A man walks from A to B and back in a certain | vedclass

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(D) Let the distance between $A$ and $B$ be $x 	ext{ km}$.In the first case,the total time taken is $T_1 = \frac{x}{3.5} + \frac{x}{3.5} = \frac{2x}{

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$7$

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(D) Let the distance between $A$ and $B$ be $x 	ext{ km}$.In the first case,the total time taken is $T_1 = \frac{x}{3.5} + \frac{x}{3.5} = \frac{2x}{3.5} = \frac{2x}{7/2} = \frac{4x}{7} 	ext{ hours}$.In the second case,the total time taken is $T_2 = \frac{x}{3} + \frac{x}{4} = \frac{4x + 3x}{12} = \frac{7x}{12} 	ext{ hours}$.According to the problem,the difference in time is $5 	ext{ minutes}$,which is $\frac{5}{60} = \frac{1}{12} 	ext{ hours}$.So,$T_2 - T_1 = \frac{1}{12}$.$\frac{7x}{12} - \frac{4x}{7} = \frac{1}{12}$.Taking the common denominator $(84)$:$\frac{49x - 48x}{84} = \frac{1}{12}$.$\frac{x}{84} = \frac{1}{12}$.$x = \frac{84}{12} = 7 	ext{ km}$.Therefore,the distance between $A$ and $B$ is $7 	ext{ km}$.

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