A, B and C race in three people, and after starting from the starting point at the same time, A is e

Analysis: According to the title, as shown in the figure below, when A walks to B, B walks to D, and C goes to C, A is 3 kilometers faster than B, and B is 2 kilometers faster than C, so DB is 3 parts, and CD is 2 parts, B takes 10 minutes to walk 3 parts, and C takes 20 minutes to walk 5 parts, so the speed ratio of B and C is 6:5;

B's speed is 1 part more than C's speed and 2 kilometers per hour, so B's speed is 12 kilometers per hour, C's speed is 10 kilometers per hour, C's time is 10 (6-5) 6=60 (minutes) = 1 hour, and the race distance is 10 1=10 (kilometers).

Answer: Solution: The velocity ratio of B and C is:

The velocity of C is:

10 (6-5) 6,10 1 6,60 (min),1 hour;

Race distance: 10 1 = 10 (km).

A: The race is 10 kilometers

So the answer is: 10

Comment: The key to solving this question is to find C's speed and time, and then solve the problem according to "speed time = distance".

Let the total length be l(km), the velocity of A is v(km h), the time t (h) is (v+3) (km h) of B, the time taken is (t+1 6)(h), and the velocity of Bing is (v+5) (km h), and the time taken is (t+2 6)(h).

According to the title: v*t=l

v+3)*(t+1/6)=l

v+5)*(t+2/6)=l

The above equation is solved by synthesis: v=15km h; t=2/3h;The whole journey l=10km;

No equations won't...

Let the distance be s, C time t, C velocity v, and the equation.

s=vt=(v+5)(t-20)=(v+2)(t-10)vt=vt-20v+5t-100 20v=5t-100

vt=vt-10v+2t-20 10v=2t-20 20v=4t-40

5t-100=4t-40 t=60v=10

s=vt=600

As shown in the figure, when A walks to B, B walks to D, and C goes to C. A is 3km faster than B per hour, and B is 2km faster than C, so DB is 3 parts and CD is 2 parts. It takes 10 minutes for B to walk 3 parts, and it takes 20 minutes for C to walk 5 parts, so the speed ratio of B and C is 6:5

B's speed is 1 part more than C's, 2km/h, so B's speed is 12km/h and C's speed is 10km/h. 1/6÷（1/12 - 1/15）=10

I won't do without equations, I'll make an equation, I hope it will inspire you.

Let the distance be s, the velocity of C is v, the velocity of B is v+2, and the velocity of A is v+2+3

s/(v+5)=s/(v+2)+1/6=s/v+1/3

A's velocity: B's velocity=5:4=15:

20, the velocity of B: the velocity of C = 5:4=20:

16, therefore: the speed of A: the speed of B:

C's velocity = 15:20:16, which is the distance of A's run and the key and :

The distance of the B run: the distance of the bright bend C = 15:20:

16 = 75: 100:80, that is, when B runs to the finish line, A runs 75 meters, C runs 80 meters, 100-75 = 25 (meters), 100-80 = 20 (meters), staring.

Answer: When B reaches the finish line, A is still 25 meters away from the finish line and C is 20 meters away from the finish line

10 minutes = 1 6 hours.

Let the velocities of A, B, and C be x, y, z(km) respectively, and the total distance is s(km), then there is s y-s x=1 6

s/z-s/y=1/6

x-y=3y-z=2

Solve the above 4 equations to get x=15 y=12 z=10s=10(km).

A catches up with B 500 minutes after departure. C speed is more than B speed = 4:5, C speed is more than A speed = 10:13, so A speed is better than B speed = 26:25. Let A start x minutes to catch up with B, then 26x=25(20+x), and the solution, x=500

B walks for 40 minutes is equivalent to C walks for 50 minutes, the speed ratio of B and C is 5:4 A walks for 100 minutes is equivalent to C walking 130 + 30 = 130 minutes, and the ratio of the speed of A and C is 13:10

The speed ratio of A and B is 13 10 : 5 4) =26:25 A catches up with B in 20 26 25-1) =500 minutes.

3 people have to list 2 formulas, so there must be 2 unknowns.

Let the speed of Bingjun x, and the time of Bingjun t

xt=（x+2）（t-10）=（x+2+3）（t-10-10）

The speed ratio of B and C is:

The velocity of C is:

10 (6-5) 6,10 1 6,60 (min),1 hour;

Race distance: 10 1 = 10 (km).

A: The distance of this race is 10 kilometers of Chi Zhi Qi Min.

So the answer is: 10