Two pipes A and B can fill a tank in 15 hours and 20 hours respectively while a third pipe C can empty the full tank in 25 hours. All the three pipes are opened in the begining. After 10 hours, C is closed. In how much time, will the tank be full?

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$12 hrs$

$13 hrs$

$16 hrs$

$18 hrs$

answer is A.

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Detailed Solution

Part filled in 10 hours

$= 10 (\frac{1}{15} + \frac{1}{20} - \frac{1}{25}) = \frac{23}{30}$

Remaining part

$= (1 - \frac{23}{30}) = \frac{7}{30}$

(A+B)'s l hour's work

$= (\frac{1}{15} + \frac{1}{20}) = \frac{7}{60}$

$\frac{7}{60} : \frac{7}{30} : 1 : x$

Or $x = (\frac{7}{30} \times 1 \times \frac{60}{7}) = 2 hours.$

$∴$ The tank will be full in $(10 + 2) hrs = 12 hrs$

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