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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]

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New post Updated on: 06 Oct 2014, 08:15
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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

Originally posted by dynamo on 24 Dec 2007, 04:10.
Last edited by Bunuel on 06 Oct 2014, 08:15, 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]

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New post 24 Dec 2007, 05:20
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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]

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New post 24 Dec 2007, 06:16
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]

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New post 24 Dec 2007, 16:48
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]

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New post 14 Sep 2014, 21:46
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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]

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New post 15 Sep 2014, 01:03
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?

Thanks
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 18 Jan 2017, 09:56
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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8


t1 = 2/10 hours
t2 = 2/6

t =t1 + t2 = 2/10 + 2/6 = 16/30

Average speed = Total distance/ total time = 4/ [16/30] = 15/2 mph

15/2 * 1/60 = 1/8 miles/min

Ans.: A
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 19 Apr 2017, 23:24
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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8


We can approach this problem using conversion factors

If the total distance is 2 miles and Dan runs at a rate of 10 miles per hour then he will cover in 1/5 of an hour or 1 mile in 6 minutes

1 mile/6 minutes = 2 miles/ 20 minutes

cross multiply add and then divide by 2

(16 miles /2) / 60 minutes

(16 miles/2)/ 1 minute

8 miles/ 1 minute

1 minute/ 8 miles
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 01 Aug 2017, 02:38
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.


Thank you for the explanation. I picked answer choice "B" because I did not pay attention to the unit (minutes) in the question.
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 03 Apr 2018, 10:21
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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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8


Average speed = (total distance)/(total time)

We know the total distance = 4 miles

The total time consists of two parts:
Let's begin with a "word equation"
TOTAL time = (time spent running) + (time spent swimming)

time = distance/speed
So, time spent running = 2/10 = 1/5 hours = 12 minutes
time spent swimming = 2/6 = 1/3 hours = 20 minutes

TOTAL time = (12 minutes) + (20 minutes)
= 32 minutes

AVERAGE speed = (total distance)/(total time)
= (4 miles)/(32 minutes)
= 1/8 miles/minute

Answer: A

Cheers,
Brent
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 05 Apr 2018, 16:59
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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8


We can use the average rate formula: average = total distance/total time. We see that the total distance is 4 miles. The total time for swimming is (2 miles)/(6 miles per hour) , and the total time for running is (2 miles)/(10 miles per hour). Thus, we have:

average = 4/(2/6 + 2/10)

average = 4/(1/3 + 1/5)

average = 4/(5/15 + 3/15)

average = 4/(8/15) = 60/8 mph

Since 1 hour = 60 minutes, then 60/8 mph = 60 mi/8 hr x 1 hr/60 min = 1/8 mi/min.

Answer: A
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 15 Apr 2018, 00:55
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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8


Avg. Speed when distance for two trips is same = \(\frac{2 * R1 * R2}{R1 + R2}\)

Avg. spee = 2 * 10 * 6/10 + 6

= 120/16

= 15/2 miles / hr

= 15/2 * 1/60 miles / minute

= 1/8

Hence (A)
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac  [#permalink]

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New post 04 Sep 2018, 09:49
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?

A. 1/8
B. 2/15
C. 3/15
D. 1/4
E. 3/8

\({\text{2}}\,{\text{miles}}\,\,\,\,({\text{go}}\,\,{\text{and}}\,\,{\text{back}})\)

\({V_{{\text{run}}}} = \frac{{10\,\,{\text{miles}}}}{{1\,\,{\text{h}}}}\,\,\,\,\,\,;\,\,\,\,\,{V_{{\text{swim}}}} = \frac{{6\,\,{\text{miles}}}}{{1\,\,{\text{h}}}}\)

\(?\,\,:\,\,\,\,\frac{{{\text{total}}\,\,{\text{miles}}}}{{{\text{total}}\,\,{\text{minutes}}}} = \frac{{4\,\,{\text{miles}}}}{{?{\,_{{\text{temporary}}\,\,{\text{focus}}}}}}\)

Excellent opportunity to use UNITS CONTROL, one of the most powerful tools of our method!

\(\left. \begin{gathered}
{\text{Run}}:\,\,\,2\,{\text{miles}}\,\,\left( {\frac{{1\,\,{\text{h}}}}{{10\,\,{\text{miles}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\left( {\frac{{60\,\,{\text{minutes}}}}{{1\,\,h}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,12\,\,{\text{minutes}} \hfill \\
{\text{Swim}}:\,\,\,2\,{\text{miles}}\,\,\left( {\frac{{1\,\,{\text{h}}}}{{6\,\,{\text{miles}}}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\left( {\frac{{60\,\,{\text{minutes}}}}{{1\,\,h}}\,\,\begin{array}{*{20}{c}}
\nearrow \\
\nearrow
\end{array}} \right)\,\,\, = \,\,20\,\,{\text{minutes}} \hfill \\
\end{gathered} \right\}\,\,\,\, \Rightarrow \,\,\,\,?{\,_{{\text{temporary}}\,\,{\text{focus}}}}\,\, = \,\,\,32\,\,{\text{minutes}}\)

\({\text{? = }}\frac{{\text{4}}}{{32}} = \frac{1}{8}\)

Obs.: arrows indicate licit converters.

This solution follows the notations and rationale taught in the GMATH method.
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Re: Triathlete Dan runs along a 2-mile stretch of river and then swims bac &nbs [#permalink] 04 Sep 2018, 09:49
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