Two pipes A and B can fill a tank in 15 minut | vedclass

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(D) Part of the tank filled by pipe $A$ and $B$ in $1$ minute $= \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.In the first $4$ minutes

Vedclass · Accepted Answer

$14 	ext{ min } 40 	ext{ sec}$

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(D) Part of the tank filled by pipe $A$ and $B$ in $1$ minute $= \frac{1}{15} + \frac{1}{20} = \frac{4+3}{60} = \frac{7}{60}$.In the first $4$ minutes,the part of the tank filled $= 4 	imes \frac{7}{60} = \frac{28}{60} = \frac{7}{15}$.Remaining part of the tank to be filled $= 1 - \frac{7}{15} = \frac{8}{15}$.Since pipe $A$ is turned off,the remaining part is filled by pipe $B$ alone.Let the time taken by pipe $B$ to fill the remaining part be $x$ minutes.$\frac{1}{20} 	imes x = \frac{8}{15} \implies x = \frac{8 	imes 20}{15} = \frac{8 	imes 4}{3} = \frac{32}{3} = 10 \frac{2}{3}$ minutes.$10 \frac{2}{3}$ minutes $= 10$ minutes and $\frac{2}{3} 	imes 60$ seconds $= 10$ minutes $40$ seconds.Total time taken to fill the tank $= 4$ minutes $+ 10$ minutes $40$ seconds $= 14$ minutes $40$ seconds.

Two pipes $A$ and $B$ can fill a tank in $15$ $minutes$ and $20$ $minutes$ respectively. Both the pipes are opened together but after $4$ $minutes,$ pipe $A$ is turned off. What is the total time required to fill the tank?

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