Pipe A can fill a cistern in $5$ hours and pipe B can fill in $20$ hours. Both Pipe are turned on but there is a leakage in bottom of the cistern. So the cistern is filled in $30$ min. more. In how many time will leakage empty the full cistern ?

30

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36

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24

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12

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Solution

The correct option is C $36$
A can fill cistern in $5$ hours
Per hour work of A = $100 / 5 = 20 %$
B an fill cistern in $20$ hours'Per hour work of $B = 100 / 20 = 5 %$
A and B together can fill cistern in $1$ hour $= 20 + 5 = 25 %$
They together can fill entire cistern $= 100 / 25 = 4$ hours
Due to leakage, they take $30$ min more to full the cistern
$1 / (4 * 2) \times 100 = \frac{1}{8} \times 100 = 12.5$
It means leakage is emptied $12.5 %$ of tank in $4$ hour and $30$ min
So, work per hour of leak $= (12.5 * 2) / 9 = 2.77 %$
It can empty full cistern $= 100 / 2.7 = 36$ hours (Approx)

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Pipe A can fill a cistern in 5 hours and pipe B can fill in 20 hours. Both Pipe are turned on but there is a leakage in bottom of the cistern. So the cistern is filled in 30 min. more. In how many time will leakage empty the full cistern ?

Pipe A can fill a cistern in $5$ hours and pipe B can fill in $20$ hours. Both Pipe are turned on but there is a leakage in bottom of the cistern. So the cistern is filled in $30$ min. more. In how many time will leakage empty the full cistern ?