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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 1

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by BTGModeratorVI » Mon Jul 13, 2020 7:50 am

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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

Answer: A
Source: Manhattan Prep
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Source: — Problem Solving |

BTGModeratorVI wrote: ↑
Mon Jul 13, 2020 7:50 am
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

Answer: A
Source: Manhattan Prep
Note that if the distance is constant (in this case), then the average speed of the to-and-fro trip is less than the arithmetic mean of the two speeds; actually the average speed = harmonic average of the speeds. Well, you need not know what harmonic mean is. Merely, knowing that harmonic mean is always less than the arithmetic mean is sufficient.

So, let's calculate the arithmetic mean. It is (10 + 6)/2 = 8 mph = 8/60 = 2/15 miles per minute

So, the correct answer must be less than 2/15. The only choice less than 2/15 is choice A (= 1/8): the correct answer.

Correct answer: [spoiler][/spoiler]

Hope this helps!

-Jay
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BTGModeratorVI wrote: ↑
Mon Jul 13, 2020 7:50 am
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

Answer: A
Source: Manhattan Prep
Average speed when distance for two trips is same \(=\dfrac{2R_1R_2}{R_1+R_2}\)

\begin{align*}
\text{Avg. Speed} &= \dfrac{2 \cdot 10 \cdot 6}{10 + 6} \\
&= \dfrac{120}{16} \\
&= \dfrac{15}{2} \text{ miles / hr} \\
&= \dfrac{15}{2} \cdot \dfrac{1}{60} \text{ miles / minute} \\
&= \dfrac{1}{8}
\end{align*}

Hence, A
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BTGModeratorVI wrote: ↑
Mon Jul 13, 2020 7:50 am
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

Answer: A
Source: Manhattan Prep
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
Brent Hanneson - Creator of GMATPrepNow.com
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BTGModeratorVI wrote: ↑
Mon Jul 13, 2020 7:50 am
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

Answer: A
Solution:

We can use the average rate formula: average rate = 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 rate = 4/(2/6 + 2/10)

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

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

average rate = 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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