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A bus from city M is traveling to city N at a constant speed
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Updated on: 28 Feb 2012, 00:01
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A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance between the two cities? A. 48 B. 72 C. 96 D. 120 E. 192 The source is GMATClub's diagnostic test... would look forward to see some innovative approach to this. Thanks!
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Originally posted by gmattokyo on 07 Nov 2009, 03:23.
Last edited by Bunuel on 28 Feb 2012, 00:01, edited 1 time in total.
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Re: two buses, same speed... head spinning
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07 Nov 2009, 09:02
A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance between the two cities?A. 48 B. 72 C. 96 D. 120 E. 192 Distance between the cities \(d\). First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway. Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours. On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover. They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\): \(\frac{d}{2}24=\frac{0.75d}{2}\) \(d=192\) Answer: E.
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Re: two buses, same speed... head spinning
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07 Nov 2009, 04:13
Hmmm I did this question a while ago and got it wrong so trying again Rate x time = distance. For the initial trip lets the distance to the midpoint be represented by P. EQN1: R x 2 = P For the second trip we know one bus left late and one left early. Together this is just a tricky way of saying one bus left an hour after the other. We know the total trip takes 4 hours (since getting to P is 2 hours). The second trip can be represented by: Since the trip takes 4 hours if a bus leaves one hour early, the reminaining 3 hours are split between the two buses, ie. 1 + 3/2 = 2.5 EQN2: R x 2.5 = P + 24 EQN2EQN1 : 0.5R=24 R=48 So the distance is rate x time = 48 x 4 = 192. ANS = E




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Re: two buses, same speed... head spinning
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07 Nov 2009, 04:35
Normally I goof up with distancespeed probs, but got this one right.
here, B1 and B2 are coming towards each other at the same speed, so when they meet they MUST have traveled the same distance. in 2 hours B1 has traveled 2S miles ( let S be the speed ) so, B2 has also traveled 2S miles
So we have a line segment MN having a distance 4S and PM=MN = 2S. Divide this line segment into four parts with each having distance S.
Okay, for the second trip the total time delay is 60 mins=1 hour. Suppose B2(at N) starts early. In this one hour B2 has traveled S miles. At this time B1 will start and both buses will meet at a point which is halfway of the remaining distance(3S) = 3S/2.
Now, 2S (which represents PM)  3S/2 (second trip's halfway meeting point) = 12 => S=48. So, total distance = 4S = 4*48 = 192.



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Re: two buses, same speed... head spinning
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07 Nov 2009, 06:45
I am not sure if I am right, but my ans differs from the 2 posted earlier. I think the distance between M and N should be 96.
Facts given: Both buses travel at the same speed. When they leave at exactly same time from M and N respectively, they meet at P in 2 hours. This means the total journey should be taking 4 hours for both of them individually.
Now, one bus leaves 36 min earlier and other one is delayed by 24 min, there is a total gap of 60 min or 1 hour between the two buses. If their meeting point is now shifted by 24 miles, it means their respective speeds are 24 miles an hour. This drives me to a conclusion that the totla distance between M and N should be 4 * 24 = 96 miles.
Please correct me if i'm wrong!!



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Re: two buses, same speed... head spinning
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07 Nov 2009, 08:45
kalpeshchopada7 wrote: I am not sure if I am right, but my ans differs from the 2 posted earlier. I think the distance between M and N should be 96.
Facts given: Both buses travel at the same speed. When they leave at exactly same time from M and N respectively, they meet at P in 2 hours. This means the total journey should be taking 4 hours for both of them individually.
Now, one bus leaves 36 min earlier and other one is delayed by 24 min, there is a total gap of 60 min or 1 hour between the two buses. If their meeting point is now shifted by 24 miles, it means their respective speeds are 24 miles an hour. This drives me to a conclusion that the totla distance between M and N should be 4 * 24 = 96 miles.
Please correct me if i'm wrong!! Dear Economist and yangsta8, thanks for the new approaches... i still am spinning though kalpesh, following is the official explanation which I spent some time to understand (24miles in 30mins): (I didn't know that these questions have their own thread, sorry for repost) gmatdiagnostictestquestion79354.htmlThe buses travel at the same constant speed. It would take one bus to travel 4 hours to cover the distance between the cities M and N (two buses drove for 2 hours each). We need to find the speed of the bus. If the first bus was delayed by 24 minutes and the second one left 36 minutes earlier, it makes the second bus \(24+36=60\) minutes ahead of the first bus. The meeting point was 24 miles away on this second day. We know the distance difference between the two meeting points, but we also need to find difference in time those 24 miles were covered. If the second bus drove for 1 hour before the first one departed, each of them had to go for another 1.5 hour to meet (1.5 hour + 1.5 hour + 1 hour). The second bus traveled for 2.5 hours and the first one for 1.5 hour. Therefore the meeting point on the second day was 30 minutes away from that of the previous day. So the second bus covered 24 miles in 30 minutes, which gives us the speed of the bus, 48 mph. We can calculate the distance as we already know the speed: \(4*48=192\) miles.



