1 Simple application problem

(1) Simple application problem: an application problem that only contains a basic quantitative relationship or is solved by one-step operation, usually called a simple application problem.

(2) Steps to solve the problem:

A understand the meaning of the question: understand the content of the application question and know the conditions and problems of the application question. When reading a question, read, think and understand the meaning of every sentence in the question without losing words or adding words. You can also repeat the conditions and questions to help you understand the meaning of the questions.

B selection algorithm and column calculation: this is the central work to solve application problems. Starting with what to say and ask, according to the given conditions and questions, and connecting with the significance of four operations, this paper analyzes the quantitative relationship, determines the algorithm, answers and marks the correct unit name.

C test: according to the conditions and problems of the application questions, check whether the listed formulas and calculation processes are correct and whether they meet the meaning of the questions. If mistakes are found, correct them immediately.

Two complex application problems

(1) An application problem that consists of two or more basic quantitative relations and is solved by two or more operations is usually called a compound application problem.

(2) An application problem of two-step calculation of three known conditions.

Find an application problem that is greater than (less than) the sum of two numbers.

The application problem of comparing the difference and multiple of two numbers.

(3) An application problem in which two known conditions are calculated in two steps.

Know the difference (or multiple relationship) between two numbers and one of them, and find the sum (or difference) of the two numbers.

Knowing two numbers and one of them, find the difference (or multiple relationship) between the two numbers.

(4) Solve the application problem of multiplication and division.

(5) Solve the application problem of three-step calculation method.

(6) Solving the application problem of decimal calculation: the application problem of addition, subtraction, multiplication and division of decimal calculation is basically the same as the formal application problem, except that there are decimals among the known numbers or unknowns.

D answer: according to the calculation results, answer orally first, and then gradually transition to written answer.

(3) Solve the problem of addition application:

An application problem of finding the total number: what is the known number A, what is the number B, and what is the sum of the two numbers A and B.

Find a number greater than the number. Application problem: Know what A number is, how much more B number is than A number, and find what B number is.

(4) Solve the application problem of subtraction:

A Finding the residual application problem: removing a part from the known number and finding the residual part.

The application problem of finding the difference between two numbers by -b: Given the numbers of A and B, find how much A is more than B, or how much B is less than A. ..

The application of c to find the number less than the number: what is the known number a, how much is the number b less than the number a, and how much is the number B.

(5) Solve the problem of multiplication application:

An application problem of seeking the sum of common addends: knowing the same addend and the number of the same addend, find the sum.

The application problem of finding the multiple of a number is: how many times is one number, how many times is another number, and how much is another number?

(6) Solve the problem of division application:

A divide a number into several parts on average, and find out how much each part is: know a number, divide it into several parts on average, and find out how much each part is.

B. Find an application problem, in which one number contains several other numbers: given a number, how many copies are there in each number, and how many copies can you find?

C the application problem of finding a number that is several times that of another number: given the number A and the number B, finding a larger number is several times that of a smaller number.

D know how many times a number is, and find the application problem of this number.

(7) Common quantitative relations:

Total price = unit price × quantity

Distance = speed × time

Total amount of work = working time × working efficiency

Total output = single output × quantity

Three typical application problems

Compound application problems with unique structural characteristics and specific problem-solving rules are usually called typical application problems.

(1) average problem: average is the development of equal division.

The key to solve the problem is to determine the total quantity and the corresponding total number of copies.

Arithmetic average: Given several unequal quantities of the same kind and the corresponding number of copies, find the average of each number of copies. Quantity relationship: sum of quantity ÷ quantity = arithmetic average.

Weighted average: given the average of two or more copies, what is the total average?

Sum of quantitative relations (partial average × weight) ÷ (total weight) = weighted average.

Average difference: divide the sum of parts greater than or less than the standard number by the total number of shares to get the average difference between the standard number and each number.

Quantitative relationship: (large number-decimal number) ÷2= sum of the difference between the largest number of digits and the decimal number ÷ total number of copies = sum of the difference between the largest number of digits and the decimal number ÷ total number of copies = decimal number.

