Three taps A, B and C can fill a tank in 12, | vedclass

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(D) The efficiency of tap $A = \frac{1}{12}$ tank/hour,tap $B = \frac{1}{15}$ tank/hour,and tap $C = \frac{1}{20}$ tank/hour.In the first hour,$A$ and

Vedclass · Accepted Answer

$7 	ext{ hours}$

Vedclass · Answer

(D) The efficiency of tap $A = \frac{1}{12}$ tank/hour,tap $B = \frac{1}{15}$ tank/hour,and tap $C = \frac{1}{20}$ tank/hour.In the first hour,$A$ and $B$ are open. Part of the tank filled $= \frac{1}{12} + \frac{1}{15} = \frac{5+4}{60} = \frac{9}{60} = \frac{3}{20}$ tank.In the second hour,$A$ and $C$ are open. Part of the tank filled $= \frac{1}{12} + \frac{1}{20} = \frac{5+3}{60} = \frac{8}{60} = \frac{2}{15}$ tank.Thus,in a cycle of $2$ hours,the total part of the tank filled $= \frac{3}{20} + \frac{2}{15} = \frac{9+8}{60} = \frac{17}{60}$ tank.In $3$ such cycles ($6$ hours),the part of the tank filled $= 3 	imes \frac{17}{60} = \frac{17}{20}$ tank.Remaining part to be filled $= 1 - \frac{17}{20} = \frac{3}{20}$ tank.In the $7^{th}$ hour,$A$ and $B$ are opened. They fill $\frac{3}{20}$ tank in exactly $1$ hour.Therefore,the total time taken $= 6 + 1 = 7$ hours.

Three taps $A, B$ and $C$ can fill a tank in $12, 15$ and $20$ hours respectively. If $A$ is kept open all the time,and $B$ and $C$ are opened for one hour each alternately,how long will it take to fill the tank?

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