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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: 02 Apr 2012, 23:43
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Difficulty:

95% (hard)

Question Stats:

35% (02:20) correct 65% (02:09) wrong based on 1174 sessions

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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

I didn't get how to solve this problem. Can someone explain me more detailed than solution provided by the test?

Originally posted by lucalelli88 on 04 Jan 2010, 03:44.
Last edited by Bunuel on 02 Apr 2012, 23:43, edited 1 time in total.
Edited the question and added the OA
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Re: test m08 question n18  [#permalink]

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04 Jan 2010, 03:57
56
51
lucalelli88 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?

I didn't get how to solve this problem. Can someone explain me more detailed than solution provided by the test?

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

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: $$\frac{d}{b+c}=4$$, $$d=4b+4c$$;

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

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

Hope it helps.
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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24 Aug 2012, 22:52
9
1
lucalelli88 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?

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

I didn't get how to solve this problem. Can someone explain me more detailed than solution provided by the test?

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).

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##### General Discussion
Intern
Joined: 23 Dec 2009
Posts: 21
Re: test m08 question n18  [#permalink]

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04 Jan 2010, 07:55
Thank you very much!!Great explanation!!+1 for you!

I think the question is pretty hard: do you think that is possible to find a question so hard in the real GMAT?
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29 Apr 2010, 08:03
1
neoreaves wrote:

I might be VERY wrong..But when i don't get the answers..I go off track..So I'm not sure that this is the answer..
But marking the answer with some guess is better than NO guess.!!

Let say the road is a straight line AB.
The cyclist starts frm A and Bus starts frm B.

So in 4 min, let the distance traveled by Cyc is x, so distance traveled by Bus is (AB - x)
so 4 = x/c...........................1
and 4= (AB - x)/b..................................2
where c and b are respective speed of the cyclist and bus

Also in 12 mins, distance traveled by cyc is 3x.

so 12= 3x/(b-c) and x= 4c (frm 1)
so 1= c/(b-c)
and b=2c........................3

put the value of b frm 3 in equation 2.

We get AB=12c

So now the total distance is 12c and bus speed is 2c. The bus travels in 6 min...

It's A COMPLETE WILD GUESS..!! I was not able to get answer in 2 min..Let me knw the OA and OE.
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Re: test m08 question n18  [#permalink]

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30 Apr 2010, 00:21
Bunuel wrote:

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 $$c$$.

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

Hope it helps.

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 ?
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Posts: 52390
Re: test m08 question n18  [#permalink]

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30 Apr 2010, 00:59
4
1
Fiver wrote:
Bunuel wrote:

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 $$c$$.

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

Hope it helps.

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.
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Re: test m08 question n18  [#permalink]

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02 Apr 2012, 19:05
Bunuel wrote:
Fiver wrote:
Bunuel wrote:

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 $$c$$.

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

Hope it helps.

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.

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.
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Re: test m08 question n18  [#permalink]

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02 Apr 2012, 21:44
3
2
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...!!
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Posts: 35
Re: test m08 question n18  [#permalink]

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02 Apr 2012, 23:22
2
2
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
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Re: test m08 question n18  [#permalink]

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02 Apr 2012, 23:51
1
Bunuel wrote:
Fiver wrote:
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.

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.

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

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30 Apr 2012, 12:28
Hi Bunuel,

This is regarding the solution you gave:

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

when you say distance between 2 buses is d, you mean to buses that start at opposite ends right?
ex.
after 12 mins
(POINT A)starting point for busA --------------cyclist(busA meets him here)-----------------staring point for busB
-------------------------distance d -----------------------

then how is following possible coz d is the total distance not the distance between busA staring point and meeting point with cyclist.
Every 12 minutes a bus overtakes cyclist: \frac{d}{b-c}=12, d=12b-12c;

Plus, is the question asking for interval between when 2 buses leave staring point of busA(POINT A)?

thanks,
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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30 Apr 2012, 23:36
kartik222 wrote:
Hi Bunuel,

This is regarding the solution you gave:

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

when you say distance between 2 buses is d, you mean to buses that start at opposite ends right?
ex.
after 12 mins
(POINT A)starting point for busA --------------cyclist(busA meets him here)-----------------staring point for busB
-------------------------distance d -----------------------

then how is following possible coz d is the total distance not the distance between busA staring point and meeting point with cyclist.
Every 12 minutes a bus overtakes cyclist: \frac{d}{b-c}=12, d=12b-12c;

Plus, is the question asking for interval between when 2 buses leave staring point of busA(POINT A)?

thanks,

No, the distance between the buses is $$d$$ means that $$d$$ is the distance between two subsequent buses.
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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02 May 2012, 22:02
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.
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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04 May 2012, 19:13
7
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manjeet1972 wrote:
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.

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

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15 Jul 2012, 20:09
why is the distance between buses a constant value?
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Re: A man cycling along the road noticed that every 12 minutes  [#permalink]

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15 Jul 2012, 21:38
3
teal wrote:
why is the distance between buses a constant value?

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.
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Re: test m08 question n18  [#permalink]

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15 Jul 2012, 22:49
1
Bunuel wrote:
lucalelli88 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?

I didn't get how to solve this problem. Can someone explain me more detailed than solution provided by the test?

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

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: $$\frac{d}{b+c}=4$$, $$d=4b+4c$$;

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

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

Hope it helps.

Hi,

I m not sure how this formula of Interval has been arrived?. Is it a standard formula???

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

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16 Jul 2012, 04:02
teal wrote:
why is the distance between buses a constant value?

Buses travel on schedule so, the distance between two successive buses must be some constant value.

bhavinshah5685 wrote:
Hi,

I m not sure how this formula of Interval has been arrived?. Is it a standard formula???

Interval between two successive buses is the time, so it equals time=distance/rate.
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Re: test m08 question n18  [#permalink]

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23 Jun 2013, 20:26
FTGNGU wrote:

The cyclist is moving in one direction, say due east. Every 4 mins, he meets a bus coming towards him (going due West)

Cyc ----> <-------- Bus (Bus coming towards him)

Cyc-><-Bus (Meets the bus)

<------Bus Cycl -------> (They cross each other)

This happens with a different bus every 4 mins.

Every 4 mins, he meets an oncoming bus. So he meets a bus coming towards him at 12:00 noon. Then, at 12:04 pm, he meets another bus coming towards him. Then again at 12:08 pm and so on.

Their relative speed = Speed of cyclist + Speed of Bus
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Re: test m08 question n18 &nbs [#permalink] 23 Jun 2013, 20:26

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