Problem Solving

Q.

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


MGMAT has the explanation but it is very long and time-consuming .. any other suggestion to solve this problem faster guys?

IMO A
Plug in numbers
Z=120km
X=20km/hr
Y=10km/hr
Rt=d (combine both rates)
T= 4hr
Then,
X travels 80km
Y travels 40km

Thanks ! i tried to plug in numbers but i chose inappropriate ones ... your numbers are better ! .. thanks again !

z=V1(t)+v2(t) =(t)* (z/x + z/y)--(1)

also we have to find out D1-D2 = (V1-V2)(t)=(z)*(t)*(1/x - 1/y)--(2)

from (1) and (2), eliminate t and get

D1-D2=(z^2)*(1/x - 1/y) * ( (z/x + z/y))^(-1)
=x(y-x)/(y+x)..whis is A