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

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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 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?

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

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by Anju@Gurome » Sat Apr 13, 2013 11:04 am
varun289 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 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?
Let us assume, the distance between the cities d miles.
On the first day, as both buses traveled at the same constant speed and leave the cities at the same time they meet halfway between the cities. Hence, each bus covers d/2 miles in 2 hours. So, speed of each bus is d/4 miles per hour.

On the second day, the first bus traveled alone for (36min + 24min) = 1 hour
Hence, the first bus covered d/4 miles.
So, (d - d/4) = 3d/4 miles is left.

Now, they will again meet halfway of these rest 3d/4 miles = 24 miles from d/2

So, (3d/4)/2 = d/2 - 24
--> 3d/8 = d/2 - 24
--> 4d/8 - 3d/8 = 24
--> d/8 = 24
--> d = 192

The correct answer is E.
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by CSASHISHPANDAY » Sun Apr 14, 2013 2:45 am
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.

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by GMATGuruNY » Sun Apr 14, 2013 4:00 pm
varun289 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 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?

A. 48
B. 72
C. 96
D. 120
E. 192
Let:
P1 = the distance traveled by EACH bus on Day 1.
P2 = the distance traveled by the LATER bus on Day 2.
d = the total distance between M and N.

The two buses travel at the SAME SPEED.
Thus, when they travel toward each other, each bus travels HALF the distance between them.

Day 1:
When the two buses travel toward each other, each travels (1/2)d.
Thus:
P1 = (1/2)d.
Since each bus travels for 2 hours, the total time required to travel the d miles between M and N = 2+2 = 4 hours.

Day 2:
The earlier bus leaves 36 minutes AHEAD of schedule, while the later bus leaves 24 minutes BEHIND schedule.
Since 36+24 = 60, the earlier bus travels for 60 minutes -- in other words, for 1 hour -- by itself.
Since it takes 4 hours to travel the d miles from M to N, in 1 hour -- 1/4 the time -- the earlier bus travels (1/4)d miles.
Then, when the two buses travel toward each other, each travels 1/2 of the remaining (3/4)d miles.
Thus:
P2 = (1/2)(3/4)d = (3/8)d.

Result:
P1 - P2 = (1/2)d - (3/8)d = (1/8)d.
Since the distance between P1 and P2 is 24 miles, we get:
(1/8)d = 24
d = 192.

The correct answer is E.
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by veenu08 » Thu Apr 18, 2013 8:50 am
Since the buses travel at the same constant speed, so we can conclude that they meet midway.
therefore we can say that the time taken to travel from M to N is 4 hours.

When one bus(lets say A) is delayed by 24 minutes and another one (lets say B) left 36 minutes early,and they meet at 24 miles from P, so Bus B traveled 24 miles in 1 hour (24+36= 60 minutes).
Hence speed of bus is 24 miles/hr.

Now we have speed= 24 miles/hr
Time= 4 hrs
Distance= Time*speed = 24*4= 96 miles

so can anyone suggest me where I am wrong ?

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by Anju@Gurome » Thu Apr 18, 2013 9:03 am
veenu08 wrote:... When one bus(lets say A) is delayed by 24 minutes and another one (lets say B) left 36 minutes early,and they meet at 24 miles from P, so Bus B traveled 24 miles in 1 hour (24+36= 60 minutes).
Hence speed of bus is 24 miles/hr.
No.
Bus B traveled (half of the distance between M and N + 24 miles) in 1 hour as they meet 24 miles from P and P is midway between M and N.

Hope that helps.
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by CSASHISHPANDAY » Fri Apr 19, 2013 8:52 am
[spoiler] When one bus(lets say A) is delayed by 24 minutes and another one (lets say B) left 36 minutes early,and they meet at 24 miles from P, so Bus B traveled 24 miles in 1 hour (24+36= 60 minutes). Hence speed of bus is 24 miles/hr.[/spoiler]

Anju has explain very well. I just want to add as follows:

look
1st bus travels (36+24)+1.5hour= 2.5 hour
2nd Bus travels 1.5 hour. = 1.5 hour

earlier each bus traveled 2 hour
now took any bus; on time front, there is a difference of 30 minutes, and on distance front of 24 km.
quite clear in 30 minute (not in 1hr) buses travel 24 KM, in 4 hour buses travel 192 KM. [/spoiler]
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by Brent@GMATPrepNow » Fri Dec 08, 2017 8:24 am
varun289 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 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?

A. 48
B. 72
C. 96
D. 120
E. 192
Buses traveling at SAME speeds meet in point P after driving for 2 hours.
IMPORTANT: Since the 2 buses are traveling at the SAME speed, then point P must be HALF WAY between city M and city N

Also recognize that the TOTAL travel time = 4hrs (2 hrs for each bus)

TOTAL distance traveled = (distance one bus traveled) + (distance other bus traveled)
Distance = (travel time)(rate)
Let r = the rate that EACH bus is traveling.
So, we get: TOTAL distance traveled = 2r + 2r = 4r
So, from the above fact, the distance from point P to city M (and to city N) = 4r/2 = 2r

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?
Another way to read this is: One bus leaves its city. ONE HOUR LATER, the other bus leaves its city.
Let's say Bus A leaves city M at noon, and Bus B leaves city N at 1pm.
Let t = Bus A's travel time until they meet
So, Let t-1 = Bus B's travel time until they meet

Since the TOTAL travel time = 4hrs, we can write: t + (t-1) = 4
Solve to get: t = 2.5
So, Bus A traveled for 2.5 hours, Bus B traveled for 1.5 hours,
This means Bus A traveled further. In fact, Bus A travels PAST point P (which is halfway between cities M and N) for an ADDITIONAL 24 miles
In other words, Bus A's travel distance = 2r + 24
Since Bus A traveled for 2.5 hours at a rate of r, we can write: 2.5r = 2r + 24
Multiply both sides by 2 to get: 5r = 4r + 48
Solve: r = 48

TOTAL distance traveled = 4r
= 4(48)
= 192

Answer: A

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