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Distance of trains

by RobertP » Sun Feb 21, 2016 3:27 pm
Hello everyone,

A high speed train will take X hours to run Z km which separates city A from city B. A conventional train will take y hours to travel the same distance. If the high speed train leaves city A to city B and the conventional train leaves city B to city A at the same time, how many more kms will the high velocity train have traveled than the conventional train when they pass each other.

(A) z (y-x) / (x+y)
(B) z (x-y) / (x+y)
(C) z (x+y) / (y-x)
(D) xy (x+y) / (y-x)
(E) xy (x+y) / (x-y)

Thank you!!
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by GMATGuruNY » Sun Feb 21, 2016 3:37 pm
It takes the high-speed train x hours to travel the z miles from Town A to Town B at a constant rate, while it takes the regular train y hours to travel the same distance at a constant rate. If the high-speed train leaves Town A for Town B at the same time that the regular train leaves Town B for Town A, how many more miles will the high-speed train have traveled than the regular train when the two trains pass each other?

A)z(y - x) / (x + y)
B)z(x - y) / (x + y)
C)z(x + y) / (y - x)
D)xy(x - y) / (x + y)
E)xy(y - x) / (x + y)
Plug in easy numbers for the two rates.
Then plug in a distance that is a multiple of both the two individual rates and the COMBINED rate.

Let the high speed rate = 3 miles per hour.
Let the regular rate = 2 miles per hour.
Combined rate for the trains = 2+3 = 5 miles per hour.
Let z = 30 miles.
Time for the high speed train to travel 30 miles = x = 30/3 = 10 hours.
Time for the regular train to travel 30 miles = y = 30/2 = 15 hours.
Time for the trains to meet = 30/5 = 6 hours.
Distance traveled by the high speed train in 6 hours = r*t = 3*6 = 18 miles.
Distance traveled by the regular train in 6 hours = r*t = 2*6 = 12 miles.
Distance for the high speed train - distance for the regular train = 18-12 = 6. This is our target.

Now plug x=10, y=15, and z=30 into the answers to see which yields our target of 6.

Only answer choice A works:
z(y - x)/(x+y) = 30*(15-10)/(10+15) = 6.
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by jain2016 » Sat Apr 30, 2016 10:44 pm
Hi Experts ,

Please check my solution and advise whats wrong.

Let z = 100 miles and x = 10 hours and y =20 hours

Let the speed of high speed train is S1 and speed of regular train is S2.

So the S1= 10m/h and S2 = 5m/h

Now let x be the distance that high speed train covers and 100-x distance cover by regular train.

Both leave the train at the same time, so our equation will be

x/S1 = 100-x/S2

x/10 = 100-x/5

from this x = 200/3.

So this is our target, if I put the values in option A, then I do not find my answer.

Please check and correct me.

Many thanks in advance.

SJ
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by GMATGuruNY » Sat Apr 30, 2016 11:13 pm
For the sake of clarity in your solution below, I've let d = the distance traveled by the high-speed train.
jain2016 wrote:Hi Experts ,

Please check my solution and advise whats wrong.

Let z = 100 miles and x = 10 hours and y =20 hours

Let the speed of high speed train is S1 and speed of regular train is S2.

So the S1= 10m/h and S2 = 5m/h

Now let d be the distance that high speed train covers and 100-d distance cover by regular train.

Both leave the train at the same time, so our equation will be

d/S1 = 100-d/S2

d/10 = 100-d/5

from this d = 200/3.

So this is our target
Since the question stem does not ask for the distance traveled by the high-speed train, the target is NOT 200/3.
Rather, the question stem asks for the DIFFERENCE between the high speed train's distance and the regular train's distance.
Distance traveled by the regular train = 100-d = 100 - (200/3) = 100/3.
Thus, the difference between the two distances = (200/3) - (100/3) = 100/3. This is our target.

Now we plug x=10, y=20 and z=100 into the answers to see which yields our target of 100/3.
Only A works:
z(y - x) / (x + y) = 100(20-10) / (10+20) = 100/3.

The correct answer is A.
Last edited by GMATGuruNY on Sun May 01, 2016 6:38 am, edited 1 time in total.
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by jain2016 » Sun May 01, 2016 6:33 am
Since the question stem does not for the distance traveled by the high-speed train, the target is NOT 200/3.
Rather, the question stem asks for the DIFFERENCE between the high speed train's distance and the regular train's distance.
Distance traveled by regular train = 100-d = 100 - (200/3) = 100/3.
Thus, the difference between the two distances = (200/3) - (100/3) = 100/3. This is our target.

