Problem Solving

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

Solution:

Since the trains traveled the z miles in x and y hours, their speeds can be represented as z/x and z/y respectively.

We can again use an RTD chart to evaluate how far each train travels when they move toward each other starting at opposite ends. Instead of using another variable d here, let's express the two distances in terms of their respective rates and times.



High-speed Regular Total

R z/x z/y

T t t

D zt/x zt/y z


Since the two distances sum to the total when the two trains meet, we can set up the following equation:

zt/x + zt/y = z divide both sides of the equation by z
t/x + t/y = 1 multiply both sides of the equation by xy
ty + tx = xy factor out a t on the left side
t(x + y) = xy divide both sides by x + y
t =
xy

x + y


To find how much further the high-speed train went in this time:
(ratehigh × time) - (ratereg × time)
(ratehigh - ratereg) × time

zy - zx xy
------- x ------
xy x+y


z(y - x)
-------
x + y




The correct answer is A.

Question: Why time is taken as constant?