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

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

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

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

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

Total Distance/Total time

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

in miles per minute 15/2*60 = 1/8

Can you kindly update OA for this?

Thanks

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

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

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)

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

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

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

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.

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

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

Why are the rates not additive here?