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# It takes the high-speed train x hours to travel the z miles

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Director
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It takes the high-speed train x hours to travel the z miles [#permalink]  25 Jun 2008, 21:56
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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)
Director
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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.
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Re: PS high-speed train [#permalink]  25 Jun 2008, 22:17
Great job durgesh79

The mgmat explanation is about 2 pages long. I wonder if it is realistic for their own people to solve it in 2 mins using their methods.
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Re: PS high-speed train [#permalink]  25 Jun 2008, 22:18
I got D as the answer as follows.

Suppose, the two trains pass each other after m hours.

Then mz/x + mz/y = z or m = xy/(x+y).

The difference between the distance travelled by two trains = m(x-y) = xy(x-y)/(x+y).

Am I doing anything wrong here?
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Re: PS high-speed train [#permalink]  25 Jun 2008, 22:26
scthakur wrote:
I got D as the answer as follows.

Suppose, the two trains pass each other after m hours.

Then mz/x + mz/y = z or m = xy/(x+y).

The difference between the distance travelled by two trains = m(x-y) = xy(x-y)/(x+y).

Am I doing anything wrong here?

The difference will be mz/x - mz/y
=mz(1/x - 1/y)
you have the value of m
= (xy/(x+y))*z*(y-x)/xy
=z(y-x)/(x+y)
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Re: PS high-speed train [#permalink]  25 Jun 2008, 22:35
Thanks. was in the middle of a call and tried to solve
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Re: PS high-speed train [#permalink]  25 Jun 2008, 22:41
scthakur wrote:
Thanks. was in the middle of a call and tried to solve

Dont worry, you wont be having you phone with you during actual GMAT.
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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 ==== >

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

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

ie. B...

please let me knwo if i m wrong...
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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.
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Re: PS high-speed train [#permalink]  26 Jun 2008, 05:33
oh yeas.... such a stupid mistake!! ....

thanks durgesh.!!
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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

Just plug away. I started with A:

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

Stupid problem >:-(

A
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Re: PS high-speed train [#permalink]  26 Jun 2008, 09:28
the answer is A.

the basic function is: t(pass each other) * (Vx-Vy)

when:

Vx= z/x, Vy= z/y , and t(pass each other)= z/(z/x+z/y)= xy/(x+y)

thus: t(pass each other) * (Vx-Vy) = xy/(x+y) * z/x - z/y = z * (y-x)/(y+x)
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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

Just plug away. I started with A:

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

Stupid problem >:-(

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