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A man cycling along the road noticed that every 12 minutes [#permalink]
04 Jan 2010, 03:44

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Question Stats:

35% (02:54) correct
65% (01:51) wrong based on 327 sessions

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? Thank you in advance!!

Re: test m08 question n18 [#permalink]
04 Jan 2010, 03:57

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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? Thank you in advance!!

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;

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.

Re: test m08 question n18 [#permalink]
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

Answer: 6 minutes.

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 ?

Re: test m08 question n18 [#permalink]
30 Apr 2010, 00:59

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Expert's post

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

Answer: 6 minutes.

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.

Re: test m08 question n18 [#permalink]
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

Answer: 6 minutes.

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.

Re: test m08 question n18 [#permalink]
02 Apr 2012, 21:44

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shadabkhaniet wrote:

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...!!

_________________

Practice Practice and practice...!!

If my reply /analysis is helpful-->please press KUDOS If there's a loophole in my analysis--> suggest measures to make it airtight.

Re: test m08 question n18 [#permalink]
02 Apr 2012, 21:54

kraizada84 wrote:

shadabkhaniet wrote:

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...!!

Thanks kraizada84 for explanation, but still yeah if you standing at a busstop then whats the time interval of arrival of bus.--> you may take the question this way. not making sense to me. If you are standing at a bus stop then the interval of arrival of bus will be in one direction only. But here we are calculating for both direction.

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

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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

_________________

Please give KUDOS if you find that my post adds value! Thank you!!

Re: test m08 question n18 [#permalink]
02 Apr 2012, 23:51

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Expert's post

shadabkhaniet wrote:

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}.

Re: test m08 question n18 [#permalink]
03 Apr 2012, 01:03

Bunuel wrote:

shadabkhaniet wrote:

Bunuel wrote:

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}.

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

Answer: B (6 minutes).

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)?

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
30 Apr 2012, 23:36

Expert's post

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

Answer: B (6 minutes).

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.

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

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
03 May 2012, 11:21

Expert's post

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.

No you can not assume that. Please refer to the solutions given above.

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
04 May 2012, 19:13

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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.