Discussion Forum

Two pipes $A$ and $B$ can fill a tank in $6$ hours and $14$ hours respectively. If they are opened on alternative hours and if pipe $A$ is opened first, in how many hours, the tank shall be full ?

Answer:

Explanation:

$A$’s work in $1$ hour $=\frac{1}{6}$

$B$’s work in $1$ hour $=\frac{1}{4}$

$(A+B)$’s $2$ hour’s work when opened alternatively $=\left(\frac{1}{6}+\frac{1}{4}\right)$ $=\frac{5}{12}$.

$(A+B)$’s $4$ hour’s work when opened alternatively $=\frac{10}{12}$ $=\frac{5}{6}$.

Remaining part $=\left(1-\frac{5}{8}\right)$ $=\frac{1}{6}$.

Now, it is $A$’s turn and $\frac{1}{6}$ part is filled by $A$ in $1$ hour.

$\therefore$ Total time taken to fill the tank $=(4+1)$ hrs = 5 hrs.