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A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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Updated on: 23 Jul 2015, 09:38
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59% (02:35) correct 41% (03:05) wrong based on 136 sessions
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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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Originally posted by shobuj40 on 31 Jan 2009, 10:05.
Last edited by Bunuel on 23 Jul 2015, 09:38, edited 1 time in total.
Renamed the topic, edited the question, added the OA and moved to DS forum.



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Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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31 Jan 2009, 13:53
shobuj40 wrote: 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 D. 1) 20x + 40 (8x) = 200. x = Fuel x and y = 8x x = 6 y = 2. suff. 2) 20/3 (a) + 40/5 (1a) = 50/7 a = 9/14 1a = 5/14 fraction of $ for fuel x + fraction of $ for fuel y = 1 a = fraction of $ for fuel x 1a = fraction of $ for fuel y suff...
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Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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01 Feb 2009, 21:04
GMAT TIGER wrote: shobuj40 wrote: 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 D. 1) 20x + 40 (8x) = 200. x = Fuel x and y = 8x x = 6 y = 2. suff. 2) 20/3 (a) + 40/5 (1a) = 50/7 a = 9/14 1a = 5/14 fraction of $ for fuel x + fraction of $ for fuel y = 1 a = fraction of $ for fuel x 1a = fraction of $ for fuel y suff... I'm sorry but I don't get the 2nd equation. Can you explain why 20/3(a) = fraction of $ for fuel x?



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Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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01 Feb 2009, 22:30
Great explanation GMATTIGER!
@wcgmatclub  For every dollar spent on Fuel X, you can travel 20/3 miles For every dollar spent on Fuel Y, you can travel 40/5 miles
For every dollar spent on the mixture of X and Y, you can travel 7 1/7 miles. The question is asking what is the mixture?
Therefore, a(20/ 3) + (1a)40/5 = 50/7
Cheers, Unplugged



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Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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13 Mar 2018, 02:05
Let there be x gallons of fuel X, and y gallons of fuel Y
question: cost per gallon? = (3x + 5y)/(x + y) all we need is relation between x and y, or the ratio of x and y
Statement 1: Using fuel currently in its tank, the vehicle burned 8 gallons to cover 200 miles pretty straight forward, x + y = 8, 20x + 40y = 200, two equations to find x and y => suff
Statement 2: The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank Total miles can cover / total cost = 50/7 (20x + 40y)/(3x + 5y) = 50/7 => rearranging we get, 3y = x, so we found the relation between x and y => sufficient
(D)




Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any &nbs
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13 Mar 2018, 02:05






