A primary school problem, a fifth grade primary school problem

The above methods are all algorithms with equations, 75=2100 (meters) It is also the deformation method of equations, which is not practical for elementary school students who are not good at equations, and the general primary school students feel that the reasoning learning method is easier to use and easy to understand than equations, so I will use a reasoning method to solve this problem!

Since the students in grades 4 and 5 have already learned the greatest common divisor and the least common multiple, I use this method to deduce the correct answer:

From the question, it can be seen that whether the speed of 60 meters minutes or 75 meters minutes is the same distance, (that is, the distance from the addition to the school), so as long as the same distance is found, and the time is in line with the topic (one is 2 minutes early, one is 5 minutes late, and the difference is 7 minutes), the distance can be found, so the greatest common divisor of 60 and 75 is required: 60*5=300;75*4=300；

So the answer is as follows:

When the distance between school and home is 300*1 meter, the former takes 5 minutes and the latter takes 4 minutes, a difference of 1 minute, which is not in line with the topic.

When the distance between school and home is 300*2 meters, the former takes 10 minutes and the latter takes 8 minutes, a difference of 2 minutes, which is not in line with the topic.

When the distance between school and home is 300*3 meters, the former takes 15 minutes and the latter takes 12 minutes, a difference of 3 minutes, which is not in line with the topic.

When the distance between school and home is 300*4 meters, the former takes 20 minutes and the latter takes 16 minutes, a difference of 4 minutes, which is not in line with the topic.

When the distance between school and home is 300*5 meters, the former takes 25 minutes and the latter takes 20 minutes, a difference of 5 minutes, which is not in line with the topic.

When the distance between school and home is 300*6 meters, the former takes 30 minutes and the latter takes 24 minutes, a difference of 6 minutes, which is not in line with the topic.

When the distance between school and home is 300*7 meters, the former takes 35 minutes and the latter takes 28 minutes, a difference of 7 minutes, which is in line with the topic.

Therefore, the distance between school and home is 300*7=2100 meters.

The mathematical method can be written like this:

Solution: According to the problem, the greatest common divisor of 60 and 75 is 300, and the time difference between the two velocities is 5+2=7 (minutes).

So, the total distance is 300*7=2100 (meters).

A: The distance from home to school is 2100 meters.

Because walking 60 meters per minute, you have to be 5 minutes late; Walking 75 meters per minute, you can advance by two minutes, the difference between the front and back is 2 + 5 = 7 (minutes), and the speed difference is 75-60 = 15 meters; If you walk at a speed of 75 meters as much time to get to school as 60 meters, then 75*7=525 meters will take a total of 525 15=35 (minutes) and the total distance will be 60*35=2100 (meters).

2100 meters.

60*5=300 meters (300 meters from time to time).

75 * 2 = 150 meters (if the time is up to walk 150 meters) 300 + 150 = 450 meters (difference of 450 meters).

75-60 = 15 meters (15 meters more per minute) 450 15 = 30 minutes (if you just arrive, it will take 30 minutes) You can choose: 30 + 5) * 60 = 2100 meters.

Or. 30-2)*75=2100 meters.

Column equations. Solution: Set the time to x minutes.

60*(x+5)=75*(x-2)

60*+300=75x-150

15x=450

x = 3060 * (30 + 5) = 2100 (m) or 75 * (30-2) = 2100 (m).

2100 meters.

Solution: Set the time to x minutes.

75*（x-2）=60*（x+5）

75x-150=60x+300

75x-60x=300+150

15x=450

x = 3060 * (30 + 5) = 2100 (m).

The time of first encounter between A and B is 30 divided by (10-5) = 6 seconds and every 90 divided by (10-5) = 18 seconds from now on.

The time of the first encounter of a b is 6, 24, 42, 60, 78, 96, and the time of the first encounter of b and c is 30 divided by (5-3) = 15 seconds and every 90 divided by (5-3) = 45 seconds from now on.

b c The meeting time is 15,60,105,150,195, so the 3 reptiles arrive at the same location for the first time 60 seconds after departure.

90 divided by 10 for 9 seconds;

90 divided by 5 for 18 seconds;

90 divided by 3 for 30 seconds;

The least common multiple of is 90 seconds;

So it takes 90 seconds to reach the same location for the first time.

It takes 90 10 = 9 seconds for reptile A to crawl in one circle; Crawler B climbs 1 lap to 90 5 = 18 seconds; Crawler C climbs 1 lap to 90 3 = 30 seconds. The least common multiple of 9 seconds, 18 seconds, and 30 seconds is 90 seconds.

A: It takes 90 seconds for them to climb to their original position for the first time.

Let the three points A, B, C, and the three points be divided into equal circumferences, point A in the front, point B in the center, point C behind the temple, point B in the front 30 cm of point C, point A in the direction of 30 cm behind point C, and the three insects climb forward clockwise at the origin point at the same time. After 30 seconds: the C-worm crawls in a full circle (90 cm) to return to point C; The B worm climbed 2 laps, missing 30 cm (150 cm), and also climbed to point C; The worm crawled 3 times and 30 centimeters (300 centimeters) more, and also climbed to point C.

So the answer to this question is: it takes 30 seconds for them to reach the same location for the first time, and then they meet again at point C every 90 seconds.

If 3 starts in the order of 1st, 5nd, 2nd, 10th, 3rd, your logic is problematic.

Because they set off at the same time, it took the same amount of time, and let the three unknown quantities represent the distance they would have traveled when they met, and the equation was solved.

This is a problem for finding the least common multiple. Find the least common multiple of [8,10], the common multiple is 40, so at least another 40 minutes will depart at the same time, which is 6:40.

This question requires the least common multiple of 8 and 10, and the least common multiple of 8 and 10 is 40, that is, the train departs at the same time after 40 minutes, which is 6:40.

8=2*2*2

The least common multiple of 8 and 10: 2*2*2*5=406:00+40 points=6:40

After another 40 minutes, it was 6:40

[8,10]=40

6 hours + 40 minutes = 6 hours and 40 minutes.

At least 40 minutes later, the train departs at the same time, at 6:40 a.m.

The least common multiple of 8 and 10 is 40, which means that the train departs at the same time after 40 minutes, which is 6:40.

It is the least common multiple of 8 minutes and 10 minutes, and the least common multiple of 8 and 10 is 40, which is 6:40

8x=10y (xy is an integer) can be obtained when x=4, y=5 the equation holds, so that 8*5=40 minutes later at the same time.

8=2*4

The common multiple is 2*4*5=40

40 minutes later, the bus departs at the same time.

It was 6:40 a.m.

Easily, let's look at the general first.

1/（n-1）n(n+1)=1/2(n-1)+1/2(n+1)-1/n

So the original formula 1 2(1 1 3) 1 2 1 2(1 2 1 4) 1 3 . 1/2（1/28＋1/30）－1/29＝1/2－1/4－1/58＋1/60＝4/15－1/58

What is the equal? I do not know.

I only know that the result is less than 1

After selling 50 kg of rice, the remaining rice and millet are 1 in total, of which millet occupies 6 11, rice station 5 11, after selling 55 kg of millet, the remaining millet is 1 6 of rice, 1 of 5 11 * 1 6, indicating that 55 kg of millet sold occupies 6 11-5 11 * 1 6 = 31 66

The original total amount of rice and millet was 55 (31 66) = 55 * 66 31, and the original warehouse was 55 * 66 31 + 50

Zhao = money - 20-10

Sun = (money-20) 2

Lee = (money-20) 2

Money = (370 + 20 + 10 + 20 2 + 20 2) (1 + 1 + 1 2 + 2).

100 (trees).