Two pipes A and B can fill a cistern in 4 min | vedclass

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Question

(A) Pipe $A$ fills $\frac{1}{4}$ of the cistern in $1$ minute.Pipe $B$ fills $\frac{1}{6}$ of the cistern in $1$ minute.In a cycle of $2$ minutes (alt

Vedclass · Accepted Answer

$4 \, 	ext{min} \, 40 \, 	ext{s}$

Vedclass · Answer

(A) Pipe $A$ fills $\frac{1}{4}$ of the cistern in $1$ minute.Pipe $B$ fills $\frac{1}{6}$ of the cistern in $1$ minute.In a cycle of $2$ minutes (alternating $A$ then $B$),the part filled is $\frac{1}{4} + \frac{1}{6} = \frac{3+2}{12} = \frac{5}{12}$.In $4$ minutes (two such cycles),the part filled is $2 	imes \frac{5}{12} = \frac{10}{12} = \frac{5}{6}$.The remaining part to be filled is $1 - \frac{5}{6} = \frac{1}{6}$.In the $5^{th}$ minute,pipe $A$ operates. Since pipe $A$ fills $\frac{1}{4}$ in $60$ seconds,it will fill $\frac{1}{6}$ in $\frac{1/6}{1/4} 	imes 60 = \frac{4}{6} 	imes 60 = 40$ seconds.Total time = $4$ minutes + $40$ seconds = $4 \, 	ext{min} \, 40 \, 	ext{s}$.

Two pipes $A$ and $B$ can fill a cistern in $4$ $minutes$ and $6$ $minutes,$ respectively. If these pipes are turned on alternately for $1$ $minute$ each,then how long will it take for the cistern to fill?

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