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Triathlete Dan runs along a 2-mile stretch of river and then swims bac

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Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 24 Dec 2007, 03:10
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Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8
[Reveal] Spoiler: OA

Last edited by Bunuel on 06 Oct 2014, 07:15, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 24 Dec 2007, 04:20
Dan travels 4 miles round trip.

Running part: (2/10 = 1/5*60 = 12 minutes)
Swimming Part: (2/6 = 1/3*60 = 20 minutes)

4 miles in (12+20) minutes
4/32 = 1/8 mile per minute

Answer: 1/8 mile per minute
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 24 Dec 2007, 05:16
dynamo wrote:
Triathlete Dan runs along a 2-mile stretch of river and then swims back along the same route. If Dan runs at a rate of 10 miles per hour and swims at a rate of 6 miles per hour, what is his average rate for the entire trip in miles per minute?


Total Distance/Total time

=4/ [( 1/5) + (1/3)]= 15/2

in miles per minute 15/2*60 = 1/8
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 24 Dec 2007, 15:48
time to run = 1/5 hr = 12 minutes

time to swim = 1/3 hr = 20 minutes

average rate = total distance / total time
= 4 miles / 20+12 = 4/32 = 1/8 mile per minute
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 14 Sep 2014, 20:37
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 14 Sep 2014, 20:46
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If distance traveled is same but at varying rates, the average rate can be found through the shortcut \frac{2ab}{a+b} where a and b are individual rates.
In this problem a=10 and b=6 therefore average speed = \frac{15}{2} miles per hour, i.e. \frac{1}{8} miles per minute.

P.S. Added tags as saw this question in an MGMAT CAT.
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac [#permalink] New post 15 Sep 2014, 00:03
Dienekes wrote:
If distance traveled is same but at varying rates, the average rate can be found through the shortcut \frac{2ab}{a+b} where a and b are individual rates.
In this problem a=10 and b=6 therefore average speed = \frac{15}{2} miles per hour, i.e. \frac{1}{8} miles per minute.

P.S. Added tags as saw this question in an MGMAT CAT.


Can you kindly update OA for this?

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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac   [#permalink] 15 Sep 2014, 00:03
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