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
Source: — Data Sufficiency |

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by deloitte247 » Sun Feb 17, 2019 6:19 am

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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
$$answer\ is\ Option\ D$$ .