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

There is only one thing you need to understand in this question - When buses are approaching him from both the sides at a constant speed, it doesn't matter whether the man is standing still or cycling, the number of buses that he will meet will be the same. Convince yourself by imagining the case where the man is standing still. He will meet a bus from each side after every few mins. When he starts cycling in a direction, he is cycling away from buses of one side but towards buses of the other side. Since in 12 mins he meets total 4 buses (1 + 3), in 6 mins he meets 2 buses, one from each side, if he were standing still. So buses ply at a frequency of 6 mins each.

Twist: Same scenario. If a man is sitting inside one bus, at what frequency will a bus from opposite side cross him?

Also try the same question by changing the time taken by buses to meet the man to 10 min and 8 min respectively (instead of 12 mins and 4 mins)

When solving motion problems, I can't do without some drawings.
So, here is my version:

Denote by B the speed of the bus and by C the speed of the bicycle. Both are assumed to be constant.
Let T be the constant time interval between consecutive buses. It means, the distance between two consecutive buses is BT.

First scenario: buses and bicycle moving in the same direction and buses overtake the bicycle. Refer to the first attached drawing.

When bus \(B\) and bicycle \(C\) are at point \(n,\) next bus \(B^*\) is at point \(m.\)
Bus \(B^*\) will overtake bicycle \(C\) at point \(p.\)
In 12 minutes, bus \(B^*\) travels the distance \(mp\) and bicycle \(C\) travels the distance \(np.\)
We know that \(mp\) is the distance between consecutive buses, therefore \(mp=BT.\)
Translated into an equation mp=mn+np, so:
\(12B=BT+12C\) (1)

Second scenario: buses and bicycle moving in opposite directions and buses meet the bicycle. Refer to the second attached drawing.

When bus \(B\) and bicycle \(C\) are at point \(m\), next bus \(B^*\) is at point \(p.\)
Bus \(B^*\) will meet bicycle C at point \(n.\)
In 4 minutes, bus \(B^*\) travels the distance \(np\) and bicycle \(C\) travels the distance \(mn.\)
We know that \(mp\) is the distance between consecutive buses, therefore \(mp=BT.\)
Translated into an equation mp=mn+np, so:
\(BT=4C+4B\) (2)

Expressing \(BT\) from both equations, we get \(12(B-C)=4(B+C)\) from which \(B=2C\) (3)
Substituting (3) in (2) for example, we get, \(2CT=8C+4C\) from which \(T=6\) (minutes).

Answer B.

We can let b = speed of the bus and c = speed of the cyclist. Let d = the distance between two consecutive buses (in the same direction). Thus, we want to determine d/b, which is the time interval between two consecutive buses.

When the bus and the cyclist are traveling in the same direction, if we assume a bus overtakes the cyclist at a certain time, then it will be 12 minutes until the next bus overtakes him. Thus, we have:

12(b - c) = d

12b - 12c = d

When the bus and the cyclist are traveling in opposite directions, if we assume the cyclist meets an oncoming bus at a certain time, then it will be 4 minutes until the cyclist meets the next oncoming bus. Thus, we have:

4(b + c) = d

4b + 4c = d

Subtracting 4b + 4c = d from 12b - 12c = d, we have:

8b - 16c = 0

8b = 16c

b = 2c

We see that the bus is twice as fast as the cyclist. Substituting 2c for b in 12(b - c) = d (or in 4(b + c) = d), we have:

12(2c - c) = d

12c = d

Recall that the time interval between two consecutive buses is d/b; thus, we have:

d/b = 12c/2c = 6

Answer: B

Hi All,

To start, this question has some vague language in it - but we're meant to assume that all of the buses travel at the same constant speed and the cyclist is traveling at a different, constant speed. It can be solved with a mix of TESTing VALUES and TESTing THE ANSWERS - and drawing a little time-table.

To start, since the 1st bus overtakes the cyclist at the 12-minute mark, I'm going to set that distance at 1 mile. Since the cyclist traveled 1 mile in 12 minutes, then the cyclist would travel 5 miles in 60 minutes; the cyclist's speed would be 5 miles/hour.

12:00 Cyclist starts riding
12:12 1st bus passes cyclist

Since we're dealing with speeds and times AND both the times in the prompt (re: 12 minutes and 4 minutes) are EVEN numbers, it's likely that the correct answer will also be an EVEN number. Let's TEST Answer B first....

Answer B: 6 minutes

If the 'chasing' buses leave every 6 minutes, then we can add some more details to our time-table:

12:00 Cyclist starts riding and a bus leaves at the same time (we don't care about this bus though)
12:06: 1st bus leaves
12:12 1st bus passes cyclist

The 1st bus that passed the cyclist had to cover that 1 mile distance in just 6 minutes; that bus would then travel 10 miles in 60 minutes; thus, the bus's speed is 10 miles/hour. Don't forget that buses are leaving every 6 minutes, so there's more info to add to the table:

12:00 Cyclist starts riding and a bus leaves at the same time (we don't care about this bus though)
12:06: 1st bus leaves
12:12 1st bus passes cyclist; 2nd bus leaves

That second bus catches the cyclist at 12:24 - meaning at the 2-mile mark. Does that make sense with the speed of the bus? YES; that bus would travel those 2 miles in those 12 minutes!

