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

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(D) Work done by $A$ in $1$ hour $= \frac{1}{10}$.Work done by $A$ and $B$ in the second hour $= \frac{1}{10} + \frac{1}{12} = \frac{6+5}{60} = \frac{

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$4 	ext{ hours } 52 	ext{ minutes}$

Vedclass · Answer

(D) Work done by $A$ in $1$ hour $= \frac{1}{10}$.Work done by $A$ and $B$ in the second hour $= \frac{1}{10} + \frac{1}{12} = \frac{6+5}{60} = \frac{11}{60}$.Total work done in the first $2$ hours $= \frac{1}{10} + \frac{11}{60} = \frac{6+11}{60} = \frac{17}{60}$.Remaining work $= 1 - \frac{17}{60} = \frac{43}{60}$.Work done by $A, B$ and $C$ in $1$ hour $= \frac{1}{10} + \frac{1}{12} + \frac{1}{15} = \frac{6+5+4}{60} = \frac{15}{60} = \frac{1}{4}$.Time taken to complete the remaining work $= \frac{43/60}{1/4} = \frac{43}{60} 	imes 4 = \frac{43}{15} = 2 	ext{ hours } 52 	ext{ minutes}$.Total time $= 2 	ext{ hours} + 2 	ext{ hours } 52 	ext{ minutes} = 4 	ext{ hours } 52 	ext{ minutes}$.

Three pipes $A, B$ and $C$ can fill a cistern in $10$ $hours$,$12$ $hours$ and $15$ $hours$ respectively. First $A$ was opened. After $1$ $hour$,$B$ was opened and after $2$ $hours$ from the start of $A$,$C$ was also opened. Find the time in which the cistern is just full.

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