Puzzling Race

for simplification we will say

A beats B by 20m and B beats C by 20m in a 100m race

When A runs 100m B runs 80m

When B runs 100m C runs 80m

So when B will run 80m C will run 80*80/100 = 64m

Hence when A runs 100m C runs 64m

A beats C by 100 - 64 = 36m

When Lily finishes, Betty has gone 80 meters = 4/5 of the race. When Betty finishes, Louisa has gone 80 meters = 4/5 of the race. So when Lily finishes, Louisa has gone 4/5 of 4/5 = 16/25 of the race. 16/25 of 100 = 64, leaving 36 meters.

Lily beats Louisa by 36 meters.

It is tempting to guess that Lily beats Louisa by 40 meters, but when Lily finishes and is 20 meters ahead of Betty, Betty is NOT 20 meters ahead of Loiusa (she's only 16 meters ahead), and it will take a couple more seconds before Betty increases her lead over Louisa to 20 meters.

To figure out the answer, we let Lily's speed be A meter/second. So it takes her 100/A seconds to finish the race. At this point, we know that Betty has run 80 meters (since Lily beats her by 20 meters). So Betty runs 80 meters in 100/A seconds, meaning that she is running at a speed of (80/(100/A)) meters/second, or (8A/10) meters per second.

So we then know that it takes Betty 100/(8A/10) seconds to finish the race, or 125/A seconds. At this point, we know that Louisa has run 80 meters (since Betty beats her by 20 meters). So Louisa runs 80 meters in 125/A seconds, meaning that she is running at a speed of (80/(125/A)) meters/second, or 80A/125 meters per second.

Now that we know Loiusa's speed, we just need to figure out how far she had run when Lily finished the race. Since Lily finished in 100/A seconds, we can determine that Louisa had run (100/A) * (80A/125) = 8000/125 = 64 meters when Lily finished the race. And so Lily beat her by (100 - 64) = 36 meters.

z is beaten by 20 m in 100 m race.so at 80 m z will be beaten by 80*20/100=16 m so z is beaten by a by (20+16)=36 m

The race is 100 meters. So think in terms of percents. 20 meters is 20%. Assign 100 to Lily. Then reduce Lily by 20% to get Betty. Then reduce again by 20% to get Louisa. 100 (0.80)(0.80) = 64 The difference is 36

I took an arbitrary time for lily to finish in. I said it took her 10 seconds to complete the 100m.

Therefore her speed was 10m/s. Beth had travelled 80m by the time lily had finished (100m-20m) in 10 seconds, making her speed 8m/s. Therefore to travel the additional 20m it would take her 2.5s to finish after lily, thus a total time of 12.5s.

At this time Louisa is at 80m, hence her speed is 6.4m/s and she takes an additional 3.125s on top of Beth's 12.5s making a total finish time of 15.625.

Now to work out how far behind she was when lily finished I took Louisa's total time away from Lily's total time to give 5.625s (15.625-12.5) I multiplied this time by Louisa's constant average speed of 6.4m/s to give the final answer of 36m - voila.

You can use any arbitrary number u wish, I just choose 10.

If Lily lets say travels at 10 m/s then Betty must be travelling at 8 m/s because after 10 seconds Lily will be done and Betty will be 20 meters off. Using this same reasoning Louisa must be travelling at 0.8(8 m/s) = 6.4 m/s. 6.4 m/s(10 s) = 64 m, 100 m - 64 m = 36 m

Runner A finishes and Runner B is at 80m meaning runner B's pace is 4/5 the pace of runner A. Runner B Finishes and runner c is at 80m meaning runner C runs at 4/5 the pace of B. so when A is finishing C is 4/5 of B at 80m therefore C is at 64m, 36m behind Runner A.

let us consider that lily ran at 10m/sec.this means that she took 10 seconds to finish the race.since it is given that all the 3 run at a constant speed throughout,we can say that betty ran at 8m/sec (since it is given that louisa beat her by 20m.that means at 10 seconds she would have finished 80 metres making her speed 8m/sec.) now running at 8 m/sec betty will take 100/8=12.5 seconds to complete the race. since it is also given that she finishes 20 m ahead of louisa,this means that in 12.5 seconds louisa has covered only 80 meters. that makes louisa's speed=80/12.5=6.4 m/sec. now when betty finished the race in 10 seconds louisa would have covered 64 meters(since she runs at 6.4m/sec).thus betty wins by(100-64)=36 m over louisa

ratio of their speeds- 25:20:16

lily beats bety by 20m
race 1
so when distance travelled by lily =100m
distance travelled by bety= 80m
similarly
race 2
distance travelled by bety=100m
distance travelled by lousia=80m
as they run with constant speed
we will find how much they ran when bety ran 400m
ditance travelled by lily=500m ; and
distance travelled by lousia=320 m
difference=180 m
as we increased the race 5 times ,we divide it by 5
180/5=36 km

Let the speeds be v1, v2 and v3. Now v1/v2=100/80, v2/v3=100/80, hence v1/v3=100/64 i.e answer is 100-64=36

let x be A's velocity, y, be B's and z be C's velocity....... when A finishes.... as time elapsed for three is same..... 100/x = 80/y = d/z.......... (i)....... d is the distance covered by C........ now when B finishes.......... 100/y=80/z...........(ii)...... thus, 80/d = 100/80 [from (i) and (ii)].......... d = 64m..... A beat C by 36 m

S=S0+VT

Lily - S=1.t

Betty - S=(1-20%).t S=0,8.t

Louisa - S=(0,8-20%).t S=0,64.t

SLily-Slouisa=0,36.t for tx D = 36m

v=d/t use 1sec for time. On 1sec: Lily's speed is 100m/s, Betty's speed is 80m/s (since she's on the 80m-mark).

When betty finished (covering 100m also), t=(100m)/(80m/s)=1.25sec. Louisa is on the 80m mark so (80m)/(1.25sec)=64m/s.

Thus in the first sec when Lily finished first, Louisa is on the 64m-mark. Lily beats Louisa by 36m.