Now we plug x=10, y=20 and z=100 into the answers to see which yield our target of 100/3.
Only A works:
z(y - x) / (x + y) = 100(20-10) / (10+20) = 100/3.

The correct answer is A.
[/quote]

Thank you so much sir for correcting me.

SJ
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by vipulgoyal » Thu Jan 12, 2017 8:19 am
Hi Mitch,

Could you please let me know where i m wrong here


(z/x+y) ---- time travelled by either train
x(z/x+y) - yz/xy
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by GMATGuruNY » Fri Jan 13, 2017 5:05 am
vipulgoyal wrote:Hi Mitch,

Could you please let me know where i m wrong here


(z/x+y) ---- time travelled by either train
x(z/x+y) - yz/xy
The portion in red is incorrect.

Rule:
If A takes x hours to do a job, and B takes y hours to a job, the combined rate for A and B working together = (x+y)/(xy), and the total time for A and B working together = (xy)/(x+y).

In the problem above:
When the two trains travel toward each other, they WORK TOGETHER to cover the z miles between them.
Since the time for the high-speed train = x, and the time for the regular train = y, the total time for the two trains working together = (xy)/(x+y).
Thus:
Since the rate for the high-speed train = d/t = z/x, the distance traveled by the high-speed train = rt = (z/x) * (xy)/(x+y) = (zy)/(x+y).
Since the rate for the regular train = d/t = z/y, the distance traveled by the regular train = rt = (z/y) * (xy)/(x+y) = (zx)/(x+y).
Difference between the distances = (zy)/(x+y) - (zx)/(x+y) = [(z)(y-x)]/(x+y).

The correct answer is A.
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by Scott@TargetTestPrep » Mon Jan 16, 2017 5:10 pm
RobertP wrote:Hello everyone,

A high speed train will take X hours to run Z km which separates city A from city B. A conventional train will take y hours to travel the same distance. If the high speed train leaves city A to city B and the conventional train leaves city B to city A at the same time, how many more kms will the high velocity train have traveled than the conventional train when they pass each other.

(A) z (y-x) / (x+y)
(B) z (x-y) / (x+y)
(C) z (x+y) / (y-x)
(D) xy (x+y) / (y-x)
(E) xy (x+y) / (x-y)
We have a "converging rate problem" in which we can use the formula:

distance of high speed train + distance of conventional train = total distance traveled

Since there are z km between the two cities:

distance of high speed train + distance of conventional train = z

We are given that a high speed train will take x hours to run z km and the conventional train will take y hours to travel the same distance. Since rate is distance/time, we know:

rate of high speed train = z/x

rate of conventional train = z/y

We are given that they leave at the same time, so we can denote the time traveled as t.

Thus, the distance traveled by the high speed train is (z/x)t = zt/x and the distance traveled by the conventional train is (z/y)t = zt/y.

Thus:

zt/x + zt/y = z

Multiplying the entire equation by xy, we have:

zty + ztx = xyz

ty + xy = xy

t(y + x) = xy

t = xy/(y + x)

So, the two trains meet in t = xy/(y + x) hours. In this amount of time, the high speed train travels (z/x)(xy/(y + x)) = zy/(y + x) kilometers and the conventional train travels (z/y)(xy/(y + x)) = zx/(y + x) kilometers. The difference is:

zy/(y + x) - zx/(y + x) = (zy - zx)/(y + x) = z(y - x)/(x + y).

Answer: A

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by Matt@VeritasPrep » Thu Jan 19, 2017 6:57 pm
vipulgoyal wrote: Could you please let me know where i m wrong here

(z/x+y) ---- time travelled by either train
You're right that each train travels for the same time, though, since they leave at the same time and meet at the same time.

To find that time in terms of either train's rate, use the Distance = Rate * Time equation:

High speed train:
Time for the whole trip = x
Distance for the whole trip = z
Rate for the whole trip = Distance / Time = z / x

Conventional train:
Time for the whole trip = y
Distance for the whole trip = z
Rate for the whole trip = z/y

Now let's say they each run for time t before meeting. That means that, together, the entire distance = the sum of each train's distance, or

z = High Speed's Distance + Conventional Distance

z = (z/x)*t + (z/y)*t

z / (z/x + z/y) = t

1 / (1/x + 1/y) = t

(x * y) / (x + y) = t
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