e. 22.5 lts will be 2/5th

of the solution after mixing. Thus solution will become 5/2 times of 22.5 i.e. 56.25 meaning that 26.25lts is addedMethod 4: Inverse proportion :

C1 × V1 = C2 × V2 Thus 2

75% 15 450V 30 30 56.25

40% 8 8= × = × = =


Method 5: Unitary method : Initial Milk & Water is 3k & 1k. Since milk has not changed, make thevalue of milk in the second ratio also as 3k. Thus final Milk & Water is 3k & 4.5k. Thus 3.5k litres of

water was mixed. Since initial solution (4k) is 30 lts, 3.5k is 3.5

30 =26.254

×

While the student may get a little overwhelmed, ask them to study all these processes as differentapproaches can be useful in different problems. Tell them that they should be sure of methods 3, 4 and5 as these are shortcuts and will be useful later on. E.g. In what ratio must 3 : 2 milk water solution bemixed with water to result in a 2 : 3 milk water solution. Using method 4, the ratio Vfinal : Vinitial is directlygiven by 3/5 : 2/5 i.e. 3 : 2. Thus ratio of mixing has to be 2 : (3-2) i.e. 2 : 1. This was simple becausethe ratios just got inverted.

5. Removal and Replacement (25 mins with exercise problems)Directly start with a problem. From 40 lts of milk and water solution with concentration of milk being90%, 4 litres of solution is removed and replaced with water. If this operation is done once more, whatis the amount of milk in the resultant solution and also the concentration of milk in the resultantsolution?Point to be highlighted in the begining : When we remove a part of any solution, we are not changingthe concentration.

Approach 1: Thus when 4 litres are removed, concentration of milk remains 90% and the only changeis that the amount of solution becomes 36 lts.

When water is added (it is adding a pure component to a mixture) and hence concentration of milk is

inversely proportional to the volume and the concentration of milk becomes 36

90% .40

×

Again when 4 litres of solution is removed, the concentration of milk remains at 36

90%40

× and the

volume becomes 36.Lastly when 4 lts of water is added back, volume increases from 36 to 40 and since concentration is

inversely proportional to volume, the concentration becomes 36 36

90% .40 40

× ×

Thus final concentration is 72.9%The fraction 36/40 is nothing but 9/10 and could also have been thought as the fraction of the solutionleft after 4 lts is removed.

Approach 2: When x% of a fraction is removed, x% of each of the constituents is also removed.Thus when 10% (4 out of 40) of the solution is removed, 10% of milk is also removed and thus 90% ofmilk remains. Originally 36 lts of milk was there in the solution and now only 90% of it remains. Whenwater is added back, amount of milk does not increase and it remains at 90% of 36. Again when thesolution is being removed 4 lts out of 40 are being removed and thus 10% of milk will again be removedand 90% of it will be left. When water is added, amount of milk does not increase and remains same.Thus after two full operations, 90% of 90% of 36 lts of milk remains. Since volume of solution is backto the original 40 lts when we divide this by 40 we get the concentration.


Ensure the students go back clearly understanding that Final concentration (or volume) = Initialconcentration (or volume) × (Fraction left)n where the operation is done n times. And that the concentrationor volume should be of that component which is not replaced.10 lts of 27 : 8 milk and water solution is removed and replaced with water. This operation is done thriceand the solution now has milk and water in ration 8 : 27. What is the total volume of the originalsolution? Initial concentration of milk is 27/35 and final concentration of milk is 8/35. Thus (fractionleft)3 = 8/27 and fraction left = 2/3 which means that 1/3rd was removed and since 1/3rd = 10 lts, totalsolution is 30 lts.Ask students to solve all problems of the exercise which are based on removal and replacement. Givethem the question number they have to solve and then clear any doubts that they have.[Q. 4, 5, 24, 25, 27]

6. Average speed and other unusual mixtures (15 mins with exercise problems)Suppose a person goes from Delhi to Chandigarh at a speed of 60 kmph and returns at a speed of 80kmph, what is the average speed for the round trip?Explain with this problem the concept of average speed. Link it up with weighted average with time asthe weights. Point out to the student that since weighted average is done, alligation can also be doneand the ratio that alligation would result in is the ratio of the time driven at the two speeds and not ofdistance. Take an example on this : A man covers a total distance of 200 kms in 4 hours but partly inbus @ 30 kmph and partly in car @ 60 kmph. What distance does the man travel by car? Using

alligation we have 30 50

2 : 1.50 60

− =− This ratio is of time driven at various speeds and not distance

covered. Thus he travels by car for 2/3rd of time i.e. for 2

23

hours and covers a distance of 160 kms.

Take another example of unusual mixtures as follows : A man borrows a total of Rs. 10,000 but partfrom one bank @ 6% and part from another bank @ 9%. If he pays a total interest of Rs. 750 per year,what part the other was borrowed at 9%? Solve it using alligation?[Q. 6, 7]

7. Class Exercise (20 mins)

Tell the students to solve all the undiscussed questions from class exercise in the stipulated timelimit.


1. Alloy A is made by mixing metal and X and Y in the ratio of 3 : 2, by weight. Alloy B is made bymixing metals Y and X in the ratio 3 : 5, by weight. An alloy C is to be made by mixing the alloys Aand B in the ratio of 1 : 4, by weight. Find the ratio of X and Y, in alloy C.a. 21 : 10 b. 20 : 29 c. 31 : 19 d. 30 : 23

2. The ratio of the incomes of Ashok and Tina is 6 : 4 and the ratio of their expenditures is 7 : 5. If theratio of their savings is 6 : 3, what percentage of his income does Ashok save?a. 53.3% b. 50% c. 55% d. 57%

3. Three identical bottles are each half filled with alcohol solution. When the content of the third bottleis divided into two halves and each half is poured into each of the first two bottles, the concentrationof alcohol in the first bottle reduces by 10% of its volume and that in the second bottle increases by10% of its volume. What is the ratio of the initial quantity of alcohol in the first bottle to that in thesecond bottle?a. 5 : 2 b. 13 : 4 c. 13 : 7 d. 11 : 6

4. Five runners - A, B, C, D, and E - started from the same point at the same time, with speeds that arein the ratio of 1 : 3 : 4 : 5 : 7 respectively and are running a race around a circular track, A, and C runin the same direction, while the remaining run in the opposite direction. At how many distinct pointson the track does A meet any other runner?a. 13 b. 12 c. 11 d. 18

5. A student took five papers in an examination, where the full marks were the same for each paper.His marks in these papers were in the proportion of 6 : 7 : 8 : 9 : 10. In all papers together, thecandidate obtained 60% of the total marks. Then the number of papers in which he got more than50% marks isa. 2 b. 3 c. 4 d. 5 (CAT - 2001)

6. Three classes X, Y and Z take an algebra test.The average score in class X is 83.The average score in class Y is 76.The average score in class Z is 85.The average score of all students in classes X and Y together is 79.The average score of all students in classes Y and Z together is 81.

What is the average for all the three classes?a. 81 b. 81.5 c. 82 d. 84.5 (CAT - 2001)

Directions for questions 7 and 8: Answer questions on the basis of the information given below:An airline has a certain free luggage allowance and charges for excess luggage at a fixed rate per kg. Twopassengers, Raja and Praja have 60 kg of luggage between them, and are charged Rs 1200 and Rs 2400respectively for excess luggage. Had the entire luggage belonged to one of them, the excess luggagecharge would have been Rs 5400.

7. What is the weight of Praja's luggage?a. 20kg b. 25 kg c. 30 kg d. 35 kg e. 40 kg (CAT - 2006)

Question Bank: Ratio


8. What is the free luggage allowance?a. 10kg b. 5 kg d. 20 kg d. 25 kg e. 30 kg

9. A milkmaid mixed a certain quantity of milk with water and sold one third of the mixture. He thenadded pure milk and water to the remaining mixture whose quantities were half and one-third,respectively, of the initial quantity of mixture. The ratio of milk and water now is the reverse of whatit initially was. Find the original ratio of milk and water

a. 2 : 5 b. 1 : 2 c. 6 : 7 d. 7 : 9

10. There are three alloys. The first contains 45% of tin and 55% of lead, the second contains 10% ofbismuth, 40% of tin and 50% of lead and the third contains 30% of bismuth and 70% of lead. Theymust be used to produce a new alloy containing 15% of bismuth. What is the least percentage oflead that can be contained in this new alloy?a. 66% b. 55% c. 40% d. 50%

11. Two vessels A and B of equal capacities contain mixtures of milk and water in the ratios 4 : 1 and3 : 1, respectively. 25% of the mixture from A is taken out and added to B. After mixing it thoroughly,an equal amount is taken out from B and added back to A. The ratio of milk to water in vessel A afterthe second operation is:a. 79 : 21 b. 83 : 17 c. 77 : 23 d. 81 : 19 (JMET 2005)

12. A, B and C are assigned a piece of work which they can complete by working together in 15 days.

Their efficiencies (measured in terms of rate of doing work) are in the ratio 1 : 2 : 3. After 13

of the

work is completed, one of them has to be withdrawn due to budget constraint. Their wages per dayare in the ratio 3 : 5 : 6. The number of days in which the remaining two persons can complete thework (at optimal cost) is:a. 18 b. 20 c. 15 d. 12 (JMET 2005)

13. The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halvedwhile the length is doubled, then the total area of the four walls of the room willa. remain the same b. decrease by 13.64%c. decrease by 15% d. decrease by 18.75%e. decrease by 30%

14. If a, b, c, d are positive quantities such that a2 = b3 = c5 = d6 then logd (abc) equalsa. 5.8 b. 6.0 c. 6.2 d. 6.4 (ATM)

15. The length, breadth and height of a room are in the ratio 3 : 2 : 1. If the breadth and height are halvedwhile the length is doubled, then the total area of the four walls of the room will(a) remain the same (b) decrease by 13.64% (c) decrease by 15%(d) decrease by 18.75% (e) decrease by 30%


Directions for question 16: Each question is followed by two statements A and B. Indicate your responsebased on the following directives.Mark (a) if the questions can be answered using A alone but not using B alone.Mark (b) if the question can be answered using B alone but not using A alone.Mark (c) if the question can be answered using A and B together, but not using either A or B alone.Mark (d) if the question cannot be answered even using A and B together.

16. The average weight of a class of 100 students is 45 kg. The class consists of two sections, I and II,

each with 50 students. The average weight, IW , of Section I is smaller than the average weight IIW ,

of the Section II. If the heaviest student say Deepak, of section II is moved to Section I, and thelightest student, say Poonam, of Section I is moved to Section II, then the average weights of the

two sections are switched, i.e., the average weight of Section I becomes IIW and that of Section II

becomes IW . What is the weight of Poonam?

A: II IW – W 1.0= .

B: Moving Deepak from Section II to I (without any move I to II) makes the average weights of the two sections equal.

17. In a cricket match, Team A scored 232 runs runs without losing a wicket. The score consisted ofbyes, wides and runs scored by two opening batsmen: Ram and Shyam. The runs scored by thetwo batsmen are 26 times wides. There are 8 more byes than wides. If the ratio of the runs scoredby Ram and Shyam is 6 : 7, then the runs scored by Ram is(a) 88 (b) 96 (c) 102 (d) 112 (e) None of the above.

