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

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

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16 Jun 2014, 03:25
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Expert's post
Another perspective:

Consider this way: from one end the buses start every 4 minutes and from the other end they start every 12 minutes. We can see that the buses, on the average, start every 6 min and this is the actual time between two consecutive buses.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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19 Jun 2014, 02:54
Whats the level of this question?

I got it wrong the first place
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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19 Jun 2014, 02:56
PareshGmat wrote:
Whats the level of this question?

I got it wrong the first place

This is definitely a 700+ question.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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16 Jan 2015, 04:34
I have found somebody ask for answering by tabulating the measurement. I do it by this provision. Please find the attach file for it.
Hope it help
Note:
Catch up: same work, diff rate and time
Meet up: same time, diff rate, together finish work
Attachments

Catch up - Meet up.docx [27.35 KiB]

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

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16 Jan 2015, 10:05
I'm trying to "think like the testmaker" and look at really tough questions like logical-reasoning problems rather than math problems.
I looked at the 12 and 4 and then at the answer choices. The buses and cyclist are traveling at constant speeds, so it seemed like 6 had the most in common with 12 and 4 since it's even and offers the least common multiple. Is there any mathematical logic to this or did I just get lucky?

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

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16 Jan 2015, 10:45
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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

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

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

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01 Feb 2015, 01:08
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Hi. I thought I'd write this, just in case someone's unaware of it.
I see a few recurring questions, here and in the other thread of the same question
The direction of the bus- This is a very legit question. We are thinking "Hey, so the buses that he meets up front, started from the opposite point, so how is d common?" Firstly, the buses are round trip buses, so, the ones that overtake the cyclist are among 'oncoming' too. And, when the question asks about consecutive buses, it wants to know the frequency of the buses. So, do not get distracted by the idea of 1 bus meeting the other.
Secondly , the reason the speeds are added and subtracted is to shorten the process, using Relative speeds concept.
(My experience- The gmat club math book is really helpful. Do read it before math forums. Almost all answers that look amazing here, in the forums, are conceptually summarized there)
But I have a question regarding this. According to relative speed (add opposite-direction speeds and subtract same-direction speeds) , the distance d has to be between bodies whose speeds we are adding or subtracting, right? Time t is time taken to cover such a gap. So then how can I consider d (distance between consecutive buses) with relative speeds (sum and difference of bus and cyclists speeds)??

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

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18 Jan 2016, 10:44
I stared at the question for 1 min when decided just to triage it.

8 is out immediately because 12-4 is too easy. 10 is out because it is too round. 5 is also likely out because it looks like an oddball.

well we have 12 and 4. those multiples of 3, 2 and 2 respectively. 9 has two factors of 3 so I dont like it. chose 6 because it nicely has 3 and 2 and passes my triage =))

how do u like it?
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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11 Feb 2016, 05:51
can this question be solved using the method shown here: distance-speed-time-word-problems-made-easy-87481-20.html ....
or do i have to learn another method just to solve this type of problems....

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

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28 Oct 2016, 05:12
VeritasPrepKarishma wrote:
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.

Hi Karishma,
Could you tell me why the distnace between the rear bus and the cyclist, the oncoming bus and the cyclist and the distance between both the buses same??????
By your explaination the distance between buses doubt is clear but what about the other two?

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

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03 Nov 2016, 04:22
Hi
I am having problems with this question..
This is my solution..
Please let me know where I am mistaken...
I am getting 12 as the answer
I am not able to get that how most of the people have equalled the distance between the 2 buses to the distance between the cyclist and oncoming bus and following bus..

I apologise for my bad handwriting

SINCE I DON'T HAVE MUCH EXPERIENCE IN POSTING ATTACHMENTS ON GMAT CLUB ,PLEASE NOTE THE 1ST PIC IS OF THE LAST PAGE OF MY SOLUTION , THE MIDDLE ONE IS THE CONTINUATION OF SOLUTION AND THE LAST PIC IS OF THE STARTING OF THE SOLUTION
Attachments

File comment: End of Solution

pg 3.jpg [ 73.58 KiB | Viewed 382 times ]

File comment: Continuation

pg 2.jpg [ 39.25 KiB | Viewed 381 times ]

File comment: start of solution

pg 1.jpg [ 103.72 KiB | Viewed 381 times ]

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

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23 Feb 2017, 15:52
Consider the cyclist's speed to be negligible. Of course he is encountering oncoming buses more frequently than he is being passed by buses, because he is heading toward oncoming buses. All buses travel at the same speed. So, consider that he is standing still. How many buses pass him, from either direction, every 12 minutes? Well, 1 comes from 1 direction and 3 come from the other direction, so 4 buses pass every 12 minutes. This averages out to 1 every 3 minutes. So, if 1 buses passes from either direction every 3 minutes, then from 1 direction 1 bus will pass every 6 minutes.
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A man cycling along the road noticed that every 12 minutes [#permalink]

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23 Apr 2017, 13:47
Hello Experts,
msk0657, Vyshak, abhimahna, Engr2012, Abhishek009, Skywalker18, VeritasPrepKarishma, Bunuel, mikemcgarry, ehsan090, gracie, ScottTargetTestPrep, rohit8865, alpham, tiklimumita, Nunuboy1994, maheshaero20

Can you someone please clarify what i am thinking/understanding from the question is correct or not:
I am not able to understand that how come within 12 or 4 minutes entire distance d is being traversed?
So after going over all the explanation on this thread and veritas page (url: http://www.veritasprep.com/blog/2012/08 ... -concepts/) i think i understand the given question as follows:

