Discussion of Question with ID = 013 under Pipes-and-Cisterns

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Question

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?

A

$18{58/75}$ mins.

B

$20{3/74}$ mins.

C

$17{24/77}$ mins.

D

$21{16/77}$ 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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