Solving mathematical problems. "Arithmetic ways of solving text problems"

Solve a math problem- it means to find such a sequence general provisions mathematics, applying which to the conditions of the problem we get what you want to find - the answer.

The main methods for solving word problems are the arithmetic and algebraic method, as well as the combined one.

Solve a problem arithmetic method - means to find the answer to the requirement of the task by performing arithmetic operations over the numbers given in the problem. The same problem can be solved in different arithmetic ways. They differ from each other in the logic of reasoning in the process of solving the problem.

Solve a problem algebraic method - means to find the answer to the requirement of the problem by compiling and solving an equation or system of equations.

The algebraic method is solved according to the following scheme:

1) allocate quantities about which in question in the text of the task, and establish the relationship between them;

2) introduce variables (letters denote unknown quantities);

3) with the help of the introduced variables and data, the tasks compose an equation or a system of equations;

4) solve the resulting equation or system;

5) check the found values according to the condition of the problem and write down the answer.

Combined the solution method includes both arithmetic and algebraic methods of solution.

AT primary school tasks are divided by the number of actions when solving for prime and compound. Problems in which only one action is required to answer a question are called simple. If two or more actions are required to answer a question, then such tasks are called composite.

A compound problem, just like a simple one, can be solved using various methods.

Task. The fisherman caught 10 fish. Of these, 3 bream, 4 perch, the rest - pike. How many pike did the fisherman catch?

practical way.

Label each fish with a circle. Let's draw 10 circles and denote the caught fish.

L L L O O O O

To answer the question of the problem, you can not perform arithmetic operations, since the number of pikes caught corresponds to unmarked circles - there are three of them .

Arithmetic way.

1) 3+4=7(p) - caught fish;

2) 10 - 7 \u003d 3 (p) - caught pikes.

Algebraic way.

Let x be the pike caught. Then the number of all fish can be written as: 3 + 4 + x. According to the condition of the problem, it is known that the fisherman caught only 10 fish. This means: 3 + 4 + x = 10. Having solved this equation, we get x = 3 and thus answer the question of the problem.

Graphical way.

bream perch pike

This method, as well as the practical one, will allow you to answer the question of the problem without performing arithmetic operations.

In mathematics, the following is generally accepted division of the problem solving process :

1) analysis of the text of the problem, schematic representation of the problem, study of the problem;

2) finding a way to solve the problem and drawing up a solution plan;

3) implementation of the found plan;

4) analysis of the found solution of the problem, verification.

Methods for finding a solution to the problem can be called the following:

1) Analysis: a) when in reasoning they move from the desired to the data of the problem; b) when the whole is divided into parts;

2) Synthesis: a) when moving from the data of the problem to the desired ones;
b) when the elements are combined into a whole;

3) Reformulation of the problem (clearly formulate intermediate tasks that arise in the course of the search for a solution);

4) inductive method problem solving: on the basis of an accurate drawing, see the properties of the figure, draw conclusions and prove them;

5) Application of analogy (remember a similar task);

6) Forecasting - anticipation of the results that the search may lead to.

Let's consider in more detail problem solving process:

Movement task. The boat traveled along the river the distance between two piers in 6 hours, and back - in 8 hours. How much time will pass the distance between the piers a raft floated along the river?

Task analysis. The problem deals with two objects: a boat and a raft. The boat has its own speed, and the raft and the river along which the boat and the raft float have a certain speed of flow. That is why the boat makes its way down the river in less time. (6h) than against the current (8h). But these speeds are not given in the problem, just as the distance between the piers is unknown. However, it is required to find not these unknowns, but the time during which the raft will cover this distance.

Schematic notation:

A boat 6 h

raft boat

Finding a way to solve the problem. We need to find the time it takes for the raft to cover the distance between the piers. BUT and B. In order to find this time, you need to know the distance AB and the speed of the river. Both of them are unknown, so we denote the distance AB by the letter S (km), and the flow rate and km/h. To relate these unknowns to the task data, one needs to know the boat's own speed. It is also unknown, suppose it is equal to V km/h From here, a solution plan arises, which consists in compiling a system of equations with respect to the introduced unknowns.

Implementation of problem solving. Let the distance be S (km), river speed a km/h, own boat speed V km/h, and the required time of the raft movement is equal to x h.

Then the speed of the boat downstream is (V+a) km/h. Behind 6h a boat traveling at this speed has traveled a distance of S (km). Therefore, 6( V + a) =S(1). This boat is moving against the current at a speed of ( V-a)km/h and this path she goes for 8 h, so 8( V-a) =S(2). A raft sailing at the speed of a river a km/h, swam the distance S (km) behind x h, hence, Oh =S (3).

