Kevin drove from \(A\) to \(B\) at a constant speed of \(60\) mph. Once he reached \(B,\) he turned right around without

[M7MBA]

Kevin drove from \(A\) to \(B\) at a constant speed of \(60\) mph. Once he reached \(B,\) he turned right around without pause and returned to \(A\) at a constant speed of \(80\) mph. Exactly \(4\) hours before the end of his trip, he was still approaching \(B,\) only \(15\) miles away from it. What is the distance between \(A\) and \(B?\)

A. 275 mi
B. 300 mi
C. 320 mi
D. 350 mi
E. 390 mi

Answer: B

Source: Magoosh

[Scott@TargetTestPrep]

Kevin drove from \(A\) to \(B\) at a constant speed of \(60\) mph. Once he reached \(B,\) he turned right around without pause and returned to \(A\) at a constant speed of \(80\) mph. Exactly \(4\) hours before the end of his trip, he was still approaching \(B,\) only \(15\) miles away from it. What is the distance between \(A\) and \(B?\)

A. 275 mi
B. 300 mi
C. 320 mi
D. 350 mi
E. 390 mi

Answer: B

Solution:

Let the distance between A and B = d. Therefore, the time from A to B is d/60, and the time from B to A is d/80, and thus the total time for the round trip is d/60 + d/80 = 4d/240 + 3d/240 = 7d/240. Since we are given that exactly 4 hours before the end of his trip, he was still approaching B (and was thus still traveling at 60 mph), only 15 miles away from it, we can create the equation:

60(7d/240 - 4) = d - 15

7d/4 - 240 = d - 15

7d - 960 = 4d - 60

3d = 900

d = 300

Alternate Solution:

Let’s concentrate only on the 4-hour time period. During this time, he was still going from A to B at a rate of 60 mph (or 1 mile per minute), and he traveled 15 miles to actually get to B. Thus, he traveled 15 miles in 15 minutes. This means that the remaining 3 hours and 45 minutes of the 4-hour travel time was all used to get from B back to A. During these 3.75 hours, he traveled at a rate of 80 mph; thus, the distance from B to A = 3.75 x 80 = 300 miles.

Answer: B