Discussion

Explanation:

First pipe can fill the tank in 10 mins while second pipe can fill the tank in 15 mins. 

Together, both pipes can fill the tank in $\frac{1}{{\displaystyle \frac{1}{10}}+{\displaystyle \frac{1}{15}}}$ = 6 mins

Now, during the first 6 mins drain pipe was also open which was closed and it takes another 6 mins for the tank to be completely filled.

∴ 6$\left(\frac{1}{10}+\frac{1}{15}-\frac{1}{x}\right)$ + 6$\left(\frac{1}{10}+\frac{1}{15}\right)$ = 1

[Where x = Time taken by drain pipe to empty the tank]

⇒ $\left(\frac{1}{10}+\frac{1}{15}-\frac{1}{x}\right)$ + $\left(\frac{1}{10}+\frac{1}{15}\right)$ = $\frac{1}{6}$
⇒ $\frac{2}{10}$ + $\frac{2}{15}$ - $\frac{1}{x}$ = $\frac{1}{6}$
⇒ $\frac{1}{3}$ - $\frac{1}{x}$ = $\frac{1}{6}$
⇒ x = 6

Alternately,
Time taken for both filling pipes to fill the tank is 6 mins.

Due to drain pipe it taken another 6 mins to fill the tank. Hence, during the first 6 mins effectively no work was done.

∴ Work done by filling pipes in first 6 mins = Work done by the draining pipe in first 6 pipes.

In 6 mins filling pipes would’ve completely filled the tank, hence the draining pipe must have completely emptied the tank in first 6 mins.

∴ It takes 6 mins for the draining pipe to completely empty the tank.

Hence, option (e).

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