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

Question

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

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

eschn3am · Accepted Answer

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

Dienekes · Answer

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.

LM · Answer

Total Distance/Total time

=4/  = 15/2

in miles per minute 15/2*60 = 1/8

pmenon · Answer

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

PareshGmat · Answer

Can you kindly update OA for this?

Thanks

roadrunner · Answer

t1 = 2/10 hours
t2 = 2/6

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

Average speed = Total distance/ total time = 4/   = 15/2 mph

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

Ans.: A

Nunuboy1994 · Answer

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

nawaf52 · Answer

Thank you for the explanation. I picked answer choice "B" because I did not pay attention to the unit (minutes) in the question.

BrentGMATPrepNow · Answer

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

JeffTargetTestPrep · Answer

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

AkshdeepS · Answer

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)

fskilnik · Answer

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

{V_{{	ext{run}}}} = \frac{{10\,\,{	ext{miles}}}}{{1\,\,{	ext{h}}}}\,\,\,\,\,\,;\,\,\,\,\,{V_{{	ext{swim}}}} = \frac{{6\,\,{	ext{miles}}}}{{1\,\,{	ext{h}}}}

?\,\,:\,\,\,\,\frac{{{	ext{total}}\,\,{	ext{miles}}}}{{{	ext{total}}\,\,{	ext{minutes}}}} = \frac{{4\,\,{	ext{miles}}}}{{?{\,_{{	ext{temporary}}\,\,{	ext{focus}}}}}}

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

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

{	ext{? = }}\frac{{	ext{4}}}{{32}} = \frac{1}{8}

Obs.: arrows indicate licit converters.

This solution follows the notations and rationale taught in the GMATH method.

davidbeckham · Answer

Can anyone tell me how do we solve this question if the distance is not given?

RastogiSarthak99 · Answer

Really the only trap here is to convert to MINUTE. Don't fall in to the trap of going with hour figures.

Avg speed = total distance/total time

total time = 12 mins + 20 mins

2 + 2 / 12 + 20 = 1/8

A

NoviceBoy · Answer

Why are the rates not additive here?

bumpbot · Answer

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

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