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# A man cycling along the road noticed that every 12 minutes

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A man cycling along the road noticed that every 12 minutes  [#permalink]

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Updated on: 21 Sep 2013, 07:53
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: a-man-cycling-along-the-road-noticed-that-every-12-minutes-88723.html

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Originally posted by noboru on 05 Nov 2009, 13:24.
Last edited by Bunuel on 21 Sep 2013, 07:53, edited 2 times in total.
Edited the question and added the OA
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06 Nov 2009, 11:11
6
1
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

Good question. +1.

Let's say the distance between the buses is $$d$$. We want to determine $$Interval=\frac{d}{b}$$, where $$b$$ is the speed of bus.

Let the speed of cyclist be $$c$$.

Every 12 minutes a bus overtakes cyclist: $$\frac{d}{b-c}=12$$, $$d=12b-12c$$;

Every 4 minutes cyclist meets an oncoming bus: $$\frac{d}{b+c}=4$$, $$d=4b+4c$$;

$$d=12b-12c=4b+4c$$, --> $$b=2c$$, --> $$d=12b-6b=6b$$.

$$Interval=\frac{d}{b}=\frac{6b}{b}=6$$

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08 Nov 2009, 10:05
Correct me Bunuel if I am wrong. I am attempting this question in a different manner

If the cyclist is static then the time interval between the two buses should be 8.
But as the cyclist is travelling but has a velocity less than that of buses the time interval should be less than 8 and greater than 4. So choice C, D and E are out

As the time intervals of the buses are even and as the cyclist velocity as a net effect on each of the buses velocity. The time interval should be even

That leaves us with choice B
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08 Nov 2009, 10:47
Correct me Bunuel if I am wrong. I am attempting this question in a different manner

If the cyclist is static then the time interval between the two buses should be 8.
But as the cyclist is travelling but has a velocity less than that of buses the time interval should be less than 8 and greater than 4. So choice C, D and E are out

As the time intervals of the buses are even and as the cyclist velocity as a net effect on each of the buses velocity. The time interval should be even

That leaves us with choice B

About the first red part: we determined that the interval is 6 min, which IS exactly the interval when cyclist is static. So 8 is not correct. I believe you just divided 12+4 by 2.

About the second red part: if we had 16min instead of 12, than the answer would be 77/5min, which is not even, so it's also incorrect (by the way 77/5 is more than (16+4)/2=10). I'd say that we can not determine whether the interval is odd or even if we knew only that the intervals of overtaking and meeting buses are even or odd.

So, I think it's a little bit more complicated than this.
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08 Nov 2009, 12:16
Thanks Bunuel
As for the first part I subtracted 4 from 12 which is 8. As the cyclist velocity is also affecting the time interval (he is adding his velocity to the buses that are coming from the opposite side and they are more numerous than the buses which are overtaking him). The time interval should be less than 8.

And if the time interval is less than 4 => the cyclist velocity is more than that of the buses, which is not true.

So the time interval lies between 4 min and 8 min.

As for the second part
I guess it is more complicated for assumptions without doing calculations
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01 Dec 2009, 10:39
1
Let x be the speed of the bus and y be the speed of man
Since the man meets oncoming buses 3 times as fast as they overtake him. We have an equation

x+y = 3(x-y) =>y=x/2

When the bus and the man travel in the same direction the speed of the buses relative to the man is 2y-y=y, while the actual speed of the bus is 2y.
If men is stationary the buses would pass him every 12*1/2=6 min

Or

When the bus and the man travel in the opposite direction the speed of the buses relative to the man is 2y+y=3y, while the actual speed of the bus is 2y.
If the men is stationary the buses would pass him every 4*3/2=6 min
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24 Jan 2011, 07:55
Hi Bunuel

Could you please elaborate on how you got these two equations :

Every 12 minutes a bus overtakes cyclist: d/b-c = 12;

Every 4 minutes cyclist meets an oncoming bus: d/b+c = 4;

What I want to know is, how is the denominator 3 here ?

