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Speed

by MI3 » Sun May 29, 2011 12:06 am
Q. 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

[spoiler] I solved the problem by finding out individual time date in minutes, then finding the average rate by summing the total distance and dividing by the calculated total time taken (sum of individual time taken) and got 2/15 as the answer. But, I was wondering whether there was a easier way to solve this?

There is another formula to calculate the average speed 2XY/(X+Y) and this gives the answer as 15/2 (but in MPH)..So ideally both approaches should work?[/spoiler]

Thanks,
M
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by manpsingh87 » Sun May 29, 2011 12:14 am
MI3 wrote:Q. 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

[spoiler] I solved the problem by finding out individual time date in minutes, then finding the average rate by summing the total distance and dividing by the calculated total time taken (sum of individual time taken) and got 2/15 as the answer. But, I was wondering whether there was a easier way to solve this?

There is another formula to calculate the average speed 2XY/(X+Y) and this gives the answer as 15/2 (but in MPH)..So ideally both approaches should work?[/spoiler]

Thanks,
M
average speed= total distance/ total time;
since D here is constant and is equal to 2;
therefore we have;
total distance =2+2;
total time= 2/v1+ 2/v2;
v1 is the speed while dan runs, and v2 is the speed while dan swims..!!!

hence average speed = 2*2/(2/v1+2/v2);
2*v1*v2/(v1+v2); subsituting v1=10; v2=6;
we have 15/2 in mph; to convert it into milesperminute; divide it by 60 we have;
15/2*(1/60)= 1/8 hence A
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by smackmartine » Sun May 29, 2011 12:20 am
IMO A
Time taken while running = 2/10 hrs= (1/5)*60 min =12 min
Time taken while swimming = 2/6 hrs= (1/3)*60 min =20 min

Avg Rate = Total Distance / Total time = 2*2 /(12+20) = 4/32 = 1/8
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by sivaelectric » Sun May 29, 2011 12:21 am
Thanks manpsingh87
If I am wrong correct me :), If my post helped let me know by clicking the Thanks button ;).

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by amar66 » Sun May 29, 2011 12:35 am
sivaelectric wrote:Thanks manpsingh87
Instead of writing Thanks as a reply in a post, you can press just Thank button(in top right corner of the poster's reply) ;-)
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by GMATGuruNY » Sun May 29, 2011 1:08 am
MI3 wrote:Q. 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
Very little math is needed here.

The faster rate is 10 miles per hour, the slower rate is 6 miles per hour.
The average of the 2 rates is (10+6)/2 = 8 miles per hour.
Since the distance traveled at each rate is the same, the time spent traveling at the slower rate is longer than the time spent traveling at the faster rate.
Thus, the average rate for the whole trip will be a little less than 8 miles per hour, since more time is spent traveling at the slower rate.

The answer choices represent the speed in miles per minute.
Thus, when the correct answer choice is multiplied by 60, the result will be a little less than 8.

Only answer choice A works:
(1/8)*60 = 7.5

The correct answer is A.
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by edvhou812 » Sun May 29, 2011 9:33 pm
Trip is 4 miles total:

First 2 miles:
2miles/10mph = 1/5. 60/1*1/5 = 2 miles in 12 minutes

Second 2 miles:
2miles/5mph = 2/3. 60/1*1/3 = 2 miles in 20 minutes

Total:
4miles in 32 minutes=> 32/4 = 8. 1 mile takes 8 minutes. Miles per minutes: 1/8
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by mcdesty » Sat Jul 12, 2014 2:13 pm
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