A cistern has 2 pipes. One can fill it with water in 10 hours and the other can empty it in 6 hours. In how many hours will the cistern be emptied if both the pipes are opened together when 3th4 of the cistern is already full of water ?

# A cistern has 2 pipes. One can fill it with water in 10 hours and the other can empty it in 6 hours. In how many hours will the cistern be emptied if both the pipes are opened together when $\frac{{3}^{th}}{4}$ of the cistern is already full of water ?

1. A

11 hrs 15 min

2. B

12 hrs 22 min

3. C

5 hrs 3 min

4. D

2 hrs 12 min

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### Solution:

Part of  the tank filled by pipe 1 hour $=\frac{1}{10}$

Part of  the tank emptied by pipe 2 in hour $=\frac{1}{6}$

Cistern emptied when both are together opened  $⇒\frac{1}{6}-\frac{1}{10}=\frac{10-6}{60}=\frac{4}{60}=\frac{2}{30}$

$\frac{2}{30}$ part is emptied in 1 hour

$⇒\frac{3}{4}th$ of  the tank

$\therefore \text{\hspace{0.17em}\hspace{0.17em}}\frac{30}{2}×\frac{3}{4}⇒\frac{90}{8}=11\frac{1}{4}=11\text{\hspace{0.17em}\hspace{0.17em}}hrs\text{\hspace{0.17em}\hspace{0.17em}}15\text{\hspace{0.17em}\hspace{0.17em}minutes}$

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