Spring is a good season for planting trees. Students, can you also participate in afforestation activities every year? Beautify and green your home, have you found that there is a lot of knowledge in planting trees? Because the route of planting trees is different, the situation of planting trees is different. Do you want to know about tree planting and learn how to solve the problem of tree planting? Welcome to join the column of Mathematics Garden to study the problems you want to solve. Please look at the following example.

Example 1: There is an expressway with a length of 1000 meters. How many weeping willows can be planted every 5 meters on one side of the expressway?

Analysis: First of all, the distance between two weeping willows should be taken as the cutting standard. The total length of the highway can be divided into several sections, that is, 1000m contains how many 5m, and 1000m ÷ 5 = 200 (sections). Because trees need to be planted at both ends of the highway, the number of trees planted is more than that of the divided sections 1, that is, 200 trees.

Solution: (1) Take 5m as a section, and the total length of the highway can be divided into: 1000÷5=200 (section).

(2) The number of weeping willows is 200+ 1=20 1 (plant).

Comprehensive formula:1000 ÷ 5+1= 201(tree)

Answer: 20 1 weeping willows can be planted

The distance between the two buildings is 56 meters. How many cedars can be planted every 4 meters?

Analysis: Taking the distance between two cedars as the standard of segmentation, the length between two buildings can be divided into several segments, that is, how many 4 meters are included in 56 meters, 56÷4= 14 (segment).

The difference between this problem and the example 1 is that trees don't need to be planted at both ends (because trees can't be planted at the roots of buildings), so the number of cedars to be planted is less than that of the divided road sections 1, 14- 1 = 13 (trees).

Solution: (1) Take 4m as a segment, and the number of segments that 56m should be divided into is: 56÷4= 14 (segment).

(2) The number of cedar plants: 14- 1= 13 (plants)

Comprehensive formula: 56÷4- 1= 13 (tree)

A: You can plant 13 cedars.

Example 3: The circumference of a freshwater lake is 1, 350 meters. Plant a willow every 9 meters, and plant two peaches between two willows. How many willows can you plant? How many peaches can you grow? How many meters is there between two peach trees?

Analysis: when planting trees in a circle, the number of plants that can be planted is equal to the number of segments, because the first species will overlap with the last one in turn; Because two willows are planted at the same distance, 2&127; The number of trees with peach branches is equal to the product of 2 times the number of nodes; How many meters should be separated between two peach trees? You should know that two peach trees are planted at equal distances between two willows, that is, there are three equal distances between four trees.

Solution: (1) Divided into a section of 9 meters, the number of sections that can be divided around the water lake, that is, the number of willows planted:

1350÷9= 150 (strain)

(2) Number of peach trees: 2× 150=300.

(3) The total number of carambola in each section is: 2+2=4 (plants)

(4) Four plants were planted at a distance of 9 meters, and three plants were equally spaced. The distance between every two plants is:

9(4- 1)= 3 (m)

Comprehensive formula: (1)1350 ÷ 9 =150 (plant)

(2)2×( 1350÷9)=300 (plant)

(3)9÷(2+2- 1)=3 (m)

Answer: 150 can plant willows; Can plant 300 peach trees; Every two peaches are 3 meters apart.

Example 4: Grade 3 students of Guanghua Road Primary School 125 participated in the entrance ceremony of the sports meeting. There are five of them in a row, the distance between them is 2 meters, and the chairman is 42 meters. How many minutes does it take them to pass the rostrum at a speed of 45 meters per minute?

Analysis: On the surface, this problem is completely different from the previous example, but in essence it is the inverse problem of planting trees. According to the total number of people participating in the sports meeting in grade three and the number of people in each line of the topic, how many lines can be arranged in grade three? Each row is equivalent to the known number of trees, and the interval between each row is 2 meters, which is equivalent to the distance between every two trees, so that the total length of the team can be calculated; Then add the length of the team to the length of the rostrum, that is, the distance that everyone has traveled through the rostrum, and then divide the distance by the speed of travel, and you can find out the time needed to pass through the rostrum.

Solution: (1) The number of lines of the third grade March is: 125÷5=25 (lines).

(2) The total length of the three-level entrance team is 2×(25- 1)=48 (m).

(3) The total length of the third-grade admission team plus the length of the podium, that is, the distance that each person walks through the podium is: 48+42=90 (meters)

(4) The distance through the rostrum is 90÷45=2 (minutes).

Comprehensive formula: [2× (125 ÷ 5-1)+42] ÷ 45 = 2 (minutes).

A: It takes 2 minutes to pass the rostrum.

Example 5: A carpenter saws a 24-meter-long piece of wood into 3-meter-long pieces, and each saw takes 5 minutes. How many minutes * * *?

Analysis: To saw a 24-meter-long piece of wood into a 3-meter-long piece, it is required to saw several pieces first, that is, 24-meter-long bread contains several 3 meters, and 24÷3=8 (pieces). Because the last piece didn't need to be sawed, the carpenter sawed only 8- 1=7 (times), each time for 5 minutes, one * * *.

Solution: (1) The number of sawable segments of a 24-meter-long wooden strip: 24÷3=8 (segments)

(2) The number of times the saw is divided into 8 sections is 8- 1=7 (times).

(3)* * * Time required: 5×7=35 (minutes)

Comprehensive formula: 5×(24÷3- 1)=35 (minutes).

A: * * It takes 35 minutes.

Like the problems involved in the above five examples, we are used to calling them planting trees.

Key points to solve the problem of planting trees:

(1) When planting trees on non-closed lines (such as straight lines, semi-circles of broken lines, etc.). ), since a tree can be planted at both ends, it should be more than the number of segments to be divided 1, and the number of plants = number of segments+1= total length ÷ plant spacing+1.

(2) If trees have been planted at both ends (or there is no need to plant trees at both ends) and then trees are planted between trees, then the number of trees planted should be 1 less than the divisible number of segments, and the number of trees = number of segments-1= total length-1.

(3) Planting trees on closed lines (such as circles, squares, rectangles, closed curves, etc.). ), because the head and tail overlap, so the number of trees is equal to the number of separable segments. Number of trees = Number of segments = Total length ÷ Plant spacing

practice

1. There is a 2000-meter-long highway. How many poplars do you need to plant from beginning to end?

2. A circular flower garden is 40 meters long, and a red flag is inserted every 4 meters around it, and a yellow flag is inserted between every two red flags. How many red flags and yellow flags have been planted around the garden?

There is an 800-meter-long walkway in a university, from the gatepost at the school gate to the wall of the teaching building. How many buttonwood trees do you need to plant every 5 meters?

There is a lamp post 50 meters apart on one side of the road. Xiaojun took a trolley bus for 2 minutes and saw a 2 1 lamppost on one side of the road. How many kilometers does the trolley bus travel per hour?

To celebrate New Year's Day, there are 52 floats inspected, each of which is 4m long and 6m apart. They all travel at a speed of 50 meters per minute. How many minutes does it take for this train to pass through a 536-meter-long parade ground?

The answer to the exercise:

(1) (2000 ÷ 5+1) × 2 = 802 (tree)

(2) 40÷4= 10 (surface) red flag, and every other red flag is inserted with a yellow flag, so there are as many yellow flags as red flags, which is 10.

(3) (800 ÷ 5-1) × 2 = 318 (tree)

(4)50×(2 1- 1)÷2×60 = 30000(m)= 30k m。

(5) [4× 52+6× (52-1)+536] ÷ 50 = 21(minutes)