Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

It takes the high-speed train x hours to travel the z miles [#permalink]
25 Jun 2008, 21:56

1

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct
0% (00:00) wrong based on 0 sessions

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? z(y – x) / (x + y)

Re: PS high-speed train [#permalink]
25 Jun 2008, 22:08

The answer is z(y-x)/(x+y)

after i did it by normal method, i found an even shorter method .. POE

We are looking for distance.... the unit of measure of answer should match with distance... For options D and E it doesnt match ... eliminate

We know y>x (slower train takes more time to cover the same distance) Distance can not be more than Z, eleiminate C Distance can not be -ve, eliminate B

Answer is A. If this is too confusing i'll post the original method.

Re: PS high-speed train [#permalink]
26 Jun 2008, 04:46

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? z(y – x) / (x + y)

z(x – y) / (x + y)

z(x + y) / (y – x)

xy(x – y) / (x + y)

xy(y – x) / (x + y)

i feel the option is B...

The time taken bythe trains to meet is z/(x+y). (x > y )

The distance covered by the first train(faster train) is x * z/(x+y)

and the distance covered by the slower train is y * z/(x+y)

so the difference between the distance covered by the trains is

The distance covered by the faster train - The distance covered by the slower train ==== >

Re: PS high-speed train [#permalink]
26 Jun 2008, 05:22

dipenambalia wrote:

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? z(y – x) / (x + y)

z(x – y) / (x + y)

z(x + y) / (y – x)

xy(x – y) / (x + y)

xy(y – x) / (x + y)

i feel the option is B...

The time taken bythe trains to meet is z/(x+y). (x > y )

The distance covered by the first train(faster train) is x * z/(x+y)

and the distance covered by the slower train is y * z/(x+y)

so the difference between the distance covered by the trains is

The distance covered by the faster train - The distance covered by the slower train ==== >

x * z/(x+y) - y * z/(x+y) ====?

(x-y) * z/(x+y)

ie. B...

please let me knwo if i m wrong...

you are thinking that x and y are the speeds of fast and slow trains.... its not. the respective speeds are z/x and z/y.

Re: PS high-speed train [#permalink]
26 Jun 2008, 09:18

gmatnub wrote:

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? z(y – x) / (x + y)

z(x – y) / (x + y)

z(x + y) / (y – x)

xy(x – y) / (x + y)

xy(y – x) / (x + y)

Haha wow... this one is nuts to try and solve with variables. I sat at my desk for about 15min... not proud of that.

I'm not the biggest fan of using numbers for rate problems, but this is def one of those problems.

Took me 1min 15 sec when I did it this way.

z=100 x=2 y=5

Thus rate of the fast train is 50 and that of the slow train is 20.

t= time for both

50t=100-d 20t=d

d=200/7 100-d = 500/7

Answer is 300/7.

Just plug away. I started with A:

100(5-2)/(5+2) --> 300/7 Heyyyyyyyyy look at that theres our answer.

Re: PS high-speed train [#permalink]
26 Jun 2008, 10:39

haha..you and me both..at the end i just set screw it..and plug numbers..

GMATBLACKBELT wrote:

gmatnub wrote:

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? z(y – x) / (x + y)

z(x – y) / (x + y)

z(x + y) / (y – x)

xy(x – y) / (x + y)

xy(y – x) / (x + y)

Haha wow... this one is nuts to try and solve with variables. I sat at my desk for about 15min... not proud of that.

I'm not the biggest fan of using numbers for rate problems, but this is def one of those problems.

Took me 1min 15 sec when I did it this way.

z=100 x=2 y=5

Thus rate of the fast train is 50 and that of the slow train is 20.

t= time for both

50t=100-d 20t=d

d=200/7 100-d = 500/7

Answer is 300/7.

Just plug away. I started with A:

100(5-2)/(5+2) --> 300/7 Heyyyyyyyyy look at that theres our answer.

Stupid problem >:-(

A

gmatclubot

Re: PS high-speed train
[#permalink]
26 Jun 2008, 10:39

Wow...I'm still reeling from my HBS admit . Thank you once again to everyone who has helped me through this process. Every year, USNews releases their rankings of...

I had an interesting conversation with a friend this morning, and I realized I need to add a last word on the series of posts on my application process. Five key words:...