A bus from city M is traveling to city N at a constant speed while ano

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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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.

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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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

EQN2-EQN1 : 0.5R=24
R=48

So the distance is rate x time = 48 x 4 = 192.
ANS = E

Normally I goof up with distance-speed 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.

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)
gmat-diagnostic-test-question-79354.html
The 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.

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(T-2/5)+24
Therefore, R(T+3/5)-24=R(T-2/5)+24
Solving for R, we get Rate=48 m/h
Plug in value in Chart#1: p=96
So, D=192

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!!

This is the only explanation that made sense to me! Thanks!! +1

M----------x------ P-----(d-x)--------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/2-24)/v

=> 48 = d/2 - d/4

=> d = 4 * 48 = 192

Answer - E
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Thanks Bunuel.

Your explanation is very clear ... I also has trouble undersatnding the official answer..
Kudos to you!

@Bunuel - Very neat explanation. Kudos to you..!!
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quite confusing question but very good explained by bynnel and karishma
thanks to both of you

good explanation bunuel.......... thanks

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/2-24=0.75d/2 (greater value minus 24 equals to smaller value).

Hope it's clear,
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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.

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?

let d=distance (what the question is looking for)

On the first day, bus a and bus b travel at the same constant speed. Because they travel at the same constant speed, when they meet at point P they have each traveled for 2 hours. Therefore, the total time (at that constant speed) from A to B is 4 hours and P is the mid distance between those two points.

On the second day each bus travels at the same constant speed. The bus that leaves first spends a total of one hour on the road before the second bus leaves (if bus a and bus b normally leave at the same time and today one leaves 36 minutes earlier and 24 minutes later respectively)

Here is where I get thrown off (using Bunuel's explanation to structure mine)

The wording is a bit ambiguous (at least to me) regarding the constant speed of bus A and bus B. On the second day, do they travel the same constant speed they did the day before or did they each travel the same constant speed that day that is different from the day before?

If they each traveled at the same rate they did before (which is 1/4 of the distance every hour) and today one bus traveled for an hour before the other set off, then today one bus traveled .25d of the way

(d/2) - 24 = (.75d/2)

(I'm not sure as to why that is)

Maybe this is why:

(d/2) represents the speed each bus travels and the midpoint at which bus A and B meet on the first day. (d/2)-24 would represent the bus reaching a point 24 miles before the midpoint. This is equal to a distance that's 3/4ths normal???? Help!!!

Bunuel, I read the link you posted offering an explanation of how that formula represents the point at which each bus meets the second day but I am still confused. Why is the halfway .75d?


Here is another way to look at it. Let's say bus A starts first and B, second. When bus A travels one hour bus B starts. When bus A has traveled for two hours, bus B has traveled for one. When bus A and B meet, A has been on the road for 2.5 hours and B, for 1.5 at a place 24 miles away from the mid distance of the journey.

We can break the journey up into four identical blocks, each representing an hour long portion of the journey (see attached image) when bus A and B meet they meet in the exact middle of one of those 4 identical blocks meaning that 24 miles represents 1/2 of one block. This means there are two 24 mile portions in each of the four blocks. 2*24*4=192.

A. 48
B. 72
C. 96
D. 120
E. 192

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

r=speed of each bus
2r=distance from each city to point P
4r=distance between two cities
2nd day early bus distance to meeting=2r+24
2nd day late bus distance to meeting=2r-24
2r+24-(2r-24)=48 miles difference between distances covered
in exactly one additional hour early bus covers 48 miles more than late bus
thus, r=48 mph
4r=192 miles distance between two cities
E

I think this could be solved with the help of relative speed & unitary method concept:
Since speed of both the bus is the same, let's assume it as "X"
Moving in opposite direction (with same speed), gives the relative speed as 2x

In 1 hour (24 min + 36 min), distance moved from point P = 24 miles
Using unitary method,
For relative speed 2x, distance moved = 24 miles (i.e. each bus reduced the distance by 24 miles)
Therefore, total distance moved by both bus = 24 miles + 24 miles = 48 miles (in 1 hour)

In 4 hours the distance travelled = 48*4 = 192 Miles