Primary Math Problem ,。。。。。。 This is a primary school math problem

Answer] Typical pick-up questions.

Plan 1] Send a part of the people to 13 kilometers first, then return to pick up the other half of the people, and then both teams arrive at the same time.

When the car returns to pick up the other half of the people, the pedestrian line is 1 copy, and the car line is 60 5 12 copies. Instructions for car dealers (12 1) 2 copies returned.

The first to get off the bus also went 1 copy, indicating that the 15 kilometers were a total of 15 kilometers, and each was 15 kilometers.

It shows that the car traveled for kilometers and the people traveled for 2 kilometers, and the shared time was 13 60 2 5 37 60 hours 42 60 hours.

Plan 2] Send half of the other half of the person on foot, and then return to pick up the other half.

The journey takes 15 60 hours each way, and 15 60 3 45 60 hours if the other half does not walk.

People walked 1 distance, and it took 2 13 2 15 60 45 60 42 60 to travel back and forth, so it was feasible.

10 60 = 1 6 hours = 10 minutes.

It only takes 10 minutes to get the car to take four people to the train station and then return to pick up the other four.

The other four walked for ten minutes and were still 9 and 1 6 kilometers from the train station.

This distance is shared by cars and people, and it takes about a minute, which can be considered 9 minutes.

In other words, it takes exactly 9 minutes for cars and pedestrians to meet. It also takes 9 minutes for the bus to return to the train station.

Exactly 28 minutes, understand?

People can catch the train, as follows:

The remaining car carries four people to the railway station at a speed of 60 kilometres (i.e. 1 kilometre minute), stops at a distance of 2 kilometres from the railway station, disembarks the four persons, and then walks to the railway station at a speed of 5 kilometres (i.e. 1 12 kilometre minutes); The car returned the same way, carrying the remaining 4 people, the 4 people had already started walking at a speed of 5 kilometers (i.e. 1 12 kilometers minutes) when the car carried the first 4 people, and the 4 people met the car after walking 2 kilometers, i.e. 13 1 12 + 11 12 = 2 (kilometers), after which the 4 people took the car to the railway station.

According to the above:

The first 4 people arrive at the train station at (15-2) 1+2 (1 12)=13+24=37 (minutes)<42 (minutes).

The arrival time of the last 4 people at the train station is 2 (1 12)+(15-2) 1=24+13=37 (minutes) <42 (minutes).

So 8 people were able to catch the train.

Plan 1: Two cars go at the same time, 15 kilometers away from the railway station, the people on the faulty car are waiting in place, and the other car can be sent back to pick up the first 4 people, 15 * 2 60 = 05 hours = 30 minutes.

Option 2: Two cars go at the same time, 15 kilometers away from the railway station, the people in the faulty car go back at the same time, and the other car picks up after sending four people.

The answer to the previous one, the first part is sloppy. Note that "one of the cars broke down on the other side 15 kilometers from the train station," not "10 kilometers".

It only takes 15 minutes to have the car take four people to the train station (15 60 = hours = 15 minutes) and then return to pick up the other four.

The other four walked for 15 minutes and were still kilometers from the train station.

The car returns to pick up the 4 people, and this distance is shared by the car and the person, hours = minutes. It takes minutes for a car to meet a pedestrian. The bus returns to the train station, which of course takes minutes.

So the total time is 15+2* minutes. Fortunately, there was no delay.

Car: 240 km, 3 hours, 80 km.

Motorcycle: 150 km, 2 hours, 75 km.

The sedan is fast, the time ratio is 3:2, the distance ratio is 8:5, and the speed ratio: 16:15

Sedan 240km 3 hours 1 hour 80km

Motorcycle: 150km, 2 hours, 1 hour, 75km

The denominator is 37*29, and the above numerators can be added, that is, 20*9+9*17=9*(20+17)=9*37, which becomes 9*37 divided by 37*29, which is equal to 9/29

20 37*9 27+9 37*17 29 Idea: Treat the denominator 37*29 as an integer, so that the denominator on both sides of the plus sign is the same! You only need to add the numerator to solve this problem.

Answer: Primitive formula = (20*9+9*17) (37*29) = "9*(20+17)" (37*29).

You can change the positions of the numerators 9 and 17 after the plus sign, and the multiplication of the numerators does not affect the result, so that there are 9 29 in the two formulas before and after the plus sign, and it becomes (20+17) 37=1, so the final result is 9 29

After the denominator is formed, the denominator on both sides of + is the same.

The denominator of the left and right parts is 37x29, and the left side of the numerator is 20x9 and the right side is 17x9, so the original formula = (20x9+17x9) 37x9=37x9 37x29=9 29

The line is required to complete the journey; 3 hours.

47（412-365）

3.Minimum 82 pieces (respectively) 4Minimum of 5 boxes.

Adding a 2 to the end of the 2 is the largest number.

The first five numbers are 315, and the sum of the five numbers is 315, plus 50 = 365, so the smallest number is 50

(1)(5x51)+3x(2+3+4+..52)=4386(2)531 47

3) Any three bags and 60, i.e. at least 20 per bag, so a total of 80 (4) 81 balls in 10 boxes, 8 in each box, and 1 more ball, with the same number of balls, at least 10, 10, 9, 9, 8, 8, 8, 7, 6, 6

Is this sure to be an elementary school topic? I'm already learning the number sequence, and oh yes, I'm not out of date.