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# If a car traveled from Townsend to Smallville at an average

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Director
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If a car traveled from Townsend to Smallville at an average [#permalink]  20 Jul 2008, 09:35
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If a car traveled from Townsend to Smallville at an average speed of 40 mph and then returned to Townsend later that evening, what was the average speed for the entire trip?
(1) The return trip took 50% longer than the trip there.

(2) The distance from Townsend to Smallville is 165 miles.
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Re: MGMAT RTD DS [#permalink]  20 Jul 2008, 09:52
IMO E.

Because it is not given that way of going is same as way of coming or not.

However as I have seen in quite a few MGMAT problems. Asnwer can be A as well. Here is reasoning for same.

Distance D, (It is same distance as going and coming back).
T1 = Time going for first time
T1 = D/40 => 40 = D/T1

T2 = Time coming back
T2 = 1.5T1

Average Speed = 2D/(T1+T2) = 2D/2.5T1 = 2*40/2.5 = 32

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Re: MGMAT RTD DS [#permalink]  20 Jul 2008, 15:36
jimmyjamesdonkey wrote:
OA is A.

Please explain. My vote is for E
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Re: MGMAT RTD DS [#permalink]  20 Jul 2008, 19:17
A. Stat. 1 is SUFF.

When you are given the rate, you can make up the distance (any smart number, 80 would work in this case) and then solve for the time on the first trip. Then, with this data, setup the formula for the return trip and use the information given in Stat. 1.

Stat. 2 is clearly insufficient because it doesn't even mention the return trip.

Use this method also to solve PS questions were you are only given the rate.

Hope this helps.
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Re: MGMAT RTD DS [#permalink]  20 Jul 2008, 19:42
will go with either of the statement
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Re: MGMAT RTD DS   [#permalink] 20 Jul 2008, 19:42
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