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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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New post Updated on: 16 Aug 2013, 19:41
1. In the case of the overtaking bus, the speed of the cyclist needs to be deducted from that of the bus and in the case of the oncoming bus their
speed needs to be added.
2. These two speeds are in the ratio 1:3 because the time is in the ratio 3:1.
3. s1+s2 =3x -----(1) and s2-s1 = x --- (2) Therefore s2 = 2x --- (3) where s1 is the speed of the cyclist and s2 is the speed of the bus.
4. To find the time interval between consecutive buses, we need to assume the case when the buses alone are running and the cyclist is stationary. That is we need to consider s2 only and assume s1=0.
5. From (1) and (3) we know that the speed of (3) is 1.5 times less than the speed of (1.) Therefore the time taken will be 1.5 times more than in the case of (1)
6. In the case of (1), the time taken is 4 min. Therefore in the case of (3) it is 4*1.5=6 min. That is when the cyclist is stationary, the buses cross him every 6 min.

Therefore the time interval between 2 consecutive buses is 6 min
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Originally posted by SravnaTestPrep on 23 Jun 2013, 23:41.
Last edited by SravnaTestPrep on 16 Aug 2013, 19:41, edited 2 times in total.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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Another approach:

1. Assume the combined speed of the cyclist and the bus is s1+ s2. s1 is the speed of the cyclist and s2 is the speed of the bus.
2. From the time a bus starts till the time it meets the cyclist, assume the distance traveled is d and the time taken is m
3. In the case of overtaking the difference in the speeds needs to be taken since both are traveling in the same direction. The relative speed is s2- s1. Assume to travel the same distance d the time taken is n
5. From (2) and (3), we have, m(s1+s2) = n(s2-s1)=d. we know m is 4 min and n is 12 min.
6. d= 4(s1+s2) = 12(s2-s1)
7. We want to find d/ s2 = 12(s2-s1)/s2 = 12(s2/s2) - 12(s1/s2) = 12-6= 6 min
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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New post 13 Aug 2013, 13:03
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?

This problem took me a while to figure out but I think I got it!

The interval refers to the time (i.e. time = distance/speed)

The interval for the bus is t=d/b where b represents the constant speed of both buses.

When the bus overtakes the cyclist, it represents the bus gaining on the cyclist (i.e. the bus speed is faster than the cyclist) so every time the bus overtakes the cyclist, which is every 12 minutes, we have:

12 = d/(b-c) where b-c is the rate that the bus gains on the cyclist.
12b-12c = d

When the oncoming bus meets the cyclist, it represents their combined forward speeds towards one another. Every time the bus meets the cyclist, which is every 4 minutes, we have:

4 = d/(b+c)
4b+4c = d
Note, we use c for both bus speeds because the question tells us that they move at the same constant speed.

12b-12c = d
4b+4c = d

12b-12c = 4b+4c
8b = 16c
b = 2c

12b-12c = d
12(2c) - 12c = d

My problem is this: how do I know to solve for d? In other words, why would I know:
b = 2c, 12b-12c = d
12*(2c) - 12c = d
24c - 12c = d
12c = d isn't correct?
How would I know to convert b=2c to c=b/2 to plug in so I can get a value for d thus getting me the right answer?


c=b/2
12b-12c = d
12b - 12(b/2) = d
12b - 6b = d
6b = d

t=d/b (the interval for both buses)
t=(6b/b)
t = 6

ANSWER: B. 6 minutes
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Re: test m08 question n18 [#permalink]

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New post 16 Aug 2013, 02:20
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?
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: \(\frac{d}{b+c}=4\), \(d=4b+4c\);

\(d=12b-12c=4b+4c\), --> \(b=2c\), --> \(d=12b-6b=6b\).

\(Interval=\frac{d}{b}=\frac{6b}{b}=6\)

Answer: B (6 minutes).

Hope it helps.


Hi Bunuel
Following doubts need to be clear

d/b-c---Why it is so if objects are moving in same direction.
d/b+c---Why it is so if objects are moving in opposite direction.

Rgds
Prasannajeet
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Re: test m08 question n18 [#permalink]

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New post 16 Aug 2013, 02:23
prasannajeet 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?
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: \(\frac{d}{b+c}=4\), \(d=4b+4c\);

\(d=12b-12c=4b+4c\), --> \(b=2c\), --> \(d=12b-6b=6b\).

\(Interval=\frac{d}{b}=\frac{6b}{b}=6\)

Answer: B (6 minutes).

Hope it helps.


Hi Bunuel
Following doubts need to be clear

d/b-c---Why it is so if objects are moving in same direction.
d/b+c---Why it is so if objects are moving in opposite direction.

Rgds
Prasannajeet


When 2 objects move in the same direction their relative speed is the difference of their individual speeds.
When 2 objects move in the opposite direction their relative speed is the sum of their individual speeds.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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

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New post 16 Aug 2013, 11:08
VeritasPrepKarishma wrote:


