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?
Pick some simple numbers. Let's say x = 2 and z = 100, so the high-speed train takes 2 hours to go 100 miles. So the high-speed train goes 100/2 = 50mph.
Let's say y = 4 and the regular train takes 4 hours to go 100 miles, so the regular train goes 100/4 = 25mph.
Let's say that they're traveling for t hours. We know that the high-speed train will cover 50t miles. (Rate* Time)
We know that the regular train will cover 25t miles.
They're starting 100 miles apart, so combined, they travel 100 miles. This gives us the following equation: 25t + 50t = 100 --> 75t = 100 --> t = 100/75 = 4/3
So the high-speed train will have gone a total of 50 * (4/3) = 200/3 miles
And the regular speed train will have gone a total of 25 *(4/3) = 100/3 miles
And the high-speed train will have gone 200/3 - 100/3 = 100/3 more miles than the regular speed train.
Now we substitute x = 2, z = 100, and y = 4 into the answer choices and see which gives us 100/3.
A gives us [100(4-2)]/(4+2) = 200/6 = 100/3. Done! That's our answer.
