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At 10:00 a.m., Peter begins traveling on a certain bike path from Rive [#permalink]
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Updated on: 20 Sep 2015, 07:44
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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
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Originally posted by crejoc on 09 Aug 2009, 09:04.
Last edited by Bunuel on 20 Sep 2015, 07:44, edited 1 time in total.
Renamed the topic, edited the question and added the OA.



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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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09 Aug 2009, 09:25
Let number of hours = x, (when John catches up to Peter.)
So in x + 4 hrs Peter traveled = 10x + 40. In x hrs John traveled = 15x.
Therefore, 15x = 10x + 40. x = 8 hrs.
So @ 2:00PM + 8 hrs i.e. @ 10:00PM John catches peter.



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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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10 Aug 2009, 01:29
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 OA: Explanation plz... 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 (1510)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



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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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10 Aug 2009, 02:30
distance travelled by peter = xp distance travelled by john = xj
x = vt + x0 ( where x0 is the distance at time t=0, x is distance at time t and v is speed at time t )
Using the above, xp = 10t + 40 xj = 15 t Now, xp = xj => 40 = 5t => t = 8, i.e. 10 pm



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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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10 Aug 2009, 03:16
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 (1510) 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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05 Sep 2011, 14:21
peter is 40miles ahead when john begins at 2:00pm. john covers 5miles each hour time required to catch up = 40/5 = 8hours so 10:00 pm
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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, 06:19
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



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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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11 Nov 2016, 00:05
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 For more on relative speed, check: https://www.veritasprep.com/blog/2012/0 ... elatively/https://www.veritasprep.com/blog/2012/0 ... concepts/
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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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14 Nov 2016, 08:28
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 We can classify this problem as a “catchup” 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, 10:30
Total relative speed = 1510 = 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
So, Option (E)




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