Question

Two pipes $A$ and $B$ can fill a tank in $6$ hours and $4$ hours respectively. If they are opened on alternate hours and if pipe $A$ is opened first, then in how many hours, the tank shall be full?

• A. $4$
• B. $4\frac{1}{2}$
• C. $5$
• D. $5\frac{1}{2}$

Answer 1

The correct option is C $5$

$A^{\prime }s$ work in $1$ hour $=\frac{1}{6}$ and $B^{\prime }s$ work in $1$ hour $=\frac{1}{4}$

$\Rightarrow$ $(A+B{)}^{\prime }s$ $2$ hour's work when opened alternately $=\frac{1}{6}+\frac{1}{4}=\frac{5}{12}$.

$\Rightarrow$ $(A+B{)}^{\prime }s$ $4$ hour's work when opened alternately $=\frac{10}{12}=\frac{5}{6}$

$\Rightarrow$ Remaining part $=1-\frac{5}{6}=\frac{1}{6}$

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

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