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

Walker,

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

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.

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..

Thanks for the explanation. Please clarify the following doubts.
Aren't we calculating the interval between 2 buses that move towards each other?
If yes, then would the interval not be 6b/b+b ?

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.

Karishma,

I did not understand the TWIST qtn. Please do explain me the question and let me try to solve

2nd qtn:
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)

i take LCM of 10 and 8 that is 40
now, in 40 mins, i meet 9 buses (4+5)
==> 9buses --> 40 mins
==> 1 bus --> 40/9 mins
==> 2 buses --> 80/9 mins, that is the time interval b/w 2 buses

is this correct?

Regards,
Murali.

The twist question - now imagine that there is a man sitting in one of the buses. My question is, at what frequency (i.e. after what time interval) will buses from opposite direction cross him (or his bus).
(e.g. a bus from opposite direction crosses his bus every t mins - so I want the value of t). This information is in addition to the given original question.

and yes, your answer to the next question is correct.

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.

Thanks bunnel for the explanation.
But I have a doubt here. what is the time interval between consecutive buses?
Isn't it the time interval between consecutive buses going in one direction. If that is the case then ans should be 12 min.
Thanks.

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...!!

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

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.

Can I assume that since Cyclist is meeting an oncoming bus every 4 minutes, his speed is 4km/hr? or lets assume that a bus crosses him every 12 minutes coming from back and in 4 minutes coming from front. Thus cumulative speed is 16km/hr however cyclist himself is also moving forward @4km/hr. Thus net effect is 12km/hr. Beyond that, I am confused.