pawan203 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?

(1) z(y – x) / x + y

(2) z(x – y) / x + y

(3) z(x + y) / y – x

(4) xy(x – y) / x + y

(5) xy(y – x) / x + y

Both train spend the same time traveling = t

Velocity(X) = Z/X

Velocity(Y) = Z/Y

Distance(X) = rate X * time = (Z/X)t

Distance(Y) = rate Y * time = (Z/Y)t

Distance(X) + Distance(Y) = Z

(Z/X)t + (Z/Y)t = Z

(t/X) + (t/Y) = 1 ---------- divides the above equation with Z

t(1/X + 1/Y) = 1

t((X+Y) / XY) = 1

t = XY / (X+Y) ---------- (1)

The question want to know Distance(X) - Distance(Y)

= (Z/X)t - (Z/Y)t ---------- (2)

= (Z/X)(XY / (X+Y) - (Z/Y)(XY / (X+Y)) ---------- (1) & (2) combined

= ZY/(X+Y) - ZX/(X+Y)

= Z(Y-X)/(X+Y)

The answer is A ( or (1) in this case).