Three pipes A, B and C can fill a cistern in | vedclass

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

Vedclass · Accepted Answer

$12$

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(B) Let the total capacity of the cistern be $1$ unit.Combined rate of pipes $A, B$ and $C$ is $\frac{1}{6}$ units per hour.In $2$ hours,the part filled by $A, B$ and $C$ together is $2 	imes \frac{1}{6} = \frac{1}{3}$ units.Remaining part to be filled $= 1 - \frac{1}{3} = \frac{2}{3}$ units.This remaining $\frac{2}{3}$ part is filled by $A$ and $B$ in $8$ hours.Therefore,the rate of $(A + B) = \frac{2/3}{8} = \frac{2}{24} = \frac{1}{12}$ units per hour.Since the rate of $(A + B + C) = \frac{1}{6}$ and $(A + B) = \frac{1}{12}$,the rate of pipe $C$ is $\frac{1}{6} - \frac{1}{12} = \frac{2-1}{12} = \frac{1}{12}$ units per hour.Thus,pipe $C$ alone can fill the cistern in $12$ hours.

Three pipes $A, B$ and $C$ can fill a cistern in $6$ $hrs$. After working together for $2$ $hrs$,$C$ is closed and $A$ and $B$ fill the remaining part of the cistern in $8$ $hrs$. Find the time (in $hrs$) in which the cistern can be filled by pipe $C$ alone.

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