Three general reviews: decimals
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Three general reviews: statistics
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Three general reviews: location
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Three General Reviews: Divisions
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(distance. The formula is 284× 2 = 20 (km) 20× 2 = 40 (km) 40 ÷ (4× 2) = 5 (hours) 28 × 5= 140 (km).
(9) Reduction problem: We call it the reduction problem of finding an unknown application problem after knowing the results of four operations.
The key to solving the problem is to find out the relationship between each step change and the unknown quantity.
Law of problem solving: Starting from the final result, the original number is gradually deduced by using the operation (inverse operation) method opposite to the original problem.
According to the operation order of the original question, the quantitative relationship is listed, and then the original number is calculated and deduced by inverse operation.
Pay attention to the operation sequence when answering the restore question. If you need to add and subtract first, don't forget to write parentheses when calculating multiplication and division later.
For example, there are four classes in grade three in a primary school, 168 students. If four classes are transferred from three to three, from three to two, from two to one, and from two to four, then the number of students in four classes is equal. How many students are there in four classes?
Analysis: When the number of four classes is equal, it should be 168 ÷ 4. Take Class Four as an example. It transfers three people to Class Three and two people from Class One, so the number of people in the original four classes minus three plus two equals the average. The original number of class four is 168 ÷ 4-2+3=43 (people).
The original number of a class is 168 ÷ 4-6+2=38 (people); The original number of class two is 168 ÷ 4-6+6=42 (people), and the original number of class three is 168 ÷ 4-3+6=45 (people).
(10) Tree planting problem: This kind of application problem takes "tree planting" as its content. Any application problem of studying the four quantitative relations of total distance, plant distance, number of segments and number of plants is called tree planting problem.
The key to solving the problem: to solve the problem of planting trees, we must first judge the terrain and distinguish whether the graph is closed, so as to determine whether to plant trees along the line or along the perimeter, and then calculate according to the basic formula.
Law of problem solving: plant trees along the line.
Tree = number of segments+1 tree = total distance ÷ distance between plants+1
Plant spacing = total distance ÷ (tree-1) total distance = plant spacing × (tree-1)
Planting trees along the periphery
Tree = total distance ÷ plant distance
Plant spacing = total distance.
Total distance = plant spacing × trees
There are 30/kloc-0 poles buried along the highway, and the distance between every two adjacent poles is 50 meters. Later, it was completely revised and only 20 1 was buried. Find the distance between two adjacent ones after modification.
Analysis: this question is to bury telephone poles along the line, and the number of telephone poles is reduced by one. The formula is 50× (301-1) ÷ (201-1) = 75 (m).
(1 1) profit and loss problem: It was developed on the basis of equal share. His characteristic is to distribute a certain number of goods to a certain number of people equally. In the two distributions, one is surplus, the other is insufficient (or both are surplus), or both are insufficient). The problem of finding the right quantity of goods and the number of people participating in the distribution is called profit and loss problem.
The key to solving the problem: The key point of profit and loss problem's solution is to find the difference of the quantity of goods that the distributors did not get in the two distributions, and then find the difference of goods in each distribution (also called total difference). The final difference is divided by the previous difference to get the number of distributors, and then get the quantity of goods.
Law of solving problems: total difference ÷ per capita difference = number of people.
The solution of the total difference can be divided into the following four situations:
The first time is redundant, the second time is insufficient, and the total difference = redundant+insufficient.
The first time is just right, the second time is redundant or insufficient, and the total difference = redundant or insufficient.
First redundancy, second redundancy, total difference = large redundancy-small redundancy.
First shortage, second shortage, total difference = big shortage-small shortage.
For example, students who participate in the art group are given the same number of colored pens. If there are 10 people in the group, there will be 25 more markers. If there are 12 people in the group, there will be 5 more markers. How many cigarettes do you want per person? * * * How many colored pencils are there?
Analysis: Each student was assigned a pen of the same color. This activity group 12 people, 2 more than 10 people, the number of colored pens is (25-5) =20, 2 people are 20 more, 1 person gets 10. The formula is (25-5) ÷ (12-10) =10 (branch)10×12+5 =125 (branch
(12) age problem: the application problem is called "age problem" with the difference between two numbers as a certain value as the condition in the problem.
