Discussion Forum

A tank is filtered with two inlet pipes $A$ and $B$. Both the pipes are kept open for $10$ minutes so that the tank is $\frac{2}{3}$ full and then pipe $A$ is closed. How much time will $B$ take to fill the remaining part of the tank ?

I. Pipe $A$ is thrice as fast as pipe $B$.

II. Pipe $B$ alone can fill the tank in $60$ minutes.

Answer:

Explanation:

I. Let $B$’s $1$ min. work $=\frac{1}{x}$. Then, $A$’s $1$ min. work $=\frac{3}{x}$.

$(A+B)$’s $1$ min. work $=\left(\frac{1}{x}+\frac{3}{x}\right)$ $=\frac{4}{x}$.

$(A+B)$’s $10$ min. work $=\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.

$\therefore$ $\frac{40}{x}=\frac{2}{3}$

$\Leftrightarrow x=60$.

$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(\frac{4}{x}\times 10\right)$ $=\frac{40}{x}$.

$\therefore$ $\frac{40}{x}=\frac{2}{3}$

$\Leftrightarrow x=60$.

$\therefore$ $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.

II. $B$’s $1$ min. work $=\frac{1}{60}$.

$\frac{1}{60}$ part is filled by $B$ in $1$ min.

$\frac{1}{3}$ part is filled by $B$ in $\left(60\times\frac{1}{3}\right)$ = 20 min.

Hence, the correct answer is (C).