Example: A car travels from A to B at a speed of 100 km/h and from B to A at a speed of 60 km/h.. Find the average speed of this car.

Analysis: The formula can also be used to find the average speed of the car. In this question, the distance from A to B can be set to "1", so the total distance traveled by the car is "2" and the speed from A to B is 100. The time required is that the speed of the car from B to A is 60 kilometers, the time required is+=, and the average speed of the car is 2 \

2) Normalization problem: two quantities that are known to be interrelated, one of which changes, and the other changes with it, with the same change law. This problem is called standardization.

According to the number of steps to find a single quantity, the normalization problem can be divided into one normalization problem and two normalization problems.

According to the problem of multiplication or division, normalization can be divided into positive normalization and negative normalization.

One problem at a time can be solved in one step. Also known as "single to one".

Two-step operation can solve the problem of twice normalization. Also known as "double return to one".

Normalization problem: after finding a "single quantity" by equal division, the normalization problem of the result is calculated by multiplication.

Denormalization problem: the normalization problem of the result calculated by division after finding a "single quantity" by equal division.

The key to solving the problem: from a group of known corresponding quantities, the number of a copy (single quantity) is obtained by equal division, and then the result is calculated according to the requirements of the topic.

Quantity relationship: single quantity × copy number = total quantity (positive normalization)

Total quantity ÷ single quantity = number of copies (normalized)

A weaver weaved 4774 meters in July. According to this calculation, how many days does it take to weave 6930 meters?

Analysis: First of all, we must find out how many meters we weave on average every day, which is a single quantity. 693 0 ÷( 477 4 ÷ 3 1) =45 (days)

(3) Sum problem: the number of units and units of measurement, as well as different units (or units) are known, and the number of units (or units) can be obtained by finding the total.

Features: Two related quantities, one changing and the other changing, have opposite changing rules and are connected by inverse ratio algorithm.

Quantity relationship: unit quantity × unit quantity ÷ another unit quantity = another unit quantity × unit quantity ÷ another unit quantity = another unit quantity.

For example, to build a canal, it was originally planned to build 800 meters a day and complete it in six days. Actually, it took four days to fix it. How many meters are repaired every day?

Analysis: Because of the length of daily maintenance, we must first find out the length of the canal. Therefore, this kind of application problem is also called "inductive problem". The difference is that "normalization" first finds a single quantity, and then finds the total quantity. The general problem is to find the total quantity first, and then find the single quantity. 80 0 × 6 ÷ 4= 1200 (m)

(4) Sum and difference problem: the sum and difference of two numbers with different sizes are known, and the application problem of finding the numbers of these two numbers is called sum and difference problem.

The key to solve the problem is to convert the sum of two numbers into the sum of two large numbers (or the sum of two decimals), and then find another number.

Law of solving problems: (sum+difference) ÷2 = large number-difference = decimal.

(sum and difference) ÷2= decimal, and-decimal = large number.

For example, there are 94 workers in Class A and Class B in a processing plant, and 46 workers in Class A are temporarily transferred from Class B due to work needs. At this time, Class B has fewer 12 workers than Class A. How many workers are there in Class A and Class B respectively?

Analysis: From Class B to Class A, the total number of people has not changed. Now the number of class B is converted into two classes B, that is, 94- 12, which means that the current class B is (94- 12) ÷ 2 = 4 1 (people), and the class B should be 4650 before being transferred to 46 people.

(5) Sum and multiple problem: The sum of two numbers and the multiple relationship between them are known, which is called the sum and multiple problem.

The key to solving the problem is to find the standard number (that is, the multiple of 1). Generally speaking, whoever says it is several times the "who" in the question is determined as the standard number. Find the standard number after finding the sum of multiples. Find the number of another number (or numbers) according to the multiple relationship between another number (or numbers) and the standard number.

Law of solving problems: sum/multiple sum = standard number × multiple = another number.

Example: There are 1 15 trucks in the automobile transportation yard, of which 7 trucks are five times more than the minivan. How many trucks and cars are there in the transportation yard?

