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(A) Let the time taken by pipe $A$ be $x$ hours. Since pipe $A$ takes half the time of pipe $B$,the time taken by pipe $B$ is $2x$ hours.The combined

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(A) Let the time taken by pipe $A$ be $x$ hours. Since pipe $A$ takes half the time of pipe $B$,the time taken by pipe $B$ is $2x$ hours.The combined rate of pipes $A$ and $B$ is $\frac{1}{x} + \frac{1}{2x} = \frac{3}{2x}$.According to the problem,this combined rate is twice the rate of pipe $C$ (since they take half the time of $C$). Therefore,the rate of pipe $C$ is $\frac{1}{2} 	imes \frac{3}{2x} = \frac{3}{4x}$.Together,all three pipes take $6$ hours $40$ minutes,which is $6 + \frac{40}{60} = 6 + \frac{2}{3} = \frac{20}{3}$ hours.The combined rate of all three pipes is $\frac{1}{x} + \frac{1}{2x} + \frac{3}{4x} = \frac{1}{20/3} = \frac{3}{20}$.Solving for $x$: $\frac{4+2+3}{4x} = \frac{3}{20} \Rightarrow \frac{9}{4x} = \frac{3}{20}$.$12x = 180 \Rightarrow x = 15$ hours.

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