For CapgeminiBoat & Stream

Formulas to remember:

Downstream/Upstream:

1. In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

2. If the speed of a boat in still water is u km/hr and the speed of the stream is v km/hr, then:

Speed downstream = (u + v) km/hr.

Speed upstream = (u - v) km/hr.

3. If the speed in downstream is a km/hr and the speed in upstream is b km/hr, then:

Speed in still water = (a + b)/2 km/hr.

Rate of stream =(a - b)/2 km/hr

Question 1

A motorboat can cover 10 1/3 km in 1 hour in still water. And it takes twice as much as time to cover up than as to cover down thesame distance in running water. The speed of the current is:

a)3 4/9 km/hr b) 2 1/3 km/hr c) 4 km/hr d) none of these

Answer : a) 3 4/9 km/hr

Solution :

Let the speed of upstream be X km/hr.

Then, speed in downstream = 2X km/hr (since boat takes twice as much as time to cover up than as to cover down the same distancein running water).

Speed in still water = (2X+X)/2 km/hr. (formula 3)= 3X/2 km/hr.

Given that, boat covers 10 1/3 km in 1 hour in still water.

Therefore, 3X/2 = 10 1/3X = 62/9

So, speed in upstream = 62/9 km/hr.And, speed in downstream = 2 x 62/9 = 124/9 km/hr

Hence, speed of the current = [(124/9 - 62/9)]/2 km/hr= 62/9x2 = 34/9 = 3 4/9 km/hr.

Question 2

A man can row a certain distance downstream in 2 hours while he takes 3 hours to come back. If the speed of the stream be 6 km/hrthen the speed of the man in still water is:

a) 15km/hr b) 30km/hr c) 25km/hr d) 29km/hr

Answer : b) 30km/hr

Solution :

Let the speed of the man in still water be X km/hr.

Given that, speed of the stream = 6 km/hr.Therefore, speed in downstream = (X+6) km/hr (by using formula 2)And, speed in upstream = (X-6) km/hr

Distance covered in downstream in 2 hours = (X+6)2 km

Distance covered in upstream in 3 hours = (X-6)3 km

Therefore, (X+6)2 = (X-6)32X+12 = 3X-18X = 30km/hr.

Question 3

A man can take the same time to row 13 km downstream and 7 km upstream. His speed in still water 5 km/hr. The speed of thestream is:

a) 5/2 km/hr b) 3/2 km/hr c) 7/2 km/hr d) 2 km/hr

Answer : b) 3/2 km/hr

Solution :

Given that, the speed in still water = 5 km/hrLet the speed of the stream be X km/hr.Then speed in downstream = (5+X) km/hrAnd, speed in upstream = (5-X) km/hr

The time taken to cover 13 km downstream = 13/(5+X)The time taken to cover 7 km upstream = 7/(5-X)

Therefore, 13/(5+X) = 7/(5-X)13(5-X) = 7(5-X)65 - 13X = 35+7X30 = 20XX = 30/20 = 3/2

Hence the required answer is 3/2 km/hr.

Question 4

A boat takes 7 hours to cover 24 km distance and comes back. And, it can cover 2 km with the stream in the same time as 1.5 kmagainst the stream. The speed of the stream is:

a) 1 km/hr b) 2 km/hr c) 3 km/hr d) 4 km/hr

Answer : a) 1 km/hr

Solution :

Let the boat takes X hours to cover 2 km in downstream.Then, speed in downstream = (2/X) km/hr

and, speed in upstream = (1.5/X)km/hr

Given that, the boat takes 7 hours to cover 24 km distance and comes back.

That is, 24/(2/X) + 24/(1.5/X) = 724X/2 + 48X/3 = 7168X/6 = 7X = 42/168 = 1/4

So, speed in downstream = 2/X = 2 /(1/4) = 8 km/hrSpeed in upstream = 1.5/X = 1.5 /(1/4) = 6 km/hr.

Speed of the stream = (8-6)/2 km/hr (by using the formula 3)= 1 km/hr.

Question 1

The rate of current and boat in still water is 3 km/hr and 18 km/hr respectively.The boat starts from the point A and reaches thedestination point B and returns back to A.Find the time taken by the boat for both forward and backward journey if the distancebetween them is 210 km.

a) 12 hours b) 20 hours c) 24 hours d) 16 hours

Answer : c) 24 hours.

Solution :

Speed of the stream = v = 3 km/hrSpeed of the boat in still water = u = 3 km/hrSpeed of the boat in downstream = (u + v) = 18 + 3 = 21 km/hrSpeed of the boat in upstream = (u - v) = 18 - 3 = 15 km/hr.

Distance between A and B = 210 km.Time taken by the boat to reach B (forward journey) (in upstream) = distance/speed = 210/15 hours.

