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A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.

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A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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New post 17 Jan 2016, 12:00
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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.

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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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New post 17 Jan 2016, 12:20
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1
at 1:00pm bus is 200 miles ahead of plane
plane gains 250 mph on bus
200/250=4/5 hour=48 minutes
plane will overtake bus at 1:48pm
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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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New post 21 Oct 2016, 05:31
Use the formula distance = speed * time
from 9:00 am to 1:00 pm total is 4 hrs
In 4 hrs bus travels a distance of 50 * 4 = 200 miles
plane starts travelling in same direction so relative speed = 300 - 50 = 250 miles/hr
distance the plane needs to cover so as to overtake the bus is 200 miles
time = 200 /250 = 4/5 hrs = 48 minutes
so at 1:48 pm the plane overtakes the bus
Correct answer - D
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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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New post 30 Jan 2017, 03:19
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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.



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.

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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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New post 01 Feb 2017, 09:20
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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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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour.  [#permalink]

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Re: A bus leaves Burbank at 9:00 a.m. traveling east at 50 miles per hour. &nbs [#permalink] 13 Apr 2018, 05:00
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