Regards,
Subhash
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24 Jan 2011, 08:31
subhashghosh wrote:
Hi Bunuel

Could you please elaborate on how you got these two equations :

Every 12 minutes a bus overtakes cyclist: d/b-c = 12;

Every 4 minutes cyclist meets an oncoming bus: d/b+c = 4;

What I want to know is, how is the denominator 3 here ?

Regards,
Subhash

When cyclist meets an oncoming bus their relative speed (combined speed) is b+c so the time needed to cover the distance d will be d/(b+c);

And when a bus overtakes the cyclist their relative speed (combined speed) is b-c (so less than in previous case) so the time needed to cover the distance d will be d/(b-c).

Hope it's clear.
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24 Jan 2011, 18:20
Hi

Sorry, I made a typo error earlier. What I wanted to know is, how is "d" the numerator, please explain that.

Regards,
Subhash
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24 Jan 2011, 19:41
5
1
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

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)
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25 Jan 2011, 02:28
1
subhashghosh wrote:
Hi

Sorry, I made a typo error earlier. What I wanted to know is, how is "d" the numerator, please explain that.

Regards,
Subhash

time*rate=distance --> time=distance/rate --> in our case d is the distance between the buses, b-c is relative speed (rate) and 12 is the time needed for a bus to overtakes the cyclist, so: 12=d/(b-c).

Hope it's clear.
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25 Jan 2011, 02:29
VeritasPrepKarishma wrote:
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

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)

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.
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25 Jan 2011, 03:58
1
muralimba wrote:
VeritasPrepKarishma wrote:
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

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)

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.

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27 Jan 2011, 08:48
3
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.
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28 Jan 2011, 10:34
Karishma, thanks. +1
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28 Feb 2011, 16:59
Bunuel wrote:
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

Good question. +1.

Let's say the distance between the buses is $$d$$. We want to determine $$Interval=\frac{d}{b}$$, where $$b$$ is the speed of bus.

Let the speed of cyclist be $$c$$.

Every 12 minutes a bus overtakes cyclist: $$\frac{d}{b-c}=12$$, $$d=12b-12c$$;

Every 4 minutes cyclist meets an oncoming bus: $$\frac{d}{b+c}=4$$, $$d=4b+4c$$;

$$d=12b-12c=4b+4c$$, --> $$b=2c$$, --> $$d=12b-6b=6b$$.

$$Interval=\frac{d}{b}=\frac{6b}{b}=6$$

HI bunuel, just a quick question - Why is the "d" in both equation equal to each other. I am not able to grasp how you are able to equate the distance intervals. Kindly clarify.
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01 Mar 2011, 01:17
shasidharancsp wrote:
Bunuel wrote:
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

Good question. +1.

Let's say the distance between the buses is $$d$$. We want to determine $$Interval=\frac{d}{b}$$, where $$b$$ is the speed of bus.

Let the speed of cyclist be $$c$$.

Every 12 minutes a bus overtakes cyclist: $$\frac{d}{b-c}=12$$, $$d=12b-12c$$;

Every 4 minutes cyclist meets an oncoming bus: $$\frac{d}{b+c}=4$$, $$d=4b+4c$$;

$$d=12b-12c=4b+4c$$, --> $$b=2c$$, --> $$d=12b-6b=6b$$.

$$Interval=\frac{d}{b}=\frac{6b}{b}=6$$

HI bunuel, just a quick question - Why is the "d" in both equation equal to each other. I am not able to grasp how you are able to equate the distance intervals. Kindly clarify.

Because d, the distance between any two subsequent buses, is constant.
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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15 May 2013, 05:03
I guess this question is simple but the stem is kind of difficult to understand.
What is asked is what is " what is the time interval between buses travelling in the same direction"

And then you can think of the stem as
if the cyclist and this bus travel in the same direction he meets one every 12 mins

If the cyclist and the bus travel in the opposite direction they meet once every 4 mins

And then Bunnels explanation in his 1st post will follow ... bunnel am i thinking correctly here ?
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02 Jun 2013, 05:37
VeritasPrepKarishma wrote:
noboru wrote:
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?

5 minutes
6 minutes
8 minutes
9 minutes
10 minutes

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)

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.
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03 Jun 2013, 04:03
2
davidfrank wrote:
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.
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Re: bus 573   [#permalink] 03 Jun 2013, 04:03

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