Solve it within 10 seconds

The speed of Bruce is 100 metres divided by 15 seconds, or 6.667 m/s. When Allen reaches the finish line in 12 seconds, B would have been at (6.667 m/s)(12 seconds), or 80 m after the start. The distance between A and B would therefore be 100 m - 80 m = 20 m.

I noted that Bruce was running 20m every 3 sec, and Allen was running 25m every 3 sec. Bruce had 3 more seconds to run, therefore 20 more meters. (A little less calculator-driven. I grew up without one, so my brain looks for something easier first.)

There is a LCF( Lowest Common Factor) of 12 and 15, which is 3. (Except 1) So 15/3 is 5 and 12/3 is 4 so that means the ratio of the length of how much did allen run to the length of how much did Bruce ran is 5:4 so the 5 is 100 m and the 4 is 4/5 of 100 m which is 80m. So 100m - 80m equals 20m

After 12 seconds, Bruce finished 12/15 of the race, leaving 3/15=1/5 of the race unfinished. So he is 1/5*100=20m from the finish line.

When Allen finished the race, at 12 secs, Bruce was still 3 secs away from finishing his 15 secs race. So he still had 3/15 (20%) of the 100m to go. 20% of 100m = 20m

At 15 sec:

speed = $\frac{Distance}{time} \ = \frac{100}{15} \ = \ 6.667$

At 12 sec. $\therefore \ distance\ of \ Bruce = 100-x \\ \therefore\quad 6.667 =\frac{100-x}{12}$

$\rightarrow x=\boxed{=>20m}$

100-100/15*12 =20

When Bruce has run for 12 seconds at a constant speed, he still has 3/15 of the 100m left to go. 3/15 = 1/5 = 0.2. 0.2 * 100 = 20m

Speed of Bruce=100/15 Time difference=3 S Distance lag =100/15 X3=20M

It is way simple.. Since both are running at constant speeds, at 12 seconds, Bruce would have covered (12/15)th of the race or 80 percent. Therefore he would need to complete 20 percent of the race. Therefore he is at 20 meters distance from the finish line.

Reiterating the question, I found that first we need to calculate the speed of Bruce then the distance ran by him in 12 seconds after that the difference between the distance ran and 100m is the answer. Speed of Bruce = 100/15 m per sec (v=s/t) Distance covered by Bruce in 12 seconds = (100/15) *12=80 (s=vt)
Distance to be ran by Bruce =100-80=20m. That's it!

Allan's speed was 12"/100 mts and Bruce's 15"/100 mts. Speed difference between runners were of 3"/100 mts in favour of Allan. Therefore Allan was 20% faster than Bruce. 20% of the 100 mts track is 20 mts. So, Bruce was 20 mts below Allan at the end of the race.

Better calculate the distance traveled by Alex in 3 seconds. Which would be (100÷15)*3=20

We calculate sppeds of Allen and Bruce as 8,33 and 6.66 per second for 100 mts race.Since bruce is 3 seconds behind , so'6.66 X3 =20 meters bruce is behind Allen

100 meters divided by 15 seconds multiplied by 3 second difference = 20 meters. college math de-simplifies this question that many 3rd graders will get right.

(1-(12/15))*100 = 100/5 = 20

After Allen arrival, Bruce has to run for 15-12 = 3 other seconds, in which he will cover $\frac{100}{15} \times 3 = 20$ meters

100-[100/15*12]=20

The laggard's speed times the finisher's time gives you the distance the laggard ran when the leader crossed the line. Subtract that amount from the length of the race and that's the distance to the finish line.

If Bruce ran 100m in 15 seconds then he was running 1m every 6.67 seconds. So in 12 seconds he would have ran (6.67 * 12) 80 meters. Subtracting the 80 meters that Bruce has covered in 12 seconds from the entire distance of 100 meters means Bruce was 20 meters from the finish line.

Allen is running 5/4 (15/12) times faster than Bruce. So when Allen finish the race, Bruce is at 4/5 of the distance to finish, 100m x 4/5 = 80m, so 20m behind.

No need to calculate the speed but just the ratio of speed.