18. If 1 Japanese Yen = 0.01 US dollars, 100 Dollars = 5000 Indian Rupees (INR), how many JapaneseYens are 100 INR?(a) 20 (b) 2000 (c) 200 (d) 500

19. Aman, Baman, and Raman jointly invested Rs. 60,000 in a small services firm. They decided toshare the profits from this investment in the ratio of their investments. The firm had a very successfulfirst year and recorded profits of Rs. 1,00,000. Aman and Baman received Rs. 40,000 and Rs.25,000 as their respective share of the profit. The respective investments of Aman, Baman, andRaman were:(a) 30,000; 10,000; 20,000 (b) 21,000; 15,000; 24,000(c) 24,000; 15,000; 21,000 (d) 25,000; 12,000; 23,000

20. A trader forms a mixture of cement and sand weighing 40 kgs. In the mixture, cement and sand arein the ratio of 4 : 1 in weight terms. Later, when he adds more sand to the mixture, the new ratiobecomes 4 : 3. Given this, mark all the correct statements.(a) The second mixture formed is one and a half times heavier than the original mixture.(b) In order to arrive at the second mixture, the trader had to add a quantity of sand weighing 16 kg.(c) Had the original mixture been in the ratio of 8 : 3, the weight of the sand in the original mixture

would have been 12 kg.(d) If the trader sells 7 kg of the second mixture formed by him, and adds 11 kg of a new mixture of

cement and sand in the ratio 7 : 4 to the residual, then the new ratio of cement to sand willbecome 7 : 5.


21. Sumit works as a state contractor for PWD and supplies bitumen mix for road construction. He hastwo varieties of bituman, one at Rs. 42 per kg and the other at Rs. 25 per kg. How many kg of firstvariety must Sumit mix with 25 kg of second variety,. So that he may, on selling the mixture at 40kg, gain 25% on the outlay?(a) 30 (b) 20 (d) 25 (d) None of these

22. Ashok a master adulterator cum grosser sells haldi powder (turmeric powder), which contains fivepercent saw dust. What quantity of pure haldi should be added to two kilos of haldi (containing fivepercent saw dust) so that the proportion of saw dust becomes four percent?(a) 1 kg. (b) 23 kgs (c) 0.5 kg. (d) None of these

23. John, Mona and Gordon, three US based business partners, jointly invested in a business project to

supply nuclear fuel to India. As per their share in the investment, Gordon will receive 23

of the profits

whereas John and Mona divide the remainder equally. It is estimated that the income of John willincrease by $60 million when the rate of profit rises from 4% to 7%. What is the capital of Mona?(a) $2000 million (b) $3000 million (c) $5000 million (d) $8000 million

24. Answer the question based on the following Table.

Lactose Glucose

Saccharin

0.16 0.74

675.00

Maltose Sucrose

0.32 1.00

Relative Sweeteners of Different Substances

What is the ratio of glucose to lactose in a mixture as sweet as maltose?(a) 8 : 21 (b) 1 : 3 (c) 3 : 2 (d) 16: 9

25. A, B and C started a business by investing 1/2, 1/3rd and 1/6th of the capital respectively. After 1/3rd

of the total time, A withdrew his capital completely and after 1/4th of the total time B withdrew hiscapital. C kept his capital for the full period. The ratio in which total profit is to be divided amongstthe partners is(a) 1:2:1 (b) 4:1:4 (c) 2:1:2 (d) 1:2:2


��

1. c In alloy A, ratio of weight of metal X to the total weight

is ( )3 3 24

3 2 5 40= =

+

Ratio of weight of X to the total weight in alloy B

= ( )5 5 25

3 5 8 40= =

+

Ratio of weight of X to the total weight in the alloy C(A and B are mixed in the ratio 1 : 4 from C)

= ( ) ( )

( )2 24 4 25

40 1 4

++

= 24 100 124 31

200 200 50+ = =

∴ Ratio of X to Y in

alloy C = ( )31 31 31

50 31 50 – 31 19= =

−

2. aL 2L

A,C meet at L, ,3 3

Income – Expenditure = Saving∴ 6x – 7y = 6z ... (1)4x – 6y = 5z ... (2)6(1) – 7(2) gives

8 x = 15z = ⇒ z 8x 15

= ... (3)

∴ Ashoke saves 8

1 0 0 5 3 .3 %1 5

× =

3. cBottle I Bo ttle II Bo ttle III

x y z

Initial dataVolume of mixture v v vAlcohol only x y zWater v – x v – y v – zAfter transfer

Alcohol1

x z2

+ 1y z

2+ 0

Total volume now1

v v2

+ 1v v

2+ 0

According to the given condition, xv

becomes 9 x

10 v

and yv

becomes 11 y10 v

.

1x z 19x2

1 10v1 v2

+⇒ =

and

1y z 11 y2

1 10 v1 v2

+=

⇒ 20x + 10z = 27x → 10z = 7x ... (1)And 20y + 10z = 33y or 10z = 13y ... (2)From equation (1) and (2), 7x = 13 y

⇒ x 13y 7

=

4. d When we have five runners as mentioned, with A, C, Band C in one direction with speeds in ratio 1 : 3 : 4 andD, E in the opposite direction.

A, B meet at L

,L,2

→

L 2LA,C meet at L, ,

3 3When, D, E move opposite to “A”,

The meeting points of A, D are L L L 2L 5L

L, , , , ,6 3 2 3 6

A, E are L L 3L L 5L 3L 7L

L, , , , , , ,8 4 8 2 8 4 8

Total distinct points 12.

Note:

0

‘O’ is the starting point and L is the length of the track.The details of the meeting points is given below interms of their distance from O in the clock wise direction.In general far the conditions given and the ratio beingn:1 where is an integer greater than 2Case:1 Same direction


The distance of the meeting point from O are

( )n – 2L 2L, ,......

n – 1 n – 1 n – 1 L i.e. (n – 1) meting points.

Case 2: Opposite directionThe distance of the meeting point O are

L 2L 3L nL, , ......

n 1 n 1 n 1 n 1+ + + + and L i.e. n + 1 meeting points.

5. c10060

5x10x9x8x7x6 =++++⇒ ⇒ 8x = 0.6

⇒ x = 0.075So the marks are 0.45, 0.525, 0.6, 0.675 and 0.75.Number of times the marks exceed 50% is 4.

6. b Let the number of students in classes X, Y and Z be a,b and c respectively. ThenTotal of X = 83aTotal of Y = 76bTotal of Z = 85c

And 79ba

b76a83 =++

, i.e. 4a = 3b

Also 81cb

c85b76 =++

, i.e. 4c = 5b

Hence, b = a34

, c = 5 5 4 5

b a a4 4 3 3

= × =

Average of X, Y and Z = cba

c85b76a83++

++

=

4 583a 76 a 85 a

3 34 5

a a a3 3

+ × + ×

+ + = 5.81

12978 =

For questions 7 and 8:Let for Raja allowed luggage be A and excess luggage be EHence for Praja his luggage must be A + 2E . If all luggagebelongs to one.(A + 3E) is the excess.E corresponds to Rs. 1200.Hence, A must correspond to (5400 - 3600) = Rs. 1800If E = 2x; A = 3xSo total weight = 2(A) + 3E = 12xOr x = 5Hence, Praja's luggage weight = 7x = 35 kg

Alternate method:Let, Raja = x kg Free allowance = F kgPraja = (60 - x) kg

According to question(x - F)V = 1200 …(1){v = rate of levy on excess luggage}(60 - x - F)V = 2400 …(2)(60 - F)V = 5400 …(3)Divide (2) equation by (1) equation:

− − =−

60 x F2

x F60 – x – F = 2x – 2F3x - F = 60 …(4)Divide (3) equation by (1) equation

− =−

60 F4.5

x F60 - F = 4.5x - 4.5F4.5x - 3.5F = 60 …(5)Multiply (4) by 1.54.5x - 1.5F = 904.5x - 3.5F = 60______________ 2F = 30 F = 15Put F in (4)th equation3x = 75 ⇒ x = 25

7. d Praja has 35 kg luggage

8. 15 kg (correct answer not present among options)

9. c The data is tabulated belowMilk Water

Initial Quantity 6 m 6 wAfter selling 4 m 4 wAfter mixing 7 m + 3w 2 m + 6 w

∴ 7m 3w w2m 6w m

+ =+

⇒ 7m2 + 3wm = 2mw = 6w2

⇒ 7k2 + k – 6 = 0 where

mk

w=

⇒ (7k – 6) (k + 1) = 0

∴ 6

k7

=

10. b In 100 units of first alloy, there are 45 units tin, 55 unitslead and 0 units of bismuth.In 100 units of second alloy, there are 40 units tin, 50units lead and 10 units of bismuth.In 100 units of third alloy, there are 0 units tin, 70 unitslead and 30 units of bismuth.As obvious from the data that whatever percentageof bismuth present in final mixture is due to secondand third alloy only. So using alligation rule.

Quantity of third alloyQuantity of sec ond alloy


Percentage of bismuthin f inal mixture Percentage in secondalloyPercentage of bismuth in thirdalloy Percentage in f inal mixutre

−=−

15 10 130 15 3

−= =−

Therefore quantity of lead in mixture of second and

third alloy = 3 50 1 70

55%4

× + × =

As percentage of lead in first alloy is also 55%, sopercentage of lead in final mixture will be 55%.

11. a

12. d

13. e

14. c

15. e Area of 4 walls = ( )+ = 2x 3x 2x 5x

= =2x 7x

[7x]2 2

−2

2

2

7x5x

25x

=2

23x

30%10x

16. c Using A: II IW 45.5 and W 44.5= =Using B: Weight of Deepak = 70kg (Only after usingstatement A)This is sufficient to find weight of Poonam using thedata given in the question statement. Hence option (3)is correct choice.

17. b As per the statements in the question, we can writehereR = 26 W …(1)R + S + W + B = 232 …(2)B = 8 + W …(3)7R = 6S …(4)Subscription R, S, W, B represent the runs scored byRam, Shyam, Wides and Byes.Solving (1), (2), (3) and (4), we getR = 96Hence (b) is the correct option.

18. c Given that 1 Japanese Yen = 0.01 US DollarsTherefore, 100 US Dollars = 104 Japanese YenAlso, given that 100 US Dollars = 5000 Indian Rupees

Therefore, 100 Indian Rupees = 410

2005 10

=×

Japanese

YenHence, option (c) is the correct choice.

19. c Profit received by Raman = Rs.10000 – (Rs.40000 +Rs.25000) = Rs.35000Ratio of the profits received by them = 8 : 5 : 7Therefore, the money invested by them will also be inthe same ratio 8:5:7Hence, the money invested by Aman, Baman andRaman will be 24,000, 15,000 and 21,000 respectively.Hence, option (c) is the correct choice.

20. b, d Initial mixture = 40 kgS and (S): Cement (C) ratio is 1 : 4

⇒ S by weight 1

40 8 kg5

= × =

and C by weight 4

40 32 kg5

= × =

Let he mixes x kg of sand to the 40 kg mixture,

8 x 3x 16 kg

40 x 7+ = ⇒ =+

⇒ he mixed 16 kg of sand to the 40 kg mixture.Option (A): Weight of second mixture = 40 + 16 = 56 kg

which is 561.4

40= times heavier and not 1.5 times.

Hence (a) is incorrect.Option (b): Correct. x = 16 kg, as solved above.Option (c): If the original mixture was in 8 : 3 ratio, theweight of sand would have been

340 10.9 kg 12 kg

8 3× = ≠

+Hence, (c) is incorrect.

