Question

Pipes, $A,B$ and $C$ are attached to a cistern $A$ and $B$ can fill it in $20$ min and $30$ respectively, while $C$ can empty it in $15$ min $HA,B$ and $C$ kept in operation successively for $1$ minute each, how soon will the cistern be filled?

• A. $167$ min
• B. $160$ min
• C. $166$ min
• D. $164$ min

Answer 1

The correct option is A $167$ min

Work done in $3$ min. $=$ $(\frac{1}{20}+\frac{1}{30}-\frac{1}{15})=\frac{1}{60}$
$\therefore$ Work done in $3\times 55=165$ $=$ $\frac{55}{60}$
$\therefore$ Remaining tank $=$ $(1-\frac{55}{60})=\frac{5}{60}=\frac{1}{12}$
Now its A's turn, $\frac{1}{20}$ part of the tank is filled by $A$ in $1$ minute.
Since there is still $(\frac{1}{12}-\frac{1}{20})=\frac{1}{30}$ tank to be filled, which will be filled by $B$ in $1$ min.
Therefore, required time $=(165+2)=167$ min.