The key to solving the problem: the age problem is similar to the problems of sum and difference, sum multiple and difference multiple. The main feature is that age increases with time, but the difference between two different ages will not change. Therefore, the age problem is a "constant difference" problem. When solving problems, we should make good use of the characteristics of constant difference.
Father is 48 years old and son is 2 1 year old. A few years ago, my father was four times as old as my son.
Analysis: The age difference between father and son is 48-2 1=27 (years old). Since the father's age was four times that of his son a few years ago, we can know that the multiple difference of the father's age is (4- 1) times. In this way, we can calculate the age of father and son a few years ago, so we can find that the age of father is four times that of son a few years ago. The formula is: 21(48-21) ÷ (4-1) =12 (year).
(13) Chicken and rabbit problem: The total number of head and legs of "chicken and rabbit" is known. How many chickens and rabbits are there? It is often called "the problem of chickens and rabbits", also known as the problem of chickens and rabbits in the same cage.
The key to solving the problem: generally, the problem of chicken and rabbit is solved by hypothesis, assuming that all animals are one kind (for example, all chickens or rabbits), and then according to the different number of legs, the number of heads of a certain kind can be calculated.
Law of problem solving: (total number of legs-number of chicken legs × total number of heads) ÷ The difference between the number of legs of a chicken and a rabbit = the number of rabbits.
Number of rabbits = (total number of legs -2× total number of heads) ÷2
If we assume all rabbits, we can have the following formula:
Number of chickens =(4× total number of heads-total number of legs) ÷2
Number of rabbits = total-number of chickens
Chicken and rabbit pass 50 heads 170 legs. How many chickens and rabbits are there?
The number of rabbits is (170-2 × 50 )÷ 2 =35 (only).
The number of chickens is 50-35= 15 (only)
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(b) Application of scores and percentages
1 Fraction addition and subtraction application problem:
Fractional addition and subtraction application problems and integer addition and subtraction application problems are basically the same in structure, quantitative relationship and solving method, but the difference is that there is a fraction in the known number or unknown number.
2 Fractional multiplication application problem:
Refers to the application of knowing a number and finding its score.
Features: The quantity and fraction of the unit "1" are known, and the actual quantity corresponding to the fraction is found.
The key to solving the problem is to accurately judge the number of units "1". Find the score corresponding to the required question, and then formulate it correctly according to the meaning of multiplying a number by a score.
3 fractional division application problem:
Find the fraction (or percentage) of one number to another.
Features: Knowing one number and another, find the fraction or percentage of one number. "One number" is a comparative quantity, and "another number" is a standard quantity. Find a fraction or percentage, that is, find their multiple relationship.
The key to solving the problem: start with the problem and find out who is regarded as the standard number, that is, who is regarded as "unit one" and who is the bonus compared with the number of unit one.
A is the fraction (percentage) of B: A is the comparative quantity and B is the standard quantity. Divide a by b ..
How much is A more (or less) than B (a few percent): A minus B is more (or less) or (a few percent) than B ... Relationship (A minus B)/B or (A minus B)/A.
Given the fraction (or percentage) of a number, find the number.
Features: Knowing an actual quantity and its corresponding fraction, find the quantity with the unit of "1".
The key to solve the problem is to accurately judge the number of units "1". Take the quantity of unit "1" as an equation of X according to the meaning of fractional multiplication, or as an equation according to the meaning of fractional division, but we must find out the known real corresponding to the fraction.
Quantity.
4 Attendance rate
Germination rate = number of germinated seeds/number of experimental seeds × 100%
Wheat flour yield = flour weight/wheat weight × 100%.
Product qualification rate = number of qualified products/total number of products × 100%.
Employee attendance = actual attendance/attendance × 100%
Five engineering problems:
It is a special case of fractional application, which is closely related to the work of integers. It is an applied problem to explore the relationship among total workload, work efficiency and working hours.
The key to solving the problem: regard the total amount of work as the unit "1", and the work efficiency is the reciprocal of the working time, and then use the formula flexibly according to the specific situation of the topic.
Quantitative relationship:
Total amount of work = working efficiency × working time
Work efficiency = total workload ÷ working hours
Working hours = total amount of work ÷ working efficiency
Total workload ÷ work efficiency = cooperation time
6 pay taxes
Paying taxes means paying a part of the collective or individual income to the state at a certain tax rate according to the relevant provisions of various national tax laws.
