AAPL wrote:Manhattan Prep
Half an hour after Car A started traveling from Newtown to Oldtown, a distance of 62 miles, Car B started traveling along the same road from Oldtown to Newtown. The cars met each other on the road 15 minutes after Car B started its trip. If Car A traveled at a constant rate that was 8 miles per hour greater than Car B's constant rate, how many miles had Car B driven when they met?
A. 14 miles
B. 12 miles
C. 10 miles
D. 9 miles
E. 8 miles
OA A
Let’s let the rate of Car B be v. Then, the rate of Car A is v + 8.
Since Car A had been traveling for 30 minutes when Car B started its trip and since the two cars met 15 minutes after, the total travel time for Car A is 30 + 15 = 45 minutes = 3/4 hour, and the total travel time for Car B is 15 minutes = 1/4 hour. Since the sum of the distances traveled by the two cars must equal the total distance between the two towns, which is 62, we can create the following equation:
(v + 8)(3/4) + v(1/4) = 62
Let’s multiply each side of this equation by 4:
(v + 8)(3) + v = 248
3v + 24 + v = 248
4v = 224
v = 56
Since the rate of Car B is v = 56 mph and since Car B traveled for 1/4 hour when the two cars met, Car B traveled 56 * 1/4 = 14 miles.
Answer: A
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