Time taken by the boat to reach A (backward journey) (in downstream) = 210/21 hours.Total time = 210/15 + 210/21 hours = 14 + 10 = 24 hours.

Question 2

If the speed of the stream is 6 km/hr and the speed of a man in still water is 20 km/hr, then the distance covered by the mandownstream in 15 minutes is:

a) 3.5 km b) 4.5 km c) 6.5 km d) 5.5 km

Answer : c) 6.5 km

Solution :

Time taken by the man = 15 minutes = 15/60 hour.Speed of the stream = v = 6 km/hrSpeed of the man in still water = u = 20 km/hr.Speed of the man in downstream = 20 + 6 = u + v = 26 km/hr.Distance Covered = time x speed = 26 x 15/60 km = 6.5 km.

Question 3

The speed of a motor-boat in still water is 12 km/hr and the speed of the current is 3 km/hr. If the boat takes 3 hours to arrive at aplace and return back, then how far is the place?

a) 17 km b) 18 km c)19 km d) 20 km

Answer : a) 17 km.

Solution :

Speed of the stream = v = 3 km/hrSpeed of the motor-boat in still water = u = 12 km/hrSpeed of the boat in downstream = (u + v) = 12 + 3 = 15 km/hrSpeed of the boat in upstream = (u - v) = 12 - 3 = 9 km/hr

Time taken by the boat to arrive and return back = 3 hours.Let the required distance be X km.Time taken by the boat in downstream = X/15 hr.Time taken by the boat in upstream = X/9 hr.Total time = 3 = X/15 + X/9(3X + 5X)/45 = 3X = 135/8 = 16.875 km.Hence the distance is 17 km (approximately).

Average Problems

Question 1

The average of five consecutive odd integers is 113. What is the second smallest of them?

a) 111 b) 115 c) 109 d) none of these

Answer : a) 111

Solution :

Let the five consecutive odd numbers be X, X+2, X+4, X+6 and X+8

Given that, their average = 113.

That is, (X + X+2 + X+4 + X+6 + X+8) / 5 = 113(5X + 20) / 5 = 1135(X+4) / 5 = 113X = 113 - 4 = 109.

Second smallest number is X + 2 = 109+2 = 111.

Question 2

The average of 6 consecutive even numbers is 207. What will be the sum of the smallest and largest number?

a) 210 b) 408 c) 414 d) 208

Answer : c) 414

Solution :

Let the six consecutive odd numbers be X, X+2, X+4, X+6, X+8 and X+10

Given that, their average = 207

That is, (X + X+2 + X+4 + X+6 + X+8 + X+10)/6 = 207(6X+30) / 6 = 2076(X+5) / 6 = 207X+5 = 207X = 202.

Therefore the smallest number = 202 and the largest number = x+10 = 202+10 = 212

Required sum = 202+212 = 414.

Question 3

The average of the four consecutive even integers is 3645 less than their sum. What is the last of these numbers?

a) 1218 b) 2146 c) 3212 d) none of these

Answer : a) 1218

Solution :

Let the 4 consecutive even integers be X, X+2, X+4 and X+6.

Then, their average = (X + X+2 + X+4 + X+6)/4 = (4X+12)/4 = (X+3) ....(1)And, their sum = (X + X+2 + X+4 + X+6) = 4X+12...(2)

Given that, average is 3645 less than the sum.

That is, from (1) and (2), 4x+12 = 3645 + X+33X = 3636X = 1212

Therefore, required last number = X+6 = 1212 + 6 = 1218.

Question 4

If the average of seven consecutive odd numbers is 2117 then what will be the average of first four?

a) 2111 b) 2112 c) 2113 d) 2114

Answer : d) 2114

Solution :

Let the 7 consecutive odd numbers be X, X+2, X+4, X+6, X+8, X+10 and X+12.

We have to find the average of X, X+2, X+4 and X+6

That is, (X + X+2 + X+4 + X+6) / 4 = (4X+12) / 4 = X+3 ...(1)

Given that, the average that 7 numbers = 2117

That is, (X + X+2 + X+4 + X+6 + X+8 + X+10 + X+12)/7 = 2117(7X + 42) / 7 = 2117(X + 6) = 2117X = 2111

Putting the above X value in (1), we get, X+3 = 2111+3 = 2114

Hence the required answer is 2114.

Question 1

The average weight of 50 men was 72.The weight of the 46th person was 68 which has been wrongly entered as 46.Find thecorrected average weight.

a) 73 b) 78 c) 75 d) none of these.

Answer : d) none of these.

Solution :

Average weight of 50 men = 72.Sum of weights of these 50 men = 72 x 50.

Since the weight 68 is wrongly entered as 46.Amount of weight decreased in the sum = 68 - 46 = 22.

