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D01-24

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D01-24  [#permalink]

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New post 16 Sep 2014, 00:12
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A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). 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
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New post 16 Sep 2014, 00:12
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Official Solution:

A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). 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 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.

Alternative Explanation

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

Hence, \(\frac{d}{2}-24=\frac{0.75d}{2}\), which gives \(d=192\) miles.


Answer: E
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Re: D01-24  [#permalink]

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New post 23 Dec 2014, 16:30
Bunuel wrote:
Official Solution:

A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). 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 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.

Alternative Explanation

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

Hence, \(\frac{d}{2}-24=\frac{0.75d}{2}\), which gives \(d=192\) miles.


Answer: E


Hey Bunue,

I didn't get why the buses meet again at the halfway of \(0.75d\), which is 24 miles from \(\frac{d}{2}\). May u help to explain?
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D01-24  [#permalink]

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New post 23 Dec 2014, 21:08
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The buses are travelling at the same speed. One bus is at the point 0.75d when the other bus starts. So, the distance between them is 0.75d and they are travelling at the same speed. Therefore, they will meet halfway through the journey. Hope that helps.
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Re: D01-24  [#permalink]

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New post 26 Dec 2014, 10:40
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total distance can be covered in 4 hrs(Given). Let it be x.
After one hour: bus A covers x/4 , bus B covers 0 distance, as there is a delay of one hour between the buses.
After two hours: bus A covers x/4 + x/4=x/2( bus A is at point p), bus B covered x/4. Note, at this point of time both are equidistant from the point they met on the round trip i.e 24 kms.
In next one hour they will cover 48kms.
distance covered in one hour=48 kms,
distance covered in four hour = 48*4=192kms.
**With diagram it becomes easy to understand.
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Re: D01-24  [#permalink]

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New post 06 Jan 2015, 11:23
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5
M------------------------------N
Bus-A Bus-B
Both at constant equal speed meet at point P which is 2 hrs away so total time is 4 hours for each bus
Now on Second day Bus-A departs 36 mins early and Bus-B departs 24 minutes late
that means when A traveled for 1 hour, B just left N
1hr.
M------|-----------------N
A B

Now time left is 3 hrs. Both travel at same constant speed so they will meet after 1.5 hrs
In this time A has traveled for 2.5 hours (.5 hour more than yesterday which represents 24miles.)
So 4 hours = .5 * 8 is equal to distance of 24*8 = 192miles
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Re: D01-24  [#permalink]

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New post 19 Feb 2015, 06:18
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its given that 1 bus starts 36 minutes late and the other 24 minutes early which means effectively 1 bus had head start of 1 hr. Since they are traveling at same speed and they meet 24 miles from where they met yesterday then the bus which had head start of 1 hr traveled 24 miles in 1 hr. Thus its speed is 24 miles. The speed of 2nd bus is also 24 (since both bus travels at constant speed)..
Thus both bus travels for 4 hrs hence distance traveled by 1 bus =24*4

Total distance= 2*24*4= 192.. Answer
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Re: D01-24  [#permalink]

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New post 21 Sep 2015, 04:43
Bunuel wrote:
Official Solution:

A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). 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 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.

Alternative Explanation

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

Hence, \(\frac{d}{2}-24=\frac{0.75d}{2}\), which gives \(d=192\) miles.


Answer: E


Sorry, I am still unable to understand direction of the travel.
In beginning both buses move towards each other and meet at point P
In return journey they move away from each other...so I am not getting how they meet again?
I think I completely missed what question is talking about.
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Re: D01-24  [#permalink]

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New post 21 Sep 2015, 08:23
anupamadw wrote:
Bunuel wrote:
Official Solution:

A bus leaves city \(M\) and travels to city \(N\) at a constant speed, at the same time another bus leaves city \(N\) and travels to city \(M\) at the same constant speed. After driving for 2 hours they meet at point \(P\). 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 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.

Alternative Explanation

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

Hence, \(\frac{d}{2}-24=\frac{0.75d}{2}\), which gives \(d=192\) miles.


