A few fourth grade math problems, and a few fourth grade math problems

1。Set the length of the team x meters.

x (120-80) + x (120 + 80) = 12 to get x = 400 meters.

2.Let the sum of the speeds of the two cars be x

3x+200+300=8x

3x+500=8x

3x+500-3x=8x-3x

500=5x

500÷5=5x÷5

100 = x8x = 8 100 = 800 km.

3.Let C's workload be x in one day, then A's is 3x, and B's is 2x, and the total amount of work is considered as 1

5（x+2x+3x）=1÷3

The solution is x=1 90

A rested for 3 days, 2 rested for two days, then only C worked for the first two days, completed 2x=2 90, then B worked with C for one day, completed x+2x=3 90, and the remaining three people did it together, leaving 1-1 3-2 90-3 90=55 90, and the three of them together could complete 6x=6 90 every day, then the rest needed 55 90 divided by 6 90 to get 9 and 1 90, that is, the rest would take 10 days to complete.

Then it will take 5+3+10=18 days to complete from the beginning.

1) A is the time from the end of the team to the head of the team, B is the time from the head of the team to the end of the team, and L is the length of the team.

120-80）*a=(120+80)*b=la+b=12

Solve l=400

2) Let the combined speed of the fast train and the slow train be v and the distance between the two places l

8v=l8v-3v=200+300

Solve l=800

Question 1: (120-80) 12=480

Question 2: 200 3+300 3=1500

1) A is the time from the end of the team to the head of the team, B is the time from the head of the team to the end of the team, and L is the length of the team 120-80) * A = (120 + 80) * B = la + B = 12

Solve l=400

2) Let the combined speed of the fast train and the slow train be v and the distance between the two places l

8v=l 8v-3v=200+300

Solve l=800

1。Set the length of the team x meters, and the equation can be listed.

x (120-80) + x (120 + 80) = 12 to get x = 400 meters.

2.According to the title, the sum of the speeds of the two cars can be set as V.

8v=3v+200+300

If the solution is v=100, the distance between the two places is 8v=800 km.

3.If this question is a fourth-grade problem, it is recommended to solve it in parts.

Let's first let C's workload for one day be x, then A's is 3x, and B's is 2x, and the total amount of work is considered as 1, then 5 (x+2x+3x)=1 3 solves x=1 90, A rests for 3 days, 2 rests for two days, then only C works for the first two days, and completes 2x=2 90, and then B and C work together for one day, and completes x+2x=3 90, and the remaining three people do it together, and the remaining three people do it together, and the remaining 1-1 3-2 90-3 90=55 90 can complete 6x=6 90 every day, then the rest needs 55 90 divided by 6 90 gets 9 and 1 90, which means that the rest will take 10 days to complete.

Then it will take 5+3+10=18 days to complete from the beginning.

202 divided by 2 minus 1) Dong Jing multiplied by 5 equals 500 meters.

24-20 + 5 = 9 hours (total standing time) 9 (6 2) = 3 hours (average group standing time) 24-3 = 21 hours (rest time).

Comprehensive column: 24-(24-20+5) (6 2)=21 (hours).

The number that two people can get before each guarantee is 1+6=7

32-1) 7=4 remainder 3

So take 3 pieces at a time for the first penitential stocking.