Option (d): The mixture weighs 56 kg. After selling 7kg of it, he is left with 49 kg of the mixture. In 11 kg ofnew mixture (7 : 4 ratio)

S and is 4

11 4 kg7 4

× =+

and Cement is 7

11 7 kg7 4

× =+

In the final mixture

Cement = 4

(49) 77

× + = 28 + 7 = 35 kg

Sand 3

(49) 4 25 kg7

× + =


Cement : Sand ratio = 35 725 5

=

Hence (d) is correct.In all B and D options are correct.

21. d The problem statement misses the word “per”. Thelast sentence should have had “....mixture at Rs. 40per kg …”

Cost price of mixture 40

Rs.32 /kg1.25

= =

Let the required ratio be x : y

25

32

42Average cost

Resu ltant cost

X

Y

Now applying alligation, we have

42 – 32 y x 7 x x 250x

32 – 25 x y 10 25 25 7= ⇒ = = ⇒ ⇒ =

22. c Pure haldi has 0% saw dust and the adulterated samplehas 5% saw dust in it. By adding appropriate amountof pure haldi, the concentration of saw dust, in the 5%sample, can be reduced to 4 %. Applying allegation:

4%

0%

14

5%Pure ha ld i

4 2 kg (of 5% sample)1 Pure Haldi

⇒ =

⇒ Pure haldi = 0.5 Kg. Hence (c) is the correct an-swer.

23. a 3% change in rate of profit is equivalent to $60 million.Therefore, profit = $2000 million.Note : Ratio of stare of profits of John, Mora andGordon 4 : 1 : 1 respectively.

24. a Sweetness of Maltose, Lactose and Glucose is 0.32,0.16 and 0.74 respectively. So by allegation we get,therefore, we get the ratio of Glucose to lactose as8 : 21.

0 .74 0 .16

0 .32

0 .16 0 .42

25. c Effective investment of A, B and C per unit of time is

1 1 1, ,

2 4 3 3 6 12× × × i.e. 2, 1 ,2 respectively.

Hence their profit sharing ratio will be 2 : 1 : 2


Topic: Time, Speed and Distance Lecture Number: 01 Duration: 120 Minutes

Objectives:� Help students to discover what makes the “Time, Speed & Distance” important in almost all the

tests.� Help students to learn how to use ratios, proportions and percentages in solving the questions.� Help students to discover the importance of assuming convenient values and avoiding fractions.� Help students to understand how to go about the study material and what to do before coming

to the next lecture.

Step1: Introduction� Faculty is supposed to introduce him/herself if he/she has not done it ever before in the very

batch.� Talk something over the importance of TSD (some very easy questions from last few years’

actual test papers can be used as examples)� Make them aware of the importance of answer options.

� Tell them that their first aim will be to be thorough with the basic concepts and to develop a habitof getting right answers in the first attempt only (Talk about the importance of a good accuracyrate)


Step 2: Fundamentals - 11. Proportionality between Time, Speed and Distance2. Relative Speed3. Trains crossing

Step 3: Fundamentals - 24. Boats and Streams5. Average Speed6. Games and Races

Step 4: Class ExerciseGive them sufficient time and if possible discuss few problems from class exercise at the end of thelecture. Tell them to get the doubts solved with the faculty available at the center before the nextlecture.NOTE: Make necessary announcements..


QA Exercise 9 - Time Speed & Distance: 1


1. Proportionality between T, S, and D (20 mins)Start by stating that speed is the rate at which distance is covered i.e. S = D/T and thus units of speedis kmph or m/s. Without wasting much time explain the conversion between the units.Immediately move on to the proportionality relation between T, S and D. Tell them that the proportionalitywill be very helpful in solving problems orally.A train leaves Calcutta for Mumbai, a distance of 1600 kms and at the same time a train leavesMumbai to Calcutta. The trains meet at Nagpur which is at a distance of 700 kms from Mumbai. Whatis the ratio of the speeds of the train from Calcutta and from Mumbai?Point out that if two persons leave at the same time and meet, the time they have been travelling till themeeting is same for both and hence distance is directly proportional to speed or vice-versa.Speed is inversely proportional to time over same distance. Thus if I travel at 5/6th of my speed and amlate by 8 minutes, what is the original and new time taken? Explain using equations and then makesure you explain how to solve this orally. The ratio of speeds is 5 : 6 and of times will be 6 : 5 anddifference give is 8. Thus time has to be 48 and 40 minutes. Give the students a couple of more suchoral problems. Also take the following problem after a couple of examples : If I go to office at 30 kmph,I am late by 10 minutes and if I go at 40 kmph, I am early by 5 minutes. At what speed should I go tothe reach on time. The first thing to identify in this problem is that speeds are given and hence I knowthe ratio of speeds. Thus the problem boils down to, “travelling at 4/3rd of my speed, I save 15 minutes”.Thus ratio of original and new time is 4 : 3 and since difference is 15 minutes, they are 60 minutes and45 minutes. At 60 minutes I am late by 10 minutes so I should take 50 minutes to reach on time andagain since ratio of times is 60 : 50 : i.e. 6 : 5, speed should be 6/5 of 30 kmph to reach on time.Take the problem of train meeting with an accident and hence arriving late by 15 mins. Had theaccident occurred 30 kms further, the train would have been late by just 7 minutes. What is thedistance between the site of accident and the destination? This problem boils down to “over 30 kms,travelling at 5/6th, I am late by 8 minutes”. Thus time taken at original speed is 40 minutes to cover 30kms and original speed is 45 kmph. Next from site of accident to destination, travelling at 5/6th speed,train is late by 15 minutes. Thus time taken at original speed from site of accident to destination is 75minutes. Since we know original speed and time taken we can calculate the distance.[Q. 8, 10, 13]CAT 2001: Three runners A, B and C run a race, with the runner A finishing 12 meters ahead of runnerB and 18 meters ahead of runner C, while runner B finishes 8 meters ahead of runner C. Each runnertravels the entire distance at a constant speed. What was the length of the race?a. 36 mts b. 48 mts c. 60 mts d. 72 mts

2. Relative Speed (25 mins)Its not worthwhile to waste a lot of time in going deep in relative speed and simply taking the followingexample to elucidate why one adds speed when travelling in opposite direction and subtracts speedwhen travelling in same direction is enough : Consider two friends A and B separated by 100 kms. Ifthey start walking towards each other, A @3 kmph and B @ kmph, in each hour A would travel 3 kmsand B would travel 2 km and with respect to each other in each hour the distance between them willdiminish by 5 km i.e. a relative speed of 5 kmph. Thus to meet they would take 100/5 = 20 hours. Nowafter meeting if they continue walking, every hour they would introduce a distance of 5 kms betweenthem. Thus in opposite direction (whether towards each other or away from each other) their relativespeed will be 5 kmph i.e. addition of speeds.


Explain the same that if A starts chasing B, every hour though he covers a distance of 3 kms, he wouldbe able to “catch up” just 1 kmph with B as B would also move away by 2 kms. Thus with respect toB, A speed is just 1 kmph i.e. difference of speeds.[Q. 1, 9]

CAT 1999: Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 AM and travels at50 kmph per hour towards Baroda situated 100 kms away. At 7:00 AM Howrah-Ahmedabad expressleaves Baroda towards Ahmedabad and travels at 40 kmph. At 7:30 AM Mr. Shah, the traffic controllerat Baroda realises that both the trains are running on the same track. How much time does he have toavert a head-on collision between the trains?a. 15 min b. 20 min c. 25 min d. 30 min

CAT 2001: A train X departs from station A at 11:00 am for station B, which is 180 km away. Anothertrain Y departs from station B at 11:00 am for station A. Train X travels of an average speed of 70 kmphand does not stop anywhere until it reaches station B. Train Y travels at an average speed of 50 kmph,but has to stop for 15 minutes at station C, which is 60 kms away from station B enroute to station A.Ignoring the lengths of the trains, what is the distance, to the nearest km, from station A to the pointwhere the trains cross each other?a. 112 b. 118 c. 120 d. None of these

CAT 2002: A train is approaching a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8of the distance AB measured from entrance A. When the train whistles the cat runs. If the CAT movesto the entrance of the tunnel, A, the train catches the cat exactly at the entrance. If the cat moves tothe exit, B, the train catches the cat at exactly the exit. The speed of the train is greater than thespeed of the cat by what order?a. 3:1 b. 4:1 c. 5:1 d. None of these

3. Trains crossing (15 mins)Again, a topic where you should not spend too much time. DO NOT start with train crossing pole andother types. Directly start with train crossing a platform. Explain to them that to find the time taken tocross a platform, the time starts when the engine just enters the platform. The engine (& the train)covers the entire length of platform. But the time des not end when engine reaches the other end of theplatform. The train has to further travel such that the guard crosses the platform. For this train will haveto travel a distance equal to its length. Thus the time taken for anything crossing any other thing is

1 2L L.

Relative Speed

+

Thus if it were not a platform but another moving train, we would take relative speed. In case of platformspeed of platform is 0 and hence just speed of train will be taken. If instead of a platform, it were a poleor a man, its length would be negligible and would be taken as 0.Ask students to solve the problems of trains crossing from the exercise of TSD. Give them the problemnumbers to solve and clear any doubts if any. Not worthwhile to take any problem from your end.[Q. 4, 5]


4. Boats and Streams (15 mins)In such problems explain what downstream and what upstream means. Give D = B + S andU = B – S and also that B = (D + U)/2 and S = (D – U)/2.

Problem 1: A man travels downstream for 5 hours and again upstream for 5 hours. Yet it is at adistance of 2 kms from the place it started. What is the speed of stream? Explain the solution using anequation but make sure you explain the following : Think of this problem as a person in a moving train.If he walks for five minutes in the direction the train is moving and then reverses direction and againwalks for 5 minutes, he would come back to his original position in the train. However if he (and thetrain) was at New Delhi when he started to walk, and now he is at Faridabad (10 kms away from Delhi),is it not obvious that in the 10 minutes the train has taken him from New Delhi to Faridabad and speedof train is 1 km/min. Thus in the above problem also had the stream been stationary, after rowing 5hours in either direction he would have come back to the original spot. But he is away from the originalspot by 2 kms means the stream is moving and has taken him 2 kms downstream in 10 hours.Problem 2: Two men are at the same point in a stream and start rowing (at different speeds) awayfrom each other, one upstream and other downstream. They row for 4 hours and then reverse theirdirection. Now how long will it take for them to meet? Same as the above problem, as both of them arein the stream, we could consider the stream to be our reference point and it would take them 4 hoursto meet.Ask the students to solve the problems on Boats and Streams from the exercise. Give them theproblem numbers and solve any doubts if they have.[Q. 6, 11]

CAT 2001: At his usual rowing speed, Rahul can travel 12 miles downstream in a certain river in sixhours less than it takes him to travel the same distance upstream. But if he could double his usualrowing speed for this 24 mile round trip, the downstream 12 miles will then take only one hour less thanthe up-stream 12 miles. What is the speed of the current in miles per hour?a. 7/3 b. 4/3 c. 5/3 d. 8/3

5. Average Speed (15 mins)Just remind them the concept of average speed being the weighted average of the speeds with weightsbeing the time driven at various speeds.[Q. 2, 3]

6. Games and Races: (10 mins)Do the problems related to games and races, the ones involving – in a game of 500, A can give B 50and B can give C 50, so how many points can A give C?

