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Rate question - when will Paul pass Mary?

by mdecarbo » Sun Dec 01, 2013 8:14 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
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by Brent@GMATPrepNow » Sun Dec 01, 2013 9:02 am
mdecarbo wrote: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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by Mathsbuddy » Mon Dec 02, 2013 9:40 am
Imagine the linear graphs of the 2 drivers, with distance y on the vertical axis and time t on the horizontal. The garage is where y = 0. Paul intersects the t axis at point X, where time t = 0.25 hours. Mary's line passes through the origin O. Both drivers intersect at point I. A lovely triangle OXI is born!

Anyway, the equations of each line are:

Paul: y = 60(t - 0.25) = 60t -15
Mary: y = 50t

where t is the absolute time from when Mary first passed the garage.

At point I, these are equal:

60t - 15t = 50t

So t = 1.5 hours since Mary passed the garage.

However Paul passed the gargage 0.25 h after Mary.

So answer = 1.5 - 0.25 = 1.25 h

D) 1 hour 15 mins.
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by theCodeToGMAT » Mon Dec 02, 2013 9:52 am
Mary's Distance in 15 minutes = 50 * 15/60 = 12.5

Time to catch = 12.5/(10) = 1.25 ==> 1.25 * 60 = 75 minutes

Answer [spoiler]{D}[/spoiler]
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by mdecarbo » Mon Dec 02, 2013 1:17 pm
Thanks team. When I missed this question I let the wording confuse me. I forgot to include the 2.5 miles that Mary drove, making the distance 12.5 miles as opposed to 10 miles. I chose C - 1 hour at first, but now understand that I needed the extra 15 mins to catch up.
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