Analysis: There are 7 trucks that are more than 5 times of the minivan, and these 7 trucks are also within the total 1 15. In order to make the total number correspond to (5+ 1), the total number of vehicles should be (1 15-7).

The formula is (115-7) ÷ (5+1) =18 (vehicle), 18 × 5+7=97 (vehicle).

(6) Difference multiple problem: Knowing the difference between two numbers and the multiple relationship between two numbers, we can find the application problem of how many two numbers are.

Law of solving problems: the difference between two numbers ÷ (multiple-1) = standard number × multiple = another number.

Examples a and b have two ropes. The length of rope A is 63m, and the length of rope B is 29m. The two ropes were cut to the same length. Results The remaining length of rope A is three times that of rope B. What are the remaining lengths of rope A and rope B respectively? How many meters per person?

Analysis: Cut the same section of two ropes with the same length difference. The remaining length of rope A is three times that of rope B, but it is (3- 1) times more than rope B, and the length of rope B is the standard number. Equation (63-29) ÷ (3-1) =17 (m) … remaining length of rope b, 17 × 3=5 1 (m) … remaining length of rope a, 29-/kloc.

(7) Travel problem: About walking and driving, distance, time and speed are generally calculated, which is called travel problem. To solve this kind of problems, we must first understand the concepts of speed, time, distance, direction, speed sum and speed difference, and understand the relationship between them, and then answer them according to the laws of this kind of problems.

The key and law of solving problems;

Go in the opposite direction at the same time: distance = speed x time.

Walking in the opposite direction at the same time: meeting time = speed and x time.

Walking in the same direction at the same time (slow in front and fast in the back): catching up time = distance and speed difference.

Walk in the same direction at the same time (slow in the back, fast in the front): distance = speed difference × time.

Example A is 28 kilometers behind B, and both of them are walking in the same direction at the same time. A travels 16 km per hour, and B travels 9 km per hour. How many hours does it take for A to catch up with B?

Analysis: A drives (16-9) kilometers per hour more than B, that is, A can catch up with B (16-9) kilometers per hour, which is the speed difference.

It is known that A is 28 kilometers behind B (pursuit distance), and 28 kilometers includes several kilometers (16-9), which is the time required for pursuit. Equation 2 8 ÷ (16-9) =4 (hours)

(8) Running water problem: Generally speaking, it is to study the problem of ships sailing in "running water". It is a special type of travel problem, and it is also a sum-difference problem. Its characteristic is mainly to consider the different functions of water velocity in retrograde and anterograde.

Ship speed: the speed at which a ship sails in still water.

Speed of water flow: the speed of water flow.

Downstream speed: the speed at which a ship sails downstream.

Current speed: the speed at which a ship sails against the current.

Forward speed = ship speed+water speed

Reverse speed = boat speed-water speed

The key to solve the problem: because the downstream speed is the sum of the ship speed and the water speed, and the upstream speed is the difference between the ship speed and the water speed, the running water problem is solved as the sum difference problem. When solving problems, we should take current as a clue.

Law of solving problems: ship speed = (downstream speed+countercurrent speed) ÷2

Water velocity = (downstream velocity and countercurrent velocity) ÷2

Distance = downstream speed × time required for downstream navigation

Distance = countercurrent speed × time required for countercurrent navigation

A ship sails from place A to place B at a speed of 28 kilometers per hour. After arriving at B, it sailed against the current and returned to A ... It takes 2 hours to travel against the current, and the known current speed is 4 kilometers per hour. How many kilometers is it between Party A and Party B?

Analysis: This question first needs to know the speed and time needed to go with the water, or the speed and time needed to go against the water. It is not difficult to calculate the speed against the current, because we know the speed along the current and the speed of the current, but we don't know the time along the current and the time against the current. All we know is that it takes 2 hours less than countercurrent. Grasping this point, we can calculate the time from A to B along the ocean current, so we can calculate the distance between A and B. The formula is 284× 2 = 20 (km) 20× 2 = 40 (km) 40 ÷ (4× 2) = 5 (hours) 28× 5 =144.

(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)

(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