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j_shreyans Legendary Member Default Avatar
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Post Mon Sep 08, 2014 10:02 pm
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.

OAE

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Post Mon Sep 08, 2014 11:20 pm
Hi j_shreyans,

This is what's called a "chase down" question. We need to figure out how far ahead the first person is, then figure out how much the second person "catches up per hour" to answer this question.

Since Peter starts at 10am and John doesn't start until 2pm, Peter has a 4 hour "head start." Since Peter is traveling 10mph, he travels 40 miles before John gets started. John travels 15mph, so every hour he travels 5 miles MORE than Peter does (so he "catches up" 5 miles every hour from the moment he starts traveling). To make up the 40 miles "head start" that Peter has, John needs 8 hours to "catch up."

Final Answer: E

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j_shreyans Legendary Member Default Avatar
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Post Tue Sep 09, 2014 2:08 am
Hi Rich ,

Thanks for your help ....is there any other approach to catch these kind of questions?

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Post Tue Sep 09, 2014 2:20 am
j_shreyans wrote:
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.

OAE
I would solve as Rich did:

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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Post Tue Sep 09, 2014 12:02 pm
Using the catch up rate is the best way to go here. Understanding this method will help simplify a lot of DS problems in rates that seem to be lacking information, but are actually sufficient.

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Post Thu Dec 07, 2017 9:37 am
j_shreyans wrote:
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.
We can classify this problem as a “catch-up” rate problem, for which we use the formula:

distance of Peter = distance of John

We are given that at 10:00 a.m., Peter begins traveling on a certain bike path from Riverdale at a constant rate of 10 mph, and that at 2:00 p.m., John begins traveling from Riverdale on the same path at a constant rate of 15 mph.

Since Peter started 4 hours before John, we can let Peter’s time = t + 4 hours, and John’s time = t.

Since distance = rate x time, we can calculate each person’s distance in terms of t.

Peter’s distance = 10(t + 4) = 10t + 40

John’s distance = 15t

We can equate the two distances and determine t.

10t + 40 = 15t

40 = 5t

t = 8 hours

Since John left at 2 p.m. and caught up to Peter 8 hours later, he caught up with Peter at 10 p.m.

Answer: E

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