Question

Two pipes $A$ and $B$ can fill a cistern is $12$ minutes and $15$ minutes respectively, but a third pipe $C$ can empty the full tank in $6$ minutes. $A$ and $B$ are kept open for $5$ minutes in the beginning and then $C$ is also opened. In what time will the cistern be emptied ?

• A. $30$ min
• B. $33$ min
• C. $37.5$ min
• D. $45$ min

Answer 1

The correct option is D $45$ min

Part filled in $5$ min $=5\times (\frac{1}{12}+\frac{1}{15})$

$=5\times \frac{9}{60}=\frac{3}{4}$

Part emptied in $1$ minute when all the pipes are opened $=\frac{1}{6}-(\frac{1}{12}+\frac{1}{15})=\frac{1}{60}$.

Now, $\frac{1}{60}$part is emptied in $1$ minute.

Therefore, $\frac{3}{4}$ part will be emptied in $60\times \frac{3}{4}=45$ minutes.