Problem Solving

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?

z(y – x)/x + y

z(x – y)/x + y


z(x + y)/y – x

xy(x – y)/x + y

xy(y – x)/x + y

anjaligeorge1,

I believe that some parenthesis are missing in the solutions.

Answer is A. I'll post explanation after I know the correct expression for the solutions.

Lets assume that they meet at distance P from Town B

Normal Train..Starts from Town B
Distance Z in Y hrs
Distance P in (YP/Z) hrs

High Speed train..Start from Town A
Distance Z in X hrs
Distance Z-P in [(Z-P)X/Z] hrs

Now,
Time it takes the Normal train to cover distance P = Time it takes the Normal train to cover distance Z-P

So, (YP/Z) = [(Z-P)X/Z]
=> P= XZ/X+Y..............I

So, Z-P = YZ/X+Y.........II

Substracting I from II gives us the result

{Z(Y-X)}/(X+Y)

I think A is the correct answer.

rajbirdadhiyala,

Correct. I realised that parenthesis were missing.

Correct ANswer is "A"
Take any pt. b/w A & B.
Lets say high-speed train travel z-d distance, and low speed train travels d distance before they meet (time takes is same).
Extra mils travelled by high speed train is (z-d)-d i.e z-2d we hav to find out:

So we have two equation:
(z-d) = z/x * t
d = z/y * t

you can find var "d" from above, to be 2xz/(x+y)
Thus z-2d = z(y-x)/(x+y)...choice "A"

Hope it help you ppl.