Therefore, the corrected new sum = 72 x 50 + 22.And, required average of 50 men = (72 x 50 + 22) / 50 = 3622 / 50 = 72.44Hence the answer is option d.

Question 2

In an entrance test, a candidate's point were wrongly taken as 42 instead of 24. Because of that, the average score for the exam gotincreased by 3/4. Find the numberof candidates who have attended the exam.

a) 24 b) 12 c) 36 d) 48

Answer : a) 24.

Solution :

Let there be N candidates in the exam.And, let X be incorrect average score.

Incorrect total marks = X x NSince, the points 2.4 is wrongly entered as 4.2.Corrected total marks = (X x N) - (42 - 24) = NX - 18 ....(1)

Since the average increased by 3/4, then the correct average = X - 3/4.Correct average = (X - 3/4)N = NX - 3N/4 ....(2)

Equating (1) and (2), we get,NX - 18 = NX - 3N/43N = 18 x 4N = 18 x 4/3 = 24.Hence, the number of candidates is 24.

Question 3

The average of 30 non-negative numbers is 30. Out of these numbers, the average of 10 numbers is 28 and that of the other 18numbers is 32. The average of theremaining numbers is:

a) 44 b) 22 c) 11 d) none of these.

Answer : b) 22.

Solution :

Average of 30 numbers = 30.Sum of 30 numbers = 30 x 30 = 900

Average of 10 numbers = 28Sum of theses 10 numbers = 28 x 10 = 280.

Average of 18 numbers = 32Sum of these 18 numbers = 18 x 32 = 576.

Now, number of remaining numbers = 30 - 10 - 18 = 2.We have to find the average of these two numbers.Sum of these remaining numbers = 900 - (280 + 576) = 44.Average of remaining numbers = 44/2 = 22.

Mixture Problems

Alligation :It is the rule that enables us to find the ratio in which two or more ingredients at the given price must be mixed to produce a mixture ofdesired price.

Alligation :If two ingredients are mixed, thenQuantity of cheaper / quantity of dearer = C.P of dearer – mean price / mean price – C.P of cheaper.We present as under:

Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c).

Question 1

Two cans A and B contains milk worth Rs.7 per litre and Rs.9 per litre respectively. If the contents of A and B are transferred toanother can C in the ratio 3 : 7 then the cost per litre of the mixture in can C is:

a) Rs.9.40 b) Rs.10.10 c) Rs.9.40 d) Rs.8.40

Answer : d) Rs.8.40

Solution :

Cost of 1 litre of A = Rs.7 = cheaper quantity.Cost of 1 litre of B = Rs.9 = dearer quantity.Let the mean price be Rs.X.Applying the rule of alligation,

Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) = (9 - X) : (X - 7)Given ratio = 3 / 7 = 9 - X / X - 7

63 – 7X = 3X - 21X = 8.4.Hence, the required mean price = Rs.8.40.

Question 2

Two qualities of rice at Rs.63 per kg and Rs.67.50 per kg are mixed with another quality of rice in the ratio 2:2:3. The final mixture soldat Rs.76.50 per kg then the rate of third quality rice per kg was:

a) Rs.87.50 b) Rs.91.50 c) Rs.81.50 d) Rs.99.50

Answer : b) Rs.91.50

Solution :

Given that, first and 2nd varieties are mixed in equal proportion.(2:2)Their average price = Rs.(63 + 67.50) / 2 = Rs.65.25The new mixture is formed by mixing two varieties, one at Rs.65.25 per kg and the other at Rs.X per kg in the ratio 4:3. (note that,2:2:3 becomes 4:3).Given that, mean price of new mixture = Rs.76.50We have to find X.Applying the rule of alligation,

Therefore, (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c) =X - 76.50 / 11.25 = 4/33X – 229.50 = 45X = 91.50Hence, the required price of 3rd quality rice = Rs.91.50.

Question 3

In what ratio, a liquid A of cost Rs.31 per litre should be mixed with liquid B of cost Rs.36 per litre, so that the cost of the liquid ofmixture is Rs.32.35 per litre?

a) 2:1 b) 3:1 c) 3:2 d) 4:3

Answer : b) 3:1

Solution :

Cost of 1 litre of liquid A = Rs.31 = cost of cheaper quantity.Cost of 1 litre of liquid B = Rs.36 = cost of dearer quantity.Given, mean price = Rs. 32.25Applying the rule of alligation,

Therefore, required ratio = (Cheaper quantity) : (Dearer quantity) = (d - m) : (m - c)= 3.75 : 1.25 = 3:1.Hence, the answer is option b.

Area Problems

Question 1

The dimension of a rectangular shaped wall was X x Y. If the breadth is increased by 40% and the length is decreased by 30% thenthe % of area of the new wall compared with previous one is:

a) 88% b) 98% c)68% d)78%

Answer : b) 98%

Solution :

Dimensions of the wall = X x Y.Without loss of generality, assume that, X is length and Y is breadth.Area of original wall = XY unit2.

