Pipes and Cisterns Quiz Set 005

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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.


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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


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