military vehicle

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military vehicle

by clock60 » Sat Dec 25, 2010 2:33 pm
hi guys,
spend too much time on this problem and not sure that i got it completely, if possible provide your rationale

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 is D
Source: — Data Sufficiency |

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by anshumishra » Sat Dec 25, 2010 3:00 pm
clock60 wrote:hi guys,
spend too much time on this problem and not sure that i got it completely, if possible provide your rationale

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 is D
X -> 3$/gallon -> 20 miles per gallon -> 20/3 miles/$
Y -> 5$/gallon -> 40 miles per gallon -> 40/5 miles/$ or 8 miles/$

cost per gallon of the mixture (x+y) = ?
We need the composition of x and y in the mixture = ?

Statement 1:
Let n gallons of X is in the mixture

20n + 40(8-n) = 200 -> Sufficient, as we can calculate n

Statement 2 :

Let z part of X is in the mixture
So,

z*20/3 + (1-z)*8 = 7+1/7 = 50/7 -> Sufficient, as we can calculate z.

Hence D.
Thanks
Anshu

(Every mistake is a lesson learned )