A boy started from his house by bicycle at 10 | vedclass

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(C) The boy started at $10:00 \, am$ and was caught at $1:30 \, pm$. The total time taken by the boy is $3 \, hours \, 30 \, minutes = 3.5 \, hours =

Vedclass · Accepted Answer

$18 \frac{2}{3}$

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(C) The boy started at $10:00 \, am$ and was caught at $1:30 \, pm$. The total time taken by the boy is $3 \, hours \, 30 \, minutes = 3.5 \, hours = \frac{7}{2} \, hours$.Distance covered by the boy = $	ext{Speed} 	imes 	ext{Time} = 12 	imes \frac{7}{2} = 42 \, km$.The elder brother started $1 \, hour \, 15 \, minutes$ later,so his travel time is $3 \, hours \, 30 \, minutes - 1 \, hour \, 15 \, minutes = 2 \, hours \, 15 \, minutes = 2.25 \, hours = \frac{9}{4} \, hours$.Since the distance covered by both is the same,let the speed of the scooter be $x \, km/h$.$x 	imes \frac{9}{4} = 42$.$x = \frac{42 	imes 4}{9} = \frac{14 	imes 4}{3} = \frac{56}{3} = 18 \frac{2}{3} \, km/h$.

$A$ boy started from his house by bicycle at $10:00 \, am$ at a speed of $12 \, km/h$. His elder brother started after $1 \, hour \, 15 \, minutes$ by scooter along the same path and caught him at $1:30 \, pm$. The speed of the scooter was (in $km/h$):

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