7. Exercise (20 mins)If time permits discuss any remaining problems from the exercise. Since this is a lengthy class,please solve the relevant problems of the exercise while explaining the relevant part. It will help insaving some time. If the exercise does not get completed, ask everyone to get the exercise in the nextclass on Work and discuss the problems at the start of the session of Work.

Announcement : Ask the students to solve all problems of TSD and get their fundabooks and thisexercise of TSD to the class in the QA session of work. In that session, take up all doubts of TSD.



Objectives:� Help students to discover what makes the “Time, Speed and Distance” important in almost all

the tests.� Help students to learn how to deal with questions based on Circular motion, clocks, time and

work etc..� Help students to understand how to go about the study material and what to do before coming

to the next lecture.� Help students to make use of LOGIC in place of conventional methods.


time, speed and distance.� To check there progress as well as to be assured of the above stated fact you must throw some

questions and discuss with them.� “Revision” always helps in assessing the level of the class and to identify whether they (or some


Step 2: Fundamentals - 11. Circular motion2. Clocks

Step 3: Fundamentals - 23. Concept of per day’s work4. LCM approach5. Problems of simultaneous working6. Working on alternate days7. Negative Work





Start the session by discussing any question that was left in the TSD exercise.

1. Circular Motion (25 mins with exercise problems)Start with an example: Two persons A and B are a the same point on a circular track of length 500mts. They start moving in the same direction A @ 20 m/s and B @ 50 m/s. Ask the students how longwill each take to complete a round. 25 seconds & 10 seconds. Ask when next will the two guys meetat the starting point again. Remind them that this is a problem on LCM. So the guys will meet after 50secs. Now ask them when would they meet for the first time? As soon as the race starts, B startsgaining some distance over A and as soon as the distance gained is 1 round, he has met A. For B tomeet A for the second time, B has to gain one full round again. The first time that they meet is also

given by track length

.relative speed

Next try to make them understand what is going on in a circular motion. With the same values asabove, they meet at the starting point for the first time in 50 sec. How many rounds would each haverun in 50 sec? A would have run 2 rounds and B would have run 5 rounds. Now explain to the studentsthat when A runs 2 rounds, he is at the starting point and B is 3 rounds ahead of him. So three roundsahead of him is back to the starting point itself and hence they are together. All this is common sense.Since ratio of speeds are 2 : 5, ratio of rounds covered is also 2 : 5 which explains when A ran 2 rounds,B ran 5 rounds. Now consider when A runs 2/3 rounds, B would have run 5/3 rounds. How far aheadwould B be with respect to A? (5/3 – 2/3 rounds = 1 round) i.e. wherever A is, B is one round ahead thanhim and thus B is along with A i.e. they have met. This is the first time they have met. This all shouldbe clear from the following :A runs 2 rounds, in same time B runs 5 rounds i.e. 3 rounds more. They meet for 3rd time.Multiplying everything by 2/3,A runs 4/3 rounds (1.33 rounds), in same time B runs 10/3 rounds (3.33 rounds) i.e. 2 rounds more.They meet for 2nd time.Multiplying original values by 1/3A runs 2/3rd round (0.66 rounds), in same time B runs 5/3 rounds (1.66 rounds) i.e. 1 round more. Theymeet for 1st time.Thus points to be noted :

1. When running in same direction on a circular track, the faster one meets the slower one whenever hegains one full round over the slower one

2. If the faster one has gained n rounds over the slower one, they have met n times. Hence if faster onehas run x rounds and slower one has run y rounds, they have met [x-y] times where [ ] refers to theintegral part.Now that students are clear about what is happening in a circular motion, When running in oppositedirection matters are more easy. For them to meet ‘together’ they must cover one round. Thus if fasterone covers x rounds and in same time slower one covers y rounds, they have met [x + y] times.Ask the students to solve all problems of circular motion that appears in the exercise. Tell them thequestion numbers and then clear any doubt if they have.


CAT 2003 (Leak): In a 4000 meter race around a circular stadium having a circumference of 1000meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, forthe first time after start of the race. All the runners have the same starting point and each runnermaintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of theslowest runner, what is the time taken by the fastest runner to finish the race?a. 20 min b. 15 min c. 10 min d. 5 min[Q. 1, 6, 7]

2. Clocks (20 mins)DO NOT mention that you are going to do clocks. Just keep going with circular motion. If on a circulartrack of length 360 mts there are two runners who start from the same place in same direction but onewith a speed of 6 mts/min and other with a speed of 0.5 mts/min. After how much time will they meet

again? After circular motion this should be very easy 3605.5

=5

6511

min. If initially they don’t start from

same position but faster one is 120 mts behind the slower one, when would they meet? When thefaster one gains 120 mts over the slower one, they would meet and thus they will meet in 120/5.5 i.e.

in 9

2111

minutes.

Now mention to them that this is exactly what happens in a clock. The track length is of 360° and thespeed of the minute hand is 6°/min and that of hour hand is 0.5°/min. At 12 O’clock they are together

and after this they will be together every 5

6511

minutes. Also the second problem we did was after 4

pm, in how much time will the hour hand and minute hand be together and the answer is after 9

2111

minutes.IN 12 hours how many times will the hour and minute hand be opposite to each other? Consider thestart at 12 O’clock. The hands are together. They will be 180° apart after 180/5.5 = 360/11 minutes.Now onwards they will be opposite to each other only when minutes hands gains 360° over hour handi.e. in 720/11 minutes. From now in 12 hours this would happen (12 × 60)/(720/11) times i.e. 11 times.Thus in any interval of 12 hours the hour hand and minute hand will be opposite to each other 11 times.DO NOT do any tougher problem of clocks. Do not spend a lot of time on Clocks. In CAT rarely has aproblem related to clocks come.[Q. 2, 11, 12 and 13]

3. Concept of per day’s work (10 mins)Directly start with an example. If A can build a wall independently in 10 days and if B can build thesame wall independently in 12 days, in how many days can they build the wall if they worksimultaneously? Explain to them that if A can build a wall in 10 days, in 1 day he builds 1/10th of thewall. Please explain to the students that the work done is additive and not the number of days.Similarly if a person can do 1/xth part of a work in 1 day, he will complete the full work (whole part) in 1/(1/x) = x days.


4. LCM approach (15 mins)Take the same problem again and explain it the LCM way. The LCM way does not reduce any amountof calculations, each and every calculation of ‘per days work approach’ is also done in the LCMapproach, however fractions are eliminated.

5. Problems of simultaneous working (15 mins)Take just one or two problems of simultaneous working : A, B and C can independently finish a work in10, 12 and 15 days respectively. A starts the work and is joined by B after 2 days and then the two arejoined by C again after 3 more days. The work got over in x days. However 6 days before the actualcompletion of the work A had left the group and 2 days before completion of work B left C. What is thevalue of x? Explain to them that any story can be build but essentially the format remains the same.

Thus, A, B, C work for x-6, x-2-2, x-5 days respectively. Thusx 6 x 4 x 5

1.10 12 15− − −+ + = Explain why the

RHS is 1 if any one is not clear. Also solve it using LCM approach.If A is twice as fast as B is and if B is twice as fast as C is and if they all together can finish a work in10 days, in how many days will A complete the work independently? In problems of this type, explainto students that C can be substituted by ½ B and that B can be substituted by ½ A. Thus A & B and

C is nothing but 1 1

12 4

+ + A i.e. 7/4 A. Thus 7/4 A can finish work in 10 days and A can finish it in 7/

4 × 10 = 17.5 days.[Q 3, 4, 5, 8, 9, 10]

CAT 2001: A can complete a piece of work in 4 days. B takes double the time taken by A. C takesdouble that of B and D takes double that of C to complete the same task. They are paired in groups oftwo each. One pair takes two-third the time needed by the second pair to complete the work. Which isthe first pair?a. A, B b. A, C c. B, C d. A, DExplain to students that pipes and cistern are also the same concept but a pipe may be an outlet pipewhich is equal to negative work. Ask the students to solve the problems of simultaneous working(including pipes) from the exercise (you will have to give them the question numbers from the exercise)and then clear any doubts if they have.

6. Working on alternate days (10 mins)A can build an entire wall in 30 days and B can demolish and entire wall in 40 days. If they work onalternate days with A starting on the first day, after how many days will the wall be built for the firsttime? It would be obvious to students that in 2 days 1/30 – 1/40 = 1/120th of the wall will be built.Explain to the students why 120 × 2 = 240 will not be the answer. Just leave it by saying that on the120th day, B will be working which means that he breaks the wall and it gets completed! Ask thestudents that if in 2 days 1/120thof the wall is built, in 116/120th of the wall will be built in how manydays? (232 days). Now on 233rd day 116/120 + 1/30 = 1 i.e. full wall is built. This explains why thequestion had “built for the first time”. What is crucial is for you to explain to students how to arrive atthe nearest day before completion without any iterations. All one has to do is to subtract from 1 thework done by A in one day i.e. 1 – 1/30 = 29/30 = 116/120 because we have to express it withdenominator as 120. Now whenever 116/120th of work is done, it will just take 1 more day.


If things are not clear give the example of a frog climbing a pole of 20 mts. In one minute he climbs upby 3 mts and in next minute, slips down by 2 mts. In how many minutes will the frog reach the top?Ask students to solve the problem on alternate working in the exercise.

7. Negative work (5 mins)There is a pole whose height is 100 metres. A monkey is trying to climb up the pole. But it so happensthat in the first minute the monkey climbs up by 5 metres but in next minute slips back by 3 metresand so on. In how much time the monkey will be at the top of the pole for the first time?

8. Exercise (20 mins)Ask students to solve the remaining problems from the exercise and clear the doubts that they have.



Objectives:� Help students to identify the effective way solving a test paper as well as what they have learnt

in the first two lectures of time, speed and distance.� Help students to identify the requirement of periodic Self-assessment.� Help students to boost their confidence up and to identify which areas they need to focus more

upon.NOTE: Testing process is not to compare the students with each other rather it is to judge one’sown weaknesses and strengths.

Step1: Review Test� Give them stipulated time but just before that make it very clear that they have to try to solve

“RIGHT” questions with good accuracy. (To be selective is to be effective)� Tell them to focus on the accuracy first and then on speed.

Step 2: Discussion� Give the students some time to calculate their own scores and to be aware of their own perfor-

mance and mistakes.� Don’t discuss about what should have been their score or what one was expected to do.� Tell them how to select the questions, how to proceed, how to scan the question paper and how

to decide which questions are to be attempted in the first go, second go etc and which ques-tions are not to be attempted at all.

� Solve as many questions as possible with “shortcut/alternative” methods and teach them tomake use of answer options.

Step 3: Doubt Solving- Solve the doubts.- Always solve the questions with fastest method first, tell them alternative ways later on.NOTE: Make necessary announcements.



1. Revision (20 mins)

Quickly revise all the concepts discussed in first two lectures of Time, Speed and Distance.[Q. 1, 2, 3, 4, 5, 6, 7 and 8]

2. Miscellaneous Questions (30 mins)

Explain how to solve the questions of Time, Speed and Distance applying logic.[Q. 9, 10, 11, 12, 13, 14 and 15]

3. Review Test (25 mins)

Tell them to solve all the question of the class exercise in the stipulated time limit.

4. Discussion (45 mins)

Discuss the questions of Review Test. Also discuss the questions from the sheets of last twolectures of TSD.


