When Alisha goes by car at 50 , km/h,she reac | vedclass

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Question

(A) Let $T$ be the scheduled time to reach the office.When traveling by car at $50 \, km/h$,the time taken is $T_1 = \frac{25}{50} = 0.5 \, hours = 30

Vedclass · Accepted Answer

$68$

Vedclass · Answer

(A) Let $T$ be the scheduled time to reach the office.When traveling by car at $50 \, km/h$,the time taken is $T_1 = \frac{25}{50} = 0.5 \, hours = 30 \, minutes$.Since she is $5 \, minutes$ late,the scheduled time $T = 30 - 5 = 25 \, minutes$.When traveling by motorbike,she reaches $3 \, minutes$ early,so the time taken is $T_2 = 25 - 3 = 22 \, minutes$.Convert $T_2$ to hours: $T_2 = \frac{22}{60} \, hours = \frac{11}{30} \, hours$.The speed of the motorbike $S = \frac{	ext{Distance}}{	ext{Time}} = \frac{25}{11/30} = \frac{25 	imes 30}{11} = \frac{750}{11} \approx 68.18 \, km/h$.Rounding to the nearest integer,the speed is $68 \, km/h$.

When Alisha goes by car at $50 \, km/h$,she reaches her office $5 \, minutes$ late. But when she takes her motorbike,she reaches $3 \, minutes$ early. If her office is $25 \, km$ away,what is the approximate average speed at which she rides her motorbike? (in $km/h$)

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