A big tanker can be filled by two pipes A and | vedclass

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(B) Let the total time required to fill the tank be $x$ minutes.The rate of tap $A$ is $\frac{1}{60}$ tank per minute.The rate of tap $B$ is $\frac{1}

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) Let the total time required to fill the tank be $x$ minutes.The rate of tap $A$ is $\frac{1}{60}$ tank per minute.The rate of tap $B$ is $\frac{1}{40}$ tank per minute.The combined rate of tap $A$ and $B$ is $\frac{1}{60} + \frac{1}{40} = \frac{2+3}{120} = \frac{5}{120} = \frac{1}{24}$ tank per minute.According to the problem,tap $B$ is used for $\frac{x}{2}$ minutes and both taps $A$ and $B$ are used for the remaining $\frac{x}{2}$ minutes.Work done by tap $B$ + Work done by $(A+B) = 1$ (full tank).$\frac{x}{2} 	imes \frac{1}{40} + \frac{x}{2} 	imes \frac{1}{24} = 1$$\frac{x}{80} + \frac{x}{48} = 1$Taking the $LCM$ of $80$ and $48$,which is $240$:$\frac{3x + 5x}{240} = 1$$\frac{8x}{240} = 1$$\frac{x}{30} = 1$$x = 30$ minutes.

$A$ big tanker can be filled by two pipes $A$ and $B$ in $60$ and $40$ $min$ respectively. What time will it take to fill an empty tanker if tap $B$ is used for half of the total time and both taps $A$ and $B$ together are used for the remaining half of the time?

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