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(C) Let the speed of the buses be $v_b = 20 	ext{ km/h}$ and the speed of the man be $v_m = x 	ext{ km/h}$.The distance between two consecutive buse

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$5$

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(C) Let the speed of the buses be $v_b = 20 	ext{ km/h}$ and the speed of the man be $v_m = x 	ext{ km/h}$.The distance between two consecutive buses is the distance covered by a bus in $10 	ext{ minutes}$.Distance $D = v_b 	imes 	ext{time} = 20 	imes (10/60) 	ext{ km} = 20/6 	ext{ km} = 10/3 	ext{ km}$.When the man moves in the opposite direction,the relative speed of the man with respect to the bus is $(v_b + v_m) = (20 + x) 	ext{ km/h}$.The man meets the buses at intervals of $8 	ext{ minutes}$,which means the distance $D$ is covered by the relative speed in $8 	ext{ minutes}$.$D = (v_b + v_m) 	imes 	ext{time interval}$.$10/3 = (20 + x) 	imes (8/60)$.$10/3 = (20 + x) 	imes (2/15)$.Multiply both sides by $15/2$:$(10/3) 	imes (15/2) = 20 + x$.$5 	imes 5 = 20 + x$.$25 = 20 + x$.$x = 5 	ext{ km/h}$.

Buses start from a bus terminal with a speed of $20 \text{ km/h}$ at intervals of $10 \text{ mins}$. What is the speed of a man coming from the opposite direction towards the bus terminal if he meets the buses at intervals of $8 \text{ mins}$? (in $\text{km/h}$)

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