Come to those who know the Mathematical Olympiad

1. Suppose the distance between the trams is 1, V1 is the speed of Xiao Wang, and V2 is the speed of Xiao Zhang. Then

V1+V car=1/9 V2+V car=1/8 2V car=1/6 V car=1/12

V1=1/9-1 /12=1/36 V2=1/24

S total=45×(1/12)=15/4 (15/4)÷(1/36+1/24)=54 points

2.400÷(400-360)=10

(10000-400×10)÷(400+18)≈14.3540

(10000- 360× 10)÷(360×5/4)≈14.2222

∴B comes first

The second question: B reaches the finish line first When the game started for 10 minutes in, A. took the lead. When B makes a complete circle, A has traveled 4,000 meters. After B has traveled 3,600 meters, A's speed is 418 m/min. The speed of B is 450 m/min. B can catch up with A in 12.5 minutes. can't finish the race in 12.5 minutes. The person who travels 6,000 m downhill, so B reaches the end point first.

The third question assumes that A's original speed is x and the return timeour is t, then the original speed of B is the speed is x+2

xt+( x+2)(24-t)=200

(x+2 )t+x(24-t)=200

Solve for x=14

So the original speed of A is 14 m/s

For the fourth question, suppose A's fastest speed is x, B's slowest speed is y, and the total distance is m, then

12x =12y+ m

4x+4y=m

Solving for x=2y means that the time it takes for y to run is 12 minutes (they meet once every 12 minutes), then x runs once. The time is 6 minutes

A and B are the two starting points of the tram. There is a tram every 12 minutes. The tram travels at 25 kilometers per hour. Please ask: (1) If Xiao Ming takes the tram from point A

11, how much is the rest?

A: (1-2/5)*(1- 2 /3)*(1-2/5)a

B: (1-2/3)*(1-2/5)*(1-2/3)b

12. If the trams pass each other every 4 minutes, wecan find out by drawing the chase diagram:

Both places must leave at the same time and 8 minutes apart

Suppose the total distance is 1, then:

We can know that the speed of the tram is 1/56, so the distance between the two oncoming trams is 1/7

At this time, Xiao Zhang met an oncoming tram every 5 minutes

That is: 1/7 of the distanceIt took 5 minutes for the tram and Xiao Zhang to complete the journey, the speed of Xiao Zhang can therefore be calculated: (1/7)/5-1/56=3/280

Similarly: Xiao Wang every 6 minutes Meet an oncoming tram

So where are Xiao Zhang and Xiao Wang Time? they had been walking when they met on the way

1/(3/280+1/168)=840/14=60 minutes

Here is my explanation of this idea and question I don't think I'm the first

But I hope the poster can see it clearly

Consider it appropriate

I hope it will be adopted

I hope that this will help you

Both places A and B will send a tram to the opposite side at the same time at the same speed, Mo

(1) 25×

=5÷50×60

=6 (minutes)

Answer: He sees a tram coming towards him every 6 minutes.

(2) 25×

=5÷30 ×60

=10 (minutes)

25×

=5÷20×60

=15 (minutes)

Answer:He will see a tram coming towards him every 10 minutes. A tram will pass. him from behind in 15 minutes.

Over the same distance, it takes 6 minutes for two trams to face each other. The trams and Mingming face each other. It takes 8 minutes to drive, and it takes 9 minutes for the tram and Lingling to face each other.

Let this distance s, the speed of the tram be d, the obvious speed is m, and the Lingling speed is L

Then

s/(d+d)=6 ①

s/(d+m )=8 ②< /p>

s/(d+l) = 9 ③

①/② Get m/d = 1/4

①/③ Get l/d = 1 /6

Therefore, Lingling and Mingming are together. The climb speed is 1/4+1/6=5/12

so the time it takes for the two to meet is 12. /5 times the total distance of the tram< /p>< p> =45*12/5 = 108 minutes.