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Cost of the trip

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by gmatblood » Sun Nov 06, 2011 3:22 pm
Ms. Jiminez plans an automobile trip of 7,000 to 9,000 miles. The cost of gasoline will be 85 to 95 cents per gallon, and her automobile will average 20 to 30 miles per gallon. What is the maximum possible cost of the gasoline for the trip?

(A) $485.00
(B) $427.50
(C) $382.50
(D) $297.50
(E) $256.00
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by GmatMathPro » Sun Nov 06, 2011 4:04 pm
Cost will be maximized if we maximize distance, maximize cost of gas, and minimize fuel efficiency:

Distance=9000 miles
Cost of gas=95 cents per gallon
fuel efficiency=20 miles per gallon

Total gallons of gas required=9000/20=450

Total cost =.95*450=450-(1/20)*450=450-22.5=427.50

Ans: B
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by neelgandham » Sun Nov 06, 2011 4:44 pm
Number of gallons used during the trip = Total distance travelled during the trip / The mileage
Cost of the trip = Number of gallons used during the trip * Cost per gallon

So, If the total distance travelled during the trip is D, the mileage M, and Cost per Gallon C
Then the Cost of the trip, CT = D*C/M

For maximum CT, D,C should be Maximum and M should be Minimum.i.e. D = 9000 (Maximum(9000,7000)), C = $0.95 (Maximum($0.85,$0.95), M = 20 (Minimum(20,30))

CT = 9000*0.95/20 = 450-(1/20)*450=[spoiler]450-22.5=427.50 Ans B[/spoiler]
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Re: Cost of the trip

by Scott@TargetTestPrep » Sun May 24, 2020 4:50 am
gmatblood wrote: ↑
Sun Nov 06, 2011 3:22 pm
 Ms. Jiminez plans an automobile trip of 7,000 to 9,000 miles. The cost of gasoline will be 85 to 95 cents per gallon, and her automobile will average 20 to 30 miles per gallon. What is the maximum possible cost of the gasoline for the trip?

(A) $485.00
(B) $427.50
(C) $382.50
(D) $297.50
(E) $256.00
The maximum possible cost of the gasoline for the trip occurs when Ms. Jiminez drives the longest possible distance, the cost of gasoline is the priciest possible per gallon, and fuel efficiency is the lowest possible. Therefore, the maximum possible cost of the gasoline for the trip is:

9000/20 x 0.95 = 450 x 0.95 = $427.50

Answer: B

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