New length (after decreasing 30%) = X – 30X/100 = 70X/100New breadth (after increasing 40%) = Y = Y + 40Y/100 = 140Y/100.New area = 70X/100 x 140Y/100 = 49XY/50.

Therefore, required % = (new area / original area ) x 100 = ((49XY/50) / XY) x 100= 98%.Hence, the answer is 98%.

Question 2

If the area of a square shaped field decreases by 36%, then the % of each side of the field decreases by:

a) 15% b) 12% c) 20% d) 19%

Answer : c) 20%

Solution :

Let us assume that the original area of the field is 100 unit2.Then, its original sides = sqrt(100) = 10 units.% decrease in area = 36%New area = 100 - 36 = 64 unit2.

Therefore, new side of the field = sqrt(64) = 8 units.i.r., decrease on 10 units = (10 - 8) = 2 units.Decreasing % = (units of decreasing / original units) x 100 = 2 x 100/10 = 20%Hence, the answer is 20%.

Question 3

The dimension of a black board is L m x B m. If l and b decreases by 20% and 40% respectively, then the area of the board beforealteration exceeds the area of new one by:

a) 48% b) 52% c) 39% d) 62%

Answer : b) 52%

Solution :

Let length = L m and breadth = B m.Original area = (LB) m2.

New length (after decreasing 20%) = (100 - 20)% of L= 80% of L = 80L /100 m = 4L / 5 m.New breadth (after decreasing 40%) = (100 - 40)% of B = 60% of B = 60B / 100 = 3B / 5 m.

New area = 4L/5 x 3B/5 = 12(LB)/25 m2.Difference in area = original area - new area = Lb - 12(LB)/25 = 13(LB)/25 m2.Therefore, required % = difference / original area x 100 = [(13(LB)/25) / (LB) ] x 100 = 13 x 4 = 52%.

Arithmetic calculations

Question 1

A fruit seller has 16 apples and cost of each apple is Rs.12. Three persons A,B and C decided to buy that apples. If A and B givesRs.84 and Rs.48 to the seller for some apples, then how many apples C can buy?

a) 8 b) 11 c) 5 d) 6

Answer : c) 5

Solution :

Given that the seller has 16 apples and he sells each apple at Rs.12Then total cost of 16 apples = 16 x 12 = Rs.192A and B buys for Rs.84 and Rs.48.Then the cost of remaining apples = Rs.(192 - (84+48))= Rs.60.

TC can buy for Rs.60Since each apple costs Rs.12, then C can buy 60/12 = 5 apples.Hence the answer is 5.

Question 2

Thomas bought X number of sports goods for Rs.9000. If each item was cheaper by Rs.30 then with the same amount he could havebought 50 more items than X. Find the number of items bought by Thomas ?

a) 100 b) 150 c) 75 d) 125

Answer : a) 100

Solution :

Transaction I

Thomas have bought X items for Rs.9000.Then the cost of each item = Rs.9000/X ...(1)

Transaction II

If each item was cheaper by Rs.30, he could have bought 50 more items.Therefore, with Rs.30 discount, the amount of each item = Rs.9000 / X + 50 ...(2)

We know that the cost of each item in transaction II will be lesser than that of transaction I by Rs. 30.

i.e (1) - (2) = 30

Since he bought 50 more items when cost of each item is less by Rs.30 then we have9000/X - 9000/X + 50 = 301/X - 1/X + 50 = 30/900050 / X(X + 50) = 1/300X(X + 50) = 15000X^2 + 50X = 15000X^2 + 50X - 15000 = 0(X+150)(X-100) = 0X = -150 or X = 100X cannot be a negative value. Therefore, X = 100.Hence the answer is 100.

Question 3

On Republic day, chocolates were to be distributed among 350 kids in a play school. But 140 kids were absent on that day and eachkid present got 3 chocolates more. Find the total number of chocolates bought for the distribution.

a) 3200 b) 2750 c) 2950 d) 1575

Answer : d) 1575

Solution :

Let the total number of chocolates be X.Total number of kids in the school(when everyone is present) = 350.If everyone is present, number of chocolates per kid = X/350 chocolates ...(1)

But, 140 kids are absent.Therefore, total number of kids present = 350 - 140 = 210Therefore, number of chocolates per present kid = X/210 chocolates ...(2)

210 kids were present and each kid gets 3 extra.

i.e (2)-(1)=3

Or X/210 - X/350=3

350X - 210X / (210 x 350)=3

140X / 210 x 350=3

X = 3 x 210 x 350 / 140 = 1575

Then the required answer is 1575.

