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

Hi, Camach700,

YES - you can TEST THE ANSWERS here, but before you start doing that work, you have to hash out all of the information that the prompt gives you to work with. We're given a number of facts in this wordy prompt (and I'm going to summarize the information into my own words):

1) Bus A and Bus B are approaching each other while they each drive at the same constant speed. The buses are both driving the same route.
2) The buses meet at point P after driving for 2 hours each.
3) The following day the buses do the return trip at the same constant speed. One bus (Bus A) is delayed 24 minutes and the other (Bus B) leaves 36 minutes earlier.
4) The NEW meeting point is 24 miles from point P

We're asked for the distance between the two cities.

To start, the 24-minute delay for one bus (Bus A) and a 36-minute early-start for the second bus (Bus B) means that Bus B travels for 1 hour MORE than Bus A does. That 1-hour "head start" has to move the MEETING POINT 24 miles. You might assume that the busses were moving 24 miles/hour, but that would NOT be correct; keep in mind that BOTH Buses will still be moving when they meet (so that 24-mile difference still has to accommodate 2 moving busses)....

Let's start by TESTing one of the larger Answers....

Answer C: 96 miles.

IF... the two buses met after 2 hours of travel, then the 4 total hours of travel would have covered 96 miles and Point P is at the 48-mile 'mark.'
D = (R)(T)
96 miles = (R)(4 hours)
96/4 = 24 miles/hour = R

On the second day, Bus B has a head start, so it travels 24 miles before Bus A gets going. Thus, the two buses would then have to travel the remaining 96-24 = 72 total miles together at 24 miles/hour each. Each hour, the TOTAL distance traveled would be 48 miles.... 72/48 = 1.5 total hours of travel, so each bus would travel 1.5 hours.

In this situation, the buses meet after Bus A has traveled 1.5 hours at 24 miles/hour.... (1.5)(24) = 36 miles.... the 36-mile 'mark.' This is NOT 24 miles away from point P.... it's only 48 - 36 = 12 miles away. This is exactly HALF the distance that it needs to be.... so I'd look to DOUBLE the total distance...

Let's TEST Answer E: 192 miles

IF... the two buses met after 2 hours of travel, then the 4 total hours of travel would have covered 192 miles and Point P is at the 96-mile 'mark.'
D = (R)(T)
192 miles = (R)(4 hours)
192/4 = 48 miles/hour = R

On the second day, Bus B has a head start, so it travels 48 miles before Bus A gets going. Thus, the two buses would then have to travel the remaining 192 - 48 = 144 total miles together at 48 miles/hour each. Each hour, the TOTAL distance traveled would be 96 miles....144/96 = 1.5 total hours of travel, so each bus would travel 1.5 hours.

In this situation, the buses meet after Bus A has traveled 1.5 hours at 48 miles/hour.... (1.5)(48) = 72 miles.... the 72-mile 'mark.' This is 96 - 72 = 24 miles from Point P, so this MUST be the answer.

Final Answer:

GMAT assassins aren't born, they're made,
Rich

I named the buses A and B. Let us refer to the diagram below:



Let speed of each bus = s miles/hr
=> total distance = 2s + 2s = 4s miles

Let P = original meeting point - midpoint of MN - say they would have met at 12:00 if they started together
P' = New meeting point when they started at different times (one early and the other late)

In the return journey: (in the diagram, scenario 3)
B was 36 min early and A was 24 min late => To simplify, let us just assume that B was 60 min early ans A was on time
Since their usual meeting time at P was supposed to be 12 noon, what actually happened is:

B reached P already at 11:00 (1 hour earlier) and then A was possibly at some point P" (at 11:00 am)
He would have taken 1 more hour to reach P (if he actually managed to do so); implying that the distance P-P" = 1 x s = s miles

However, the buses meet at the point P'.
Since both buses have same speeds, P' must be midway between P and P"

=> Distance P-P' = Distance P"-P' = s/2 miles

But P-P' = 24 miles (given) => s/2 = 24 => s = 48

=> Distance MN = 4s = 192 miles

Let the distance between m and n be D.
Both the buses move D/2 first day
Next day one bus started 1 hour earlier so covered D/4 before other bus starts because on first day both the buses covered D/2 in
2 hours.
Now out of remaining distance 3D/4 , each will cover 3D/8
So the bus which started an hour earlier has covered D/4+3D/8=5D/8
So second day distance cover extra by this bus is 5D/8-D/2= D/8
Now D/8 =24
So D =192

Posted from my mobile device

It takes 4 hours for each bus to travel the whole distance.

