An empty tank can be filled by pipe A in 4 h | vedclass

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(C) Pipe $A$ fills $\frac{1}{4}$ of the tank in $1$ $h$.Pipe $B$ fills $\frac{1}{6}$ of the tank in $1$ $h$.In a cycle of $2$ $h$ (opening $A$ then $B

Vedclass · Accepted Answer

$4 \frac{2}{3} 	ext{ h}$

Vedclass · Answer

(C) Pipe $A$ fills $\frac{1}{4}$ of the tank in $1$ $h$.Pipe $B$ fills $\frac{1}{6}$ of the tank in $1$ $h$.In a cycle of $2$ $h$ (opening $A$ then $B$),the part of the tank filled is $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$.In $4$ $h$ (two cycles),the part of the tank filled is $2 	imes \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$.Remaining part of the tank $= 1 - \frac{5}{6} = \frac{1}{6}$.Now,it is pipe $A$'s turn to fill the remaining $\frac{1}{6}$ part.Time taken by pipe $A$ to fill $\frac{1}{6}$ part $= \frac{1/6}{1/4} = \frac{1}{6} 	imes 4 = \frac{2}{3} 	ext{ h}$.Total time taken $= 4 	ext{ h} + \frac{2}{3} 	ext{ h} = 4 \frac{2}{3} 	ext{ h}$.

An empty tank can be filled by pipe $A$ in $4$ $h$ and by pipe $B$ in $6$ $h$. If the two pipes are opened for $1$ $h$ each alternately,starting with pipe $A$,then in how much time will the tank be filled?

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