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GMAT Diagnostic Test Question 24 Field: word problems (rate) Difficulty: 700

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 same trip again 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?

Say the distance between the cities is \(d\) miles.

Since both buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway, so the first meeting point P, is \(\frac{d}{2}\) miles away from M (and N).

Next, since the buses meet in 2 hours then the total time to cover \(d\) miles for each bus is 4 hours.

Now, on the second day one bus traveled alone for 1 hour (36min +24min), hence covered \(0.25d\) miles, and \(0.75d\) miles is left to cover.

The buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\):

Bunuel I don't get the last step. However I did get the catch that the diff between the distance traveled by both the buses on the second day is not 24 but 48 miles.

p + 24 = v(2+0.6) ------I p - 24 = v(2-0.4) ------II Subtracting we get v = 48 d = 4v = 4*48 = 192

Bunuel I don't get the last step. However I did get the catch that the diff between the distance traveled by both the buses on the second day is not 24 but 48 miles.

p + 24 = v(2+0.6) ------I p - 24 = v(2-0.4) ------II Subtracting we get v = 48 d = 4v = 4*48 = 192

Can you explain your last step?

Bunuel wrote:

\(\frac{d}{2}-24=\frac{0.75d}{2}\) \(d=192\)

Quote:

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 \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

You should visualize this problem:

M--------X----P-------------N

First meeting point P=d/2; Second meeting point X=0.75d/2, which is 24 miles from point P: \(\frac{d}{2}-24=\frac{0.75d}{2}\) --> solving for \(d\) --> \(d=192\). _________________

Bunuel I don't get the last step. However I did get the catch that the diff between the distance traveled by both the buses on the second day is not 24 but 48 miles.

p + 24 = v(2+0.6) ------I p - 24 = v(2-0.4) ------II Subtracting we get v = 48 d = 4v = 4*48 = 192

Can you explain your last step?

Bunuel wrote:

\(\frac{d}{2}-24=\frac{0.75d}{2}\) \(d=192\)

Quote:

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 \(\frac{d}{2}\):

\(\frac{d}{2}-24=\frac{0.75d}{2}\)

\(d=192\)

You should visualize this problem:

M--------X----P-------------N

First meeting point P=d/2; Second meeting point X=0.75d/2, which is 24 miles from point P: \(\frac{d}{2}-24=\frac{0.75d}{2}\) --> solving for \(d\) --> \(d=192\).

Sorry to bother you, please explain - hence covered .25d , and 0.75d is left cover. Is it because the bus traveled for one hour - (36+24) and the trip took four hours the day before hence 1/4 d.

Sorry to bother you, please explain - hence covered .25d , and 0.75d is left cover. Is it because the bus traveled for one hour - (36+24) and the trip took four hours the day before hence 1/4 d.

Thanks for your continuos support

Yes. As the buses travel at the same constant speed, leave the cities same time and meet in 2 hours then total time to cover \(d\) for a bus is 4 hours. So bus needs 4 hours to cover \(d\). On the second day first bus traveled alone 1 hour (36min +24min), hence covered 1/4th of \(d\) and 3/4th of \(d\) is left to cover. _________________

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. Please note the text in red. It is very important to understand that the meeting point on the second day was only 30 minutes away from the meeting point from the first day. I really hope it helps because I can't explain it better than it is already explained in the second post of this thread.

sorry but i am not being able to get it all... is it that second one has travelled 1 hour more but that 24 miles were coverved by both both so they met 30 minutes apart...? could you please elaborate that thanks

Maybe I'm just misreading this, but something seems off to me.

The first trip takes x amount of time at s speed for both buses from opposite directions. If P is not city M - then why would they ever meet outside the city on one of the buses sides? It seems like the wording means they are going to the city N and stopping.

Same with the second part, how will they meet away from city N if they are coming from opposite directions and not passing through?

M >-----------------N----------------X(second bus home)

given the wording, it seems to me P lies somewhere in between M and N or N and X. Am I misreading it?

Assumed distance between M and N to be = 2D miles Distance travelled by each bus = D miles Time taken to cover the distance = 2 hours Speed of each bus = D/2 miles per hour

Since the meeting point has now moved 24 miles away from point P, hence Distance travelled by one bus would be (D+24) miles and the Distance travelled by the other bus would be (D-24) miles. The time difference between the two buses is one hour (24+36). Time = Distance/Speed [(D+24)/(D/2)] - [(D-24)/(D/2)] = 1 Solving the equation gives D = 96 miles

The distance between M and N is thus 2D = 192 miles.

Assumed distance between M and N to be = 2D miles Distance travelled by each bus = D miles Time taken to cover the distance = 2 hours Speed of each bus = D/2 miles per hour

Since the meeting point has now moved 24 miles away from point P, hence Distance travelled by one bus would be (D+24) miles and the Distance travelled by the other bus would be (D-24) miles. The time difference between the two buses is one hour (24+36). Time = Distance/Speed [(D+24)/(D/2)] - [(D-24)/(D/2)] = 1 Solving the equation gives D = 96 miles

The distance between M and N is thus 2D = 192 miles.

This problem has driven me crazy for more than 1 hr now, I am trying to find what is the issue in my understanding..

Worst, I am getting 3 diff answers.. I assume total dist i.e. dist bet cities = D. Total time = 240 min. Constant speed for both buses = D/240.

RTD Chart

T R D 240 D/240 D 156 (i.e.120+36) D/240 (D/2) + 24 For bus leaving early, no need to care abt other bus 96 (i.e. 120-24) D/240 (D/2) - 24 For bus leaving late, no need to care abt other bus

1st Linear equation for D (bus leaving early): (156D)/240 = (D/2) + 24=>D=160 2nd Linear equation for D (bus leaving late): (96D)/240 = (D/2) - 24=>D=240 3rd case, same as tingle15 above: ( (D/2) + 24 ) /(D/240) - ( (D/2) - 24 ) /(D/240) = 156 - 96=>D=192 (OE).

