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Re: A man cycling along the road noticed that every 12 minutes [#permalink]
16 Aug 2013, 19:11

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 _________________

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. _________________

Re: test m08 question n18 [#permalink]
28 Aug 2013, 03:37

1

This post received KUDOS

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

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 ?

Re: test m08 question n18 [#permalink]
28 Aug 2013, 07:01

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

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. _________________

Re: test m08 question n18 [#permalink]
29 Aug 2013, 19:40

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. _________________

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
31 Dec 2013, 07:54

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

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
16 Jun 2014, 02:25

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. _________________

Re: test m08 question n18 [#permalink]
17 Jun 2014, 23:33

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

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
14 Jan 2015, 21:52

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? 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: \(\frac{d}{b+c}=4\), \(d=4b+4c\);

Re: A man cycling along the road noticed that every 12 minutes [#permalink]
16 Jan 2015, 03:34

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

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