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Re: two buses, same speed... head spinning
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29 Mar 2011, 20:29
Bunuel wrote: Distance between the cities \(d\).
First meeting point \(\frac{d}{2}\), as both buses travel at the same constant speed and leave the cities same time they meet at the halfway.
Total time to cover the \(d\) 4 hours, as the buses meet in 2 hours.
On the second day first bus traveled alone 1 hour (36min +24min), hence covered \(0.25d\), and \(0.75d\) is left cover.
They meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):
\(\frac{d}{2}24=\frac{0.75d}{2}\)
\(d=192\) This is the only explanation that made sense to me! Thanks!! +1



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Re: two buses, same speed... head spinning
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29 Mar 2011, 21:41
Mx P(dx)N 2v + 2v = d x = 2v (they meet at halfway) Next day they meet at x  24 miles from city M d/2v = 2 > On day 1 Next day, in 1 hr, B2 travels v distance = d/4 miles, and then it travels (24 miles + d/4) from there for t hrs at speed v where it meets B1 (B1 has also traveled for t hrs by then) (24+ d/4)/v = (d/224)/v => 48 = d/2  d/4 => d = 4 * 48 = 192 Answer  E
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Re: two buses, same speed... head spinning
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25 Oct 2011, 15:47
Thanks Bunuel.
Your explanation is very clear ... I also has trouble undersatnding the official answer.. Kudos to you!



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Re: two buses, same speed... head spinning
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27 Oct 2011, 06:15
gmattokyo wrote: A bus from city M is traveling to city N at a constant speed while another bus is making the same journey in the opposite direction at the same constant speed. They meet in point P after driving for 2 hours. The following day the buses do the return trip at the same constant speed. One bus is delayed 24 minutes and the other leaves 36 minutes earlier. If they meet 24 miles from point P, what is the distance between the two cities?
* 48 * 72 * 96 * 120 * 192
The source is GMATClub's diagnostic test... would look forward to see some innovative approach to this. Thanks! A little bit of visualization can help you solve this question in a moment. Look at the diagram below. Since they travel at same speed for 2 hrs, point P must be right in the middle of M and N. Attachment:
Ques3.jpg [ 13.29 KiB  Viewed 44093 times ]
On the return trip, the following is what happens. One bus starts at 11:24 and reaches P at 1:24. The other bus starts at 12:24 and reaches mid way between P and N at 1:24 (in one hour, it covers half of the distance it usually covers in 2 hrs). At 1:24, both buses have a fourth of the total distance between them. They will both cover equal distances and meet at a point which is 1/8th of the total distance from P. This distance is given as 24 miles. Hence total distance = 24*8 = 192 miles
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Re: two buses, same speed... head spinning
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27 Oct 2011, 12:33
@Bunuel  Very neat explanation. Kudos to you..!!
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Re: two buses, same speed... head spinning
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09 Dec 2011, 04:39
Fro those of you who are dependent on MGMAT RTD chart (as i am) can try the following: From the attached chart#2 P=R(T+3/5)24 P=R(T2/5)+24 Therefore, R(T+3/5)24=R(T2/5)+24 Solving for R, we get Rate=48 m/h Plug in value in Chart#1: p=96 So, D=192
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Re: two buses, same speed... head spinning
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19 Dec 2011, 03:04
quite confusing question but very good explained by bynnel and karishma thanks to both of you



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Re: two buses, same speed... head spinning
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30 Dec 2011, 02:14
good explanation bunuel.......... thanks



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Re: two buses, same speed... head spinning
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01 Jan 2012, 10:28
suppose total distance = 2d and point p is at d distance from M and N. for bus M d+24 = v(1+36/60) = 1.6v for bus N d24 = v(124/60) = 0.6v subtract them, we get v=48 final distance = 2v*t = 4v =192 ( final distance = 2vt because relative speed = v1+v2 = 2v)
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Re: two buses, same speed... head spinning
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27 Feb 2012, 23:03
Bunuel wrote: \(\frac{d}{2}24=\frac{0.75d}{2}\)
why do you assume 24 instead of +24?



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Re: two buses, same speed... head spinning
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27 Feb 2012, 23:36
T740qc wrote: Bunuel wrote: \(\frac{d}{2}24=\frac{0.75d}{2}\)
why do you assume 24 instead of +24? You can derive this either my reasoning or simply by noticing that d/2>0.75d/2, so it should be d/224=0.75d/2 (greater value minus 24 equals to smaller value). Hope it's clear,
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Re: A bus from city M is traveling to city N at a constant speed
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12 Apr 2012, 06:51
Help me please to realize whats wrong with my solution? @what point exactly I'm making a mistake?
form the first part of data i can get (1 bus) X*2=2X (2 bus) X*2=2X >the total distance is 4X
Taking into account the following assumptions, 1)the 1st bus began to drive later for 24 min 2)both buses have the same rate > the 1st has to travel less distance than the 2d, since it started to travel later but the second earlier 3)the total distance is 4X i set up two equations: (1 bus) X*(t+24)=2X24 (2 bus) X*(t36)=2X+24 since 2X is a half of the distance i can rewrite the two equations: X(t+24)+24=X*(t36)24 >60X=48 what makes me crazy and i don't know where is the flaw in my logic (excluding the fact, that I'm a woman)



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Re: A bus from city M is traveling to city N at a constant speed
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08 Jul 2013, 01:10
Bumping for review and further discussion*. Get a kudos point for an alternative solution! *New project from GMAT Club!!! Check HERE
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Re: A bus from city M is traveling to city N at a constant speed
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09 Jul 2013, 04:40
Let s be the speed of the buses. Thus total distance betn the two cities is 4s. On the first day, thus they meet P, which is at a distance of 4s/2= 2s from their starting points. The next day , in effect one bus has travelled for 1 hour before the other starts. Distance covered in 1 hour =4s/4= s. Remaining distance is 3s. Point at which the two buses meet now, is at a distance of 3s/2 which is also a distance of 24 from P. Thus, 3s/2 = 2s 24 or s =48 . therefore distance = 4 * 48 =192.




Re: A bus from city M is traveling to city N at a constant speed
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