Question 4

8 friends planned to go to hotel for dinner and to share the bill amount equally. If one of them have forget to bring the wallet, then whatwill be the extra amount contribute by each to pay the bill of Rs.1904 ?

a) Rs.28 b) Rs.54 c) Rs.34 d) Rs.38

Answer : c) Rs.34

Solution :

Given that the bill amount = Rs.1904Actual share of each = Rs.1904/8 = Rs.238If one of 8 is left, then the sharing amount = Rs.1904/7 = Rs.272Then the extra amount given by each = Rs.(272 - 238) = Rs.34.Hence the answer is Rs.34

Look And Say

Note :

A look-and-say sequence is a sequence of integers, expressed in decimal notation, where each successive term is generated bydescribing the previous one.

For instance, if x1 (the first term of the sequence) is 1, the next term is the description of this term, 11 ("one 1"), which is described by21 ("two 1's"), which is described by 1211 ("one 2 one 1"), etc.; the series continues 111221, 312211, 13112221, ...

Question 1

Here is a sequence of numbers: 64 1614 1116114 31163114 132116132114. It seems to be strange sequence, but yet there is asystem behind it. What is the number of digits in the next term in this sequence?

a) 20 b) 12 c) 16 d) 18

Answer : a)20

Solution :

Given sequence is 64 1614 1116114 31163114 132116132114.It is a look-and-say sequence.

The first term is 64.The next term is the description of the first term, 1614 (one 6 one 4).2nd term is described in the 3rd term as 11161114, (one 1 one 6 one 1 one 4).Then the 4th term is 31163114 ( three 1 one 6 three 1 one 4).And the 5th term is 132116132114 (one 3 two 1 one 6 one 3 two 1 one 4)Then the required term is 11131221161113122114 (one 1 one 3 one 2 two 1 one 6 one 1 one 3 one 2 two 1 one 4).Therefore the number of digits in the required term is 20.

Question 2

If 9 is the 1st term of a look-and-say sequence then what will be its 6th term?

a) 1113122119 b) 1113222119 c) 1113122119 d) 1113421119

Answer : c)1113122119.

Solution :

The 1st term is 9.Then the successive terms are:2nd term = 193rd term = 11194th term = 31195th term = 132119Then the 6th term = 1113122119.Hence the answer is 1113122119.

Question 3

What will be the sum of digits in 5th term of a look-and-say sequence generated by 81?

a) 20 b) 18 c) 28 d) 24

Answer : d)24

Solution:

The look-and-say sequence for 81 is,1st term of the sequence = 81.2nd term of the sequence = 1811.3rd term of the sequence = 111821.4th term of the sequence = 31181211.5th term of the sequence = 132118111221.Then the required sum = 1 + 3 + 2 + 1 + 1 + 8 + 1 + 1 + 1 + 2 + 2 + 1 = 24.Hence the answer is 24.

Profit & Loss Problems

Question 1

Find the profit % of a sales man, if he had sold the emergency lamp at a marked price. Note that, the cost price of the emergencylamp is Rs.640 and he can make a profit of 30% even after reducing the marked price of the emergency lamp by Rs.64.

a) 32% b) 40% c)12% d) 42%

Answer : b) 42%

Solution :

Cost price (C.P) of an emergency lamp = Rs.640Profit made by him = 30%i.e., if the C.P is 100 then it's selling price(S.P) = 130.Here, the S.P of emergency lamp = 130/100 x 640 = Rs.13 x 64 = Rs.832

After reducing the marked price by Rs.64, he sold the lamp for Rs.832.Then it's marked price = Rs.832 + 64 = Rs. 896Now marked price = Rs.896 and C.P = Rs.640

If he sells at marked price then profit = Rs.896 - Rs.640 = Rs.256Then required profit % = 256/640 x 100 = 40%Hence the answer is 40%.

Question 2

A man bought a bike at the marked price of Rs.65,000 with a discount of 3% on its marked price. If the salesman makes 30% of profiton this sale then what will be the cost price of the bike?

a) Rs.48,500 b) Rs.39,300 c) Rs.52,400 d) Rs.51,100

Answer : a) Rs.48,500

Solution :

Marked price of the bike = Rs.65000Discount on marked price = 3%

Then the selling price (S.P) of the bike = 97% of Rs.65000ie., S.P = 97/100 x 65000 = Rs.(97 x 650) = Rs.63050

The sales man made a profit of 30%.i.e., if the S.P is Rs.130 then its C.P = Rs.100Therefore the C.P of the bike = Rs. 100/130 x 63050 = Rs.(100x485) = Rs.48500Hence the required answer is Rs.48500.

