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(D) Part of the tanker filled by pipe $A$ in $1$ minute $= \frac{1}{60}$.Part of the tanker filled by pipe $B$ in $1$ minute $= \frac{1}{40}$.Let the

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(D) Part of the tanker filled by pipe $A$ in $1$ minute $= \frac{1}{60}$.Part of the tanker filled by pipe $B$ in $1$ minute $= \frac{1}{40}$.Let the total time taken to fill the tank be $2x$ minutes.According to the problem,pipe $B$ works alone for $x$ minutes,and pipes $A$ and $B$ work together for $x$ minutes.Therefore,the total work done is: $\frac{1}{40} 	imes x + (\frac{1}{60} + \frac{1}{40}) 	imes x = 1$.Simplifying the equation: $\frac{x}{40} + (\frac{2+3}{120})x = 1$.$\frac{x}{40} + \frac{5x}{120} = 1$.$\frac{x}{40} + \frac{x}{24} = 1$.Taking the least common multiple $(120)$: $\frac{3x + 5x}{120} = 1$.$\frac{8x}{120} = 1$.$x = \frac{120}{8} = 15$ minutes.Total time taken $= 2x = 2 	imes 15 = 30$ minutes.

$A$ large tanker can be filled by two pipes $A$ and $B$ in $60$ $minutes$ and $40$ $minutes$ respectively. How many minutes will it take to fill the tanker from an empty state if pipe $B$ is used alone for half the time and pipes $A$ and $B$ are used together for the other half?

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