Bunuel wrote:

It costs 10¢ a kilometer to fly and 12¢ a kilometer to drive. If you travel 200 kilometers, flying x kilometers of the distance and driving the rest, then the cost of the trip in dollars is

(A) 20

(B) 24

(C) 24 – 2x

(D) 24 – .02x

(E) 2,400 – 2x

An alternate approach would be to model the question, if you don't know how to set up the algebra.

Pick an easy number for x, that is not an exact split (e.g. 100) or a number from the problem, such as 50.

If x = 50, then it costs 50 x .10 = $5 for the flight portion of the trip.

Then, there are 200 - 50 = 150 kilometers left for the drive portion of the trip, so it costs 150 x .12 = $18 for the drive portion of the trip.

Add the two values to find that when x = 50, the cost of the trip is $23. Now, eliminate choices A and B, which are defined numeric values that do not equal $23.

Then, plug in x = 50 into the remaining choices seeking one that yields $23.

(C) 24 - 2(50) = -76 ; eliminate choice C

(D) 24 - .02(50) = 23 ; keep choice D

(E) 2,400 - 2x = 2,300 ; eliminate choice E

The correct answer is choice D.

_________________

M.S. Journalism, Northwestern University '09

Director of Online Tutoring, MyGuru

http://www.myguruedge.com

Find out more about online GMAT tutoring with MyGuru

http://www.myguruedge.com/gmat-prep/online-gmat-tutoring