Discussion of Question with ID = 017 under Pipes-and-Cisterns

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Question

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.


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