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Question
A man leaves a town at 8 a.m. on his bicycle moving at 10 km/hr. Another man leaves the same town at 9 a.m. on his scooter moving at 30 km/hr. At what time does he overtake the man on the bicycle?
A man leaves a town at 8 a.m. on his bicycle moving at 10 km/hr. Another man leaves the same town at 9 a.m. on his scooter moving at 30 km/hr. At what time does he overtake the man on the bicycle?
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Solution
The correct option is D 9.30 a.m.
Let 't' be the time taken for the scooter to travel the distance, so as to overtake the bicycle. The time taken by the bicycle to cover the same distance will be (t+1). The speed of the bicycle is 10 Km/hr.and that of the scooter is 30 km/hr.
Now, using the strategy, the distance travelled by the scooter as well as the bicycle till the point of overtake will be the same.
i.e. 30×t=10×(t+1)
⇒30t=10t+10
⇒30t−10t=10
⇒20t=10
⇒t=12 hour
= 30 mins
Hence, the scooter will overtake the bicycle at 9:30 a.m.
Let 't' be the time taken for the scooter to travel the distance, so as to overtake the bicycle.
The time taken by the bicycle to cover the same distance will be (t+1). The speed of the bicycle is 10 Km/hr.and that of the scooter is 30 km/hr.
Now, using the strategy, the distance travelled by the scooter as well as the bicycle till the point of overtake will be the same.
i.e. 30×t=10×(t+1)
⇒30t=10t+10
⇒30t−10t=10
⇒20t=10
⇒t=12 hour
= 30 mins
Hence, the scooter will overtake the bicycle at 9:30 a.m.
Now, using the strategy, the distance travelled by the scooter as well as the bicycle till the point of overtake will be the same.
i.e. 30×t=10×(t+1)
⇒30t=10t+10
⇒30t−10t=10
⇒20t=10
⇒t=12 hour
= 30 mins
Hence, the scooter will overtake the bicycle at 9:30 a.m.
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