The taxes paid are called taxes payable.
The ratio of taxable amount to various incomes (sales, turnover, taxable income) ...) is called tax rate.
* Interest
Money deposited in the bank is called principal.
The extra money paid by the bank when withdrawing money is called interest.
The ratio of interest to principal is called interest rate.
Interest = principal × interest rate× time
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Chapter II Weights and Measures
A length
What is length?
Length is a measure of one-dimensional space.
(2) Common length units
* kilometers (km) * meters (m) * decimeters (dm) * centimeters (cm) * millimeters (mm) * microns (um)
(3) Conversion between units
* 1mm = 1000mm * 1cm = 10mm * 1cm = 1cm * 1m = 1000mm * 1km。
Two areas
(1) What is the area?
Area is the size of the plane occupied by an object. The measurement of the surface of three-dimensional objects is generally called surface area.
(2) Public area unit
* square millimeter * square centimeter * square decimeter * square meter * square kilometer
(3) conversion of area units
* 1 cm2 =100mm2 * 1 cm2 =1cm2 * 1 m2 =100mm2.
* 1 ha = 10000 m2 * 1 km2 = 100 ha.
Three volumes and volumes
(1) What are volume and volume?
Volume is the size of the space occupied by an object.
Volume, the volume of objects that can be accommodated in boxes, oil drums, warehouses, etc. , usually called their volume.
(2) Common units
1 unit of volume
* cubic meter * cubic decimeter * cubic centimeter
2 unit of volume * L * mL
(3) Unit conversion
1 unit of volume
* 1 m3 = 1000 cubic decimeter
* 1 cubic decimeter = 1000 cubic centimeter
2 unit of volume
* 1L = 1000ml
* 1 l = 1 m3
* 1 ml = 1 cm3
Four qualities
A length
What is length?
Length is a measure of one-dimensional space.
(2) Common length units
* kilometers (km) * meters (m) * decimeters (dm) * centimeters (cm) * millimeters (mm) * microns (um)
(3) Conversion between units
* 1mm = 1000mm * 1cm = 10mm * 1cm = 1cm * 1m = 1000mm * 1km。
Two areas
(1) What is the area?
Area is the size of the plane occupied by an object. The measurement of the surface of three-dimensional objects is generally called surface area.
(2) Public area unit
* square millimeter * square centimeter * square decimeter * square meter * square kilometer
(3) conversion of area units
* 1 cm2 =100mm2 * 1 cm2 =1cm2 * 1 m2 =100mm2.
* 1 ha = 10000 m2 * 1 km2 = 100 ha.
Three volumes and volumes
(1) What are volume and volume?
Volume is the size of the space occupied by an object.
Volume, the volume of objects that can be accommodated in boxes, oil drums, warehouses, etc. , usually called their volume.
(2) Common units
1 unit of volume
* cubic meter * cubic decimeter * cubic centimeter
2 unit of volume * L * mL
(3) Unit conversion
1 unit of volume
* 1 m3 = 1000 cubic decimeter
* 1 cubic decimeter = 1000 cubic centimeter
2 unit of volume
* 1L = 1000ml
* 1 l = 1 m3
* 1 ml = 1 cm3
Four qualities
What is quality?
Mass refers to the weight of an object.
(2) Common units
* tons * kilograms * grams
(3) General conversion
* one ton = 1000 kg
* 1 kg =1000g
quintic
(1) What is time?
Refers to a period of time with a starting point and an ending point.
(2) Common units
Century, year, month, day, hour, minute and second.
(3) Unit conversion
* 1 century = 100 year
* 1 year = average number of years of 365 days.
* One year =366 days leap year
* January, Wednesday, Friday, July, August, October and December are big months, with 3 1 day.
* Four, six, nine and eleven are abortions. Abortion lasts for 30 days.
* There are 28 days in February in a normal year and 29 days in February in a leap year.
* 1 day = 24 hours
* 1 hour =60 points
* One minute =60 seconds
Six currencies
(1) What is money?
Money is a special commodity, which acts as the equivalent of all commodities. Money is a general representative of value and can buy any other commodity.
(2) Common units
* Yuan * Jiao * fen
(3) Unit conversion
* 1 yuan = 10 angle.
* 1 angle = 10 point
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s=(a+b)h/2=mh
I hope these are useful to you.