Answer: E


Sorry, I am still unable to understand direction of the travel.
In beginning both buses move towards each other and meet at point P
In return journey they move away from each other...so I am not getting how they meet again?
I think I completely missed what question is talking about.


I hope alternative solutions help: a-bus-from-city-m-is-traveling-to-city-n-at-a-constant-speed-86478.html
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Re: D01-24  [#permalink]

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New post 08 Dec 2015, 19:25
Alternative solution:
as 2 buses meat at the middle by driving 2 hours each. It means that the whole time each buses spends is 4 hours.
If one bus delayed 24 min and the other leaves 36 min earlier, it means that the 2nd bus drives 1 hour alone.
In 1 hour it will cover 1/4 of the distance. Then the rest of the distance 2 buses travel together and they will meet in the middle of the rest distance, thus 1-1/4=3/4 middle is 3/4/2=3/8. The distance between two points will be 1/2-3/8=1/8 which equals to 24 miles. Thus the whole distance is 8*24=192 miles.
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Re D01-24  [#permalink]

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New post 15 Dec 2015, 19:46
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I think this is a high-quality question and I don't agree with the explanation. In second scenario, when one bus is delayed for 24 minutes and another bus leaves 36 minutes , it is clear that one bus has traveled for one hour extra before they meet. Had they traveled for same time they would meet at P but they meet 24 miles from point P. This means this distance is traveled in one hour by one of the bus that has traveled one hour more than other by the time they meet. So speed of bus would be 24 miles per hour. since it takes 4 hours so distance would be 96 miles.
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Re: D01-24  [#permalink]

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New post 29 Dec 2015, 01:08
Dear Buffaloboy,

Please note that 24 kms from point p are not covered in 01 hour. See the reasoning below:

- Difference in time between two buses 1 hour

- Total time required to cover the distance: 04 hours

Y Y Y Y Y Y X X X X (Here X X X X reflects the distance covered in 01 hour by bus leaving earlier; Please note its just for understanding the reasoning)

- Not pending distance to be covered is - Y Y Y Y Y Y -

- As the speed is constant for both buses, each bus would take similar time to cover half distance (point where they meet; 3 Ys in our example)

- Now remember from question stem that 04 hours required to travel the distance

- Hence total time for buses would be 1.5 (half distance of Y Y Y Y Y Y by one bus) + 1.5 Y Y Y Y Y Y (half distance of Y Y Y Y Y Y by other bus) + 1 hour (extra one)

From there on, I think explanation is self explanatory.

Hope it helps!
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D01-24  [#permalink]

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New post 08 Jul 2016, 07:00
I think this is a high quality question .

To solve it I used a straight approach, converting a S*T=D problem into a R*T=W problem.
The distance between the 2 cities becomes the work => W=MN
We know that the 2 bus are driving toward each other a constant speed v therefore in 1h their summed speed is gonna be 2v
We know that the time is 2h
r-------t------w
2v-----2------MN => basically we are answering the question: " how long will it take to a constant rate of 2v km/h to close a gap of MN km?"

so we can deduce that 4v=MN => v=MN/4

r---------t------ w
MN/4----2------.5MN => we know the speed and the time => we can calculate the work done = distance covered (by one bus) => .5MN=P

Then if one bus leaves 36 min earlier than scheduled, while the other leaves 24 min later than scheduled therefore for 1 h one of the two bus moves while the other doesn't

r-------- t-------w
MN/4----1------MN/4 => in this hour one bus covers MN/4 therefore when the second bus depart so that the two bus are driving toward each other there still 3MN/4 to be covered which is gonna be the work.

r---------t------ w
MN/2----1.5----3MN/4 => speed is MN/2 because the rate at which the gap (3MN/4) is closing is 2v

r------------t---------w
MN/4------1.5-------3MN/8 the new meeting point P' will be 3MN/8


in conclusion MN/2= (3MN/8) +24 => MN=24*8= 192
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Re: D01-24  [#permalink]

