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sorry if this problem has already been posted, but I do not

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djhouse81
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sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 06:47
(sorry if this problem has already been posted, but I do not feel like hunting for it)

Maria takes a 40 mile trip. Her average speed for y miles is x mph. Her average speed for the remaining 40-y miles is 1.25x mph. The time it took Maria to travel the 40 miles was what percent of the time it would have taken her if her average speed for the entire trip was x mph.

1) X = 48
2) Y =20




OA is B



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amitsamel
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Re: sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 07:27
The answer is C for me.

Both X = 48 and Y =20 will be essential to find the total time required during Maria's travel with two different speeds.
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Re: sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 08:00
C is incorrect, but that is what I came up with during the test. According to GMAT Prep the answer is B. Can anyone explain?
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ridhimagupta
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Re: sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 13:03
My first instinct was also C but I think you can get B.

You have to develop the equation for the first 40 mile trip with the two different speeds.
Then for the second which will be T = 40/x

Since it is asking for a percentage you can further solve them by expressing in the percent form which should cancel x and then you substitute for y and getet the answer.

I haven't solved it completely yet.

I might be completely wrong but sitting at work and it is past 4pm and I am unable to think straight right now :)
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Re: sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 13:42
I got B...here's how:

Given:
1) avg rate = y/T = x
2) avg rate = 40-y/t = 1.25x

Looking for t/T.

1) y/x=T
2) (40-y)/t = 1.25x --> (40-y)/1.25x= t

((40-y)/1.25x) / (y/x) = (40-y)/1.25x * x/y = (40-y)/1.25y

So, therefore, we only need to know y to solve the problem.
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Re: sorry if this problem has already been posted, but I do not [#permalink] New post   19 Jun 2007, 13:55
Brilliant! Thank you!

I had the separate equations set up, but I did not take it a step further and put Maria's real trip over the 40/x trip. I would have seen that the x's cancelled out.

Those simple mistakes will get ya.



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Re: sorry if this problem has already been posted, but I do not [#permalink]
19 Jun 2007, 13:55
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