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

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Re: A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 09 Jul 2013, 04:40
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.

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Re: A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 05 Aug 2013, 11:04
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
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Re: two buses, same speed... head spinning [#permalink]

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New post 31 Dec 2013, 06:18
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Bunuel wrote:
T740qc wrote:
Bunuel wrote:
\(\frac{d}{2}-24=\frac{0.75d}{2}\)



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,


Here's the way I did it. Let's call X the total distance

Let's focus on the second part of the problem

One bus had a head start of 1 hour. In that hour we traveled 1/4 of the distance. (Since he traveled 1/2 of the distance in 2 hours from question stem)

Now, there are 3/4x of the distance to be traveled

Since they both have the same rate, they will meet at half the distance of 3/4x, so they will meet at 1/4x + (3x/4/2) = 5/8x

Now remember from the first part of the question that point P is exactly the mid point (Same rates, same time)

So we are now at point 5/8x. The question also says that this distance is equal to 24 miles.

So we have 5/8x - 1/2x =1/8x = 24 miles

And so, we get 24*8 = 192 miles for the total distance 'x'

Hope it helps!
Cheers!

J :)

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A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 12 Oct 2015, 14:12
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R=speed of each bus
2R=distance from each city to P
4R=distance between two cities
bus1 distance to meeting 2nd day=2R+24
bus2 distance to meeting 2nd day=4R-(2R+24)=2R-24
2R+24-(2R-24)=48 miles
bus1 drives 48 miles more than bus2 in 1 extra hour
R= 48 mph
4R=192 miles distance between two cities

Last edited by gracie on 03 Apr 2016, 11:10, edited 1 time in total.

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A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 19 Dec 2015, 05:58
gmattokyo 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 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 source is GMATClub's diagnostic test... would look forward to see some innovative approach to this.
Thanks!


I just don't know, why my brain is in the standby mode during the cat, now I've solved this one very quickly ~ 1m 20sec )))
P is the midpoint or Total \(\frac{Distance}{2}\), so one of them have traveled 24 miles more than the half of the distance, and guess which bus has made it (the one, that started his trip 36+24=60 min earlier, let's say it was Bus B)
Bus A:\(r*t=p-24\)
Bus B \(r*(t+1)=p+24\)
=> \(r*t+24=r*(t+1)-24\)
Rate = 48, so if each of them traveled 2 hours on the first day, than the whole distance can be covered in 4 hours by each of them --> \(4*48=192\)
Answer E
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Re: A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 16 Mar 2016, 03:51
T740qc wrote:
Bunuel wrote:
\(\frac{d}{2}-24=\frac{0.75d}{2}\)

why do you assume -24 instead of +24?

You can write it as following instead -

d/4 + [(3d/4)/2] = d/2 + 24 ..... (1)
= d/4 + 3d/8 = d/2 +24
Solve for d=192

The reason you can write (1) is because d/4 was already covered by Bus A before even Bus B started, and half of the remaining distance was covered together by both buses. So total distance covered by Bus A (which traveled for an hour more) is exceeding there previous mid point by 24mi

If you see both Bunuel's and above explanation, simplify to the common solution.

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Re: A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 16 Mar 2016, 04:45
gmattokyo 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 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 source is GMATClub's diagnostic test... would look forward to see some innovative approach to this.
Thanks!

Let the distance be 4x
Together they take 4 hours to meet at constant speed.
In the first case they meet at the mid point.
In the second case, one bus start 1 hour earlier than the other. So, it must have covered x distance.
In the remaining distance of 3x, each bus covers 1.5x
It implies one bus covers 2.5x and another 1.5x
That is from the mid point .5x distance, which is given as 24 miles.
Therefore, total distance is 8*24=192 miles

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A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 03 Apr 2016, 08:15
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gmattokyo 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 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 source is GMATClub's diagnostic test... would look forward to see some innovative approach to this.
Thanks!


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

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Re: A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 15 Jun 2017, 09:22
gmattokyo 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 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 source is GMATClub's diagnostic test... would look forward to see some innovative approach to this.
Thanks!


Hi..
Q can be solved by taking it the following way..
Since it is constant speed let's take that only one bus travels from m to N.. so that bus will take 2*2=4 hrs..

Now in SECOND case 24 minutes delay and 36 minutes early means difference of 24+36=60 min or 1 HR..
So remaining distance will be covered in 4-1=3 HR..
The remaining distance will be travelled half way that is 3/2 HR by each bus. So the bus which has traveled 1 HR already travels another 1.5 HR.. so total 2 1/2 HR..
But P is half way or 2 hrs away..
Therefore the new point is 2 1/2-2=1/2 HR away..
This 1/2 HR means 24 miles ..
So total distance of 4 hrs corresponds to 4/(1/2) *24=8*24=192 miles
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A bus from city M is traveling to city N at a constant speed [#permalink]

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New post 07 Oct 2017, 21:57
I solved like this,

Let v (m/hr)- velocity, d - distance between two cities.
First meeting point P = (d/2) (since both travel with same velocity)

Now return trip, one bus yields (36 + 24) = 1hr lead to another bus
so in 1 hr, the second bus would have travelled = v miles.

Now meeting point = (d - v)/2 (since both travel with same velocity, they meet at midpoint)

given: d/2 (Point P) - (d-v)/2 = 24 => solving this we get v = 48 m/hr

so given 2 hours for first meeting point , bus covers (d/2) distance in 2 hours = (d/2) = velocity (distance in one hour) * 2 = 96
d/2 = 96 => d = 192 miles

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A bus from city M is traveling to city N at a constant speed   [#permalink] 07 Oct 2017, 21:57

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