1. Two motorist drive on two concentric tracks separated by 5 m. Their speeds are in the ratio 3 : 8. Ifthey start racing on their respective tracks they take the same time to complete one round. If theyinterchange tracks, the slower motorist takes 5sec to complete one round around his track. Findthe speed of the faster motorist

a. 12 m/sπ b. 16 π m/s c. 4.57 π m/s d. 13 π m/s

2. Tripti and Deepti walk back and forth between the town hall and the county station at respectivespeeds of 2 kmph and 3 kmph. They start simultaneously - Tripti from the town hall and Deepti fromthe county station .If they croses each other for the first time 60 minutes from the start, at whatdistance from the county station will they cross each other fro the fifth time?a. 3 km b. 4 km c. 2.5 km d. None of these

Directions for questions 3 to 5: Answer the questions on the basis of the information given below.New Heaven Couriers (NHC) sends a person on a motorcycle everyday to the airport to collect the courier.This person reaches the airport exactly when the plane lands at the scheduled time. One day BHC cameto know that the flight would be late by 30 minutes. So, the NHC person also started from his office 50minutes later that the usual time. However, the plane landed earlier than was anticipated by the NHC. So,the airport authorities immediately dispatched the courier through a cyclist to BHC. The cyclist met themotorcyclist of BHC after travelling for exactly 10 minutes and handed over the courier to the motorcyclist.As a result, the BHC person returned to his office 6 minutes earlier than excepted. Assume that themotorcyclist and the cyclist travel at their own uniform speeds.

3. What is the delay in the landing of the plane when compared to the normal schedule?a. 10 minutes b. 15 minutes c. 20 minutes d. 17 minutes

4. If the cyclist covered a distance of 2 km before he met the motorcyclist, What is the speed of themotorcyclist?a. 60 kmph b. 45 kmph c. 40 kmph d. Cannot be determined

5. If on the particular day, when the cyclist covered 2 km before he met the motorcyclist, the motocyclistspent a total of 30 minutes to complete his journey, what is the distance from NHC’s office to theairport?a. 20 km b. 15 km c. 12 km d. Cannot be determined

Question Bank: Time, Speed & Distance


Directions for questions 6 and 7: Answer the questions based on the following information.Rajiv reaches city B from city A in 4 hours, driving at speed of 35 kmph for the first two hour and at 45 kmphfor the next two hours. Aditi follows the same route, but drives at three different speeds: 30, 40 and 50kmph, covering an equal distance in each speed segment. The two cars are similar with petrol consumptioncharacteristics (km per litre) shown in the figure below. (CAT - 1999)

16 16

24

30 40 50

M ileagekm per litre

Speed km per hour

6. The quantity of petrol consumed by Aditi for the journey isa. 8.3 l b. 8.6 l c. 8.9 l d. 9.2 l

7. Zoheb would like to drive Aditi’s car over the same route from A to B and minimize the petrolconsumption for the trip. What is the quantity of petrol required by him?a. 6.67 l b. 7 l c. 6.33 l d. 6.0 l

8. A sprinter starts running on a circular path of radius r metres. Her average speed (in metres/minute)

is πr during the first 30 seconds, r

2π

during next one minute, r

4π

during next 2 minutes, r

8π

during

next 4 minutes, and so on. What is the ratio of the time taken for the nth round to that for theprevious round?a. 4 b. 8 c. 16 d. 32 (CAT - 2004)

9. Laxman and Bharat decide to go from Agra to Delhi for watching a cricket match and board twodifferent trains for that purpose. While Laxman takes the first train the leaves for Delhi, Bharatdecides to wait for some time and take a faster train. On the way, Laxman sitting by the window-seat noticed that the train boarded by Bharat crossed him in 12 seconds. Now the faster train cantravel 180 km in three hours, while the slower train takes twice as much time to do it. Given this,mark all the correct options.a. If the faster train has taken 30 seconds to cross the entire length of the slower train, the difference between the lengths of the two trains is 50 m.b. If the faster train had been running twice as much faster, it would have taken 10 seconds to overtake the slower train.


c. Had the faster train taken 24 seconds to cross the entire length of the slower train, the length of the slower train would have been 100 m.d. If the slower train had been running at one and a half times of its current speed, the faster train would have taken 24 seconds to overtake Laxman. (IIFT 2006)

10. A contractor takes up an assignment that 20 men can complete in 10 days. The same assignmentcould be finished by 15 women in 20 days. The contractor decides to employ 10 men and 10 womenfor the project. Given this, mark all the correct options.a. If the wage rate for men and women are Rs. 50 and Rs. 45 respectively, the total wage bill for the project will be Rs. 11,400.b. If the wage rate for men and women are Rs. 45 and Rs. 40 respectively, the total wage bill for the project will be Rs. 10,200.c. If the wage rate for men and women are equal at Rs. 40, the total wage bill for the project will be Rs. 9,100.d. If the contractor decides to employ 20 men and 30 women for the project and the wage rate for men and women are Rs. 40 and Rs. 35 respectively, the total wage bill for the project will be Rs. 9,250. (IIFT 2006)

11. Five people A, B, C, D, E do a certain job., A, B and C together complete the job in 7.5 hrs. A, C andE together complete it in 5 hours. A, C and D together complete it in 6 hours B, D, E togethercomplete it in 4 hours. If all the five people work together, then how much time will be required?a. 3.5 hour b. 3 hours c. 2.5 hours d. 1.5 hours

12. John takes twice as long as Joe to build a sand castle. Jill takes thrice as Joan to build the samesand castle. Joan, being the sister of John takes the same amount of time to build the sand castle.If all four work together, the castle can be built in 1 day. How long will Joan and Joe working togethertake to build the castle?

a. 136

b. 139

c. 1311

d. 1312

13. Shyama and Vyom walk up an escalator (moving stairway). The escalator moves at a constantspeed. Shyama takes three steps for every two of Vyom’s steps. Shyama gets to the top of theescalator after having taken 25 steps, while Vyom (because his slower pace lets the escalator do alittle more of the work) takes only 20 steps to reach the top. If the escalator were turned off, howmany steps would they have to take to walk up?a. 40 b. 50 c. 60 d. 80 (CAT - 2001)

Directions for question 14: Answer question on the basis of the information given below:The cost of fuel for running the engine of an army tank is proportional to the square of the speed and Rs. 64/hour for a speed of 16 kmph. Other costs amount to Rs. 400/hour. the tank has to make a journey of 400km at a constant speed.

14. The most economial speed for this Journey isa. 20 kmph b 320 kmph c. 35 kmph d. 40 kmph (JMET - 2006)


15. The duration of a rail journey varies directly as the distance covered and inversely as the speed ofthe train. The speed varies directly as the square root of the quantity of coal required per km andinversely as the number of carriages in the train. In a journey of 25 km, in half an hour with 18carriages, 100 kg of coal is required. How much of coal will be required in a journey of 21 km in 28minutes with 16 carriages?a. 40 kg b. 64 kg c. 54 kg d. 81 kg

Directions for question 16: Answer question on the basis of the information given below:A tortoise on a beach lays eggs and goes back to the sea where its mate is waiting. One way to reach themate is going down 12 km on a straight line perpendicular to the sea, turning 90° and swimming for 5 kmon a straight line. On its way down to the sea, the tortoise can cut the water at any point. The speed of thetortoise on land is 1 kmph and in water is 2 kmph.

16. If the tortoise takes the described route, at what distance from the point where it had laid the eggswill the tortoise cut the water, if the total time taken is 12 hrs?a. 7 km b. 8 km c. 5 km d. 4 km

17. A taxi is traveling at a uniform speed. The driver sees a milestone showing a 2-digit number. aftertraveling for an hour the driver sees another milestone with the digits in reverse order. After anotherhour the driver sees another milestone containing three digits and with the same two digits included.The average speed of the taxi isa. 45 kmph b. 36 kmph c. 54 kmph d. 42 kmph

18. A train after traveling 2 hours is detain for 12

an hour after which it proceeds at 45

its former speed

and arrives 1

12

hour late. If the detention had taken place 25 km further on, the train would have

arrived 15 minutes earlier that it did before. The speed of the train and the distance traveled is:a. 25 kmph, 200 km b. 20 kmph, 150 kmc. 20 kmph, 200 km d. 25 kmph, 150 km

19. Two cars X and Y start from two point A and B towards each other simultaneously. They meet for thefirst 40 km from B. After meeting they exchange their speeds as well as directions and proceed totheir respective starting points. On reaching their starting points, they turn back with the samespeeds and meet at a point 20 km from a. Find the distance between A and B.a. 130 km b. 100 km c. 120 km d. 110 km

Directions for question 20: Answer question on the basis of the information given below:A group of 10 workers can plough a field in 20 days. This group starts the work and after every two days,one additional worker joins the group. Assume that the capacity of each worker is the same.

20. Find the time taken top complete the work.

a. 1

12 days7

b. 1

13 days17

c. 1

14 days17

d. 1

15 days17


21. In a town, the local trains start from and arrive at the station at fixed intervals and run at a uniformspeed. A boy was walking down the railway track at a certain speed. Every half an hour, a local trainovertook him. Every 20 minutes, a local train passed him in the opposite direction. Find the timeinterval between a local train passing a certain point on the railway route and the immediately nextlocal train passing that point in the same direction.a. 26 minutes b. 24 minutes c. 27 minutes d. 25 minutes

22. A swimmer jumps from a bridge over a canal and swims 1 kilometer up stream. after that firstkilometer, he passes a floating cork. He continues swimming for half an hour and then turns aroundand swims back to the bridge. If he is swimmer and the cork reach the bridge at the same time. Theswimming with constant speed, how fast does the water in the canal flow?a. 1 kmph b. 2 kmph c. 0.5 kmph d. 1.3 kmph

Directions for question 23: Answer question on the basis of the information given below:Yogi walked down a descending escalator (a moving staircase) and took 40 steps to reach the bottomBhogi started simultaneously from the bottom, taking two steps fro every one step taken by Yogi. The timetaken by Yogi to reach the bottom from the top is same as the time taken by Bhogi to reach the top fromthe bottom.

23. How many steps more than Yogi did Bhogi take before they crossed each other on the escalator?a. 20 b. 40 c. 3 d. Cannot be determined

24. The speed of an engine is 150 kmph. By connecting coaches to the engine its speed decraeses.the decreases in speed is given by the function d = x2 + 3x – 2 (kmph), their x is the number ofcoaches connected. Find the minimum speed with which the train can run.a. 1 kmph b. 11 kmph c. 22 kmph d. 33 kmph

25. In an 800 m race, A beats B by 40 m and c by 30 m. If they run at the same respective speeds in thenext race, then by what distance will C beat B in a 400 m race?

a. 24

4477

b. 10

1477

c. 15

577

d. None of these

26. I take 4 hours less to row down a 12 mile stream than 1 take to row up. For this 24 mile roundup, ifI double my rowing speed, I would take half an hour to row downstream than to row upstream. Findthe speed of the stream, in miles/hr.a. 6 b. 8 c. 0 d. None of these

27. Three athletes run a 4 km race. Their speeds are as 16 : 15 : 11. Assuming all the three athletescover the same distance in each lap, the positions of the other two, when the winner teaches thewinning post, will bee

Starting Point

a. 250 and 1250 m behind the first. b. 200 and 1250 m behind the first.c. 200 and 1500 m behind the first. d. None of these


28. Yash Chopra’s office is 10 km from his home, which he covers in 2 hours,. One day, as he was onhis way to office, he passed his friend’s office who immediately pointed out that he had forgotten towear his shoes. Yash Chopra therefore turned back, put on his shoes and set off the office. As thepassed his friend’s office, she again pointed out that now he had forgotten his glasses. Yash chopraagain went back, collected his glasses and reached the office 3 hours late. What is the distancebetween his home and friend’s office?a. 3.75 b. 7.35 c. 8.5 d. Insufficient data

29. Points A, B and C are at the distance of 60.55 and 56 km from point M respectively. Three people leftthose points for point M simultaneously: the first person started from point A, the second from B andthe third from C. Which the first person who arrived simultaneously. The second person, havingtraveled 40 km at the same speed as the first, stopped for an hour. The rest of the way he traveledwith a speed which is less than the speed of the third person by the same amount as the speed ofthe third is less than the first. The third person covered the whole way at a constant speed. Determinethe speeds of the first and the third person.a. 5 kmph, 1 kmph b. 3 kmph, 2 kmphc. 6 kmph, 3 kmph d. None of these

Directions for question 30: Answer question on the basis of the information given below:A tunnel measuring 4 km and 636 meters is designed specifically for two trains to pass simultaneously inthe same or opposite directions. Therefore two express trains of length 400 m each, travel through thetunnel at the rate 56 kmph and 80 kmph.