The resulting equations form a system of equations for the unknowns a, x, S, V. Since we only need to find X, then we will try to eliminate the rest of the unknowns.

To do this, from equations (1) and (2) we find: V + a = , V - a = . Subtracting the second equation from the first equation, we get: 2 a= - . From here a = . Let us substitute the found expression into equation (3): x = . Where x= 48 .

Verification of the solution. We have found that the raft will cover the distance between the piers in 48 hours. Therefore, its speed, equal to speed the course of the river is . The speed of the boat along the river is km/h, but against the current km/h In order to verify the correctness of the solution, it is enough to check whether the own speeds of the boat, found in two ways, will be equal: + and
- . After doing the calculations, we get the correct equality: = . It means that the problem is solved correctly.

Answer: the raft will cover the distance between the piers in 48 hours.

Solution Analysis. We have reduced the solution of this problem to the solution of a system of three equations in four unknowns. However, one unknown had to be found. Therefore, the idea arises that this decision not the best, but simple. You can suggest another solution.

Knowing that the boat sailed the distance AB along the river in 6 hours, and against - in 8 hours, we find that at 1 hour the boat, going down the river, passes part of this distance, and against the current. Then the difference between them - = is twice the part of the distance AB, floated by the raft in 1 hour. Means. The raft will cover part of the distance AB in 1 hour, so it will cover the entire distance AB in 48 hours.

With this solution, we did not need to compose a system of equations. However, this solution is more complicated than the one above (not everyone will guess to find the difference in the speeds of the boat along the flow and against the flow of the river).

Exercises for independent work

1. A tourist, having sailed along the river on a raft for 12 km, returned back on a boat, the speed of which was standing water is equal to 5 km/h, having spent 10 hours on the whole trip. Find the speed of the river.

2. One workshop must sew 810 suits, the other for the same period - 900 suits. The first completed the execution of orders for 3 days, and the second for 6 days before the deadline. How many suits per day did each workshop sew if the second one sewed 4 more suits per day than the first?

3. Two trains left towards each other from two stations, the distance between which is 400 km. After 4 hours, the distance between them was reduced to 40 km. If one of the trains left 1 hour earlier than the other, then they would meet in the middle of the journey. Determine the speeds of the trains.

4. There are 500 tons of coal in one warehouse, and 600 tons in the other. The first warehouse releases 9 tons daily, and the second - 11 tons of coal. In how many days will the coal in the warehouses become equal?

5. The depositor took 25% of his money from the Savings Bank, and then 64,000 rubles. After that, 35% of all money remained in the account. What was the contribution?

6. Artwork two-digit number and its sum of digits is 144. Find this number if the second digit in it is greater than the first by 2.

7. Solve the following problems using the arithmetic method:

a) On the way down the river powerboat took 6 hours and Return trip- 10 hours Boat speed in still water 16 km/h. What is the speed of the river?

c) The length of a rectangular field is 1536 m and the width is 625 m. One tractor driver can plow this field in 16 days, and another in 12 days. What area will both tractor drivers plow, working for 5 days?

Problem solving arithmetic way

Math lesson in 5th grade.

“If you want to learn how to swim, then boldly enter the water, and if you want to learn how to solve problems, then solve them”.
D. Poya

Goals and objectives of the lesson:

formation of the ability to solve problems in an arithmetic way;

development creativity, cognitive interest;

development logical thinking;

education of love for the subject;

education of the culture of mathematical thinking.

Equipment: signal cards with numbers 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

During the classes

I. Organizational moment (1 min.)

The lesson is devoted to solving problems in an arithmetic way. Today we will solve problems different types, but they will all be solved without the help of equations.

II. History reference (1 min.)

Historically, for a long time, mathematical knowledge was passed down from generation to generation in the form of a list of practical problems along with their solutions. In ancient times, a person who knew how to solve problems was considered trained. certain types encountered in practice.

III. Warm up (problem solving orally - 6 min.)
a) Tasks on the cards.
Each student is given a card with a problem that he solves orally and gives an answer. All tasks for action 3 - 1 = 2.

(Students solve problems correctly, and who doesn’t. At all orally. They raise cards and the teacher sees who solved the problem; the cards should be number 2.)

b) Tasks in verse and logical tasks. (The teacher reads the problem aloud, the students hold up the card with the correct answer.

gave the ducklings a hedgehog
Who will answer from the guys
Eight leather boots
How many ducklings were there?
(Four.)

Two nimble piglets
They are so cold, they are shivering.
Count and say:
How many boots to buy them?
(Eight.)

I entered the pine forest
And I saw a fly agaric
Two honey agaric,
Two morels.
three oilers,
Two lines...
Who has the answer:
How many mushrooms did I find?
(Ten.)