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New post 19 Oct 2017, 06:20
My alternative solution:
Let's rid out of the delayed bus and continue with just one bus that left earlier. What does it give us?
The bus can handle distance between M and N for 4 hours. P is the midpoint between the cities.
According to the question stem, one bus left 36 min earlier and another delayed for 24 min. It is the same as saying that early-bird bus left 60 min earlier (24+36 = 60).
Let's say that he must have been living at 8 am, but he actually left at 7 am.
So, at 10 am.
Planned time: 2 hours
Planned distance: MP
Fact time: 3 hours
Fact distance: MP+48 miles (48 miles = 24 miles*2. Why times 2? Since we considering the speed of the second bus as 0, therefore, nobody is driving toward us and we have to drive distance of the second bus by ourselves, in addition to our own distance)
Therefore, the speed is 48 m/h. And the distance is 48*4 = 192.
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Re: D01-24  [#permalink]

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New post 21 Oct 2017, 23:34
Mind = Blown ! Brilliant explanation Bunuel
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New post 09 Oct 2018, 04:30
Let the distance between M & N be D miles.

Since both buses took two hours before they met halfway round at point P. It took 4 hours for them to cover the complete distance D. (Its given that the two buses travel at the same constant speed and leave the cities at the same time then they meet at the halfway in two hours , i.e. cover distance D/2...and meet mid way i.e. at point P)

Attachment:
Q2.jpg


On the second day one bus traveled alone for 1 hour before another bus started from its destination, hence covered D/4 of the distance in one hour. The remaining distance to be covered (3/8)D before they meet at point C (in figure above). This point is 24 miles from P. Which implies that (3/8)D+ 24 = D/2 or 3D+192 = 4D


Therefore \(D=192\) miles.


Correct Answer: Option E
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Re D01-24  [#permalink]

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New post 13 Apr 2019, 08:02
I think this is a high-quality question and I agree with explanation.
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New post 28 Aug 2019, 08:49
I think this the explanation isn't clear enough, please elaborate.
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Re: D01-24  [#permalink]

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New post 22 Sep 2019, 23:54
Consider the distance between M and N as D. As both the buses travel at same constant speed, after 2 hours, they meet at P, which is at a distance D/2 from M (and N).

Therefore, the total time required to travel from M to P(mid-point) is 2 hours and hence from M to N = M to P + P to N = 2(M to P) = 4 hrs.

The following day, first bus is delayed by 24 minutes and second bus starts 36 minutes earlier, which means that one of the bus, say A, travels for (24+36) 60 minutes (from N) before the other bus starts (from M), say B.

As we know that the total time required to cover the distance between M and N is 4 hours, A covers D/4 distance in 1 hour, i.e. A travelled a distance of D/4 from N (is at a distance of 3D/4 from M).

Now, B starts at the same constant speed from M. At this instant, the distance between A & B is 3D/4. Hence, each bus must travel half of the distance between them, i.s. 3D/8 for them to meet at a point P'.

Total distance covered by A = D/4 + 3D/8 = 5D/8, (i.e A travelled 5D/8 from N or A is at a distance of 3D/8 from M)
Total distance covered by B = 3D/8 ( i.e. B travelled 3D/8 from M or is at a distance of 5D/8 from N)

Point P' is 24 miles away from point P, which is at a distance D/2 from M (and N).

Therefore,
5D/8 = D/2 + 24 or 3D/8 = D/2 - 24
4D/8 = 24
D = 24*8 = 192 miles

Hope this explanation is elaborate and clear.
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D01-24  [#permalink]

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New post 29 Oct 2019, 06:36
The thing to keep in mind with respect to this exercise is the following.

If the objects move towards each other, the gap between them shrinks at the objects' combined speed.

We are given that one object covers 24 miles (38.62 km) in 1 hour.

Since we also now that both take 2 hours and both move at the same speed,

we know that the total distance must be \((24+24)*(2+2) = 48 * 4 = 192 \ miles \ (308.99 km)\)
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