Pipes and Cisterns Quiz Set 014 | Maths Aptitude Quiz Questions and Mock Test on Pipes and Cisterns Set 14 | mock tests for Bank PO, SSC Exams, SO Clerical Exam, GATE Exam, LIC AO, GIC AO, etc.

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Question 1

A tank is filled in $1{1/3}$ 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?

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 ${4/3}$ minutes work of all the taps should add to 1. So we have, ${4/3}$ × $(1/{1 - d} + 1/1 + 1/{1 + d})$ = 1, which is same as $2/{1 - d^2} + 1$ = ${3/4}$. Solving we get d = ±3.

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Question 2

A city tanker is filled by two large pipes, X and Y, together in 42 and 28 minutes respectively. On a certain day, pipe Y is used for first half of the time, and both X and Y are used for the second half. How many minutes does it take to fill the tank?

21 mins.

22 mins.

20 mins.

23 mins.

Soln.
Ans: a

Let the time taken be x. Y is running for x mins, and X for x/2. So $(x/28 + x/{2 × 42})$ = 1. Solving for x, we get x = 21 mins.

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Question 3

Tap M can fill a cistern in 16 mins. And, a tap N can empty it in 11 mins. In how many minutes will the cistern be emptied if both the taps are opened together when the tank is $7/15$^th already empty?

$18{58/75}$ mins.

$20{3/74}$ mins.

$17{24/77}$ mins.

$21{16/77}$ mins.

Soln.
Ans: a

1 filled cistern can be emptied in ${16 × 11}/{16 - 11}$ mins. So $1 - 7/15$ = $8/15$ filled cistern can be emptied in ${16 × 11}/{16 - 11}$ × $8/15$ = ${1408/75}$, which is same as: $18{58/75}$ mins.

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Question 4

A bucket can be filled by a tap in 5 minutes. Another tap on the same bucket can empty it in 10 mins. How long will it take to fill the bucket if both the taps are opened together?

10 mins.

11 mins.

9 mins.

$4{1/3}$ mins.

Soln.
Ans: a

Net part filling in one hour is $1/x - 1/y$ = $(y - x)/(xy)$. So complete filling occurs in $(xy)/(y - x)$ = ${5 × 10}/{10 - 5}$ = 10 mins.

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