Maths section had one error. In the below set the correct option for the question no. 6 wasn’t available.

Q.1.     ‘A’ and ‘B’ complete a work in 12 days, ‘B’ and ‘C’ in 8 days and ‘C’ and ‘A’ in 16 days. ‘A’ left after working for 3 days. In how many days more will ‘B’ and ‘C’ finish the remaining work?

1. 7 ¾
2. 3 ¾
3. 6 ⅚
4. 4 ¾

Sol 1: Let total units of work = 48 units

Then, A + B = 4 units per day

B + C = 6 units per day

C + A = 3 units per day

Now, 2(A+B+C) = 13 units per day

Or, A + B + C = 6.5 units per day

This means A + B + C  will do 19.5 units in 3 days

Work left = 48 – 19.5 = 28.5 units, which has to be done by B and C

Time taken by B and C to finish remaining work = 28.5/6 = 4.75 days

Option, (4)

 

Q.2.     Praveen has Rs. 4,662 in the form of 2, 5 and 10 rupee notes. If these notes are in the ratio of 3:5:8, the number of five rupees notes with him is:

1. 250
2. 84
3. 336
4. 210

Sol 2: Let the number of notes of Rs 2, Rs 5 and Rs 10 be 3x, 5x and 8x resp

Then, 3x * 2 + 5x * 5 + 10x * 8 = 4662

i.e. 111x = 4662

i.e. x = 42

Then, number of Rs 5 notes = 5 * 42 = 210

Option, (4)

 

Q.3.     A vessel contains a mixture of milk and water in the ratio of 5:3 respectively. How much of the mixture must be siphoned off and replaced with water, so that the mixture may be half milk and half water?

1. 1/7
2. 1/3
3. 1/4
4. 1/5

Sol 3: Final fraction of milk = initial fraction of milk * (1 – fraction of mixture taken out)ⁿ

Where ‘n’ is the number of times process is done

Thus, 1/2 = 5/8 * ( 1 – f )¹

i.e. 1 – f = 4/5

i.e. f = 1/5

Option, (4)

 

 

Q.4.     There are two urns. One contains two white balls and four red balls, the other contains three white and nine red balls. All balls are of the same shape and size. From each urn, one ball is drawn. What is the probability of getting both the balls of the same colour?

1. 7/12
2. 1/24
3. 1/2
4. 1/12

Sol 4: Required probability =  2/6 * 3/12 + 4/6 * 9/12 =  42/72 = 7/12

Option, (1)

 

Q.5.     Taps ‘A’ and ‘B’ can fill a tank in 37 ½ minutes and 45 minutes respectively. Both taps are opened and after some time tap ‘B’ is turned off. The tank is filled completely in exactly 30 minutes, if tap ‘B’ is turned off after:

1. 10 minutes
2. 12 minutes
3. 15 minutes
4. 9 minutes

Sol 5: Let total units to be filled = LCM (37.5, 45) = 900 units

Now, A can fill = 900/37.5 = 24 unit in a minute

B can fill = 900/45 = 20 units in a minute

Since, A fills for entire 30 units, it must have filled = 24 * 30 = 720 units

This, means B fills 900 – 720 = 180 units

To fill 180 units, B will take 180/20 = 9 minutes

i.e. after 9 minutes B should be turned off.

Option, (4)

 

Q.6.     In an office, 1/3 of the workers are Men, 1/2 of the men are married and 1/3 of the married men have children. If ¾ of the women are married and 2/3 of the 2/3 of the married women have children, then the part of workers without children are:

1. 4/9
2. 11/18
3. 5/18
4. 17/36

Sol 6: Let total number of workers be 900. Then, Number of males = 1/3 * 900 = 300

Now, Married men = ½ * 300 = 150

Married men with children = 1/3 * 150 = 50

Now, Number of Females = 900 – 300 = 600

Number of married women = ¾ * 600 = 450

Number of married women with children = 2/3 of 2/3 of 450 = 200

Total married worker with children = 50 + 200 = 250

Now, Number of workers without children = 900 – 250 = 650

Thus, fraction of workers without children = 650/ 900 = 13/18

(No option is available for the same).

 

Q.7.     The average weight of three men ‘X’, ‘Y’ and ‘Z’ is 75 kgs. Another man ‘A’ joins the group and the average weight now becomes 80 kgs. If another person ‘B’ whose weight is 5 kgs more than ‘A’ replaces ‘X’, then the average weight of ‘Y’, ‘Z’, ‘A’ and ‘B’ will be 85 kgs. What is the weight of ‘X’?

