A man cycling along the road noticed that every 12 minutes a bus overt

Hi,
I am not able to understand how did you arrive at 2 buese in every 6 mins. I do agree that in 12 mins there will be 4 buses but how did you come to the conclusion of 2 buses in every 6 minutes and that too from one from each side. Can you please clarify.

The buses travel at constant intervals, at constant speeds.

Imagine that the man is standing still at the center. He meets 4 buses every 12 mins - two from each side A and B. So every 6 mins, he meets a bus - one from each side. What happens if he starts walking towards A? He will meet buses from A more frequently and buses from B less frequently. Overall, he will still meet 4 buses in 12 mins.

Make a diagram to understand this:

Man M standing in the middle. Buses B at a distance of 6 mins from each other converging towards the man every 6 mins.

B............B.............B.............M.............B.............B.............B.

What happens if the man starts moving towards right at the same speed as the buses?
The bus from the left never meet him (since they will always be 6 mins away from him). But he meets a bus from the right every 3 mins. So in all, he still meets 4 buses in 12 mins. The speed of the man doesn't matter as long as it is less than or equal to the speed of the bus.
So we can imagine that he is standing still instead (to make it easier for us). The question tells us that he meets 4 buses in 12 min so he must meet 2 buses every 6 mins.

1. In the case of the overtaking bus, the speed of the cyclist needs to be deducted from that of the bus and in the case of the oncoming bus their
speed needs to be added.
2. These two speeds are in the ratio 1:3 because the time is in the ratio 3:1.
3. s1+s2 =3x -----(1) and s2-s1 = x --- (2) Therefore s2 = 2x --- (3) where s1 is the speed of the cyclist and s2 is the speed of the bus.
4. To find the time interval between consecutive buses, we need to assume the case when the buses alone are running and the cyclist is stationary. That is we need to consider s2 only and assume s1=0.
5. From (1) and (3) we know that the speed of (3) is 1.5 times less than the speed of (1.) Therefore the time taken will be 1.5 times more than in the case of (1)
6. In the case of (1), the time taken is 4 min. Therefore in the case of (3) it is 4*1.5=6 min. That is when the cyclist is stationary, the buses cross him every 6 min.

Therefore the time interval between 2 consecutive buses is 6 min

Another approach:

1. Assume the combined speed of the cyclist and the bus is s1+ s2. s1 is the speed of the cyclist and s2 is the speed of the bus.
2. From the time a bus starts till the time it meets the cyclist, assume the distance traveled is d and the time taken is m
3. In the case of overtaking the difference in the speeds needs to be taken since both are traveling in the same direction. The relative speed is s2- s1. Assume to travel the same distance d the time taken is n
5. From (2) and (3), we have, m(s1+s2) = n(s2-s1)=d. we know m is 4 min and n is 12 min.
6. d= 4(s1+s2) = 12(s2-s1)
7. We want to find d/ s2 = 12(s2-s1)/s2 = 12(s2/s2) - 12(s1/s2) = 12-6= 6 min

A man cycling along the road noticed that every 12 minutes a bus overtakes him and every 4 minutes he meets an oncoming bus. If all buses and the cyclist move at a constant speed, what is the time interval between consecutive buses?

A. 5 minutes
B. 6 minutes
C. 8 minutes
D. 9 minutes
E. 10 minutes

SOLUTION:
Bus from rear overtakes every 12 min i.e. speed = 1/12
Bus from front meets every 4 min i.e. speed = 1/4

For objects traveling in opposite direction, we SUBTRACT speeds. Therefore, the relative speed is:
(1/4) - (1/12) = 1/6
Hence, time interval = 6 min

ANSWER: B

Can someone link similar questions to iterate learnings on?

Check here: All the Different Types of Distance/Rate Problems

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Why do we assume the same d for buses in the same direction and buses in the opposite direction? Shoudn't we assume as d1 and d2 instead? Since d denoted distance covered between the cyclist and bus in same direction/opp direction.

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Check updated wording of the problem. Hope it clears things up.