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Peter and John

Expert replies
Source: — Problem Solving |

by GMATGuruNY » Wed Feb 17, 2016 7:37 am
At 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph. If, at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph, at what time will he catch up to Peter?

A)6:00 p.m.
B)6:30 p.m.
C)8:00 p.m.
D)8:30 p.m.
E)10:00 p.m.
From 10am to 2pm, the distance traveled by Peter in 4 hours = rt = 10*4 = 40 miles.

The CATCH-UP RATE is equal to the DIFFERENCE between John's rate and Peter's rate:
15-10 = 5 miles per hour.
Since John travels 15 miles per hour, while Peter travels 10 miles per hour, every hour John catches up by 5 miles.

Time for John to catch up 40 miles = (catch-up distance)/(catch-up rate) = 40/5 = 8 hours.
Since John starts to travel at 2pm, the time at which he catches up to Peter = 2pm + 8 hours = 10pm.

The correct answer is E.

Alternate approach:

From 10am to 2pm, the distance traveled by Peter in 4 hours = rt = 10*4 = 40 miles.
Every hour Peter travels 10 more miles, while John travels 15 miles.
To determine when John catches up to Peter, WRITE OUT the total distance that each has traveled, hour by hour.

2pm:
Peter = 40 miles, John = 0 miles.
3pm:
Peter = 40+10 = 50 miles, John = 0+15 = 15 miles.
4pm:
Peter = 50+10 = 60 miles, John = 15+15 = 30 miles.
5pm:
Peter = 60+10 = 70 miles, John = 30+15 = 45 miles.
6pm:
Peter = 70+10 = 80 miles, John = 45+15 = 60 miles.
7pm:
Peter = 80+10 = 90 miles, John = 60+15 = 75 miles.
8pm:
Peter = 90+10 = 100 miles, John = 75+15 = 90 miles.
9pm:
Peter = 100+10 = 110 miles, John = 90+15 = 105 miles.
10pm:
Peter = 110+10 = 120 miles, John = 105+15 = 120 miles.

The correct answer is E.

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https://www.beatthegmat.com/p-and-q-circ ... 22287.html (GMAT Prep)
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by didieravoaka » Wed Feb 17, 2016 4:44 pm
Thanks Mitch for the answers and for the similar problems. But tell me, is it that rule that you use in he first approach...
"When elements compete, determine the difference between the rates"?
Thanks.
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by Matt@VeritasPrep » Thu Feb 18, 2016 6:06 pm
didieravoaka wrote:Thanks Mitch for the answers and for the similar problems. But tell me, is it that rule that you use in he first approach...
"When elements compete, determine the difference between the rates"?
Thanks.
I'd think of it as "when one person/thing is trying to catch another, treat the rate as Faster Thing - Slower Thing".

Similarly, when two things/people are traveling TOWARD each other or working together, treat the rate as Thing 1 + Thing 2.
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by didieravoaka » Thu Feb 18, 2016 6:08 pm
Thanks Matt. I'd think the same way.
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