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Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour

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Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post Updated on: 28 Jul 2016, 08:58
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Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour. One hour into Reggie’s hike, Cassie started hiking from the same starting point on the loop trail at 3 miles per hour. What is the shortest time that Cassie could hike on the trail in order to meet up with Reggie?

(A) 0.8 hours
(B) 1.2 hours
(C) 2 hours
(D) 3 hours
(E) 5 hours

So this question is from the MGMAT ROADMAP. I understand the first part of solution provided in it, that it will take cassie 2 hours to catch up to Regan. However, the solution also calculates time if cassie would do the hike in the opposite direction. Now this is really confusing to me. There is no methodology that I take from it that how Cassie is able to hike miles 6 miles in reverse while reggie is only able to hike 2 miles for that direction.

Answer is A.

Please help.

Originally posted by jonmarrow on 28 Jul 2016, 08:51.
Last edited by Bunuel on 28 Jul 2016, 08:58, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 28 Jul 2016, 09:18
Didn't get this (assumed they were both hiking in the same direction).

\(\text{speed} = \frac{\text{distance}}{\text{time}}\)

\(\text{distance travelled in 1 hour: }2 \text{ miles } (\text{speed} \times \text{time})\).

\(\text{Distance from }\color{red}{\text{end of the trail (anticlockwise)}}: 6 - 2 = 4 \text{ miles}\).

Speed that they are coming together: \(3mph + 2mph = 5mph\).

\(\text{Time} = \frac{\text{distance}}{\text{speed}} = \frac{4}{5} = 0.8\)
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Re: Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 29 Jul 2016, 02:00
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DAllison2016 wrote:
Didn't get this (assumed they were both hiking in the same direction).

\(\text{speed} = \frac{\text{distance}}{\text{time}}\)

\(\text{distance travelled in 1 hour: }2 \text{ miles } (\text{speed} \times \text{time})\).

\(\text{Distance from }\color{red}{\text{end of the trail (anticlockwise)}}: 6 - 2 = 4 \text{ miles}\).

Speed that they are coming together: \(3mph + 2mph = 5mph\).

\(\text{Time} = \frac{\text{distance}}{\text{speed}} = \frac{4}{5} = 0.8\)



we don't need to assume here the direction.we need to find which will take less time

so there are 2 cases

1)both moving in same direction...then time would be 2 hrs

2)both moving in opp direction then time would be 0.8 hrs

as we need the shortest time it would be the second case
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Re: Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 05 Aug 2016, 09:58
jonmarrow wrote:
Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour. One hour into Reggie’s hike, Cassie started hiking from the same starting point on the loop trail at 3 miles per hour. What is the shortest time that Cassie could hike on the trail in order to meet up with Reggie?

(A) 0.8 hours
(B) 1.2 hours
(C) 2 hours
(D) 3 hours
(E) 5 hours

So this question is from the MGMAT ROADMAP. I understand the first part of solution provided in it, that it will take cassie 2 hours to catch up to Regan. However, the solution also calculates time if cassie would do the hike in the opposite direction. Now this is really confusing to me. There is no methodology that I take from it that how Cassie is able to hike miles 6 miles in reverse while reggie is only able to hike 2 miles for that direction.

Answer is A.

Please help.


I think the term "loop" trail answers the confusion. Since, it's a loop so Cassie can go in the same direction as that of Regan or in the opposite direction and we need to find out which one takes lesser time. Per the calculations, heading in opposite direction takes lesser time (0.8 hours). Hence, it's the answer.
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Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 05 Aug 2016, 10:25
let t=cassie's time
3t+2(t+1)=6 miles
t=4/5=.8 hours
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Re: Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 06 Jan 2017, 12:39
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They are walking in opposite directions around a loop. He walks alone for one hour at a rate of 2 miles per hour, which means he walks two miles. Now, she begins walking from the same starting point moving the other direction around the loop, and they are 6-2= 4 miles apart. When objects move toward each other, we add their rates. Thus, they are closing the 4 mile gap at a rate of 5 miles an hour.

Rate x Time = Work

5 x Time = 4
Time = 4/5 or .8
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Re: Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour  [#permalink]

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New post 16 Apr 2018, 16:23
jonmarrow wrote:
Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour. One hour into Reggie’s hike, Cassie started hiking from the same starting point on the loop trail at 3 miles per hour. What is the shortest time that Cassie could hike on the trail in order to meet up with Reggie?

(A) 0.8 hours
(B) 1.2 hours
(C) 2 hours
(D) 3 hours
(E) 5 hours


Since the trail is a loop, the quickest way for Cassie to meet Reggie is to hike in the opposite direction of Reggie’s path:

We can let Reggie’s time = t + 1 and Cassie’s time = t, thus:

2(t + 1) + 3t = 6

2t + 2 + 3t = 6

5t = 4

t = 4/5 = 0.8 hours

Answer: A
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Re: Reggie was hiking on a 6-mile loop trail at a rate of 2 miles per hour &nbs [#permalink] 16 Apr 2018, 16:23
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