Bunuel wrote:

A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour. At 1:00 p.m. a plane leaves Burbank traveling east at 300 miles per hour. At what time will the plane overtake the bus?

A. 12:45 p.m.

B. 1:10 p.m.

C. 1:40 p.m.

D. 1:48 p.m.

E. 1:55 p.m.

We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Plane = distance of Bus

We are given that at 9:00 a.m., a bus leaves Burbank traveling east at 50 mph, and that at 1:00 p.m., a plane leaves Burbank traveling east at 300 mph.

Since the bus started 4 hours before the plane, we can let the time of the bus = t + 4 hours, and the time of the plane = t.

Since distance = rate x time, we can calculate each distance in terms of t.

Bus’s distance = 50(t + 4) = 50t + 200

Plane’s distance = 300t

We can equate the two distances and determine t.

50t + 200 = 300t

200 = 250t

t = 200/250 = 4/5 hours = 48 minutes

Thus, the plane will overtake the bus at 1:00 p.m. + 48 minutes = 1:48 p.m.

Answer: D

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