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Time, Distance Problems

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by GMATGuruNY » Sat Jan 28, 2017 4:13 am
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph.
Let's say Mary passes the station at 1:45pm.
Implication:
At 2pm -- 15 minutes later -- Paul is AT the gas station, whereas Mary has traveled for 15 minutes BEYOND the gas station.
Since Mary's rate = 50mph, the distance traveled by Mary in 1/4 of an hour = rt = (50)(1/4) = 12.5 miles.

Paul has to CATCH-UP by 12.5 miles.
The CATCH-UP rate is equal to the DIFFERENCE between the two rates.
Since Paul's rate = 60mph, while Mary's rate = 50mph, every hour Paul travels 10 more miles than Mary, with the result that every hour Paul catches up by 10 miles -- the DIFFERENCE between the two rates:
60 - 50 = 10mph.
Thus:
Time for Paul to catch-up = (catch-up distance)/(catch-up rate) = (12.5)/(10) = 1.25 hours = 1 hour, 15 minutes.

The correct answer is D.
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by Brent@GMATPrepNow » Sat Jan 28, 2017 7:27 am
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins
Mary passes gas station and travels at 50 mph for 15 minutes
So, Mary drove for 1/4 hours
Distance traveled = (time)(speed)
= (1/4)(50)
= 50/4
= 12.5 miles

Paul passes the same gas station and travels at 60 mph
IMPORTANT: At this point, Paul is 12.5 miles behind Mary.
However, since Paul is traveling 10 mph FASTER than Mary, the gap between them SHRINKS at a rate of 10 mph.
So, at this point, we need to determine the time it takes for the 12.5 mile gap to shrink to zero.
Time = (distance)/(speed)
= 12.5/10
= 1.25 hours
= 1 hour 15 mins
= D

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Brent
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by Jay@ManhattanReview » Mon Jan 30, 2017 1:20 am
Mary passed a certain gas station on a highway while traveling west at a constant speed of 50 mph. Then, 15 minutes later, Paul passed the same gas station ]while traveling west at a constant speed of 60 mph. If both travelers maintained their speed and both remained on the highway for at least 2 hours, how long after he passed the gas station did Paul catch up with Mary?

A 35 mins
B 45 mins
C 1 hour
D 1 hour 15 mins
E 1 hour 30 mins
These kinds of questions can be solved with the application of the concept of relative speed.

1. If two objects A and B are moving in the same direction, having their individual speed as Sa and Sb, then their relative speed = |Sa - Sb|.

2. If two objects A and B are moving in the opposite direction, having their individual speed as Sa and Sb, then their relative speed = Sa + Sb.

The problem falls in the first case.

We know that Mary is 50*(15/60) = 50*(1/4) = 12.5 miles way from paul.

Since Speed of Paul (Sp = 60 mph) > Speed of Mary (Sm = 50 mph), Paul will catch her.

Since both are moving in the same direction, relative speed = Sp - Sm = 60 - 50 = 10 mph.

Thus, at a relative speed of 10 mph, Paul has to cover a distance of 12.5 miles.

Thus, time taken = 12.5/10 = 1.25 hours = 1 hour 15 min.

Relevant book: Manhattan Review GMAT Word Problems

Answer: D

Hope this helps!

-Jay
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