Two pipes can fill a cistern in 14 h and 16 h | vedclass

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(D) Part of the cistern filled by two pipes in $1 	ext{ h} = \frac{1}{14} + \frac{1}{16} = \frac{8 + 7}{112} = \frac{15}{112}$.Time taken to fill the

Vedclass · Accepted Answer

$43 \frac{19}{23} 	ext{ h}$

Vedclass · Answer

(D) Part of the cistern filled by two pipes in $1 	ext{ h} = \frac{1}{14} + \frac{1}{16} = \frac{8 + 7}{112} = \frac{15}{112}$.Time taken to fill the cistern without leakage = $\frac{112}{15} 	ext{ h} = \frac{112}{15} 	imes 60 	ext{ min} = 448 	ext{ min}$.Due to leakage,the time taken to fill the cistern = $448 + 92 = 540 	ext{ min} = 9 	ext{ h}$.Let the leakage empty the cistern in $x 	ext{ h}$.Net work done in $1 	ext{ h} = \frac{1}{9}$.So,$\frac{15}{112} - \frac{1}{x} = \frac{1}{9}$.$\frac{1}{x} = \frac{15}{112} - \frac{1}{9} = \frac{135 - 112}{1008} = \frac{23}{1008}$.$x = \frac{1008}{23} = 43 \frac{19}{23} 	ext{ h}$.

Two pipes can fill a cistern in $14$ $h$ and $16$ $h$,respectively. The pipes are opened simultaneously and it is found that due to leakage in the bottom,it took $92$ $min$ more to fill the cistern. When the cistern is full,in what time will the leak empty it?

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