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?

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)

jain2016 wrote:Hi Experts ,

Please check my solution and advise whats wrong.

Let z = 100 miles and x = 10 hours and y =20 hours

Let the speed of high speed train is S1 and speed of regular train is S2.

So the S1= 10m/h and S2 = 5m/h

Now let d be the distance that high speed train covers and 100-d distance cover by regular train.

Both leave the train at the same time, so our equation will be

d/S1 = 100-d/S2

d/10 = 100-d/5

from this d = 200/3.

So this is our target

Since the question stem does not for the distance traveled by the high-speed train, the target is NOT 200/3.
Rather, the question stem asks for the DIFFERENCE between the high speed train's distance and the regular train's distance.
Distance traveled by regular train = 100-d = 100 - (200/3) = 100/3.
Thus, the difference between the two distances = (200/3) - (100/3) = 100/3. This is our target.

Now we plug x=10, y=20 and z=100 into the answers to see which yield our target of 100/3.
Only A works:
z(y - x) / (x + y) = 100(20-10) / (10+20) = 100/3.

The correct answer is A.

vipulgoyal wrote:Hi Mitch,

Could you please let me know where i m wrong here


(z/x+y) ---- time travelled by either train
x(z/x+y) - yz/xy

vipulgoyal wrote: Could you please let me know where i m wrong here

(z/x+y) ---- time travelled by either train