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Junior | Next Rank: 30 Posts
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by topspin330 » Mon Dec 23, 2013 6:32 pm
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
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by Stuart@KaplanGMAT » Mon Dec 23, 2013 8:47 pm
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.
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by theCodeToGMAT » Mon Dec 23, 2013 9:45 pm
topspin330, assuming that it's a DS question and you are trying validate the only one Statement's solution, which gives answer as 6, here is my solution;

Time to Catch = GAP/(Difference of Speed)
= (x*3/2)/(y-x)
= 3x/2y-2x
= 3/[2(y/x) - 2]
Using (1)
= 3/(2*4/3 - 2)
= 3*3 / (8 - 6)
= 9/2
= 4.5

Question asks: "then at what time will Train Y catch Train X ?"

So, 11:30AM + 4:30hrs = 4 PM

Total time of first train = 1:30 + 4:30 = 6 hrs
Total time of second train = 4:30 hrs
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by topspin330 » Thu Jan 02, 2014 9:31 pm
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.
Hi Stuart!

It's a data sufficiency question but I primarily wanted to focus on the answer provided in the first choice (which was sufficient). So I wanted to know how to get to that. Thanks for confirming my equation!
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