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Bunuel
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.


Always better to use relative velocity (or speed) concept here:
steps:
1. Till 1:00 PM (i.e. 4 hrs) distance travelled by bus = 200 miles (since numbers are simple its better to calculate mentally i.e. in 1 hr bus travels 50 miles, therefore, in 4 hrs 200 miles)
2. Same direction travel = Subtract the speeds...therefore relative speed = 250 m/h (this speed is combined speed of both bus and plane)
3. make the bus stationary (its speed is included in step no. 2 above)
4. time taken by plane = t to travel 200 miles (remember the bus is stationary now -but only "RELATIVELY")
5. Simple formula now, 250 t = 200 i.e. t = 4/5 hrs or 4/5X 60 minutes i.e. 48 mins.
6. Add 48 mins in 1 PM

So 1:48 is the answer or "D".

Hope it helps.

Thanx
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Bunuel
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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