Question 3

If a salesman made a profit of 12% by selling an item whose marked price is 20% more than it's cost price then what will be thediscount % of the item?

a) 12/5% b) 13/3% c) 21/4% d) 20/3%

Answer : d)20/3%

Solution :

Let the C.P of the item be Rs.100Marked price is 20% more than it's cost price = Rs.120

Salesman made a profit of 12%.Then its selling price = (CP + Profit) = Rs.112

Discount = marked price - selling priceRs.120 - Rs.112 = Rs.8Required discount % = 8/120 x 100 = 20/3 %Hence the answer is 20/3%.

Question 4

A seller gives 6% discount on the labelled price and gives 2 items free for buying every 11 items and gains 20%. Then the labelledprice is above the cost price by(in percentage):

a) 25% b) 12.5% c) 45% d) 18%

Answer : a) 25%

Solution :

The seller earns a gain of 20% by selling 13 items. (11 items + 2 items free)Let the C.P of each item be Rs.100Then C.P of 13 items = 13 x Rs.100 = Rs.1300

Since 20% profit is earned, S.P of 13 items = Rs.1300 x 120/100 = Rs.1560.And S.P of each item = Rs.1560/13 = Rs.120

Given that, the seller gives discount of 6% on the labelled price.i.e., S.P is Rs.94 then labelled price = Rs.100

And if S.P is Rs.1560 then labelled price = Rs. (100/96) x 120 = Rs.125Therefore, labelled price above cost price = labelled price - C.P = 125 - 100 = 25Hence the required % is 25%.

Ratio Problems

Question 1

Ratio of the score of two teams P and Q is 5:8. If the score of team P is increased by 60% and those of Q decrease by 35%, then thenew ratio of their score becomes 9:8.Find the score of team P.

a)35 b)12 c)42 d)data inadequate

Answer : d) data inadequate

Solution :

Let the original score of P and Q be 5X and 8X respectively.New score of P after increasing 60% = 160% of 5X160 / 100 x 5X = 8XNew score of Q after decreasing 35% = 65% of 8X65 / 100 x 8X = 26X / 5Therefore 8X : 26X / 5 = 9 : 8 (given)40X / 26X = 9 / 8This cannot give X valueHence the given data is inadequate.

Question 2

805 ml of mixture1 is mixed with 700ml of mixture2. If mixture1 has acid1 and acid2 in the ratio 4:3 and mixture2 has acid2 and acid3in the ratio 2:5 then the amount of acid2 in the new mixture is:

a)125 ml b)545 ml c)345 ml c)625 ml

Answer : b)545 ml

Solution :

Acid1 and acid2 is in the ratio 4:3, thenQuantity of acid2 in 805ml of mixture1 = 805 x 3 / 7 = 345 mlAnd Acid2 and acid3 in the ratio 2:5.Quantity of acid2 in 700ml of mixture2 = 700 x 2 / 7 = 100 x 2 = 200mlTherefore quantity of acid2 in new mixture = 345 + 200 = 545ml.Hence the required answer is 545 ml.

Question 3

A chemical solution of 2430 ml is poured into 3 bottles namely P,Q and R such that 5 ml,10 ml and 15 ml are taken out and theremaining solution in P, Q and R will be in the ratio 3 : 4 : 5. Find the initial amount of solution in Q.

a)810 ml b)270 ml c)970 ml d)720 ml

Answer : a) 810 ml

Solution :

Total amount of solution = 2430 mlAfter taking 5 ml,10 ml and 15 ml out, the remaining amount of Solution = 2430 - (5 + 10 + 15) ml= 2400 mlTherefore 2400 ml of solution is in the ratio = 3 : 4 : 5Amount of solution present in Q now = 4/12 x 2400 = 800 mlInitial amount of solution in bottle Q = New amount of solution in Q + Solution taken out from Qi.e., 800 + 10 = 810 ml contained in bottle Q initially.

Question 4

The ratio of length of sides of a machine is 1/6:1/4:1/3 and the sum of length of sides is 104,. If we increase each side by 3 m then thelength of largest side will be:

a)45 m b)46 m c)49 m d)50 m

Answer : c)49 m

Solution :

The sides of machine are in the ratio 1/6 : 1/4 : 1/3 = 4:6:8Let the sides of the machine be 4X,6X and 8XGiven that the sum of the sides = 104 mi.e., 4X + 6X + 8X = 104 mX = 104/18 = 52/9

Then the largest side 8X = 8 x 52/9 = 416 / 9The sides are increased by 3 m.Now the length of the largest side = 416 / 9 + 3 = 443 / 9= 49.22 = 49 m (approximately)Hence the answer is 49 m

Age Problems

Question 1

The sum of the 1/2 of Sidharth's age two years from now and 1/3 of his age three years ago is twenty years, then how old is he now?

a)20years b)2years c)24years 4)4years

Answer : c)24years

Solution :

This problem refers, Sidharth's age two years in the future and three years in the past.Let Sidharth's age now be S.His age after two years from now is S + 2 and three years ago is S - 3.

