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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. 24 Mar 2021, 05:30
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85% (hard)

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43% (03:16) correct 57% (02:53) wrong based on 14 sessions

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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

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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. 26 Mar 2021, 00:00
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Avg speed of A - 60 + 60*(20% of 60) = 72 + 72*(20% of 72) = 86.4 + 86.4*(20% of 86.4) = 103.68,
i.e 60+72+86.4+103.68/4 = 80.52mph, so if he travels at speed of 80.52 miles in 60 mins, he will take to cover 40 miles = 30.30 mins or 30.30*60 = 1818 seconds.

Avg speed of B is 80mph, so if he travels at speed of 80 miles in 60 mins, he will take to cover 40 miles = 30mins or 30*60 = 1800 seconds

Therefore, A will reach 1818-1800 = 18 seconds earlier. so B, i hope i am right
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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. 26 Mar 2021, 02:09
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I think that’s a wrong explanation

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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. 31 Mar 2021, 09:13
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Bunuel
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
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Solution:

In seconds, biker A reaches the finish line in:

3600[10/60 + 10/(60 * 1.2) + 10/(60 * 1.2^2) + 10/(60 * 1.2^3)]

= 3600[1/6(1 + 1/1.2 + 1/1.2^2 + 1/1.2^3)]

= 600(1 + 1/1.2 + 1/1.2^2 + 1/1.2^3)

≈ 600 + 500 + 417 + 347

= 1,864 seconds

Meanwhile, biker B reaches the finish line in 40/80 x 3600 = 1,800 seconds.

So, biker A takes approximately 64 seconds longer to finish the finish line than biker B.

Answer: 64
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Two bikers participate in a 40 mile race with 4 laps of 10 miles each. 01 Apr 2021, 00:24
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Bunuel
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
Show more


Biker A - Distance covered is the same for each speed.

60 mph for 10 miles means time taken = 10/60 * 60 = 10 mins = 600 secs
Next, speed becomes 20% more so ratio of speed1:speed2 = 5:6 so ratio of time taken = 6 : 5. Since 600 secs were taken in case1, now 600 * 5/6 = 500 secs will be taken.
Next, speed again becomes 20% more so speed2:speed3 = 5:6 so ratio of time taken = 6 : 5. Since 500 secs were taken in case1, now 600*(5/6)*5/6 secs will be taken.
Next, 600*(5/6)*(5/6)*(5/6) secs will be taken.

Total time taken by biker A = 600 (1 + 5/6 + (5/6)^2 + (5/6)^3) = 600 * (1 - (5/6)^4)/(1/6) = 600 * (61*11/216) = 1864

Total time taken by Biker B for 40 miles at 80 mph is 30 mins = 1800 secs

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