Two friends A and B are 30 , km apart and the | vedclass

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(D) Let the speed of $B$ be $v_B = x \, km/min$ and the speed of $A$ be $v_A = 3x \, km/min$.Since they move towards each other,the rate at which the

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$30$

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(D) Let the speed of $B$ be $v_B = x \, km/min$ and the speed of $A$ be $v_A = 3x \, km/min$.Since they move towards each other,the rate at which the distance between them decreases is $v_A + v_B = 3x + x = 4x \, km/min$.Given that the distance decreases at $2 \, km/min$,we have $4x = 2$,so $x = 0.5 \, km/min$.Thus,$v_B = 0.5 \, km/min$ and $v_A = 1.5 \, km/min$.In the first $10 \, minutes$,the distance covered by both is $10 	imes (v_A + v_B) = 10 	imes 2 = 20 \, km$.The remaining distance between them is $30 - 20 = 10 \, km$.After $10 \, minutes$,$A$ stops,so $B$ must cover the remaining $10 \, km$ at his speed $v_B = 0.5 \, km/min$.Time taken by $B$ to cover the remaining distance is $\frac{10}{0.5} = 20 \, minutes$.The total time from the start until $B$ meets $A$ is $10 \, minutes$ (initial travel) $+ 20 \, minutes$ (time for $B$ to reach $A$) $= 30 \, minutes$.

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