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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, 10:38
Updated on: 23 Jul 2015, 10:38
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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
(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
Joined: 02 Apr 2014
Posts: 369
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Location: India
Schools: XLRI"20
GMAT 1: 700 Q50 V34 
GPA: 3.5
Re: A certain military vehicle can run on pure Fuel X, pure Fuel Y, or any
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13 Mar 2018, 03:05
13 Mar 2018, 03:05
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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)
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
[#permalink]
31 Jan 2009, 14:53
31 Jan 2009, 14:53
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shobuj40
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
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 (8-x) = 200.
x = Fuel x and y = 8-x
x = 6
y = 2. suff.
2) 20/3 (a) + 40/5 (1-a) = 50/7
a = 9/14
1-a = 5/14
fraction of $ for fuel x + fraction of $ for fuel y = 1
a = fraction of $ for fuel x
1-a = fraction of $ for fuel y
suff...
