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Time Speed and Distance- Bus and A car- Part 2

Expert replies
Source: — Problem Solving |

by [email protected] » Sun May 11, 2014 11:42 pm
Hi s91arvindh,

The information that we determined in "Part 1" of this question is required to answer Part 2, so I cut-and-pasted it here...

----------START
This questions requires that you use the Distance Formula and keep track of 2 moving vehicles.

The bus is moving 60mph and is in motion for 20 minutes....

D = R x T
D = 60mph x 1/3 hour

D = 20 miles

So we know the bus traveled 20 miles.

We also know that the bus started 5 miles behind a car, but ended 10 miles ahead of that car. So the bus "made up" 15 miles on the car, even though the bus traveled 20 miles. This means that the car had to have been moving; since the bus "made" up 15 miles instead of 20 miles, the car must have traveled 5 miles during that time.

D = R x T
5 miles = R x 1/3 hour
R = 15mph
----------------FINISH

Now that we know the speed of the car (15mph), we can answer the question. This is what's called a "chase down" question. Knowing the two rates of the vehicles, we can figure out how long it takes the bus to overtake the car.

Since the bus is traveling 60mph and the car is traveling 15mph, that means that the bus will travel 45 MORE miles per hour than the car.

The bus starts 5 miles BEHIND the car, so the bus has to "make up" those 5 miles to overtake the car.

D = R x T
5miles = 45mph x T

[spoiler]T = 1/9 hour = 60/9 = 6 2/3 minutes[/spoiler]

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by GMATGuruNY » Mon May 12, 2014 2:45 am
s91arvindh wrote:A bus moving at 60 mph is 5 miles behind a car. After 20 minutes, it is 10 miles ahead of the car.
Bus overtakes car after what time?:
[spoiler]
6.66 mins[/spoiler]
In 20 minutes, the bus catches up from 5 miles BEHIND the car to 10 miles AHEAD of the car, for a total catch-up distance of 15 miles.
To REACH the car, the bus must catch-up 1/3 of the total catch-up distance (5 miles).
Thus, it will take 1/3 the time:
(1/3)(20 minutes) = 20/3 minutes.
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