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A certain military vehicle can run on pure Fuel X, pure Fuel

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A certain military vehicle can run on pure Fuel X, pure Fuel

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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

OA D

Source: Gmat Prep

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Let gallons of fuel X = x
Let gallons of fuel Y = y
What is the cost per gallon of the full mixture currently in the vehicle tank
$$\frac{\left(3x+5y\right)}{x+y}=\ ratio\ of\ x:y$$
Statement 1
Using fuel currently in its tank , the vehicle burned 8 gallons to cover 200 miles
x + y = 8
y = 8 - x
20x + 40y = 200
20x + 40(8x) = 200
$$-\frac{20x}{20}=\frac{\left(200-320\right)}{-20}$$
$$x=\frac{-120}{-20}=6$$
$$from\ x+y=8$$
$$\ 6+y=8$$
$$y=8-6=2$$
$$\cos t\ per\ gallon=\frac{3x+5y}{x+y}$$
$$=\frac{3\left(6\right)+5\left(2\right)}{8}$$
$$=\frac{18+10}{8}$$
$$=\frac{28}{8}$$
$$=3.5\ dollars$$
Statement 1 is INSUFFICIENT

Statement 2
The vehicle can cover 7 hours 1/7 miles for every dollar of fuel currently in the tank
$$\frac{\left(Total\ miles\ it\ can\ cover\right)}{Total\ \cos t}=\frac{50}{7\ }=7\ \frac{1}{7}$$
$$\frac{\left(20x+40y\right)}{3x+5y}=\frac{50}{7\ }$$
$$7\left(20x+40y\right)=50\left(3x+5y\right)$$
$$7\left(20x+40y\right)=50\left(3x+5y\right)$$
$$\left(140x+280y\right)=\left(150x+250y\right)$$
$$\left(280y-250y\right)=\left(150x-140x\right)$$
$$3y=x$$
For every dollar spent of the mixture of ratios 3y : 1x, the vehicle can travel a distance of 11 1/7 miles
Statement 2 is SUFFICIENT.
Hence both statement together are SUFFICIENT
$$answer\ is\ Option\ D$$ .

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BTGmoderatorDC 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
Statement 1:
X = 20 miles per gallon.
Y = 40 miles per gallon.
Since the mixture of X and Y in the tank allows 200 miles to be covered using 8 gallons, we get:
X+Y = 200/8 = 25 miles per gallon.

To determine how X and Y must be combined to yield a rate of 25 miles per gallon, use ALLIGATION.

Step 1: Plot the 3 rates on a number line, with the rates for X and Y on the ends and the rate for the mixture in the middle
X 20------------25-------------40 Y

Step 2: Calculate the distances between the percentages.
X 20-----5-----25-----15-----40 Y

Step 3: Determine the ratio in the mixture.
The ratio of X to Y is equal to the RECIPROCAL of the distances in red.
X:Y = 15:5 = 3:1.

Of every 4 gallons in the tank, X= 3 gallons and Y = 1 gallon.
Since X = $3 per gallon and Y = $5 per gallon, the cost for 3 gallons of X and 1 gallon of Y = (3*3) + (1*5) = $14.
Since 4 gallons costs $14, the cost per gallon = 14/4 = 7/2 = $3.50.
SUFFICIENT.

Statement 2:
The cost of the 8 gallons used in Statement 1 to travel 200 miles = (8)(7/2) = $28, implying that the miles per dollar = 200/28 = 100/14 = 50/7 = 7 1/7 miles.
Implication:
Statement 2 -- which requires that 7 1/7 miles be covered for every dollar -- conveys the SAME INFORMATION as Statement 1.
Since Statement 1 is SUFFICIENT, so must be Statement 2.
SUFFICIENT.

The correct answer is D.

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BTGmoderatorDC 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

OA D

Source: Gmat Prep
Say there be x gallons of fuel X, and y gallons of fuel Y in the tank.

Cost per gallon = (3x + 5y)/(x + y)

We have to get the value of (3x + 5y)/(x + y).

Let's take each statement one by one.

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

=> x + y = 8; and 20x + 40y = 200 => x + 5y = 10. We have two unique linear equations, we can get the unique values of x and y, thus, of (3x + 5y)/(x + y). Sufficient.

(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 = 7 and 1/7 miles = 50/7 miles/dollar

=> (20x + 40y)/(3x + 5y) = 50/7
(2x + 4y)/(3x + 5y) = 5/7
14x + 28y = 15x + 25y
x = 3y

Thus, (3x + 5y)/(x + y) => (3*3y + 5y)/(3y + y) = 14y / 4y = 7/2. Sufficient

The correct answer: D

Hope this helps!

-Jay
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