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Running on a 10-mile loop in the same direction, Sue ran at

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Running on a 10-mile loop in the same direction, Sue ran at [#permalink] New post 13 Mar 2018, 09:06
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Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles per hour and Rob ran at a constant rate of 6 miles per hour. If they began running at the same point on the loop, how many hours later did Sue complete exactly 1 more lap than Rob?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
[Reveal] Spoiler: OA

Last edited by Carcass on 14 Mar 2018, 12:53, edited 1 time in total.
Edited by Carcass
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Re: Question on work/rates - please help! [#permalink] New post 14 Mar 2018, 01:25
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Shrija Roy wrote:
2. Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles per hour and Rob ran at a
constant rate of 6 miles per hour. If they began running at the same point on the loop, how many hours later did
Sue complete exactly 1 more lap than Rob?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


Rewriting the question: How many hours later was Sue 10 miles ahead of Rob.

Say this happens x hours later.

So;
distance covered by Sue in x hours -distance covered by Rob in x hours = 10 miles

or \(8 \times x - 6 \times x =10\). Solving for x; \(x=5\) hours.

Hence option C is correct!
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Re: Question on work/rates - please help! [#permalink] New post 14 Mar 2018, 07:29
Sorry, one question here -

Sue has a speed of 8m/hr and Rob has the speed of 6 m/hr. So how are we taking the same x time?
Ideally, speed is inversely proportional to time so Sue should complete the lap early, no?
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Re: Running on a 10-mile loop in the same direction, Sue ran at [#permalink] New post 15 Mar 2018, 02:29
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For every hour sue has a lead of \(\frac{2miles}{hr}\) over Rob.
To get an exact 1 lap lead sue has to have total lead of \(10\) miles as the lap is 10 miles long
Hence \(\frac{10}{2} = 5\). Time that is required to put a one lap lead on Rob
option c
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Re: Running on a 10-mile loop in the same direction, Sue ran at [#permalink] New post 16 May 2018, 09:23
Expert's post
Shrija Roy wrote:
Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles per hour and Rob ran at a constant rate of 6 miles per hour. If they began running at the same point on the loop, how many hours later did Sue complete exactly 1 more lap than Rob?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


When Sue completes exactly 1 more lap than Rob, she will have traveled exactly 10 miles more than Rob (since the loop is 10 miles long). We can let both of the times = t and create the equation:

8t = 6t + 10

2t = 10

t = 5

Answer: C
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Re: Running on a 10-mile loop in the same direction, Sue ran at [#permalink] New post 16 May 2018, 10:10
Expert's post
Shrija Roy wrote:
Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles per hour and Rob ran at a constant rate of 6 miles per hour. If they began running at the same point on the loop, how many hours later did Sue complete exactly 1 more lap than Rob?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7


1 lap = 10 miles
So, if Sue completes 1 lap more than Rob completes, then we know that Sue has traveled 10 miles more than Rob

So, let's start with a word equation:
(Sue's travel distance) = (Rob's travel distance) + 10 miles

Distance = (speed)(time)
We know each person's speed, but we don't know their travel times.
Let t = Sue's travel time
So, t = Rob's travel time also

Now take our word equation and plug in the necessary values:
10t = 8t + 10
Solve to get: t = 5

Answer: C

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Re: Running on a 10-mile loop in the same direction, Sue ran at [#permalink] New post 08 Jun 2018, 04:34
I still have not been able to understand why the time would taken the same for both Sue and Rob, could anybody kindly advise?
Re: Running on a 10-mile loop in the same direction, Sue ran at   [#permalink] 08 Jun 2018, 04:34
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Running on a 10-mile loop in the same direction, Sue ran at

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