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Intern
Joined: 25 Dec 2008
Posts: 41

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16 Jan 2009, 01:28
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any mixture of X and Y. Fuel X costs \$3 per gallon; the vehicle can go 20 miles on a gallon of Fuel X. In contrast, Fuel Y costs \$5 per gallon, but the vehicle can go 40 miles on a gallon of Fuel Y. What is the cost per gallon of the fuel mixture currently in the vehicle’s tank?

1) Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles.

2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank
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Status: Stanford GSB
Joined: 02 Jun 2008
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16 Jan 2009, 02:21
This problem has to do with weighted averages. To find the cost per gallon of the fuel mixture currently in the vehicle’s tank, we need to know the ratio (by gallons) of Fuel X to Fuel Y in that mixture.

Statement 1: SUFFICIENT. The statement tells us that the vehicle covered 200 miles on 8 gallons of the fuel mixture; that is, the fuel mixture delivers 25 miles per gallon. As a result, we can find the ratio of Fuel X (20 mpg) to Fuel Y (40 mpg) in the mixture (this ratio turns out to be 3:1, but we do not need to find the exact ratio; we simply need to know that there will be one unique ratio). Finally, we can use this ratio to find the weighted average cost per gallon of the fuel mixture (this weighted average turns out to be \$3.50 per gallon, but again, we do not need the exact number).

Statement 2: SUFFICIENT. The statement tells us that \$1 of fuel “buys” 7 and 1/7 miles. Multiplying through by 7, we can rephrase the ratio as \$7 for every 50 miles. Let’s now express the weighted average cost per gallon as \$3w + \$5(1-w) = 5 – 2w, where w represents the percent of Fuel X in the mixture (and 1-w represents the percent of Fuel Y in the mixture). The number of gallons bought by the \$7 is then \$7 divided by the average cost per gallon, or 7/(5 – 2w). Likewise, we can express the weighted average fuel efficiency (miles per gallon) as 20w + 40(1-w) = 40 – 20w. The number of gallons burned to cover 50 miles is then 50 divided by the average miles per gallon, or 50/(40 – 20w). We can now set these numbers of gallons equal to each other:

7/(5 – 2w) = 50/(40 – 20w)
280 – 140w = 250 – 100w
30 = 40w
3/4 = w

Knowing that the mixture is 75% Fuel X and 25% Fuel Y, we can now in theory calculate the average cost per gallon of the mixture. (In fact, we should have stopped before calculating w – simply knowing that we could calculate that percentage is sufficient.)

The answer is D: EACH statement ALONE is sufficient to answer the question.
Re: sourceGMATPrep   [#permalink] 16 Jan 2009, 02:21
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