topspin330 wrote:Can someone please check my solution and point out the flaw:

Train X leaves Los Angeles at 10:00AM and travels East at a constant speed of x miles per hour. If another Train Y leaves Los Angeles at 11:30AM and travels East along the same tracks at speed y, then at what time will Train Y catch Train X ?

(1) y = 4x/3

Solution:

Distance traveled by X by 11:30: 1.5x
Difference in rate: y-x
Distance for Y to travel to catch up: 1.5x
Time it takes for y to catch up: 1.5x / (y-x)
Using (1), I get 4.5 when the OA is 6.

What am I doing wrong?

Thanks

Stuart Kovinsky wrote:

topspin330 wrote:Can someone please check my solution and point out the flaw:

Train X leaves Los Angeles at 10:00AM and travels East at a constant speed of x miles per hour. If another Train Y leaves Los Angeles at 11:30AM and travels East along the same tracks at speed y, then at what time will Train Y catch Train X ?

(1) y = 4x/3

Solution:

Distance traveled by X by 11:30: 1.5x
Difference in rate: y-x
Distance for Y to travel to catch up: 1.5x
Time it takes for y to catch up: 1.5x / (y-x)
Using (1), I get 4.5 when the OA is 6.

What am I doing wrong?

Thanks

Hi!

You haven't provided any answer choices, so I'm a bit confused - is this a problem solving or a data sufficiency question?

You said that the OA is "6", but since we only have variables, how can the answer be an actual number?

Your equation for time to catch up is dead on, since:

Catch up time = (distance to catch up)/(difference in rates)

so that's not the issue.