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 efficiency of pipes $A, B,$ and $C$ is $(A+B+C) = \frac{1}{6}$ units/hr.In $2$ hours,th

Vedclass · Accepted Answer

$12$

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(B) Let the total capacity of the cistern be $1$ unit.Combined efficiency of pipes $A, B,$ and $C$ is $(A+B+C) = \frac{1}{6}$ units/hr.In $2$ hours,the work done by $A, B,$ and $C$ is $2 	imes \frac{1}{6} = \frac{1}{3}$ units.Remaining work $= 1 - \frac{1}{3} = \frac{2}{3}$ units.This remaining work is done by $A$ and $B$ in $8$ hours.So,the efficiency of $(A+B) = \frac{2/3}{8} = \frac{2}{24} = \frac{1}{12}$ units/hr.Now,the efficiency of pipe $C = (A+B+C) - (A+B) = \frac{1}{6} - \frac{1}{12} = \frac{2-1}{12} = \frac{1}{12}$ units/hr.Therefore,the time taken by pipe $C$ to fill the cistern alone is $\frac{1}{1/12} = 12$ $hrs$.

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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