1. 82 kgs.
2. 80 kgs.
3. 78 kgs.
4. 84 kgs.

Sol  7:  Average of X, Y and Z = 75 kgs, i.e. total weight of X, Y and Z = 75 *3 = 225

Average of X, Y, Z and A = 80 kgs, i.e. total weight of X, Y, Z and A = 80 * 4 = 320

Thus, weight of A = 320 – 225 = 95

Now, weight of B = 95 + 5 = 100 kgs

Total weight of Y, Z, A and B = 85 * 4 = 340

Thus, total weight of Y, Z and A = 340 – 100 = 240

Weight of X = 320 – 240 = 80 kgs

Option, (2)

 

Q.8.     The difference between simple interest and compound interest at the same rate for rupees 5,000 for two years is rupees 98. The rate of interest is:

1. 12%
2. 10 ½%
3. 10%
4. 14%

Sol 8: P(R/100)^2 = Difference between CI and SI for 2 years

i.e. 5000 * (R/100)^2 = 98

i.e. (R/100)^2 = 98/5000

i.e. (R/100)^2 = 49/2500

i.e. (R/100) = 7/50

i.e. R = 14%

Option, (4)

 

Q.9.     The Banker’s discount on a sum of money for 18 months is Rs. 600 and the true discount on the same sum for 3 years is Rs. 750/-. The rate percentage is:

1. 10%
2. 15%
3. 12%
4. 20%

Sol 9: Banker’s discount (BD) for 1.5 years = Rs 600

i.e. BD for 3 years = Rs 1200

True Discount (TD) for 3 years = Rs 750

Sum = (BD * TD)/ (BD – TD) = (1200 *  750)/450 = Rs 2000

Thus, Rs 1200 is the SI on Rs 2000 for 3 years

Or, Rs 400 is the SI on Rs 2000 for 1 year

i.e. Rate = (400/2000) * 100 = 20%

Option, (4)

Q.10.   A man rows to a place 45 kms distant and back in 12 hours. He realizes that he can row 5 kms downstream in the same time as 3 kms against the stream. The velocity of the stream is:

1. 1 km/hr
2. 2 km/hr
3. 4 km/hr
4. 1.5 km/hr

Sol 10: Speed of man = x km/hr

Speed of stream = y km/hr

Upstream speed = (x – y) km/hr

Downstream speed = (x + y) km/hr

Given, 5/(x+y) = 3/(x-y)

i.e. 5 x – 5 y = 3x + 3y   =>  2x = 8y  => x = 4y

i.e.  Also, 45/(x-y) + 45/(x+y) = 12

i.e. 45/3y + 45/5y = 12

i.e. y = 2

Option, (2)

 

Q.11.   A train ‘X’ leaves station ‘A’ at 3 p.m and reaches station ‘B’ at 4.30 p.m., while another train ‘Y’ leaves station ‘B’ at 3.00 p.m and reaches station ‘A’ at 4.00 p.m. These two trains cross each other at:

1. 3.40 p.m.
2. 3.36 p.m.
3. 3.30 p.m.
4. 3.20 p.m.

Sol 11: Train A takes = 1.5 hours to go from A to B

Train B takes = 1 hour to go from A to B

Let distance between AB = 3 kms

Speed of train from A = 2 km/hr

Speed of train from B = 3 km/hr

Time taken by trains to meet = 3 / (2 + 3) = 0.6 hrs i.e. 36 minutes after 3

Option, (2)

Q.12.   A can do a piece of work in 8 days and B alone can do the same work in 10 days. A and B agreed to do the work together for Rs. 720. With the help of C, they finished the work in 4 days. How much C is to be paid?

1. Rs. 80
2. Rs. 70
3. Rs. 82
4. Rs. 72

Sol 12:  Let total units of work = 40 units

A’s 1 day work = 5 units

B’s 1 day work = 4 units

(A+B+C)’s 1 day work = 10 units

i.e. C’s 1 day work = 1 unit

Now, money is distributed in ratio of 1 day’s work.