12:24 2nd bus passes cyclist

Thus, these calculations 'match up' with what we were told in the prompt. Now, what about the 'approaching' buses...?

Since the cyclist is traveling 5 miles/hour and the buses each travel 10 miles/hour, when they APPROACH one another, we ADD their speeds to find the total distance traveled. In one hour, the total distance traveled between the cyclist and the approaching bus would be 5+10 = 15 miles; so their combined rate is 15 miles/hour. That's 15 miles in 60 minutes or 1 mile every 4 minutes. That 'fits' the other piece of information that we're given; since all of this fits everything that we're told, this MUST be the answer.



GMAT assassins aren't born, they're made,
Rich

I am not sure how you are assuming speeds, but if you are looking for a relatively theoretical solution, you can think in this way:

Say the cyclist is stationary at a point. Buses are coming from opposite directions (same speed, same time interval). A bus will meet the cyclist every t minutes from either direction. Let's say, a bus from each direction just met him. After t minutes, 2 more buses from opposite directions will meet him again and so on...

Now if the cyclist starts moving, (cyclist speed = c and bus speed = b), the ratio of the relative speeds of the buses is the inverse of the ratio of time taken i.e. it will be 4:12

(b-c):(b+c) = 4:12 which gives you c = (1/2)b

This means that the bus travelling at a relative speed which is half its usual speed (b-c = b/2) takes 12 minutes to meet the man. If it were travelling at its usual speed, it would have taken 12/2 = 6 mins to meet the man.

d = distance the bus and cyclist togather run
c = cyclist's speed
b = bus's speed

d/(c+b) = 4
d/(b-c) = 12
d/(c+b) = d/3(b-c)
3c+3b = b-c
4c = 2b
b = 2c, or
c = c/2

we need bus's time to complet d distance. so replace c in terms of b:

d/(c+b) = 4
d/(b/2 + b) = 4
d/b = 4x3/2
d/b = 6

Thats the time taken to pass d for a bus. after every 6 minuets another bus comes.

So it is B.

I know this is another hard question but when the concept is clear, it wont be difficult to get the result..

yeah if you standing at a busstop then whats the time interval of arrival of bus.--> you may take the question this way.

use relative velocity:

vel of bus=a
vel of cyclist=b
distance between them=d

now,
d=12(a-b)=4(a+b)

hence d=6a
so time interval is d/a=6 min

hope this clarifies...!!

Not sure I understood your question...

Anyway: question asks "what is the time interval between consecutive buses". Or time intervals between subsequent bus arrivals to a given bus stop (some static point). Which is: constant distance between two subsequent buses divided by the constant rate of these buses d/b. After some calculations we've gotten that d=6b, hence d/b=6b/b=6.

Same scenario. If a man is sitting inside one bus, at what frequency will a bus from opposite side cross him?

Ok let me answer the twist question. Let's say the bus in which the man is sitting is not moving. Then a bus from opposite direction crosses him every 6 mins because that is the frequency of the buses. Now, since his bus is also actually moving with the same speed towards the buses coming from opposite direction, he will meet those buses in 3 mins (half the distance will be covered by his bus and half by the bus coming from opposite direction). So he will meet a bus every 3 minutes.

You see in ONE DIRECTION, the distance between two buses is "d", the speed of the buses is "b"(lets just say its km/minute).
So to find this interval, we take this distance divided by speed of bus(all in ONE DIRECTION). so d/b.

Now through algebra 12(a-b)=4(a+b) , we find that distance "d" is equal to "6b". remember d is distance between two buses in ONE DIRECTION. And so we do the division and get the answer 6 minutes.

Just because we do this algebra equation d=12(a-b)=4(a+b), doesn't mean that the nature of b,c or d has changed. It stays true to its original meaning.

No idea what else can clear up your confusion :S

Yes. we have to assume. If the distance is not the same, the buses will collect in the one of the end stations and we have many possible solutions.



1. I wrote two equations for the time intervals of both the oncoming and overtaking buses using formula: t=L/V
2. I divided one equation by other one in order to exclude the distance between buses.
3. I found relationship between the speed of the buses and the speed of the cyclist
4. I used the finding and got L/Vbus (the time interval between consecutive buses) from first equation.

Assume that starting from a bus station, all buses run at the same speed of 50 mph.
Say a bus starts at 12:00 noon. Another starts at 1:00 pm i.e. exactly one hr later on the same route. Can we say that the previous bus is 50 miles away at 1:00 pm? Yes, so the distance between the two buses initially will be 50 miles. The 1 o clock bus also runs at 50 mph. Will the distance between these two buses always stay the same i.e. the initial 50 miles? Since both buses are moving at the same speed of 50 mph, relative to each other, they are not moving at all and the distance between them remains constant.

The exact same concept is used in this question.

Walker,

If possible can can you dumb it down a little for us lowly earthlings . Thanks in advance.

Consecutive buses mean consecutive buses in one direction, how else? So, if the distance between two consecutive buses is \(d\) and the rate of the bus is \(b\) then \(Interval=\frac{d}{b}\).

Hope it helps.

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.

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