30. Assuming that both the trains enter the tunnel at the same point of time (t = 0) from the two differentends, then the minimum value of ‘t’ such that both the trains have cleared the tunnel will be.a. 2 min 45 sec b. 2 min 24 secc. 2 min 36 sec d. None of these

31. Assuming that the guard starts walking immediately as he enters the tunnel and the train is travelingat the speed of 80 kmph, what is the time by taken a guard walking at the rate of 5 kmph along thecorridor towards the engine to clear the tunnel?a. 3 min 46 sec b. 3 min 53 secc. 4 min d. None of these

32. Dr. Desai always goes walking to the clinic and takes the same time to come back. One day henoticed something. When he left home, the hour hand and the minute hand were exactly oppositeto each other and when he reached the clinic, they were together. Similarly, when he left the clinic,the hour hand and the minute hand were together and when he reached home, they were exactlyopposite to each other. The least time Dr. Desai takes to reach home from the clinic isa. 32 minutes 43 seconds b. 35 minutesc. 30 minute d. 30 minutes 45 seconds


33. A 14.4 kg. gas cylinder runs for 104 hours when the smaller burner on the gas stove is fully openedwile it runs for 80 hours when the larger burner on the gas stove is fully opened. Which of thesevalues are the closest to the percentage difference in the usage of gas per hour, of the smallerburner over the larger burner?(a) 26.23% (b) 30% (c) 32.23% (d) 23.07%

34. A cyclist drove one kilometer, with the wind in his back, in three minutes and drove the same wayback, against the wind in four minutes. If we assume that the cyclist always puts constant force onthe pedals, how much time would it take him to drive one kilometer without wind?

(a) 132 (b) 3

73 (c) 373 (d) 7

123

35. Two people, A and B, need to cross a bridge. A can cross the bridge in 10 minutes and B can crossin 5 minutes. There is also a bicycle available and any person can cross the bride in 1 minute withthe bicycle. Then the shortest time that both men can get across the bridge is nearest toa. 7.23 minutes b. 3.77 minutes c. 2.72 minutes d. 2.12 minutes (ATM)

36. Arun, Barun and Kiranmala start from the same place and travel in the same direction at speedsof 30, 40 and 60 km per hour respectively. Barun starts two hours after Arun. If Barun andKiranmala overtake Arun at the same instant, how many hours after Arun did Kiranmala start?(a) 3 (b) 3.5 (c) 4 (d) 4.5 (e) 5

Directions for Questions 37 and 38: Answer the following questions based on the information given below:Cities A and B are in different time zones. A is located 3000 km east of B. The table below describes theschedule of an airline operating non-stop flights between A and B. All the times indicated are local and onthe same day.

Departure Arrival

City Time City Time

B 8:00 am A 3:00 pm

A 4:00 pm B 8:00 pm

Assume that planes cruise at the same speed in both directions. However, the effective speed is influencedby a steady wind blowing from east to west at 50 km per hour.

37. What is the time difference between A and B?(a) 1 hour and 30 minutes (b) 2 hours (c) 2 hours and 30 minutes(d) 1 hour (e) Cannot be determined

38. What is the plane’s cruising speed in km per hour?(a) 700 (b) 550 (c) 600(d) 500 (e) Cannot be determined.


39. Rahim plans to drive from city A to station C, at the speed of 70 km per hour, to catch a train arrivingthere from B. He must reach C at least 15 minutes before the arrival of the train. The train leaves B,located 500 km south of A, at 8:00 am and travels at a speed of 50 km per hour. It is known that Cis located between west and northwest of B, with BC at 60° to AB. Also, C is located between southand southwest of A with AC at 30° to AB. The latest time by which Rahim must leave A and stillcatch the train is closest to(a) 6 : 15 am (b) 6 : 30 am (c) 6 :45 am (d) 7 : 00 am (e) 7 : 15 am

40. Rajiv is a student in a business school. After every test he calculates his cumulative average. QTand OB were his last two tests.83 marks in QT increased his average by 2. 75 marks in OBfurther increased his average by 1. Reasoning is the next test, if he gets 51 in Reasoning, hisaverage will be ___?(a) 59 (b) 60 (c) 61 (d) 62 (e) 63

41. A train left station X at A hour B minutes. It reached station Y at B hour C minutes on the same day,after travelling for C hour A minutes (clock shows time from 0 hours to 24 hours). Number of possiblevalue (s) of A is ___.(a) 3 (b) 2 (c) 1 (d) 0 (e) None of the above.

42. Rajesh walks to and fro to a shopping mall. He spends 30 minutes shopping. If he walks at speedof 10 km an hour, he returns to home at 19.00 hours. If he walks at 15 km an hour, he returns tohome at 18.30 hours. How fast must he walk in order to return at 18.15 hours?(a) 17.5 km/hour (b) 17.5 km/hour (c) 18 km/hour(d) 19 km/hour (e) None of the above.

43. Sangeeta and Swati bought two wristwatches from Jamshedpur Electronics at 11.40 A.M. IST.After purchasing they found that when 60 minutes elapses on a correct clock (IST), Sangeeta’swristwatch registers 62 minutes whereas Swati’s wristwatch registers 56 minutes. Later in theday Sangeeta’s wristwatch reads 10 P.M., then the time on Swati’s wristwatch is:(a) 8:40 PM (b) 9:00 PM (c) 9:20 PM (d) 9:40 PM (e) Cannot be calculated.

44. An artist has completed on fourth of a rectangular oil paining. When he will paint another 100 squarecentimeters of the painting, he would complete three quarters of the painting. If the height of the oilpainting is 10 centimeters, determine the length (in centimeters) of t he oil painting(a) 15 (b) 20 (c) 10 (d) 25

45. Pavan builds an overhead tank in his house, which has three taps attached to it. While the first tapcan fill the tank in 12 hours, the second one takes one and a half times more than the first one to fillit completely. A third tap is attached to the tank which empties it in 36 hours. Now one day, in orderto fill the tank, Pavan opens the first tap and after two hours opens the second tap as well. However;at the end of the sixth hour, he realizes that the third tap has been kept open right from he beginningand promptly closes it. What will be the total time required to fill the tank?(a) 8 hours 48 minutes (b) 9 hours 12 minutes(c) 9 hours 36 minutes (d) 8 hours 30 minutes


46. The digging work of the DMRC on the Adohini-Andheriamore stretch requires Twenty-four men tocomplete the work in sixteen days. As a part of the task if DMRC were to hire Thirty-two women,they can complete the same work in twenty-four days. Sixteen men and sixteen women startedworking and worked for twelve days. Due to time bound schedule the work had to be completed inremaining 2 days, for which how many more men are to be employed?(a) 48 (b) 24 (c) 36 (d) 16

47. The Ghaziabad-Hapur-Merut EMU and the Meerut-Hapur-Ghaziabad EMU start at the same timefrom Gahaziabad and Meerut and proceed towards each other at 16 km/hr and 21 km/hr respec-tively. When they meet, it is found that one train has traveled 60 km more than the other. Thedistance between two stations is:(a) 445 km (b) 444 km (c) 440 km (d) 450 km

48. A boat goes 30 km. upstream and 44 km. downstream in 10 hours. In 13 hours, it can go 40 kmupstream and 55 km down-stream. The speed of the boat in still water is:(a) 3 km/hour (b) 4km/hour (c) 8km/hour (d) None of the above

49. Delhi Metro Corporation engaged 25,000 workers to complete the project of IP state to DwarkaMetro Line in 4 years. At the end of the first year 10% workers were shifted to the other projectsof Delhi Metro. At the end of second year again 5% workers were reduced. However, the numberof workers increased by 10% at the end of the third year to complete the above project in time.What was the size of work force during the fourth year?(a) 23145 (b) 23131 (c) 23512 (d) 23513

50. A flight of Jet Airways from Delhi to Mumbai has an average speed of 700 kilometres per hourwithout any stoppage, whereas a flight of Kingfisher from Delhi to Mumbai has an average speedof 560 kilometers per hour with stoppage at Baroda. What is the average stoppage time per hourof Kingfisher flight if both the planes fly at the same speed?(a) 8 minutes (b) 12 minutes (c) 16 minutes (d) 24 minutes

51. Two sea trawlers left a sea port simultaneously in two mutually perpendicular directions. Half anhour later, the shortest distance between them was 17 km, and another 15 minutes later, one seatrawler was 10.5 km farther from the origin than the other. Find the speed of each sea trawler.(a) 16 km/h, 30 km/h(b) 18 km/h, 24 km/h(c) 20 km/h, 22 km/h(d) 18 km/h, 36 km/h


��

1. b Let the radius of the inner circular track be ‘r’ m.∴ (r + 5) m is the radius of the outer circular track.

Ratio of speeds of the cyclists is 23

.

Since they both complete one round on their respectivetracks in the same time, the slower cyclist must be onthe inner track and the faster on the outer track.

Time Dis tance

Speed=

( )2 r 2

r 10 m2 r 5 3

π∴ = ⇒ =π +

The slower cyclist takes 5 sec to complete one roundof the longer track.Let v m/s be the speed of the slower cyclist then,

( )2 10 5 2 15V 6 m/ s

5 5

π + π ×= = = π

∴ Speed of the faster cyclist = 8

6 16 m/ s3

× π = π

2. a Relative speed of Tripti with respect to Deepti= 2 + 3 kmph = 5 kmph

Distance = speed × time = 60

5 5 km60

× =

∴ the distance covered by both of them, when theycross each other for the first time is 5 km.

5 km .

2 km phTripti

3km phDeepti

A B

Let the distance covered by Vicky and Nicky whenthey cross each other for the first time be AB.For such a situation, when the ratio of Speeds = 2 : 3it can be observed that distance covered by both 07them put together, between and two consecutivemeetings is always twice the distance AB.When they cross each other fro the 2nd time, theymust have covered a distance of AB + 2ABWhen they cross each other for the 3rd time. Theymust have crossed AB + 4AB.∴ Total distance covered by them when they crosseach other for the fifth time is = AB + 4 (2AB) = 9 AB= 5 (9) = 45 km.

