# Pipes and Cisterns Quiz Set 005

### Question 1

A tank is filled in 13 minutes by three taps running together. Tap A is twice as fast as tap B, and tap B is twice as fast as tap C. How much time will tap A take to fill the tank?

A

91 mins.

B

92 mins.

C

90 mins.

D

93 mins.

Soln.
Ans: a

Let the time taken by tap A be x mins. Then 13 minutes work of all the taps should add to 1. So we have, \$13 × 1/x + 13 × 2/x + 13 × 4/x\$ = 1, which is same as \$13 × 7/x\$ = 1. Solving, we get x = 91 mins.

### Question 2

A tank is filled in \$1{2/33}\$ minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?

A

6.

B

7.

C

5.

D

8.

Soln.
Ans: a

Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then \${35/33}\$ minutes work of all the taps should add to 1. So we have, \${35/33}\$ × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${33/35}\$. Solving we get d = ±6.

### Question 3

Two pipes, A and B, can fill a bucket in 15 and 19 mins respectively. Both the pipes are opened simultaneously. The bucket is filled in 5 mins if B is turned off after how many minutes:

A

\$12{2/3}\$ mins.

B

\$20{1/2}\$ mins.

C

\$11{2/3}\$ mins.

D

\$9{2/5}\$ mins.

Soln.
Ans: a

Let B be closed after it has been filling for x minutes. Work done by pipes A and B should add to 1. So \$5/15\$ + \$x/19\$ = 1. Solving, we get x = \${38/3}\$, which is same as: \$12{2/3}\$.

### Question 4

A tank is filled in 3 minutes by three taps running together. Times taken by the three taps to independently fill the tank are in an AP[Arithmetic Progression]. If the first tap is a leakage tap and the second tap takes 1 minute to fill the tank, then, the common difference of the AP can be?

A

2.

B

3.

C

1.

D

4.

Soln.
Ans: a

Let the times taken by the three taps be 1 - d, 1 and 1 + d. The time taken by the first tap will be negative because it is a leakage tap. Then 3 minutes work of all the taps should add to 1. So we have, 3 × \$(1/{1 - d} + 1/1 + 1/{1 + d})\$ = 1, which is same as \$2/{1 - d^2} + 1\$ = \${1/3}\$. Solving we get d = ±2.

### Question 5

A tap can fill a tank in 2 hours. Because of a leak it took 3 hours to fill the tank. When the tank has been completely filled, the tap is closed. How long will the water last in the tank?

A

6 hrs.

B

7 hrs.

C

5 hrs.

D

3 hrs.

Soln.
Ans: a

Work done by the leak in one hour is \$1/2 - 1/({18/6})\$ = \$1/2 - 6/18\$ = \$6/36\$. So the leak will complete the whole task in 6 hours.

This Blog Post/Article "Pipes and Cisterns Quiz Set 005" by Parveen (Hoven) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
Updated on 2017-04-07.

Posted by Parveen(Hoven),
Aptitude Trainer