Question

Pipe A can fill a tank in $4$ hours and pipe B can fill it in $6$ hours. If they are opened at alternate hours and if pipe A is opened first, in how many hours will the tank be filled?

• A. $5$ hour
• B. $5\frac{2}{3}$ hours
• C. $4\frac{2}{3}$ hours
• D. 4 hours

Answer 1

Answer: (C)

Part of tank filled by $A$ in $1hr:\frac{1}{4}$

Part of tank filled by $B$ in $1hr:\frac{1}{6}$

$\because$ pipe $A$ and pipe $B$ are opened alternately.

$\therefore$ Part of tank filled by $A$ and $B$ respectively in $2hr=\frac{1}{4}+\frac{1}{6}$

$=\frac{3+2}{12}=\frac{5}{12}$

Part of tank filled by $A$ and $B$ respectively $=1-\frac{5}{6}=\frac{6-5}{6}=\frac{1}{6}$

Now $\frac{1}{6}$th part of tank filled by $A$

$\therefore$ time taken to fill $\frac{1}{6}$th part by $A$

$=\frac{\frac{1}{6}}{\frac{1}{4}}=\frac{1}{6}\times \frac{4}{1}=\frac{2}{3}$

$\therefore$ total time taken $=4+\frac{2}{3}=4\frac{2}{3}$hr