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by ieeyorei » Mon Feb 16, 2009 11:01 am
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)


The answer is A.
Is there another way to solve this without picking numbers?
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by Bidisha_800 » Mon Feb 16, 2009 12:13 pm
speed of high speed train = z/x
speed of regular train = z/y

relative speed of these two trains when they are coming towards each other = z/x+z/y

time it will take to meet T = z/(z/x+z/y)
=xy/(x+y)

highspeed train traveled in time T = zy/(x+y)
regular train traveled in time T = zx(x+y)

difference = z(y-x)/(x+y)

(A)
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by evanr2 » Sat Jul 24, 2010 7:24 am
I just got this problem on MGMAT CAT 3, the official answer was B, which I disagree with.

Question:

"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)

Note from the stem that x < y since x corresponds with the high speed train. All other numbers are positive. Therefore if you choose answer B, which is suggested by the given solution, you will result in a negative distance which is impossible.

Combined speed, call it C = (z/x + z/y) = z(x+y)/xy
Time of meeting, call it T = z/C = z / (z*(x+y)/xy) = xy / (x+y)

High speed distance = T * z/x
Low speed distance = T * z/y

Difference = T*z(y-x) / xy

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

A.
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by rahul goyal » Tue Aug 10, 2010 11:50 pm
evanr2 wrote:I just got this problem on MGMAT CAT 3, the official answer was B, which I disagree with.

Question:

"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)

Note from the stem that x < y since x corresponds with the high speed train. All other numbers are positive. Therefore if you choose answer B, which is suggested by the given solution, you will result in a negative distance which is impossible.

Combined speed, call it C = (z/x + z/y) = z(x+y)/xy
Time of meeting, call it T = z/C = z / (z*(x+y)/xy) = xy / (x+y)

High speed distance = T * z/x
Low speed distance = T * z/y

Difference = T*z(y-x) / xy

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

A.
Thank you very much evanr2.
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by lunarpower » Wed Aug 11, 2010 2:57 am
evanr2 wrote:I just got this problem on MGMAT CAT 3, the official answer was B, which I disagree with.
you might want to check that exam again -- i just looked this problem up in our database and the answer is (a), as required. we haven't edited this problem in a long time and your post is dated last week, so you must have read the answer incorrectly (or perhaps looked at the question on the previous line, or on the next line).

please double-check. thanks.
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by mcdesty » Sat Jul 12, 2014 1:00 pm
You just have to stay organized to tackle this one. See Img below.
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I am experienced - I have tutored calculus and linear algebra for over two years.
For a very modest fee, I will ensure that your GMAT journey is a smooth one: Daily assignments and careful micro management.
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