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

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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 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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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!

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

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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 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

Originally posted by gracie on 12 Oct 2015, 14:12.
Last edited by gracie on 08 Apr 2018, 22:06, edited 2 times 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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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 --> $$4*48=192$$
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Re: A bus from city M is traveling to city N at a constant speed  [#permalink]

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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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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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GMAT 1: 730 Q50 V38 A bus from city M is traveling to city N at a constant speed  [#permalink]

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

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Not sure if I can post a similar question to this old thread.. a bit of a necro:
I've just changed the times and distances according to my question but I can't for the life of me solve it:

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 1 hour and 30 minutes. The following day the buses do the return trip at the same constant speed. One bus leaves 25 minutes early and the other is delayed by 15 minutes. If they meet 15 kilometers from point P, what is the distance between the two cities?

I've tried similar formulae as presented here, but I end up with terrible fractions and numbers:
$$\frac{d}{2}-15 = \frac{0.7778d}{2}$$

Am I on the right track?
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A bus from city M is traveling to city N at a constant speed  [#permalink]

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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

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

Cheers,
Brent
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Originally posted by GMATPrepNow on 08 Dec 2017, 09:23.
Last edited by GMATPrepNow on 11 Jul 2019, 05:58, edited 1 time in total.
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Re: A bus from city M is traveling to city N at a constant speed  [#permalink]

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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!

Let the distance between the two cities be X miles. Since they travel at the same constant speed, p is X/2 miles away from the cities. The difference between the start times on the following day is 1 hour. Now, the relative distance is only 3X/4; the mid point is 3X/8. The difference X/8 is 24 miles. Therefore, the total distance is 8*24=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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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!

D = Distance between 2 cities, S = Speed of buses, T = Time for the journey between the two cities
D = S x T
D = S x 4
2P = 4S
P = 2S

Slow bus: (P-24)= S x 1.6 (1)
Early bus: (P+24) = S x 2.6 (2)
48 = S (subtract equation 1 from equation 2)

P = 96
D = 2P
D = 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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the difference of time: 36-(-24) = 60min or 1h
the difference of distance 24-(-24) = 48m
since s1 = s2 = 48/1 = 48 m/h
the total distance = 48 * 4h = 192
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A bus from city M is traveling to city N at a constant speed  [#permalink]

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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

After reading how other people solved this question I can't help but feel I do not have a proper grasp yet on rate questions. Here is my solution:

$$4s = d$$

$$s* (t - \frac{24}{60}) + s* (t + \frac{36}{60}) = 4s$$

$$t = \frac{19}{10}$$

$$2s - s*(\frac{19}{10} - \frac{24}{60}) = 24$$

$$s = 48$$

$$d = 192$$
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Re: A bus from city M is traveling to city N at a constant speed  [#permalink]

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Let x be the speed of each bus. Then relative speed will be x+x = 2x miles per hour
They meet at mid-point after 2 hours
Therefore total distance between points M and N = speed * time= 2x*2= 4x miles

Mid-point P where they met is 4x/2= 2x on day1
Next day bus A has a lead of 60 minutes(36+24) / 1 hour over other bus. In 1 hour that bus travels x miles .
Thus actual distance between both buses today is 4x-x= 3x mile on day 2
Mid- Point where they met on day 2 is 3x/2

And the difference has been given as (2x-3x/2)= 1/2 x= 24 miles
solving x = 48 miles and the total distance i.e 4x = 4*48= 192 miles

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

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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 took a different approach than most, but got the answer faster than many.

Let the speed of each bus be X, so distance travelled by each bus on Day 1 = 2X

Total distance between the cities, 2X + 2X = 4X --- (1)

Now, we know one of the buses have left an hour earlier (relatively (36+24 mins.)), and has travelled an additional distance of 24 miles from the mid point (the point where both the buses met on Day 1)

Now, let the time taken by Bus 1 on Day 2 = Distance Travelled / Time = (2X + 24) / X

Time taken by Bus 2 on Day 2 = (2X - 24) / X

Now, since Bus 1 travelled an hour more than the Bus 2, Time taken by Bus 1 = Time taken by Bus 2 + 1

(2X + 24) / X = ((2X - 24) / X) + 1

Solving this, we get X = 48, and from (1), we have total distance between the cities = 4X, so the answer here is 4*48 = 192 (E)

HIT KUDOS IF YOU LIKE THE 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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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?

Finally i solved this with below reasoning

to start with, the key is both buses are travelling at same constant speed. they are travelling opposite direction and they meet after 2 hours.
If A & B are walking , 1 step at a time, and meet after 2 hours.
Assuming A can walk 1 step in 1 hr and B walk 1 step in hr. then in 2 hours they both will walk 2 steps , so total distance is 4 steps in total 4 hrs this solves the first analogy. now note P is the midpoint . Let's say total distance is d , so P = d/2

Next step - one bus is delayed by 24, and other delayed by 36, so there is a gap of 60 mins, 1 hr. if A and B are walking in 1 hr , they cover d/4 distance.
so one of the bus , lets say A, has covered d/4 distance. remaining distance is $$d-\frac{d}{4}$$ = $$\frac{3}{4}$$d
As A and B still walk same speed, their meeting point will also be equidistance from when B start (note A has already started an hour ago)
Equidistance of $$\frac{3}{4}$$d = $$\frac{3}{4}$$ * d * $$\frac{1}{2}$$ = $$\frac{3}{8}$$ * d
Now its given equidistance of now meeting point and earlier meeting point p is 24
$$\frac{d}{2}$$ - $$\frac{3}{8}$$ * d = 24
d = 192 miles  Re: A bus from city M is traveling to city N at a constant speed   [#permalink] 30 Jul 2019, 11:10

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