From the problem,1/2 of age two years from now is (1/2)(S + 2) = S/2 + 11/3 of age three years ago is (1/3)(S - 3) = S/3 - 1

The sum of these two numbers is twenty,i.e., S/2 + 1 + S/3 - 1 = 20S/2 + S/3 = 203S + 2S = 120S = 120/5 = 24

Hence, Sidhaeth is 24 years old now.

Question 2

Jo is 3 years elder than Rahul. Naren was 28 years of age when Mahi was born while Sheela was 26 years of age when Rahul wasborn. If Mahi was 4 years of age when Jo was born, then what was the age of Naren and Sheela respectively when Jo was born ?

a)32 & 23years b)42 & 24years c)23 & 28years d)24 & 32years

Answer : a)32 & 23years

Solution :

Given that, " Naren was 28 years of age when mahi was born"If Mahi was 4 years of age then Naren's age is 28 + 4 = 32 years.Also given that, If mahi was 4 years of age when Jo was born.From the above two statements we would have Naren's age when Jo was born is 32 years.

Also given that, Jo is 3 years elder than Rahul and sheela was 26 years of age when Rahul was born.Sheela's age when Jo was born = 26 - 3 = 23 years

Hence, the answer is 32 & 23 years

Question 3

My father was 28 years older than my brother and my mother was 23 years elder than my sister when I was born. If my brother is 10years older than me and my mother is 4 years younger than my father, how old was my sister when I was born?

a)10yrs b)11yrs c)12yrs d)13yrs

Answer : b)11yrs

Solution :

When I was born, my brother was 10 year old,My father's age is 10 + 28 = 38 yrs. oldMy Mother's age is 38 - 4 = 34 years old and my sister's age is 34 - 23 = 11 year oldHence the answer is 11 years.

Problems On Probability

Question 1

If a number selected at random from a set which contains only 2 digit numbers, find the probability that the selected number will be amultiple of '8'.

a)1/10 b)4/45 c)1/15 d)none of these

Answer : b)4/45

Solution :

There are a total of 90 two digit numbers.Therefore, the set contains 90 numbers.We know that the first two digit number that is a multiple of 8 is 16 and every 8th number from 17 will be a multiple of 8. (16 is 7thnumber of the set)Therefore, there are 11 (integral value of 90/8 = 11) of those numbers that are divisible by '8'.(alternatively the two digit multiples of 8are 16,24,32,40,48,56,64,72,80,88 and 96)Then, the required probability is 8/90 = 4/45

Question 2

What is the probability that a number selected at random from the set of 3 digit numbers will be a multiple of '9' and a divisor of 900?

a)1/450 b)5/36 c)1/180 d)7/900

Answer : c)1/180

Solution :

The three digit numbers are 100,101,102,...,999There are a total of 900 three digit numbers.We know that the number 99 preceded by 100 is divisible by 9 and then every nine-th number from 101 will be divisible by '9'.Therefore, there are (900/9) 100 of those numbers that are divisible by '9'.Now, the 3 digit divisors of 900 are 900/1, 900/2, 900/3, 900/4, 900/5, 900/6 and 900/9.i.e., the 3 digit divisors of 900 are 900, 450, 300, 225, 180, 150 and 100.Here, except 150 and 100 other numbers are multiple of 9.

The remaining 5 numbers are both multiple of 9 and a divisor of 900.Hence, the required ratio = 5/900 = 1/180.

Question 3

What is the probability that a two digit number selected at random will be a divisor of '96' and not a divisor of 24?a)2/45 b)1/30 c)1/45 d)1/15

Answer : a)2/45

Solution :

There are totally 90 two digit numbers.The divisors of 96 are 1,2,3,4,6,8,12,16,24,32,48 and 96.The two-digit divisors of 96 are 12,16,24,32,48 and 96.Here, the two digit divisors of 24 are 12 and 24 only.Therefore, 16,32,48 and 96 are the required numbers.Hence the required probability = 4/90 = 2/45

Question 4

Find the probability that a number selected at random from the set of two digit numbers will not be a multiple of 12 and not a multipleof 4?

a)29/90 b)7/90 c)61/90 d)34/45

Answer : d)34/45

Solution :

The two digit numbers are 10,11,12,...,99There are 90 two digit numbers.Since the set of multiples of 4 contains multiple of 12 then it's enough to find the number of multiples of 4.The first two digit number that is divisible by 4 is 12 and every 4th number from 13 is multiple of 4.Here, 22 (integral value of 90/4) numbers of multiples of 4 & 12 together.Then the number of two-digit non-multiple of 12 and non-multiple of 4 is 90-22 = 68.Hence the required ratio is 68/90 = 34/45.

