Discussion of Question with ID = 073 under Pipes-and-Cisterns

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Question

Tap M can fill a cistern in 18 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 $13/17$th already empty?

A

$6{78/119}$ mins.

B

$7{85/118}$ mins.

C

$5{68/121}$ mins.

D

$9{60/121}$ mins.

Soln.
Ans: a

1 filled cistern can be emptied in ${18 × 11}/{18 - 11}$ mins. So $1 - 13/17$ = $4/17$ filled cistern can be emptied in ${18 × 11}/{18 - 11}$ × $4/17$ = ${792/119}$, which is same as: $6{78/119}$ mins.


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