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(C) Let $V$ be the speed of the bus. The speed of the cyclist is $v = 20 \; km/h$.When the bus moves in the direction of the cyclist,the relative spee

Vedclass · Accepted Answer

$9 \; min$ and $40 \; km/h$

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(C) Let $V$ be the speed of the bus. The speed of the cyclist is $v = 20 \; km/h$.When the bus moves in the direction of the cyclist,the relative speed is $(V - 20) \; km/h$. The time interval between buses passing the cyclist is $18 \; min = 18/60 \; h$. The distance between two consecutive buses is $V 	imes (T/60)$. Thus,$(V - 20) 	imes (18/60) = V 	imes (T/60) \implies (V - 20) 	imes 18 = VT \quad \dots (i)$.When the bus moves in the opposite direction,the relative speed is $(V + 20) \; km/h$. The time interval between buses passing the cyclist is $6 \; min = 6/60 \; h$. Thus,$(V + 20) 	imes (6/60) = V 	imes (T/60) \implies (V + 20) 	imes 6 = VT \quad \dots (ii)$.Equating $(i)$ and $(ii)$:$(V - 20) 	imes 18 = (V + 20) 	imes 6$$3(V - 20) = V + 20$$3V - 60 = V + 20$$2V = 80 \implies V = 40 \; km/h$.Substituting $V = 40$ in $(ii)$:$(40 + 20) 	imes 6 = 40 	imes T$$60 	imes 6 = 40T$$360 = 40T \implies T = 9 \; min$.

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