Three pipes A, B and C can fill a tank in 6 h | vedclass

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(C) Let the total capacity of the tank be $1$ unit.Combined efficiency of pipes $A, B$, and $C$ is $\frac{1}{6}$ tank per hour.In $2$ hours, the part

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(C) Let the total capacity of the tank be $1$ unit.Combined efficiency of pipes $A, B$, and $C$ is $\frac{1}{6}$ tank per hour.In $2$ hours, the part of the tank filled by $A, B$, and $C$ is $2 	imes \frac{1}{6} = \frac{1}{3}$.Remaining part of the tank $= 1 - \frac{1}{3} = \frac{2}{3}$.This remaining $\frac{2}{3}$ part is filled by $A$ and $B$ in $7$ hours.Therefore, the combined efficiency of $A$ and $B$ is $\frac{2/3}{7} = \frac{2}{21}$ tank per hour.Efficiency of pipe $C = (	ext{Efficiency of } A+B+C) - (	ext{Efficiency of } A+B)$.Efficiency of $C = \frac{1}{6} - \frac{2}{21} = \frac{7 - 4}{42} = \frac{3}{42} = \frac{1}{14}$ tank per hour.Thus, pipe $C$ alone can fill the tank in $14$ hours.

Three pipes $A, B$ and $C$ can fill a tank in $6$ $hours$. After running them together for $2$ $hours$, the tap $C$ is closed and $A$ and $B$ fill the remaining part in $7$ $hours$. How much time (in $hours$) would $C$ take to fill the tank alone?

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