In the second trip, the bus that starts early travels for 1 hour before the other bus starts and so covers 1/4 of the distance.

Then the other bus starts and each travels (3/4)/2 = 3/8 of the distance.

The bus that started early totally travelled 1/4 + 3/8= 5/8 of the total distance

We know that 5/8-1/2 of the total distance = 24 miles

Total distance = 24*8 = 192 miles

Posted from my mobile device

call the 1st Bus A leaving from City M to City N

call the 2nd Bus B leaving from City N to City M

Let the Gap Distance between the 2 Buses (and thus the 2 Cities) = d


The Key Fact is that they travel the Same Speed.

After 2 hours they meet at point P. Since they are both traveling over the Same Time (Time is Constant) and they have the Same Speed (their Ratio of Speeds = 1 : 1) ---- they each will travel 1/2 of the Total Gap Distance between them when they meet at Point P.

Thus, the Mile Marker at Point P will = 1/2 * d

Speed of A = can travel 1/2 * d in 2 hours = 1/2d / 2 = d/4
Speed of B = Speed of A = d/4


Now they each have made it to the opposite City they were heading towards and are coming back on the round-trip.

Let Bus B be the bus that is delayed 24 minutes. This means Bus A is the other bus that leaves 36 minutes earlier.

Thus, the Total Head Start that Bus A gets is 1 hour. In 1 hour at the Speed of d/4, Bus A will have covered 1/4th of the total Gap Distance = 1/4*d

Now, when Bus B is ready to leave, there is only 3/4 of d remaining between them. Again, since Both buses run at the Same Speed, they will meet at the half way point. Thus, they will meet after each traveling = (3/4*d) * 1/2 = 3/8*d

Since A already traveled 1/4*d when he had the head-start, A will have traveled a Total of ---- 3/8*d + 1/4*d = 5/8*d. This is the New Mile Marker Point that the 2 buses Meet at (Call it Point Q)


Point Q is at mile marker 5/8*d away from City N where A started on the 2nd Day.
Point P is at mile marker 1/2*d away from City N where B started on the 1st Day.

We are told that when they meet at Point Q on the 2nd Day, that this is 24 miles away from Point P on the 1st Day.

Point Q - Point P = 5/8*d - 1/2*d = 5/8d - 4/8d = 1/8d = 24 miles distance between the 2 Points

1/8*d = 24

d = 24 * 8 = 192 miles

The entire distance between the 2 Cities = 192 miles

A.C. -E-

Bunuel,

Could you please advise whether there is a variation of this question for additional practice?

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Let the total distance be 4D.
It takes the bus 4 hours to go from M to N.
So in 1 hour it covers \(\frac{4D}{4} = D\) -------- (1)
P is the point where they meet after 2 hours, so P = 2D

0(M)------D------2D(P)-------3D-------4D (N)

Now, one bus is 36 earlier (Bus A) and the other (Bus B) is 24 minutes late.
Therefore, Bus A has a head start of one hour
Bus A has started D distance (From equation (1))
Now Bus A and B have a distance of 3D between them.

(M)------D(Bus A)------2D(P)-------3D-------4D (N)(Bus B)

Mid point of 2D and 3D is where they will meet as their speed is the same.
\(\frac{D}{2}\) = midpoint between 2D and 3D
We are told that this point is 24 miles away from P.

So \(\frac{D}{2}=24\)

\(D = 48\)

\(Total Distance = 4D = 4*48 = 192\)

Answer. E

First with this problem you have to understand when you have two objects moving at the same speed from opposite direction wherever they meet that's the midpoint.

