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# Jeff takes 20 minutes to jog around the race course one time, and 25

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Math Expert
Joined: 02 Sep 2009
Posts: 59722
Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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14 May 2016, 03:59
00:00

Difficulty:

15% (low)

Question Stats:

80% (01:59) correct 20% (02:22) wrong based on 90 sessions

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Jeff takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

A. 6
B. 8
C. 10
D. 12
E. 14

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Re: Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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14 May 2016, 08:08
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Bunuel wrote:
Jeff takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

A. 6
B. 8
C. 10
D. 12
E. 14

Total distance covered is 6 miles ( 3 miles + 3 miles )
Total time taken is 3/4 hours ( 45 minutes )

$$Average$$ $$speed$$ = $$\frac{{ Total \ Distance }}{{ Total \ Time \ taken }}$$ => $$\frac{6}{(3/4)}$$ = 8

Hence correct answer will be B. 8

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Math Expert
Joined: 02 Aug 2009
Posts: 8309
Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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14 May 2016, 08:32
Bunuel wrote:
Jeff takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

A. 6
B. 8
C. 10
D. 12
E. 14

various methods to do it--

I)
speed first time =$$\frac{3}{20}*60 = 9 mph$$..
speed second time = $$\frac{3}{25}*60 = 7.2mph$$..

so average speed has to be between 7.2 and 9..
ONLY 8 is given
ans B..

II) the two speed are 9mph and 7.2 mph..
so average speed by formula when distance covered is same on both occasions = $$\frac{2ab}{(a+b)} = \frac{2*9*7.2}{(9+7.2)}= \frac{2*7.2*9}{9(1+0.8)} = \frac{2*7.2}{1.8} = 2*4 = 8$$..
ans B

III)
distance = 3+3 = 6 miles
time taken = 20+25 = 45 min = \frac{45}{60} hour = \frac{3}{4} hour
avg speed = $$6/\frac{3}{4}= 8$$
B
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Re: Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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15 May 2016, 03:14
Average Speed = Total Distance Covered / Total Time

= (3+3) Miles / (20+25) mins

= 9/45 *(60) = 8 mph
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Re: Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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15 May 2016, 09:29
Avg Speed= total distance /total time

3+3/20+25
6/45

Since we have to find the speed in miles/hr, multiply the above number by 60

6*60/45= 8 miles/hr

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Re: Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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19 Nov 2019, 02:48
Avg Speed = Total Dist./ Total Time

= 3+3/(1/3 + 5/12) = 8.

Ans B
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Re: Jeff takes 20 minutes to jog around the race course one time, and 25  [#permalink]

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22 Nov 2019, 12:59
Bunuel wrote:
Jeff takes 20 minutes to jog around the race course one time, and 25 minutes to jog around a second time. What is his average speed in miles per hour for the whole jog if the course is 3 miles long?

A. 6
B. 8
C. 10
D. 12
E. 14

The average speed is:

6/(20/60 + 25/60) = 6/(45/60) = 360/45 = 8 mi/hr

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Re: Jeff takes 20 minutes to jog around the race course one time, and 25   [#permalink] 22 Nov 2019, 12:59
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