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# A certain military

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A certain military [#permalink]  31 Jan 2009, 10:05
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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Re: A certain military [#permalink]  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 (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...
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Re: A certain military [#permalink]  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 (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...

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 [#permalink]  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) + (1-a)40/5 = 50/7

Cheers,
Unplugged
Re: A certain military   [#permalink] 01 Feb 2009, 22:30
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