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

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24 Aug 2012, 23:52

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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?

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!!

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).

Answer B.

Attachments

BusBicycle1.jpg [ 7.78 KiB | Viewed 2388 times ]

BusBicycle2.jpg [ 7.75 KiB | Viewed 2386 times ]

_________________

PhD in Applied Mathematics Love GMAT Quant questions and running.

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

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

what does " every 4 minutes he meets an oncoming bus. " refers? please help

FTGNGU, it means every 4 minutes, the cyclist meets a bus moving in opposite direction. So, the relative speed between them is the absolute value of [speed of cyclist - speed of the bus]

Hope it helps,
_________________

Please +1 KUDO if my post helps. Thank you.

"Designing cars consumes you; it has a hold on your spirit which is incredibly powerful. It's not something you can do part time, you have do it with all your heart and soul or you're going to get it wrong."

what does " every 4 minutes he meets an oncoming bus. " refers? please help

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
_________________

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
_________________

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
_________________

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

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08 Aug 2013, 17:13

Hello.

While I understand the algebra behind the problem, I'm not sure I understand the logic. Consecutive buses are passed every four minutes and the cyclist is passed every 12. I'm unable to rationalize how the time between consecutive buses is 6 minutes? I'm not even sure I understand the context of "consecutive" in this problem.

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

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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

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

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/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.
_________________

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.
_________________

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 ?

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.
_________________

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]

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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