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M08 #18 - Bus intervals : Retired Discussions [Locked] - Page 2

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Re: M08 #18 - Bus intervals [#permalink]
03 Jun 2010, 03:46

2

This post received KUDOS

Expert's post

1

This post was BOOKMARKED

ykaiim wrote:

In every 2 min, cyclist is meeting bus from opposite side. So, at the end of the 12th min, he will meet the 6th bus from opposite end. That means there are 5 intervals passed at the end of 12th min. So, frequesncy = 12/5 = 2 min 24 sec

Bunuel wrote:

ykaiim wrote:

My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.

What would be the answer with your approach if it were 2 minutes instead of 4?

When I asked the above question I wanted to show that your approach is not correct. It worked for the given numbers but it won't work for another ones. Answer to my question would be 24/7 minutes not 12/5 minutes.

This question was posted in PS forum as well below is my solution for it:

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: M08 #18 - Bus intervals [#permalink]
03 Jun 2010, 03:58

Bunuel,

I dont deny this mathematical approach but how the answer is 24/7 in that case? I tried this by consedring an interval of 12 min for overtake and 2 min for cross-over. _________________

Re: M08 #18 - Bus intervals [#permalink]
03 Jun 2010, 04:22

Expert's post

ykaiim wrote:

Bunuel,

I dont deny this mathematical approach but how the answer is 24/7 in that case? I tried this by consedring an interval of 12 min for overtake and 2 min for cross-over.

ykaiim, please do the math: substitute new numbers (12 min and 2 min) in the above approach. _________________

Re: M08 #18 - Bus intervals [#permalink]
18 Aug 2010, 19:36

1

This post received KUDOS

Here is some additional explanations that might help:

We are looking for the time between two conseutive buses. Since the bus speed is constant, the time between two consecutive buses will be constant and so will be the distance between two consecutive buses as distance = time*speed.

Let's l be the distance between two buses. The time between two buses is equal to l/b where b is the speed of the bus. We need to express l in terms of b in order to have the result of this ratio.

The information provided allows us to do so. 1/we know that the cyclist crosses a bus every four minutes. It means that the bus plus the cyclist run the distance between two buses in 4 minutes. Hence we have l=4(b+c) (1) where c is the speed of the cyclist. (The relative speed of U and V with speed equal u and v going in the same direction is u+v)

2/ we know that the cyclist is caught up by a bus every twelve minutes. During these twelve minutes the bus will do more than the distance between two buses as the cyclist is going forward. We know that the relative speed between U and V running in opposite direction is u-v (where u>v). Hence to get the distance l we need to calculate the distance covered by the bus in twelve minutes and minus the distance covered by the cyclist. This means l =12b-12c--> l=12(b-c) (2)

From 1 and 2 we get 4b+4c=12b-12c-->16c=8b-->c=1/2b. We can now replace c by 1/2 b in (1) and we get l=4b+2b=6b. The time between two buses is l/b=6b/b=6

Re: M08 #18 - Bus intervals [#permalink]
19 Aug 2010, 01:21

This whole question seems a bit confusing to me. I really think the current wording is not clear. I mean we can assume that we have equal distance, really that is the only way to solve this problem.

I want to recommend the following question as a replacement for this question:

A man cycling around a circular track notices that every 12 minutes a bus overtakes him while every 4 minutes he meets an oncoming bus. If both buses and the cyclist move at constant speeds, what is the time interval between the two buses?

Re: M08 #18 - Bus intervals [#permalink]
30 Aug 2010, 09:19

using the MGMAT "RTD overtake" method, v = 1/4 - 1/12 = (3-1)/12 = 2/12 = 1/6 therefore 6 minutes.... don't even udnerstand how it works in this case but seems to be working.........please someone explain if understand.....

Re: M08 #18 - Bus intervals [#permalink]
20 Oct 2010, 09:51

if the relative speeds are in AP then the relative time intervals are in HP

Since the speeds b+c , b , b-c are in AP

The relevant times are in HP.......

speed is b when the cyclist is standing. We are asked to find the time interval between the buses. That time interval will be the time when cyclist is stationary. Time interval when bus is moving on the same side is 12 and on the other side is 4.

Thus the times 12 , t, 4 are in HP

=> t = 2*12*4/ (12+4) = 6

The above result is very useful and time saving. _________________

Re: M08 #18 - Bus intervals [#permalink]
23 Aug 2012, 05:49

Bunuel wrote:

ykaiim wrote:

My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.

What would be the answer with your approach if it were 2 minutes instead of 4?

Bunuel,

Can you please explain this problem. I get so confused with the speed problems... _________________

Regards, Harsha

Note: Give me kudos if my approach is right , else help me understand where i am missing.. I want to bell the GMAT Cat

Re: M08 #18 - Bus intervals [#permalink]
23 Aug 2012, 05:58

Expert's post

harshavmrg wrote:

Bunuel wrote:

ykaiim wrote:

My approach was different.

If in 12 min a bus overtakes the cyclist, while in 4 min they both cross each other. So, at the end of 12th min, the overtaking bus will meet the 3rd bus coming from the opposite side, which means there are 2 buses gone in the 12 min time period.

We are given speeds of buses and cyclist constant. So, in 12 min, if two buses passes at an interval of 6 min.

What would be the answer with your approach if it were 2 minutes instead of 4?

Bunuel,

Can you please explain this problem. I get so confused with the speed problems...

Re: M08 #18 - Bus intervals [#permalink]
29 Aug 2012, 23:26

I did end up getting the answer as 6 but I am not sure if my approach was correct and can be used again for similar problems.

Personally I did not prefer approaching the equations way of solving this problem. They might be way more easier but instead I took a particular scenario wherein the total time was from zero to 12 mins. In this time duration, one bus overtook the cyclist and he faced 3 oncoming buses.

i.e. total time = 12 mins + 12 mins = 24 mins Total buses passed = 1 (overtaking) + 3 (oncoming)

Interval = 24/4 = 6.

There was particular response bu Bunuel where he asked what would be the solution if the interval was 2 mins for the oncoming buses. In that case the interval between the buses is 24/7 as he has mentioned.

Will dig in further to see if my approach is correct or whether it was just apt for this particular set of numbers and was a fluke. _________________

My attempt to capture my B-School Journey in a Blog : tranquilnomadgmat.blogspot.com

There are no shortcuts to any place worth going.

gmatclubot

Re: M08 #18 - Bus intervals
[#permalink]
29 Aug 2012, 23:26