Thus, A : B : C = 5 : 4 : 1

C is paid = 1/10 * 720 = Rs 72

Option, (4)

 

Q.13.   Age of father 10 years ago was three times the age of his son. After 10 years, father’s age is twice that of his son. The ratio of their present ages is:

1. 7:4
2. 9:5
3. 7:3
4. 11:7

Sol 13: Let 10 years ago, age of father = 3x and that of son = x

Now, present age of father = 3x +10 and present age of son = x + 10

10 years later, age of father = 3x + 20 and son = x +20

Also, 3x + 20 = 2* (x + 20)

i.e. x = 20

Thus, present ages of father and son = 70 and 30 respectively

i.e. ratio = 7 : 3

Option, (3)

 

Q.14.   A boat travels upstream from A to B and back from B to A in 5 hours. The speed of the boat in still water is 8 km/hour and the speed of the current is 4 km/hour. Then, the distance from A to B is:

1. 9 kms.
2. 10 kms.
3. 12 kms.
4. 15 kms

Sol 14: Let distance = D

Then, D/4 + D/12 = 5

D = 15 kms

Option, (4)

 

Q.15.   A trader sells rice at a profit of 20% and uses weights which are 10% less than the correct weight. The total gain earned by him is:

1. 35%
2. 30%
3. 33 ¹/₃%
4. 22 ²/₉%

Sol 15: Let CP of 1 gram = Rs 1

Now, 1 kg costs = Rs 1000 but as he sells only 900 gm, thus, CP = Rs 900

Further, he sells at markup of 20%, thus SP = Rs 1200

So, profit% = 300/900 * 100 = 33.33%

Option, (3)

 

Q.16.   Two men and seven boys can do a work in 14 days. Three men and eight boys can do the same work in 11 days. Further eight men and six boys can do three times the amount of this work in:

1. 24 days
2. 18 days
3. 30 days
4. 21 days

Sol 16: 2 men + 7 boys —–> 14 days   => 28 men + 98 boys  —-> 1 day – (1)

3 men + 8 boys ——> 11 days  => 33 men + 88 boys ——> 1 day – (2)

Thus, from (1) and (2), we get : 1 man = 2 boys

Now, 8 men + 6 boys = 22 boys

Also, we know from (1): 154 boys can do the work in 1 day.

i.e. 1 boy can do the work in 154 days

i.e. 22 boys can do the work in 154/22 = 7 days

Thus, 22 boys can do thrice the work in 7 * 3 = 21 days

Option, (4)

 

Q.17.   A piece of cloth costs rupees 75. If the piece is four meters longer and each meter costs rupees 5 less, the cost remains unchanged. What is the length of the piece?

1. 12 meters
2. 10 meters
3. 8 meters
4. 6 meters

Sol 17: Let cost be Rs x per metre

Then, length of piece in 1^st case = 75/x

Also, in 2^nd case: length of piece is (75/x) + 4 and cost per piece = (x -5)

But as total cost is same, thus, [ (75/x) + 4 ] * (x – 5) = 75

Solving, 75x /x – 375/x + 4x – 20 = 75

i.e. 375/x + 4x -20 = 0

i.e. 4x^2 – 20 x + 375 = 0

i.e.  4x^2 – 50 x + 30 x + 375 = 0

i.e. 2x ( 2x – 25) + 15(2x – 25) = 0

i.e. (2x + 15) (2x – 25) = 0

i.e. x = Rs 12.5 would satisfy (as other value of cost would be negative)

Thus, length of piece = 75/12.5 = 6 metres

(Alternatively, for quicker approach use options and verify)

Option, (4)

 

Q.18.   Gold and copper are as heavy as water by 19 and 9 times respectively. The ratio in which these two metals be mixed so that the mixture is 17 times as heavy as water is:

1. 4:1
2. 2:3
3. 3:4
4. 3:2

Sol 18: Use alligation rule:   19 —— 17 —— 9

Gap = 2  : 8

Thus, ratio = 8 : 2 = 4 : 1

Option, (1)

 

Q.19.   Keerthi’s father gave him some money to buy books. He spent half of the money equally to buy books and entertaining his friends. Whatever amount left with him, he deposited half in his savings account and gave Rs. 5 to a poor person as charity. Finally, Keerthi was left with Rs. 20 which he returned to his father. What amount did his father give him initially?

1. Rs. 160
2. Rs. 120
3. Rs. 100
4. Rs. 200

Sol 19: Use the backward approach. Total money will be = [(20 + 5)* 2 ] * 2 = Rs 100

(Alternatively, use options rather than equations).

Option, (3)

 

Q.20.   A clock was set correct at 12 O’ clock. It loses 10 minutes per hour. What will be the angle between the hour and minute hands of the clock after one hour?

1. 75º
2. 105º
3. 90º
4. 85º

Sol 20: Since the clock loses 10 minutes per hour, the time shown by it after an hour after 12 o clock will be 12:50

Now, Angle (made at 12:50) = mod [11/2 * 50 – 30 * 12] = 85

Option, (4)