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)

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.

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.

You just have to stay organized to tackle this one. See Img below.