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

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

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New post 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

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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New post 09 Aug 2009, 09:25
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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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New post 10 Aug 2009, 01:29
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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

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 (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
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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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New post 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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New post 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 (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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New post 05 Sep 2011, 14:21
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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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New post 21 Sep 2015, 06:19
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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
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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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New post 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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New post 14 Nov 2016, 08:28
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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


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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New post 07 Jan 2018, 10:30
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

So, Option (E)
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Re: At 10:00 a.m., Peter begins traveling on a certain bike path from Rive &nbs [#permalink] 07 Jan 2018, 10:30
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