renanmenezes wrote:A high-speed train takes x hours to cover z km that separate city A from city B. A conventional train takes y hours to cover the same distance. If the high-speed train leaves the city A towards B and conventional train leaves from B toward A at the same time, how many kilometers the high speed train will have traveled over conventional train when they pass each other?

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I posted a solution to this problem here:
https://www.beatthegmat.com/distance-of- ... tml#766111

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)

Rate = Distance/Time

The rate of the high speed train is z/x

The rate of the conventional train is z/y.

Time = Distance/Rate

The time the trains travel to meet each other is Distance/Combined Rates = z/[(z/x) + (z/y)].

Distance = Rate * Time

During that time, the high speed train will go distance z/x * z/[(z/x) + (z/y)].

The conventional train will go distance z/y * z/[(z/x) + (z/y)].

So the difference between the distances the two trains will have gone will be the following.

(z/x - z/y) * z/[(z/x) + (z/y)]

Simplify.

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

The correct answer is A.