AAPL wrote:GMAT Prep

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.

OA D

Given for Fuel X:

"¢ Cost: 3 $/gallon;
"¢ Mileage: 20 miles/gallon;
=> Miles/dollar: 20/3 miles/$

Given for Fuel Y:

"¢ Cost: 5 $/gallon;
"¢ Mileage: 40 miles/gallon;
=> Miles/dollar: 8 miles/$

We have to find out the mix of fuels in the tank.

Let's take each statement one by one.

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

=> Milage = 200/8 = 25 m/g

We see that mileage (25 m/g) is between 20 m/g and 40 m/g; thus, the ratio of mix of Fuel X : Fuel Y :: (40 - 25) : (25 - 20) => Fuel X : Fuel Y :: 3 : 1

Thus, Fuel X and Fuel Y are mixed in the ratio of 3 : 1. Thus, the cost of mix = (3*3 + 1*5)/(3 + 1) = 3.5 $/gallon. Sufficient

2) The vehicle can cover 7 and 1/7 miles for every dollar of fuel currently in its tank.

=> Miles/dollar = 7 1/7 miles/$ = 50/7 miles/$

We see that miles/dollar (50/7 miles/$) is between 20/3 miles/$ and 8 miles/$, thus, the ratio of mix of Fuel X : Fuel Y :: (8 - 50/7) : (50/7 - 20/3) => Fuel X : Fuel Y :: 9 : 5

Thus, Fuel X and Fuel Y are mixed in the ratio of 9 : 5. Thus, the cost of mix = (9*3 + 5*5)/(9 + 5) = 52/14 = ~3.5 $/gallon. Sufficient

The correct answer: D

Hope this helps!

-Jay
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