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Two bikers participate in a 40 mile race with 4 laps of 10 miles each.
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08 Mar 2021, 04:03
08 Mar 2021, 04:03
Expert Reply
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Question Stats:
100% (03:18) correct
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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. If Biker A starts at 60 miles per hour in the first lap, and on each lap he goes 20% faster than the previous lap, and if Biker B maintains a uniform speed of 80 miles throughout the race then approximately how many seconds earlier would Biker A reach the finish line?
A. 21
B. 18
C. 15
D. 12
E. 9
A. 21
B. 18
C. 15
D. 12
E. 9
Re: Two bikers participate in a 40 mile race with 4 laps of 10 miles each.
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08 Mar 2021, 07:10
08 Mar 2021, 07:10
1
Solution:
For A,
When we calculate the speed with a 20 % increase we get, 60+72+86.4+103.68=\(\frac{322.08}{4}=\)80.52 is the average speed for each lap
B has an avg speed of 80 for each lap.`
Thus the time taken by A=\(\frac{40}{80.52}\)=0.49677 hours= 29 mins 80 sec
Time taken by B=\(\frac{40}{80}\)=0.5 hour= 30 mins
Thus, 30-29.80=20 seconds.
The closest value is 21.
IMO A
Hope this helps!
For A,
When we calculate the speed with a 20 % increase we get, 60+72+86.4+103.68=\(\frac{322.08}{4}=\)80.52 is the average speed for each lap
B has an avg speed of 80 for each lap.`
Thus the time taken by A=\(\frac{40}{80.52}\)=0.49677 hours= 29 mins 80 sec
Time taken by B=\(\frac{40}{80}\)=0.5 hour= 30 mins
Thus, 30-29.80=20 seconds.
The closest value is 21.
IMO A
Hope this helps!
Re: Two bikers participate in a 40 mile race with 4 laps of 10 miles each.
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28 Feb 2022, 07:10
28 Feb 2022, 07:10
It should be D, due to 1 minute is 60 seconds, so 0.496h = 29.8 min = 29 min + 48 sec (0.8x60=48s)
So 30min - 29min 48sec = 12 sec. Thanks
So 30min - 29min 48sec = 12 sec. Thanks
