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

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(D) Pipe $A$ fills the tank in $15$ minutes,so its rate is $\frac{1}{15}$ tank per minute.Pipe $B$ fills the tank in $20$ minutes,so its rate is $\fra

Vedclass · Accepted Answer

$14 \min 40 	ext{ sec}$

Vedclass · Answer

(D) Pipe $A$ fills the tank in $15$ minutes,so its rate is $\frac{1}{15}$ tank per minute.Pipe $B$ fills the tank in $20$ minutes,so its rate is $\frac{1}{20}$ tank per minute.Both pipes are opened together for $4$ minutes. The work done by both in $4$ minutes is $4 	imes (\frac{1}{15} + \frac{1}{20}) = 4 	imes (\frac{4+3}{60}) = 4 	imes \frac{7}{60} = \frac{7}{15}$.The remaining part of the tank to be filled is $1 - \frac{7}{15} = \frac{8}{15}$.Pipe $B$ fills this remaining part at a rate of $\frac{1}{20}$ per minute.Time taken by pipe $B$ to fill the remaining part $= \frac{8/15}{1/20} = \frac{8}{15} 	imes 20 = \frac{32}{3} = 10 	ext{ minutes and } 40 	ext{ seconds}$.Total time taken $= 4 	ext{ minutes} + 10 	ext{ minutes } 40 	ext{ seconds} = 14 	ext{ minutes } 40 	ext{ seconds}$.

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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