Also, since speed is constant, even using simple direct proportion, we get D=160: 36->24 240->D =>D=160

Can Bunuel/somebody pls explain the mistake here/if the no. 24 in the question should be changed..Worst, I hope there is no calculation mistake on my part..

Sorry to dig out this question... I tried to simplify it to the greatest degree. To me, since the other bus has 1 hour advantage and travels 24 miles farther, alludes that 1 hour of trip=24 miles. Since they traveled 4 hours total, it would equal 96 miles. I do understand above explanations, but just wanted to expose the pitfall some may have fell into. _________________

[highlight]Monster collection of Verbal questions (RC, CR, and SC)[/highlight] http://gmatclub.com/forum/massive-collection-of-verbal-questions-sc-rc-and-cr-106195.html#p832142

[highlight]Massive collection of thousands of Data Sufficiency and Problem Solving questions and answers:[/highlight] http://gmatclub.com/forum/1001-ds-questions-file-106193.html#p832133

So, bus one (we'll call M) drove 1.6 hours to get to p-24 Bus two (bus N) drove 24 minutes + 36 minutes + the 1.6 hours that M was driving = 2.6 hours.

1.6*M = P - 24 2.6*N = P + 24 So, 2N = P and 2M = P...

2.6N - P - 24 = 2N - P yields: .6N = 24 > N = 40 mph

1.6M - P + 24 = 2M - P yields: 24 = .4M > M = 60 mph

2h * 60 mph = 120 miles (bus M) 2h * 40mph = 80 miles (bus N)

= 200 miles?

The biggest issue I am having is how they are getting 1.5, 2.5, etc. 36 minutes / 60 minutes = .6 hours not .5 HOURS!

Bus M was scheduled to leave at 12, but was delayed 24 min (.4 hours...24min / 60min). So, this left at 12:24 p.m.

Bus N was scheduled to leave at 12 p.m. as well, but left 36 minutes early at 11:24 a.m. Bus N takes at LEAST 2 hours to get to point P. Thus, it cannot arrive before 1:24 p.m., however, we know that bus N travels farther since we are going past point P (or you can look at it as M can at best reach half way to point P since it has only been going for an hour at 1:24 p.m.). Thus, you go to 2 p.m. which is the original area they would have met at point P, but bus M is not to point P since delayed (it is 24 miles short) and bus N has been going longer than 2 hours which is how it gets to be P+24.

Let me know if that helps or further confuses..i realize that was incredibly wordy.

Also, in case that equation wasn't clear... Bus M only gets to drive 1.6 hours (delayed 24 min, 24 min / 60 = .4, 2-.4 = 1.6 hours). And 36 min early departure + 24 min. the delay that bus N didn't experience = 1 hour extra that bus N gets to drive over bus M. Thus:

Bus M: 1.6x = P - 24 (24 miles shorter than point P) and... Bus N: 2.6x = p + 24 (went 24 miles past point P in the 1 hour extra it had compared to bus M)

Bus M was scheduled to leave at 12, but was delayed 24 min (.4 hours...24min / 60min). So, this left at 12:24 p.m.

Bus N was scheduled to leave at 12 p.m. as well, but left 36 minutes early at 11:24 a.m. Bus N takes at LEAST 2 hours to get to point P. Thus, it cannot arrive before 1:24 p.m., however, we know that bus N travels farther since we are going past point P (or you can look at it as M can at best reach half way to point P since it has only been going for an hour at 1:24 p.m.). Thus, you go to 2 p.m. which is the original area they would have met at point P, but bus M is not to point P since delayed (it is 24 miles short) and bus N has been going longer than 2 hours which is how it gets to be P+24.lost you here

Let me know if that helps or further confuses..i realize that was incredibly wordy.

Also, in case that equation wasn't clear... Bus M only gets to drive 1.6 hours (delayed 24 min, 24 min / 60 = .4, 2-.4 = 1.6 hours). And 36 min early departure + 24 min. the delay that bus N didn't experience = 1 hour extra that bus N gets to drive over bus M. Thus:

Bus M: 1.6x = P - 24 (24 miles shorter than point P) and... Bus N: 2.6x = p + 24 (went 24 miles past point P in the 1 hour extra it had compared to bus M)

M leaves @ 12:24PM N leaves @ 11:24AM

As you rightly said: N travels P/2 distance until 12:24PM, when M starts its journey.

Let's try to understand this using a line. Let the two extreme points be X and Y.

I. X(M)---------------------------P---------------------------(N)Y @11:24AM, where XP=YP

II. X(M)---------------------------P------------(N)R--------------Y @12:24PM, because N left early and covered half of YP; PR=RY=1/4*XY

III. X--------------------(M)(N)T-------P-------------R-------------Y @Some time after 1:24PM. The important thing to note here is that the point, let's call it T, where "MN" have met is again midway from X and R in the immediately previous line(II). Why? Because M and N travels with same constant speed.

PR+PT=XT [:Note: T is midway from X and R] PR+24=XT PR=1/2*PY=1/2*1/2*(XY) [:note: P is midway from X and Y] XT=PR+24=1/4*XY+24 --------------------------1

See it another way; XT= 3/8*XY M travels XY in 4 hours M would travel 3/8*XY in (3/8)*4=1.5 hours Thus, M and N meet 1.5 hours after M starts its journey i.e. @1:54PM ********************************************************

Please let me know if something is unclear. _________________

Its a blinder.In the second journey, 1 bus should be ahead of the other as it has 1 hr lead.More so, the speeds have not changed too. _________________

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