Grade 7 Mathematics uses equations to solve travel problems

Answer: Minutes and hours.

The car breakdown point is 15km away from the airport, and only 5 people can be sent each time, and the average speed of the car is 60km h

1) After the car sends off the first batch of people, the second group waits for the car to return to pick up and drop off;

It took 15 to 60 hours for the car to take the first group of people to the airport;

The time it takes for the car to return to pick up the second group of people and then to the airport is 15 2 60 hours;

Shared time hours hours.

Therefore, it is not possible to make all 8 fans rush to the airport within the specified time.

2) While the car sends off the first group of people, the second group walks towards the airport at an average speed of 5km h, and then gets on the bus when they meet the returning car on the way.

It took 15 to 60 hours for the car to take the first group of people to the airport;

At this time, the second group of people was 15 5 km from the airport;

The time it takes for the car to return to meet the second group of people hours;

At this time, the second group is km away from the airport;

The time it takes for the car to pick up the second group of people and then get to the airport is 165 13 60 11 52 hours;

Shared time hours hours.

Therefore, all 8 fans can rush to the airport within the specified time.

Therefore, only the second method can get all 8 fans to the airport within the allotted time.

2. (1) It takes 9 seconds for two trains to travel in opposite directions, and it takes 9 seconds to stagger from meeting each other (two locomotives meet each other and leaving at the tail of two trains).

The length of the two trains is 144 meters and 180 meters respectively.

The distance traveled by the two trains is 144 180 324 meters.

The speed of two trains and 324 9 36 meters in seconds.

Car A travels 4 meters more per second than Car B.

A car speed (36 4) 2 20 m sec.

B car speed (36 4) 2 16 m sec.

2) If you are traveling in the same direction, the front of vehicle A will chase from the rear of vehicle B to vehicle A and all of them will exceed vehicle B.

At this time, it is quite stationary with car B, and car A walks at a speed of 4 meters and seconds the length of car A and car B.

Time taken (144 180) 4 81 seconds.

3. The speed of setting the armor is x kilometers per hour.

45 minutes hours.

The distance between the two places is 1 kilometer, A arrives at B and stays for 45 minutes (B has not yet reached B), and the speed of car A is 3 times the speed of car B and 1 km more.

Then they returned from B to A and met B on the way, which was 3 hours after their departure.

A walking time of 3 hours;

B has a speed of (x 1) 3 km/h.

The distance traveled by A and B when they meet is twice the distance between A and B.

x 16 km h.

x 1) 3 (16 1) 3 5 km hours.

Therefore, the speed of A is 16 kmh, and the speed of B is 5kmh.

1:15 60=,15 scheme 1:15+15+15=45, so no.

15/60=,15m.5*, 13m 15+13+13=41 Option 2 is OK.

2: Let B be x, then A x+4, (144+180) (x+x+4)=9 , x=16Then A is: 20

144+180) 4=81 seconds.

3: Let B speed x, then A's 3x+1; 3-[ Touch x is sufficient.

Drawing a line segment diagram, a mathematician said that drawing a line segment diagram is a bridge to solve the problem During the eleventh day, the engineering team was going to spend three days to repair the road, and on the first day it repaired three-quarters of the road, and the second day completed the remaining one-third, and there were still four problems left. Q, how many meters is the road?

4/（1-3/1）=6m 6/ (1-4/3) =24m

Itinerary question: the sum of the distances = the total distance.

Given the unknowns, it will be much easier to solve them with equations

Drawing a line segment diagram, a mathematician said that drawing a line segment cave travel diagram is a bridge to comic and solve problems.

During the eleventh day, the engineering team was ready to spend three days repairing the road, and on the first day, three-quarters of the road was repaired, and the remaining third was completed on the second day, leaving four questions. Q, how many meters is the road?

4/（1-3/1）=6m24m

First: (50 + 70) x2 = 240 (m) Calculate the distance between A and B when C meets B, and use 240 (60-50) = 24 (minutes), which is the time taken by C and B to walk a whole journey, 24 (70+60）=

Solution: Let the time taken by C and B to meet be t hours.

70t+60t=50(t+2 60)+70(t+2 60) The solution is t=

So the distance between a and b is multiplied by 70 = 28 meters.

Answer...

The whole journey is x kilometers and takes t t hours.

20*t = x

18 = x

The equation is solved by itself, too lazy to calculate.

Solution: If you start to travel 18xkm, the remaining distance is (.

18x-（

Tailwind: x 24 2 hours 50 minutes = 17 6 hours (x+24)*17 6

Headwind: X-24 3 hours (X-24)*3

Equation, (x+24)*17 6= (x-24)*3, solution x=8402In this question, if the total workload is considered as 1

Students in the first year of junior high school can complete the total workload by working alone for 1 hour 1 5 Students in the second year of junior high school can complete the total workload by working alone for 1 hour.

The amount of work that students in the first and second years of junior high school work together for 1 hour can be expressed as 1.

The remaining amount of work completed by students in the second year of junior high school for x hours alone is expressed as x 5.

Solution: Let the second junior high school students complete the remaining part of the equation for x hours alone: 1