And based on this we get d = 12(b - c) = 4(b + c)
=> 12b - 12c = 4b + 4c
=> 8b = 16c
=> b = 2c

d = 12b−6b = 6b
Interval x = (between 2 buses leaving the same location/station)
Interval x = d/b = 6 | B

NOTE: speed of bus = b = s2 & speed of cyclist = c = s1
Attachments

b-c.jpg [ 817.95 KiB | Viewed 288 times ]

b+c.jpg [ 605.34 KiB | Viewed 286 times ]

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

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24 Apr 2017, 01:07
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manishtank1988 wrote:
Hello Experts,
msk0657, Vyshak, abhimahna, Engr2012, Abhishek009, Skywalker18, VeritasPrepKarishma, Bunuel, mikemcgarry, ehsan090, gracie, ScottTargetTestPrep, rohit8865, alpham, tiklimumita, Nunuboy1994, maheshaero20

Can you someone please clarify what i am thinking/understanding from the question is correct or not:
I am not able to understand that how come within 12 or 4 minutes entire distance d is being traversed?
So after going over all the explanation on this thread and veritas page (url: http://www.veritasprep.com/blog/2012/08 ... -concepts/) i think i understand the given question as follows:

And based on this we get d = 12(b - c) = 4(b + c)
=> 12b - 12c = 4b + 4c
=> 8b = 16c
=> b = 2c

d = 12b−6b = 6b
Interval x = (between 2 buses leaving the same location/station)
Interval x = d/b = 6 | B

NOTE: speed of bus = b = s2 & speed of cyclist = c = s1

And based on this we get d = 12(b - c) = 4(b + c)

From where did you get 4(b + c)?

The oncoming bus takes 6 mins to meet so it should be

d = 12(b - c) = 6(b + c)
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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24 Apr 2017, 07:53
VeritasPrepKarishma wrote:
manishtank1988 wrote:
Hello Experts,
msk0657, Vyshak, abhimahna, Engr2012, Abhishek009, Skywalker18, VeritasPrepKarishma, Bunuel, mikemcgarry, ehsan090, gracie, ScottTargetTestPrep, rohit8865, alpham, tiklimumita, Nunuboy1994, maheshaero20

Can you someone please clarify what i am thinking/understanding from the question is correct or not:
I am not able to understand that how come within 12 or 4 minutes entire distance d is being traversed?
So after going over all the explanation on this thread and veritas page (url: http://www.veritasprep.com/blog/2012/08 ... -concepts/) i think i understand the given question as follows:

And based on this we get d = 12(b - c) = 4(b + c)
=> 12b - 12c = 4b + 4c
=> 8b = 16c
=> b = 2c

d = 12b−6b = 6b
Interval x = (between 2 buses leaving the same location/station)
Interval x = d/b = 6 | B

NOTE: speed of bus = b = s2 & speed of cyclist = c = s1

And based on this we get d = 12(b - c) = 4(b + c)

From where did you get 4(b + c)?

The oncoming bus takes 6 mins to meet so it should be

d = 12(b - c) = 6(b + c)

Thanks VeritasPrepKarishma, you are right that the sum in the veritas article uses 6 min value for oncoming bus. But this question uses value of 4 minute. I think i should have included the question to remove any kind of confusion. But still please let me know if my understanding of this question is correct or not?

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

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

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10 Jun 2017, 09:52
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.

Bunuel Why is the distance when the bus overtakes the bicycle in 12 minutes and the distance when the bicycle is met by the opposing bus in 4 minutes same??? Please clarify. Thanks.

Posted from my mobile device

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

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10 Jun 2017, 10:16
Romannepal wrote:
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.

Bunuel Why is the distance when the bus overtakes the bicycle in 12 minutes and the distance when the bicycle is met by the opposing bus in 4 minutes same??? Please clarify. Thanks.

Posted from my mobile device

In once case a man is moving towards the bus and thus the relative speed is higher, which gives smaller time intervals and in another case a man is moving away from the bus and thus the relative speed is lower, which gives greater time intervals.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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17 Jun 2017, 23:06
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.[/quote]
Bunuel Why is the distance when the bus overtakes the bicycle in 12 minutes and the distance when the bicycle is met by the opposing bus in 4 minutes same??? Please clarify. Thanks.

Posted from my mobile device[/quote]

In once case a man is moving towards the bus and thus the relative speed is higher, which gives smaller time intervals and in another case a man is moving away from the bus and thus the relative speed is lower, which gives greater time intervals.[/quote]

HI Bunuel... first we consider the distance between two buses as d..

and every 12 mins a bus overtakes the cylcist... which means int he formula d/b-c,,,,d should indicate the distance between bus and cycle,,correct me if im rong

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

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18 Jun 2017, 00:31

B2 B1
----------------
C
The distance between two buses in same direction will be (speed of bus)*( time interval between two buses)= b*t

At T=0
B1 paseed C. Now B2 will be behind B1 by a distance of b*t.
B2 passes C at T=12 min
B2 travelled a distance of b*t with a relative speed of (b-c) in 12 mins

bt/(b-c)= 12
Similarily bt/(b+c)=4

Solve the two equations to get t=6 mins

Hope this helps

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

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20 Jun 2017, 20:13
Hi experts/Bunuel,

I did not understand as to how the overtake distance 'd' is equal to the meet distance 'd'. If it takes 12 and 4 minutes, does it not mean that the distance covered by overtake bus is different from the oncoming bus? Please clarify my not so profound doubt.

Warm Regards,
Karthik

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A man cycling along the road noticed that every 12 minutes   [#permalink] 20 Jun 2017, 20:13

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