1 ) v(120) = d/2
--> v (240) = d

2) v (t - 24) = d/2 - 24 (This bus travelled less, because it was delayed)
3) v (t+ 36) = d/2 + 24 (This bus got to leave 36 min early. Thus it travelled more)


--> From (2) and (3) v = 48/60
--> plug the value into (1) we get (48/60)*240 = 192

Since the buses were going the same rate over the same time, they must have traveled the same distance.

Since the sum of the distances is the total distance, D, each then traveled D/2 over 2 hrs, the location of point P.

Each buses rate is then (D/2)/2 = D/4 mph.

The next day one bus started effectively 1 hr before the other, 36 min early + 24 min delay.

The first bus then had traveled D/4 * 1 hr before the second bus started.

Using the same logic from the first day, each bus started then (3/4)D apart and therefore met in the middle, (3/8)D.

This point is 24 miles from point P.

Since point P is D/2 or 4D/8 and the point on the second day is 3D/8, the difference between these two, D/8,= 24 miles as stated.

Therefore D/8=24 and D=192

Posted from my mobile device

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?

I got tripped up by the language - for example when the Q says one bus is delayed by 24 minutes while the other leaves 36 minutes earlier. I understood that 'the other' leaves 36 minutes earlier than when the other bus leaves, but it seems they meant 36 minutes earlier than it usually does. Wording could be improved to reduce ambiguity.

A small doubt here.

We all agree that the total time taken to cover the distance between the cities is 4 hours. And on the second day one bus gets a headstart of 1 hour and and covered 24 miles in it. As the question states that the bus is travelling with a constant speed, so we can assume that the bus is travelling at 24 km/hr. So the can't the total distance be 24*4= 96km?

Please help me figure out the flaw in the above logic. Thanks!

Hi All,

I am having trouble understanding the language: "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?"

The 24 min delay and the 36mins are relative to what??? For example, it says "the other leaves 36min earlier"... earlier than Bus 1 or earlier than Bus 2 usually does??

This isn't a problem from the OG, is it??

Assuming the distance between the two cities is D.
The speed of the two buses is S.
Given that both of them start at the same time with the same speed meet exactly after 2 hours. Hence in the period of two hours, each bus travels a distance of \(\frac{D}{2}\) in two hours. Hence the speed of each bus is \(\frac{D}{4}\).
Hence the junction point is at a distance \(\frac{D}{2} \)from both M and N.
Assuming the bus starting from M starts earlier and the other bus delays by 24 minutes. This is effectively equivalent to the point that one bus travelled a distance equivalent to an hours time in the first 1 hour when the other started.
Hence they need to travel a distance of \(\frac{3D}{4}\). Since \(\frac{D}{4}\) was covered by one of them in the hour. The meeting point is
\(\left(\frac{3D}{8}\right)\). The distance from M to this point is \(\left(\frac{3D}{8}\right)+\ \frac{D}{4}=\ \frac{5D}{8}\) from M.
Hence the difference in the meeting points is \(\ \frac{5D}{8}-\ \frac{D}{2}=\ \frac{D}{8}\)
This is mentioned to be equal to 24miles. Hence the distance D = 192 miles.

GK002

It is implied that they both start at the same time on day 1, say at 12 noon.
Next day on the return journey, they don't start at the same time. Compared with the previous day, one leaves 36 min earlier (36 mins before 12 noon) and the other leaves 24 mins late (at 12:24)
No, it is not an OG question though the concept tested is very much within the scope of GMAT.

Here is a solution:

Posted from my mobile device

KarishmaB Vinit800HBS

Reading the first half we get:

Let total distance = D
Let Rate of each bus = R

Since they meet at D/2 in 2 hours we can infer that the total time = 4hrs

Now, reading the latter we get

Bus 1 had an early start of 1hr. So for 1hr, the bus drove at a rate of R thereby covering D/4 the distance.

But we see that the extra 1 hour results in the extra 24 miles. So why can't we say that Bus 1 drove at a rate of 24 miles/hour? and hence:

D = R x T
D = 24 x 4 --> 96 miles.