swerve 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?
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
The OA is A
Source: Manhattan Prep
We have a converging rate problem in which:
Distance(1) + Distance(2) = Total Distance
We are given that 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. Thus, we know the following:
rate of the high-speed train = z/x
rate of the regular-speed train = z/y
We are also given that they leave at the same time, so we can let the time that has elapsed when both trains pass each other be t. We can substitute our values into the total distance formula.
Distance(1) + Distance(2) = Total Distance
(z/x)t + (z/y)t = z
zt/x + zt/y = z
We can divide the entire equation by z and we have:
t/x + t/y = 1
Multiplying the entire equation by xy gives us:
ty + tx = xy
t(y + x) = xy
t = xy/(y + x)
Now we can calculate the distance traveled by both trains for time t, using the following formula:
distance = rate x time
distance of high-speed train = (z/x)[xy/(y + x)] = zy/(y + x)
distance of regular-speed train = (z/y)[xy/(y + x)] = zx/(y + x)
Now we need to calculate the difference between the distance of the high-speed train and the distance of the regular-speed train:
difference of distance traveled = distance of high-speed train - distance of regular-speed train
difference of distance traveled = zy/(y + x) - zx/(y + x)
difference of distance traveled = (zy - zx)/(y + x) = z(y - x)/(x + y)
Answer: A
