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At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]

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09 Aug 2009, 08:04

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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

Re: At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]

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10 Aug 2009, 00:29

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crejoc 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? 6:00 p.m. 6:30 p.m. 8:00 p.m. 8:30 p.m. 10:00 p.m

since Peter had early start of 4 hrs, in this 4 hrs he covered 40miles. after that John starting in the same path. the differences in speed is (15-10)mph = 5mph, in order to catch up Peter, john has to cover the above 40miles. =40/5 = 8 hrs i.e 8 hrs from 2:00 pm is equal to 10:00p m

Schools: NUS, NTU, SMU, AGSM, Melbourne School of Business

Re: At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]

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10 Aug 2009, 02:16

Ans : 10 pm Peter covered 40 miles when John started at 2 pm. So the distance between them is 40. Difference between the speeds of Peter and John is 5 (15-10)

Therefore, time reqd to cover that distance will be 40 / 8 = 8 hrs.

Hence, David will catch up Peter at 10 pm (2+8).

Hope this helps.
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Re: At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]

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21 Sep 2015, 05:19

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crejoc 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

My ans: E By the time John starts, Peter has already covered = 4 hr * 10 mph = 40 miles Relative speed = 15- 10 = 5mph To catch up, John needs to cover 40 miles which can be covered in = 40/5= 8 hours If John leaves at 2 pm, he will catch Peter at 10 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

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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Re: At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]

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07 Jan 2018, 09:30

Total relative speed = 15-10 = 5mph Total distance between the two at the start = 40 miles (Peter traveled for 4 hours at a speed of 10mph) sO, TIME = 40/5= 8 hrs