There is a question D of the primary school itinerary, and I am eager to explain it in detail

Solving ternary equations can be a little too difficult for elementary school students.

Now since the place where the two people meet is relatively close to point C, and the speed of the descent is the same, we might as well assume that CD is x kilometers longer than AB.

Then we discuss the origin of the 9-minute gap, we can know that it is caused by the difference in the distance between two people up and down, A walked X kilometers uphill, B walked X kilometers downhill, so there is the following equation.

x/4-x/6=9/60=3/20

x = km. Then we can talk about the length of the flat land.

Because the distance between the two sections is kilometers apart, A arrives on the flat land an hour earlier than B.

During this time, A can walk 5 * kilometers, after walking this way, B also reached the flat land, and then after a period of time the two met, after B arrived at the flat land, the distance of the two people is the same, both are 1 6 of the flat land, so the kilometer is the 2 3 of the flat land, that is to say, the distance of the flat land is kilometers.

Finally, we have to discuss the length of the slope, because it took two people an hour from the beginning to the meeting, and because the length of the flat land is already known, the length of the slope is very simple.

The distance that A walks on flat ground is kilometers, and the time it takes is hours, that is, the time it takes for him to go uphill and downhill in AB is hours, and the distance he can walk during this time is kilometers, so AB is kilometers long.

And because CD is kilometers longer than AB, the length of CD is kilometers, so the total length of the three segments is kilometers.

Let ab=x, bc=y, cd=z, then:

At the time of the encounter, the time of A = x 6 + 4 5y 5 = the time of B = z 6 + 1 5y 5 = 1 hour.

Time for B to reach point a = x 4 + y 5 + z 6

Time for A to reach point d = x 6 + y 5 + z 4

So: (x 6 + y 5 + z 4) - (x 4 + y 5 + z 6) = 9 minutes = hours.

Solve the system of equations: x=,y=1 3,z=

The final answer may be a little incorrect......The calculation part still needs to be done by yourself.

Set the three sections of AB BC CD respectively xyz

x/6+y/5+z/4)-(x/4+y/5+z/6)=9/60

This is the correct answer, the result of the calculation is unknown.

James's mistake is known that EC is 1 of 5 !!

Set the three sections of AB BC CD respectively xyz

x/6+y/5+z/4)-(x/4+y/5+z/6)=9/60

Get: x= y= z=

The whole journey is km.

It's tiring to calculate

A and B have the same speed, both of which are neutral from A to D, why does B arrive 9 minutes first?

Because the length of the distance between the ab bc cd is not the same on the first floor Also, the distance between the BC should be 6 equal parts. (for the 2nd floor) 3rd floor-supporting tooth cs awesome! Hehe.

Coaches and vans traveled in 3 hours.

Passenger cars and trucks are separated by encounters.

Journey 105 (27 20-1) = 300 (km) Here's an example Let's say the speed of a passenger car is x km h and the speed of a train is y km h.

Inspired by the title: 5x+

5x) y = (solution x = 75 km h

Let the encounter time be x, then x 5 = solution x = 6 (825 6) * [6 (5+6)] = 75

Let's assume that the speed of the passenger car is x km h and the speed of the train is y km h.

Inspired by the title: 5x+

5x)/y = (

The solution is x = 75 km h

It's a matter of encounter! It's a pity I don't know, I'm sorry! I can't help you!

Six years? Will there be fewer conditions?

Haven't you learned binary equations?

The conditions under which the bike can go the farthest are: the front and rear tires are constantly adjusted, and finally they break together.

The front tyre consumes 1,5000 per kilometre and the rear tyre 1,3000 per kilometre.

The average tire consumption per kilometer traveled by the bicycle is (1 5000 + 1 3000) 2 = 1 3750

As a result, the bike can travel up to 3,750 km

3000+5000=8000

8000 2=4000 I don't need to explain the reason.

If car A takes x hours to travel from A to B, then car A takes x+x 2 hours to make a round trip between A and B.

When car A returns to the midpoint of A and B, that is, it takes x+x 4 hours, car B arrives at place B, that is, car B takes 5x 4 hours from A to B;

It can be seen that the speed ratio of A and B is 5:4

Then the time for cars A and B to reach the midpoint and place B respectively from the phase is: x 2 (2 + 4 5) = 5 x 28

So 5x 4- 5x 28=4

x=56/15

x+x 2=28 5 (hours).

That is, it takes an hour for car A to go back and forth between A and B.

Why is it calculated that the time for cars A and B to reach the midpoint and place B respectively from the phase to the meeting is: x 2 (2 + 4 5)??

Find the velocity of ab first, let a be the velocity of a and b be the velocity of b. From the meaning of the title, it can be seen that A>B, when B arrives at place B, A is at the midpoint of return. The system of column equations 1 a+1 (2*2a)=1 b, 4b+(4-1 a)*2a=1, the simultaneous solution of the two formulas gives a=15 56, b=3 14, and the time taken for a round trip is 1 (15 56)+1 (15*2 56)=

Let the speed of the passenger car be 7x kilometers per hour and the truck speed is 9x kilometers, so (7x+9x)*5=480*(1-1 6).

The solution is x=5;So the speed of the bus is 35 kilometers per hour.

The speed ratio of the two is 60:80=3:4c occupies the whole distance from place a3 (3+4) = 3/7Distance to place B accounts for 4/7 of the whole journeyIf B does not rest, then when A returns to C

B can be used for the whole process4/7 2 4/3 = 32/21

So, the whole process is14 80 (32/21-2 3/7).

= 1120 2/3= 1680 (m).

60 14 = 840 meters.

840 (60+80) = 6 minutes.

80 6 = 480 meters.

480 2 = 240 meters.

240 (80-60) = 12 minutes.

80+60) 12=1680 meters.

A and B are 1680 meters apart.

Just make a binary system of equations.

Let ac=x, bc=y

x/60=y/80

2y/60=2x/80+14

Find x=720 y=960

AB is 1680 meters apart.

Because: average speed = total distance Total time So: the distance from the bottom to the top can be regarded as 1, the time taken to go up the mountain is 1 6 = 1 6, and the time taken to go down the mountain is 1 12 = 1 12

Column : 1+1) (1 6+1 12)=2 3 12=2x12 3=8 (hours).

Speed is equal to the distance divided by the time, i.e. v=s t=(6+6) (1+

Let the distance from the bottom of the mountain to the top of the mountain be s and the average velocity be v, then v=2s (s 6+s 12)=8 km.

If the length of this road is S km, then there is a time to go up the mountain for S 6 hours; The descent time is s 12 hours, and the average speed is (s 6 s 12) 4 kmh according to the speed distance.

The speed is equal to the distance divided by the time, i.e. v=s t=[6+12] [1+1]=18 2=9km h[uncertain].