The questions about the Olympiad itinerary need to be explained in detail by the teachers, and the a

Question 1: Analysis: This kind of question should take the time taken to travel the distance between a and b as a unit time, and according to the question, the distance between two cars and a b is 3 a b, so it takes 3 units of time.

When the two cars met for the second time, they traveled a total of 3 A B distances, and both took 3 units of time. ）

According to "the first time the two cars met at a distance of 32 kilometers from place A", that is, car A traveled 32 kilometers in one unit time, then he traveled three units of time, and must have traveled a total of 32 3 96 kilometers.

The distance between the two places is (it is easy to see from the line segment diagram): 64 96 160 km.

Naturally, the distance between a a b and two places is:

160 2 80 km.

If you don't understand it, you can draw a line diagram to cooperate with my analysis.

Question 2. 2005+2006) 4 Yu 3

2005+2007) 4 surplus 0

2005+2008) 4 surplus 1

2006+2007) 4 more than 1

2006+2008) 4 Yu 2

2007+2008) 4 Yu 3

If you look closely, it's easy to see that the 2008 Athlete Race is over

1 2 3 6 (field).

When AB meets, it is point 1 and 32km away from place A, and car A and car B travel a whole way, and car A can only travel 32 Suppose that car A and car B start from point 1 to AB and then return and meet at point 2, this is the two cars A and B have traveled 2 whole journeys, that is, car A has traveled 64 kilometers, and car B has traveled 96 kilometers (he has traveled from point 1 to place A 32, and from place A to point 2 64 kilometers) (64+96) 2=80

From the first encounter and the second encounter, it can be concluded that the two people walked a total of three courses, A walked 32 kilometers in one course, and the three courses walked 32*3=96 kilometers. 64 km for the last two whole journeys = 32 + 16 * 2 km. So a whole journey is 32 + 32 + 16 = 80 kilometers.

Solution: Let the velocity of A x m seconds, B y m seconds, then:

5（x-y）=10

4(x-y)=2y

Solution: x=6

y=4A: A6, B4

Let the velocity of A be x and the velocity of B be y

A system of binary linear equations can be listed by using the problem.

5x=5y+10

4x=6y can be solved to x=6y=4

So the velocity of A is 6m s and the velocity of B is 4m s

First walk 4 parts, 40 parts of the car line, the difference is 40-4 = 36 parts, then walk another 36 (50 + 4) 4 = 8 3 parts of walking, a total of 4 + 8 3 = 20 3 parts.

Full length 40 + 20 3 = 140 3 parts.

The ratio of walking distance to the whole journey is 20 3:140 3 = 1:7

The fuel used by the truck on the way should be as much oil as it poured into the other car as it would use when it returned. So it can only go up to 150 3 = 50 km. In other words, the fuel it poured into another car could be used for 50 kilometers.

Therefore, the farthest distance from the headquarters of the detachment may be 150 + 50 = 200 kilometers.

Set car A as a document car and car B as an oil delivery vehicle.

1.For A, the problem needs to be considered is the fuel consumption A on the road and the fuel supply B of car B. Since the maximum volume of a mailbox is fixed at 150, the equation can be obtained:

150-A+B <= 150, B<=A, when B=A is the maximum value, that is, the fuel tank is full of 150 during driving

2.For B, the problem that needs to be considered is that the one-way fuel consumption c on the road and the amount of fuel B to supply A, at the same time, the fuel consumption of B needs to be considered round trip, so, the fuel consumption of B is 2C, and the equation is obtained

2c+b<=150.

3.From the meaning of the title, it can be seen that B supplies nail polish at a certain point, so a=c, we get the equation:

150-a+b=150

2a+b=150

A=b=50, so the farthest distance from the detachment is 150+50=200 kilometers.

After supplying fuel halfway and returning safely to the station, he can only supply 150 3 = 50 km of oil;

The farthest distance from the headquarters to the detachment station may be 150 + 50 = 200 kilometers.

Let the car that sends the document be A, the refueling car is B, the distance that car A can travel is Z, and B will give A the amount of fuel that can travel Y kilometers after driving x kilometers, then there is for B: 2X+Y=150;For a: 150+y=z

Finding the maximum value of z is to find the maximum value of y, because when both cars are driving x kilometers b to refuel a, they cannot add more fuel than the distance of x kilometers that can be traveled, so x>=y, from 2x+y=150, the maximum y = x=50, so z is 200 maximum

The distance from Mr. Liu's home to his unit is x kilometers.

Then there is: [(x 3) 8+(2x 3) 16-1 2]*(3 8*8+5 8*16)=x

x/12-1/2=x/13

x=12*13/2

x=78km

Solution: The speed ratio of cycling to riding is 1:2

Going to the unit is less than going home and riding the whole journey (1-5 8)-1 3=1 24 car, so the ratio of cycling time to riding on the bus during this journey is 2:1, [note that the distance remains the same, and the time is inversely proportional to the speed].

During this time, I took a car when I went and rode a bicycle when I came back, so the time ratio before and after this journey was 1:2, so it took 2 minutes to take a car when I went to this distance, [obtained by the proportional relationship (2 (2-1))*1].

So this journey takes 2 (1 24) = 48 minutes, so the total distance is 16 * 48 60 = km.

3/8-1/3=1/24 ……The distance to work is less than that to go home (i.e., 1 24 This section is a ride when you go to work, but you ride a bike when you go home).

8:16=1:2 ……The speed ratio of riding to riding.

2:1 ……The ratio of the time spent riding to the car when traveling the same distance.

2 (2-1) = 2 minutes ......The time of each serving.

2*2=4 minutes ......The time it takes to ride a bicycle for this 1 24 distance.

8*4 60 (1 24) = km ......The distance from Mr. Liu's home to the unit.

or 16 * 2 60 (1 24) = km.