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
I followed this approach: Since trains are approaching each other, relative speed =x+y
Let t be the time taken then t= z/(x+y)
Now distance travelled by high speed train = t*x=zx/(x+y)
distance travelled by average speed train = t*y=zy/(x+y)
Hence difference between both = z(x-y)/(x+y)
But the OA is z(y-x)/(x+y)
What is wrong in my approach?