Time taken 4 5

9h r5

= =

∴Distance from the county station = 3 × 9 = 27 km.= 5 × 4 + 2

= 2. km from the town hall.i.e. 3 km from the county station

12 km ph

Tripti

18 km ph

Deepti

A B

P 1 P 2 P 3 P 4

(1 ) (2 )(Town Ha ll) (County s ta tion )

(5 ) (4 )

Tripti’s and Deepti’s speeds are in the ratio of 2 : 3,Dividing the total distance between A and B into 5equal parts by

( )1 2 3 4 1 1 2 2 3 2 4 4P ,P ,P ,P AP P P P P P P P B x say= = = = =we see that Tripti covers 4 parts and Deepti 6 partsbefore they meet again. Their meetings are tabulatedbelow.

2

4

4

2

Dis tance Dis tanceMeeting Meeting

Covered CoveredNumber Point

by Tripti by Deepti

1 2x 3x P

2 6x 9x P

3 10x 15x A

4 14x 21x P

5 18x 27x P

( ) 60AB 2 3 km 5 km.

60 = + =

( )23

BP 5 km 3km.5

∴ = =

Solutions for questions 3 to 5:

P R Q

N H C Airport

Let R the point where the cyclist hands over the mail to themotorcyclist.

The motorcyclist saved 6 minutes by not traveling the distanceRQ and QR (both of which are equal). Hence the motorcyclistis 3 minutes away from the airport when he met the cyclist.The cyclist has travelled for 10 minutes.This means that by the time motorcyclist met the cyclist, it was10 minutes since the flight landed.

3. d Had the motorcyclist continued towards the airport,after he met the cyclist he would have reached theairport in another 3 minutes.


When he met the cyclist, it was already 10 minutessince the plane landed, so, when he reaches theairport, it will be 13 minutes (10 +3) since the flightlanded.On that particular day, the flight was expected to bedelayed by 30 minutes and the landing was expectedto take place at the time the motorcyclist reached theairport.Since the flight landed 13 minutes prior to themotorcyclist reaching the airport, the plane landed 13minutes (30 – 13) later than normal schedule.

4. c RQ = 2 km which would have been covered by themotorcyclist in 3 minutes.

Hence, in 1 hour, he covers (2) 603

So, his speed is 40 kmph

5. c Time taken to cover PR 30

15 m in utes2

= =

Time taken for RQ = 3 minutesTime taken for the total journey from office to airport =18 minutesSpeed = 40 kmph (as per previous question)

Hence, distance PQ = 18

40 12 km60

=

6. c Distance between A and B = (35 × 2) + (45 × 2)= 160 km.Distance covered by Aditi in each speed segment

= 160

3Hence, total petrol consumed

= 160 1 160 1 160 1

8.9l3 16 3 24 3 16

× + × + × =

7. a For minimum petrol consumption, Zoheb should drive

at 40 kmph, petrol consumption = 16024

= 6.67 l.

8. c As options are independent of n Let n = 2

Time taken for first round = 1

1 2 4 7.5 minutes2

+ + + =

Time taken for second round = 8 + 16 + 32 + 64 = 120minutes

120Ratio 16

7.5= =

9.(a,c,d)Laxman takes the first train which is the slower one.Bharat takes the faster train.

Let the trains be A and B respectively.Speed of the faster train, train B =

180 km60 km/hr

3 hr=

BV 60 km/hr=

As train A takes twice the time, so AV 30 km/hr=Speed of train B, w.r.t. Laxman (when he is sitting inthe train A) is 60 – 30 = 30 km/hrLaxman observes the train B, pass by him in 12seconds. If LB were the length of the faster train then,

BL30 km/hr

12 seconds=

B30 1000

L 123600×⇒ = × m

BL 100 m⇒ =

Option a: A BL L

30 km /hr30 sec onds

+=

{LA = Length of the slower train}

A B30 1000

L L 303600×⇒ + = ×

A BL L 250 m+ =

( )AL 250 100 m⇒ = − or LA = 150 m

So, LA – LB = 50 moption (A) is correct.

Option (b) If VB = 60 × 2 = 120 km/hrVA = 30 km/hr, as before.

To overtake, train A; train B has to cover its length, LA.As we cannot determine the length of the slowertrain, we cannot find the time taken in overtake.Hence, (b) is not correct.

Option (c): VA = 30 km/hrVB = 60 km/hr

A BL L30 km/hr

24+=

( )A1000

L 100 30 243600

⇒ + = × ×

LA = 200 – 100LA = 100 m

Option (c) is correct.

Option (d): A3

V 30 45 km/hr2

= × =

BV 60 km/hr=

BL 10015 km/hr m

t t⇒ = =


100 m 3600t

15 1000×⇒ =

×= 24 secondsHence (d) is correct.

10.(a,b,d)20 men can finish the work in 10 days and 15 womencan finish it in 20 days.From here we can say that⇒ 20 × 10 M = 15 × 20 W

⇒ 2M = 3W …(i)Hence 10 Men and 10 Women together can finish thework in N days such that{N = Number of Days taken working together.}⇒ 20 × 15 = 25 × N

⇒ N = 12Option (a) Total wage for Men = 12 × 50 × 10

= Rs. 6000Total wage for Women = 12 × 10 × 45

= Rs. 5400∴ Total wage bill = Rs. 11400Hence, option (a) is correct.

Option (b) Total wage for Men = 12 × 10 × 45 = Rs. 5400

Total wage for Women = 12 × 10 × 40 = Rs. 4800

∴ Total wage bill = Rs. 10200Hence, option (b) is correct.

Option (b) Total wage for Men = 12 × 10 × 40 = Rs. 4800

Total wage for Women = 12 × 10 × 40 = Rs. 4800

∴ Total wage bill = Rs. 9600Hence, option (b) is not correct.

Option (d) If 20 Men and 30 Women are employed.Then, together they way finish the work in N dayssuch that⇒ 20 × 15 = 60 × N {� 20 M = 30 W}

⇒ N = 5Total wage for Men = 5 × 20 × 40 = Rs. 4000Total wage for Women = 5 × 30 × 35 = Rs. 5250∴ Total wage bill = Rs. 9250Hence, option (d) is correct.

11. c According to the problem

1 1 1 1A B C 7.5

+ + = ------ (a)

1 1 1 1A C E 5

+ + = ------ (b)

1 1 1 1A C D 6

+ + = -------(c)

1 1 1 1B D E 4

+ + = ------- (d)

By (a) + (b) + (c)

1 1 1 1 1 1 1 13

A C B D E 7.5 5 6 + + + + = + +

= 1 11

7.5 30+

1 1 1 1 1A C 3 2 6

⇒ + = =

Adding to (d), 1 1 1 1 1 1 1 5A B C D E 6 4 12

+ + + + = + =

∴ Time required = 2.5 hrs

12. b Let Joe take X days to build the castle. Then:

John 2x days

Joan 2x days

Jill 6x days

Thus they together will take:

1 1 1 1 131which gives x

2x 2x 6x x 6+ + + = =

Thus Joan and Joe will take

1 136 3 9

13 13

= +

13. b If Shyam takes 1 min for every 3 steps, then he takes

31

min for every step.

For 25 steps, he takes 325

min, i.e. 8.33 min.

So Vyom takes 21

min for every step.

For 20 steps, he takes 2

20min, i.e. 10 min.

Difference between their time = 1.66 min.Escalator takes 5 steps in 1.66 min and difference innumber of steps covered = 5Speed of escalator is 1 step for 0.33 min,i.e. 3 steps per minute.If escalator is moving, then Shyam takes 25 steps andescalator also takes 25 steps.Hence, total number of steps = 50.

14. d Cost of fuel is proportional to square of the speed.E = KS2

⇒ 64 = K(16)2

∴ 1K

4=


Total cost 21S t 400t

4= +

Most economical speed, checking options we get mosteconomical speed at 40 km/hr.

Total cost at 40 km/hr 21 400(40) 400

4 40 = +

= 10 × 400 + 4000= Rs.8000.

15. b If d = distance , s = speed, t = time, n = number ofcarriages, c = quantity of coal.

then t = k d n

c

× × where k is constant.

1 1 1 1

2 2 2 1

t d n cc2 30 25 18t d n c 28 21 16 100

×= ⇒ =×

2c 64 kg⇒ = .

16. a

w ater edge

eggs

12 km

5 km

x 12–x

Total time taken x 2 – x 51 2

+= +

x 17 – x12

1 2= +

∴ 24 = 2x + 17 – x∴ x = 7.

17. a After reversing a 2 digit number and adding it to theoriginal number and getting these two digit in anothernumber, the difference between the two digits mustbe 5. So the only possible such numbers are, 16, 27,38 and 49. Now check one by one with the numbers.

16 61⇒ . (After reversing) Difference, 61 – 16 = 45.Adding 45 + 61 = 106.Hence this number is having. 1 and 6. So this is satisfied.

27 72 72 – 27 45 45 72 117⇒ ⇒ = ⇒ + ⇒ ⇒ n o tsatisfied

38 83 83 – 38 45 45 83 128⇒ ⇒ = ⇒ + ⇒ ⇒ n o tsatisfied

49 94 94 – 498 45 45 94 139⇒ ⇒ = ⇒ + ⇒ ⇒ n o tsatisfiedHence the only number satisfied is 16.Thus the average speed of taxi is 45 kmph.

18. b Let actual speed of the train = y km.

now, 2x 1 y 2x y 3

4x 2 x 2x5

++ + = +

5 5 y y 3 1y2 1 y 4x

2 4 x x 2 4x⇒ + = + + ⇒ = ⇒ = ---- (i)

Again 2x 25 1 y – 25 2x y 3 1

–4x 2 x 2 4x5

+ ++ + = =

25 1 5y 125 3 1 y2 – 2 –

x 2 4x 4x 2 4 x25 125 1y 3

–x 4x 4x 4

⇒ + + + = + +

⇒ + =

(On sub standing the value of y from equation (i) )

( )25 34 – 5 1 ,x 25 kmph

4x 4⇒ + = =

∴ y = 4 × 25 = 100 km and AB = 2x + y = 50 + 100 = 150km.

Shot cut: 25 km is covered by the train with differentspeeds in different cases and the total time taken bythe train is different in both cases.so,

25 25 15– x 25 kmph and y 4 25

4 x 60x5

100 km

= ⇒ = = × =

AB = 2 × 25 + 100 = 150 km.

19. b The exchange of speeds and directions at the firstmeeting does not make a difference to the time orplace of their meeting (when the 2 cars together coverL, the distance between A and B). Y covers 40 km, bythe 2nd meeting, the two cars together cover 3L andthe distance covered by Y upto the fist meeting plusthe distance covered by X between the 1st and 2ndmeetings is 120 km. Since the point of the 2nd meetingis 20 km from A, L = 100 km.

20. d As 10 workers complete the work in 20 days, totalmany days (units) required to complete the work in200. In the first 2 days, 10 (2) = 20 units is completedin the next 2 days, 11 workers complete 22 units. Inthe next 2 days, 12 workers complete 20 units.Proceeding this way, 182 units are completed in 14days. On the 15th day, 17 workers complete 17 unitsof the work i,.e. the balance of 1 unit of work has tobe completed on the 16 th day.On the 16th day, there will be 17 workers, and they

can complete the one unit of the work in 1

days17


∴ Total number of days required is 1

1517

+

115

17= days.

21. b Let the speed of the local shuttle train be V. Let theinterval between the start of the trains from theirrespectively stops be t. Let the speed of the boy be U.So, the distance the previous train covers by the timethe next train starts is Vt.

Vt30 minutes

V – U=

Vt30 minutes

V – U=

V U V U 1 1Vt Vt 20 30+ ++ = +

t = 24 minutes.