Nice post, thanks! I suppose you wouldn't have a post breaking down this: on-a-partly-cloudy-day-derek-decides-to-walk-back-from-work-147050.html#p1257285

monster, would you? :-D
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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New post Updated on: 28 Aug 2013, 08:14
1. If the distance between two points is d and two objects are moving with speed s1 and s2 and we are given d/(s1+s2) and d/(s2-s1), the harmonic mean gives d/s2 i.e., harmonic mean is 2 / ((s1+s2)/d +(s2-s1)/d) = d/s2
2. The present problem can be solved within the above framework. The speed of the cyclist can be taken as s1 and the speed of the buses as s2
3. If the distance is called AB and is equal to 2d , and the cyclist is midway and starts moving towards B, then buses starting from A, take d/(s2-s1) minutes to meet the cyclist after the previous bus meets him and similarly buses starting from B take d/(s1+s2) minutes to meet him after the previous bus meets him.
4. What the problem asks is effectively to find when the cyclist is stationary i.e, when s1=0, , how long does it take a bus to meet the cyclist after the previous bus meets him. i.e., it asks us to find d/s2. This can be found using the harmonic mean formula.
5. Since d/(s1+ s2) = 4 and d/(s2-s1) = 12, d/s2 = 2 / (1/4 + 1/12) = 6 minutes
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Originally posted by SravnaTestPrep on 16 Aug 2013, 20:11.
Last edited by SravnaTestPrep on 28 Aug 2013, 08:14, edited 3 times in total.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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New post 19 Aug 2013, 05:27
WholeLottaLove wrote:

Nice post, thanks! I suppose you wouldn't have a post breaking down this: on-a-partly-cloudy-day-derek-decides-to-walk-back-from-work-147050.html#p1257285

monster, would you? :-D


Well, I wrote this question for Veritas and I can tell you that it is not very easy. It's conceptual and has a trap. I haven't written a post on this question but will be willing to write one next week. Look out for the post next Monday on my blog.
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Re: test m08 question n18 [#permalink]

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New post 28 Aug 2013, 04:37
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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?
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: \(\frac{d}{b+c}=4\), \(d=4b+4c\);

\(d=12b-12c=4b+4c\), --> \(b=2c\), --> \(d=12b-6b=6b\).

\(Interval=\frac{d}{b}=\frac{6b}{b}=6\)

Answer: B (6 minutes).

Hope it helps.


Harmonic mean of the two times gives 6 as answer ? Is there a connection b/w Mean and the solution to such probs. May be yes because there is a sense of harmonics ( i.e. repetition is multilple of the first term. ) but I am not able to explain it or find a concrete analysis. brunuel what do you say ?
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Re: test m08 question n18 [#permalink]

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New post 28 Aug 2013, 08:01
ygdrasil24 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?
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: \(\frac{d}{b+c}=4\), \(d=4b+4c\);

\(d=12b-12c=4b+4c\), --> \(b=2c\), --> \(d=12b-6b=6b\).

\(Interval=\frac{d}{b}=\frac{6b}{b}=6\)

Answer: B (6 minutes).

Hope it helps.


Harmonic mean of the two times gives 6 as answer ? Is there a connection b/w Mean and the solution to such probs. May be yes because there is a sense of harmonics ( i.e. repetition is multilple of the first term. ) but I am not able to explain it or find a concrete analysis. brunuel what do you say ?


Hi,

I have corrected my previous reply and have given my thoughts on this. Kindly take a look.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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New post 28 Aug 2013, 23:01
WholeLottaLove wrote:
VeritasPrepKarishma wrote:


Nice post, thanks! I suppose you wouldn't have a post breaking down this: on-a-partly-cloudy-day-derek-decides-to-walk-back-from-work-147050.html#p1257285

monster, would you? :-D


Actually, Ron beat me to it and gave a full fledged, brilliant analysis of this question here: http://www.veritasprep.com/blog/2013/08 ... -the-gmat/

This should be helpful.
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Re: test m08 question n18 [#permalink]

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New post 29 Aug 2013, 09:11
Hi,

I have corrected my previous reply and have given my thoughts on this. Kindly take a look.[/quote]

Thanks Srinavasan, but can you explain what harmonic mean is doing here ?
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New post 29 Aug 2013, 20:40
ygdrasil24 wrote:
Hi,

I have corrected my previous reply and have given my thoughts on this. Kindly take a look.


Thanks Srinavasan, but can you explain what harmonic mean is doing here ?[/quote]

Consider two buses one traveling at 25 miles/hr and the other at 50 miles /hr. Let us assume a distance of 300 miles. When they are moving in the opposite direction the distance is covered in 300/ (50+25) = 4 hours. The slower bus covers the distance in 300/25 = 12 hours and the faster bus covers the distance in 300/50 = 6 hours You see the ratio of the combined speed and the speed of the faster bus is 1.5:1 and this is reflected in the time taken ie, 4 hours and 6 hours . Similarly the ratio of the speed of the faster bus and the slower bus is 2:1 and this is reflected in the time taken ie., 6 hours and 12 hours.

We can see that the difference in the time taken by the faster bus and the combined time (6-4= 2 hrs) and the difference in the time taken by the slower bus and the faster bus(12-6 = 6 hrs) is in the ratio of one extreme being the combined time taken and the other extreme being the time taken by the slower bus. That is the distance of the "mean" is at a distance from the two extremes that is proportionate to the ratio of the two extremes and this is what harmonic mean is.

Thus it turns out that in the above scenario the use of harmonic mean to find the time taken by the faster bus is apt.
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Re: A man cycling along the road noticed that every 12 minutes [#permalink]

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New post 31 Dec 2013, 08:54
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?
Thank you in advance!!


Let's say 'b' for bus, and 'c' for cyclist rates.

Time interval means = d / b

We have that 12(b-c) = 4(b+c) = Distance

So we get b = 2c

Now we need to find the distance, replace in second equation and we get 12c

Since b = 2c then 6b = 12c

Interval is therefore 6b/b = 6 min

Answer is B

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

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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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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

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

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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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New post 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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New post 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
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A man cycling along the road noticed that every 12 minutes   [#permalink] 23 Apr 2017, 13:47

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A man cycling along the road noticed that every 12 minutes

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