Simplification

Question 1Every day, a man spends 1/4 of his time to work, 3/8 of his time to sleep and the rest of the time in family. Then, during a week howmany hours did he spend with his family?a) 48 b) 54 c) 12 d) 63

Answer : d) 63

Solution :

Time taken to work and sleep = 1/4 + 3/8 = 5/8 partThen the remaining part = 1 - 5/8 = 3/8Remaining part of job 3/8 spend with his family.

1 part of a day means he takes 24 hourThen 3/8 part of a day takes = 3 x 24/8 = 9 hoursAnd, for a week he spends (9 x 7) = 63 hours with his family.Hence the answer is 63 hours.

Question 2

Assume that 1/6, 1/10 and 1/3 part of a human body's weight is made of bones, skins and muscles respectively. Also assume that theremaining weight is due to water and fluids. If a man's weight is 60 kg find the weight of water and fluids in his body.a)1 2 kg b) 20 kg c) 24 kg d) 36 kg

Answer : c) 24 kg

Solution:

Given that, 1/6 part made of bones,1/10 part made of skin and 1/3 part made of muscles.Then the remaining part (i.e., weight of water and fluids) = 1 - (1/6 + 1/10 + 1/3)1 - (10 + 6 + 20)/60 = 1 - 36/60 = 2/5

Given that the man's weight is 60 kg.Then water and fluid weight = 2 x 60/5 = 24 kg.Hence the required answer is 24 kg.

Question 3

A boy spends 2/5 of his study time for maths, 3/10 of his study time for science and 1/8 of his study time for arts. If he spendsremaining 30 minutes for playing with friends then find the total time taken by him to learn science and arts.a) 1 3/14 hours b) 2 1/7 hours c) 3 4/9 hours d) 1 3/10 hours

Answer : a) 1 3/14 hours.

Solution :

Part of the time left (remaining time) = 1 - (2/5 + 3/10 + 1/8)1 - 33/40 = 7/40Let the total time taken by him to study all subjects be X hours. We have just found that he had 7/40 remaining time after studyingmaths, science and arts.From the question, we can say that he spends this remaining 7/40 of his time for playing and this time equals 30 minutes.

Then 7/40 of X = 1/2 hour (we have expressed 30 minutes as half hour i.e 1/2 hour)7X / 40 = 1/2X = 20/7 hours.

Time taken to learn science = 3/10 of X = 3/10 x 20/7 = 6/7 hours. ...(1)Time taken to learn arts = 1/8 of X = 1/8 x 20/7 = 5/14 hours. ...(2)Time required to learn science and arts = (1) + (2) = 6/7 + 5/14 = 17/14 hours = 1 3/14 hours.Hence the answer is 1 3/14 hours

Consecutive Numbers

Question 1

If the average of 4 consecutive odd numbers is 92, then the least number of those numbers is:a) 89 b) 71 c) 81 d) 91

Answer : a) 89

Solution :

Let the four consecutive odd numbers be x, x+2, x+4 and x+6

We have to find the least number i.e., x

Given that the average is 92. That is, x + x+2 + x+4 + x+6 / 4 = 924x + 12 = 92(4)4x + 12 = 3684x = 368 - 12 = 356x = 356/4 = 89

Hence the required number is 89.

Question 2

If the sum of five consecutive even numbers is 1580, then find the average of the next five consecutive even numbersa) 326 b) 312 c) 325 d) 318

Answer : b) 312

Solution:

Let the five consecutive even numbers be x, x+2, x+4, x+6 and x+8

The sum of the above five numbers is 1580.i.e., x + x+2 + x+4 + x+6 + x+8 = 15805x + 20 = 15805x = 1580 - 20x = 312

Therefore, the 5 even numbers are 312, 312+2, 312+4, 312+6 and 312+8Or 312, 314, 316, 318 and 320

Then the next 5 consecutive even numbers are 322, 324, 326, 328 and 330

Now, the required average = 322+324+326+328+330 / 5 = 1630/5 = 326.

Hence the answer is 326.

Question 3

If the sum of the 4 consecutive even numbers is 4 more than three times the largest number, then the average of those numbers is:a) 32 b) 12 c) 25 d) 13

Answer : d) 13

Solution :

Even numbers can be represented to have the form 2n where n = 0,1,2,3....

Based on our above understanding, any four consecutive even numbers can be represented as,

2k, 2k + 2, 2k + 4, 2k + 6

Now, adding these we get,

2k + (2k + 2) + (2k + 4) + (2k + 6) = 8k + 12

But the above sum is 4 more than 3 times the largest number(which is 2k + 6) .

i.e., 8k + 12 = 4 + 6k + 18

2k = 10k = 5

Therefore, the sum of the numbers = 8k + 12 = 52 (by substituting the value of k)

Then the required average = 52/4 = 13