22. a It is oblivious that the cork has the same as the waterin the canal.So if the swimmer is swimming away from the corkfor half an hour (upstream), it will take him anotherhalf hour to swim back to the cork again (because thecork remains still relative to the water, and the swimmerswims with a constant speed relative to water).The water in the canal flows at a speed of 1 kmph.

23. a Let us assume the by the time Yogi took 40 steps toreach the bottom from the top, the escalator moves nsteps in escalator will be (40 + n) steps.Bhogi takes 2 steps for every 1 step of Yogi, whichmeans Bhogi moves with a speed twice as fast asYogi. The time in which Yogi takes 40 steps Bhogi willbe take 80 steps. Since Bhogi is moving up words ona descending escalator he has to cover n steps moreto reach the top because escalator makes him movedown by n steps. From this argument, we can say,the total number of steps in escalator we get 40 + n =80 – n2 n = 40 n = 20Therefore, total number of steps = 60.The time when Yogi and Bhogi cross each other Bhogi

moves 23

of total steps and Yogi moves 13

of total

steps, because their speeds are in the ratio 2 : 1.

So, number of steps covered by Bhogi = 2

603

× = 40

stepsNumber of steps covered by Ramesh = 20∴ Difference = 40 – 20 = 20.

24. c The train will not run if d = 150.∴ 150 = x2 + 3x – 2 i.e., x2 + 3x – 152 = 0.

3 9 68 –3 617x

2 2± + ±= =

617 625 25< =

–3 25x 11or – 14

2±∴ < =

We ignore – 14∴ x < 11For x = 10, d = 128Hence, the speed is 150 – 128 = 22 kmph.

25. c In a 800 m race, when a completes the race, C will be30 m behind and B will be 40 m behind.∴ when C completes 800 – 40 = 760 m.∴ 770 : 760 : : 400 : x

76 400 30400 62x m 394 m

77 77 77×∴ = = = .

C beats B by 62 15

400 – 394 577 77

= . Hence.

26. b Let the rowing speed be y and the speed of the streambe x.

2 212 15 24x

– 4 i.e. 4y – x y x y – x

= =+

... (i)

and 12 15 1

–2y – x 2y x 2

=+

i.e., 2 224x 1

24y – x=

Dividing i by ii we get,2 2

2 24y – x

8y – x

=

2 2 2 2 2 24y – x 8y – 8x 4y 7x∴ = ∴ =Putting this in ii we get,

2 2 224x 1 24x 1

2 27x – x 6x= ∴ =

∴ x = 8 .

27. a A covers 4000 m, B covers (4000 – x) m and C covers(4000 – y)m.Such that 4000 : 4000 – x : 4000 – y = 16 : 15 : 11.Working backward from options, only u (a) satisfiesthe equation.

28. a Let the distance between home and friend’s office is

x. His speed = 10

52

= kmph. Extra time taken is 3 hrs.


So the total extra distance covered in this time durationis 15 km. This distance is traveled because he goesfrom home to friends office and back twice.Thus 4x = 15 or x = 3.75.

29. a Let the sped of the first person = x kmph. Time takenby first person to complete the distance of 60 km =

60x

hours. Now, let the speed of third person be y

kmph less than first person.Hence, sped of the third person = (x – y kmph and timetaken by third person to complete the distance of 56km = 56 (x – y) hours.Now, by the condition given in question, the speed ofthe second person during last 15 km Now, again bythe conditions given in question

40 15 601 – 2

x x – 2y x+ + = ... (i)

and 40 15 56

1x x – 2y x – y

+ + = ... (ii)

By solving equations (i) an d(ii) we get x = 5 kmph andy = 1 kmph.

30. d Distance to be covered 4636 + 400 = 5036 mSpeed of the slower train = 56 kmph = 56 × (5 / 18) m∴ sec∴ time taken = 5.39 min.

31. d Time taken = ( )4636

580 5

18+ × = 196 sec. = 3 min, 16 sec.

32. a In twelve hours, the minute hand and the hour are

together for 11 times. It means that after every 1211

hours. both the hands are together, similarly in twelvehours, the minute hand and the hour hand are exactlyopposite to each other for 11 times. It means that after

every 1211

hours, both the hands are opposite. Now,

let’s take an example. We know that at 12, both thehands are together and at 6 both the hands are exactlyopposite to each other.

After 6, both the hands are in opposition at 12

611

+ hours.

126 2

11 + ×

hours, 12

6 311

+ × hours and so on.

The sixth such time is 12

6 611

+ × hours which is the

first time after 12.Thus after 12, both the hands are opposite to eachother at 12 : 32 : 43.Hence, Dr. Desai takes 32 minutes and 43 seconds toreach home from the clinic.

33. d Difference in hourly usage of gas by both the burners

14.4 14.4 14.4 380 104 1040

×= − =

Required percentage

14.4 3 80100 23.07%

1040 14.4×= × × =

34. b Let the speed of the cyclist and wind be ‘x’ and ‘y’respectively.

Effective speed with the wind = x + y = 13

Effective speed against the wind = x – y = 14

therefore, we get the value of x as 724

.

Now time taken to drive one kilometer without wind

( )1 3

37 7

24

= =

35. b

36. c Let us assume that Arun started running at 10 AMand Barun started at 12 noon. So, in these twohours distance traveled by Arun is 60 km and therelative speed of Barun w.r.t Arun is 10 km/hr. So

Barun will overtake Arun after = 6010

= 6 hours

So, Barun reaches there at 6 PM.So, Kiranmala also overtakes Arun at 6 PM.Let us assume Kiranmala takes 't' time to overtakeArun and the relative speed of Kiranmala w.r.t Arunis 30 km/hr and Arun ran for 8 hrs.So, distance travelled by Arun isWhile Kiranmala's distance traveled ist = 4 hoursSo, after 4 hrs, Kiranmala will start running.

For questions 37 and 38:Let the cruising speed of the plane and the time differencebetween A and B be y km/hr and x hours respectively.Distance between A and B = 3000 kilometers. For, the planemoving from city A to City B: 3000 = (7 – x) × (y – 50). This issatisfied for x = 1 and y = 550. These are the only values givenin the options that satisfy the above equation.

37. d 38. b


39. b As per the conditions given in the questions, we getthe following figure.

250

B

C

A

500 km

30°

90°

60°50 km /hr

425 km70 km /hr

The train leaving at B reaches C at 1:00 p.m. taking atotal time of 5 hours, which means that Rahim shouldreach C by 12:45 p.m.Now total time taken by Rahim moving with a speed of70 km/hr is ‘t’.

500 3t km/hr

70= = 6.07 hrs.

Therefore, the latest time by which Rahim must leaveA and still catch the train is closest to 6:30 a.m.Hence, option (b) is the correct choice.

40. e Let x be the average marks and n be the number oftests, then we can write here

83 – x2

n= …(i)

( )75 – x 21

n 1

+=

+ …(ii)

Solving (i) and (ii), we getx = 61 and n = 11If Rajiv get 51, his average will be 63.Hence (e) is the correct option.

41. c Train left at Ahr. B min or we can say after (60A + B)minute and Reaches at (Bhr C Min) ⇒ (60 B + C) min.Total time of journey.(60B + C) – (60A + B) = 60C + A.⇒ 59(B – C) = 61A

( )59A B – C

61= .

For A = 0 (A, B, C) < 24.Hence one value of A satisfies it as journey is completedon the same day.Hence, (c) is correct choice.

42. e As per the question, let D be the total distance and ‘t’ isthe time taken. So we haveD = 10 t = 15(t – 0.5)

t 1.5 hrs⇒ =

D 15 km⇒ =Now, for the condition given we have

315 S t

4 = −

where ‘S’ is the required speed.

3 3S

2 4 = −

⇒ S = 20 km/hrs.

43. b As per the data in the question when Sangeeta’s wrist-watch moves 62 minutes Swati’s wrist-watch movesonly 56 minutes.⇒ When Sangeeta’s wrist-watch will move 620 min-utes, Swati’s wrist-watch will move only 560 min-utes.So there will be a difference of 60 minutes betweenthe times shown by the wrist-watches of Sangeetaand Swati.⇒ If Sangeeta’s wrist-watch shows 10 p.m. Swati’swrist watch will show 9 p.m.

44. b Let the total area of the rectangular oil painting be ‘x’

Therefore, 2x 3x100 x 200 cm

4 4+ = ⇒ =

So, the length of the oil painting is 200

10 cm.10

=

Hence, option (b) is the correct choice.

45. b Here we can find after 6 hours, how much part wasfilled by the three pipes.

16 4 6 1 2 112 18 36 2 9 6

= + − = + −

9 4 3 10 518 18 9

+ −= = =

Remaining part = 1 5 4

9 9− =

Since, this part is to be filled by A and B together.

Further, A and B together can fill in one hour1 1

12 10= +

part 5

36= part.

∴49

part will be filled by A and B together in time

13

5= h.

∴ Total time required = 6h + 3h + 12 min= 9 hrs and 12 min.Hence, option (b) is correct.

46. b From the given data we can write that the total workis equivalent to (24 × 16) Man-Days which in turn isequivalent to (32 × 24)Woman-Days. Hence 1 Man-


Day is equivalent to 2 Woman-Days. Let x be the num-ber of additional men required for the last two days’work.

Total work ≡ 24 × 16 ≡ (16 Men +16 Women) ×12-Days +(16 Men +16 Women)

× 2-Days

≡ 16

16 Men Men2

+ ×12-Days +

( ) 1616 x Men Men

2 + +

×2-Days

Or, 24 × 16 = 24 × 12 + (24 + x) ×2⇒ x = 24. Hence option (b) is correct.

47. b The two trains start simultaneously. Let they meetafter a time ‘t’.

The train that has covered 60 km more must be thefaster of the two. Hence, 60 = (21 – 16) × t⇒ t = 12 hoursSince they are traveling towards each other, totaldistance is the sum of the distances traveled by thetwo trains individually.Total distance = 16 × 12 + 21 × 12 = 444 Km. Hence (b)is the correct option.

48. c If B = speed of boat in still waterR = speed of streamThen as per the question:

30 4410

B R B R+ =

− + ----(i)

40 5513

B R B R+ =

− + -----(ii)

Equation (i) x 4 – Equation (ii) x 3

120 176 120 165(40 39)

B R B R B R B R+ − − = −

− + − +

176 1651

B R B R− =

+ +

(B R) 11

(B R) 5

⇒ + =− =

⇒ B = 8 km/hr

49. d Size of work force during the fourth year

90 75 10025000

100 100 100= × × × = 23513

Option (d) is the correct choice.

50. b Let the distance be x from Delhi to Mumbai

Time taken by Jet x

700=

Time taken by Kingfisher x

560=

Stoppage time x x 10x 8x x

560 700 5600 2800−= − = =

Average stoppage time

x560 12800 12min

x 2800 5560

= = = =

51. a The two trawlers move in perpendicular directions,as shown below. Traw le r2

O

t=1/2 h r

t=3/4 h r

Let the speeds (in km/h) of the two trawlers be x and

y respectively. After 12

hours, the distances covered

by them are x

km2

and y

km2

. We have:

2 22x y

172 2

+ = Or

2 2 2x + y = 34 …(1)

After 34

hours, one of the trawlers is 10.5 km farther

from the origin than the other one. We can write:

3 x 3 y= 10.5

4 4− Or

x- y 14= …(2)

Solving the two equations, we get x and y as 16 km/hand 30km/h.

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