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. $8:30am$ B. $9:00am$ C. $9:30am$ D. $10:00am$

Hint: We have to consider the fact that the distance traveled by bicycle and scooter until the point of overtaking will be the same. The formula for distance, say d is $d = s \times t$, where s is the speed and t is the time taken.

Let the time taken by the bicycle is $t$.
As mentioned in the question the scooter leaves the same town after 1 hour.
Therefore, time taken will be $t - 1$
Now, the distance is the product of speed and time i.e., $d = s \times t$ where s is the speed and t is the time taken.

For bicycle men, speed is $10km/hr$ and time is $t$.
Therefore, the distance traveled by bicycle man will be
$\Rightarrow d = 10 \times t$
Now, for scooter man, speed is $30km/hr$ and time is $(t - 1)$
Therefore, The distance traveled for scooter man will be
$\Rightarrow d = 30(t - 1)$
As we know, the distance traveled by bicycle and scooter at the point of overtaking will be the same.
$\Rightarrow d = 10t = 30(t - 1) \\ \Rightarrow 20t = 30 \\ \Rightarrow t = \dfrac{3}{2}hr \\$
The time taken by the scooter man is $t - 1 = \dfrac{3}{2} - 1 = \dfrac{1}{2}hr$.
The scooter overtakes the bicycle after half an hour. i.e., $9:30am$

So, the correct answer is “Option C”.

Note: These types of questions come under the Meeting point questions type. A different kind of question may be, If two people travel from two points A and B towards each other, and they meet at point T. The Total Distance covered by them at the meeting will be AB. The Time taken by both of them to meet will be the same. As the Time is constant, Distances AT and BT will be in the ratio of their Speed. Say that the Distance between A and B is d. If two people are walking towards each other from A and B, When they meet for the First time, they together cover a Distance “d” When they meet for the second time, they together cover a Distance “3d”.