4. Chickens and dogs were walking in the yard. The boy counted their paws. Got ten. How many chickens and how many dogs could there be. (Two dogs and one chicken, one dog and three chickens.)

5. According to the doctor's prescription, 10 tablets were bought at the pharmacy. The doctor prescribed to take medicine 3 tablets a day. How many days will this medicine last? (Full days.)

6. Brother is 7 years old and sister 5. How old will sister be when brother is 10 years old?

7. Numbers are given: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. which is greater: their product or sum?

8. When building a fence, carpenters placed 5 posts in a straight line. The distance between the posts is 2 m. What is the length of the fence?

IV. Problem solving

(Tasks for children are given on cards - 15 min. Children solve problems at the blackboard)
Tasks a) and b) are aimed at repeating the connection of the relations "by ... more" and "by ... less" with the operations of addition and subtraction.

a) The turner's apprentice turned 120 parts per shift, and the turner turned 36 more parts. How many parts did the turner and his apprentice turn together?

b) The first brigade collected 52 devices during the shift, second?"; - 9 devices less than the first, and the third - 12 devices more than the second. How many devices did three teams collect during the shift?

With the help of task c), students can be shown the solution to the problem “in reverse”.

c) There are 44 girls in three classes, which is 8 less than boys. How many boys are in three classes?

In problem d) students can offer several solutions.

d) Three sisters were asked: “How old is each of the sisters?”. Vera replied that she and Nadia were together for 28 years, Nadia and Lyuba were together for 23 years, and all three were 38 years old. How old are each sisters?

Task e) is designed to repeat the relation "more in ..." and "less in ...".

e) Vasya had 46 marks. During the year, his collection increased by 230 stamps. How many times has his collection increased?

V. Physical education (2 minutes.)

Stay on one leg
Like you're a solid soldier.
Raise your left leg.
Look, don't fall.
Now stay on the left
If you are a brave soldier.

VI. vintage, historical tasks. Tasks with fabulous content (10 minutes.)

Problem f) to find two numbers by their sum and difference.

e)(from "Arithmetic" by L.N. Tolstoy)

Two men have 35 sheep. One has 9 more than the other. How many sheep does each have?

Movement task.

g)(An old problem.)Two trains left Moscow for Tver at the same time. The first passed at an hour of 39 versts and arrived in Tver two hours earlier than the second, which passed at 26 versts per hour. How many miles from Moscow to Tver?

(The equation makes it easier to get to the answer. But students are encouraged to look for an arithmetic solution to the problem.)

1) 26 * 2 \u003d 52 (versts) - so many versts the second train lagged behind the first;

2) 39 - 26 \u003d 13 (versts) - so many versts the second train lagged behind the first in 1 hour;

3) 52: 13 = 4 (h) - so much time was the first train on the way;

4) 39 * 4 = 156 (versts) - the distance from Moscow to Tver.

You can look in reference books to find the distance in kilometers.

1 verst = 1 km 69 m.

Part task.

h)Kikimora's task.Vodyanoy decided to marry a kikimore Ha-Ha. He planted several leeches on the kikimore's veil, and twice as many on his cape. During the holiday, 15 leeches fell off, and only 435 remained. How many leeches were on the kikimora's veil?

(The problem is given for solving using an equation, but we solve it in an arithmetic way)

VII. Live numbers (unloading pause - 4 min.)

The teacher calls 10 students to the board, gives them numbers from 1 to 10. The students receive different tasks;

a) the teacher calls the numbers; those named take a step forward (eg: 5, 8, 1, 7);

b) only the neighbors of the named number come out (eg: number 6, 5 and 7 come out);

c) the teacher comes up with examples, and only the one who has the answer to this example or task comes out (for example: 2 ´ 4; 160: 80; etc.);

d) the teacher makes several claps and also shows a number (one or two); a student should come out whose number is the sum of all the numbers heard and seen (for example: 3 claps, number 5 and number 1.);

What number is 4 greater than 4?

I conceived a number, subtracted 3 from it, I got 7. What number did I conceive?

if you add 2 to the intended number, you get 8. What is the intended number?

We must try to select such tasks so that the answers do not repeat the same numbers, so that everyone can actively participate in the game.

VIII. Summing up the lesson (2 minutes.)

- What did we do in class today?

- What does it mean to solve a problem by arithmetic?

- It must be remembered that the found solution of the problem must satisfy the conditions of the problem.

IX. Homework assignment. Grading (2 minutes.)

№ 387 (solve problems in an arithmetic way), for weak students. For medium and strong students, homework is given on cards.

1. There were 645 kg of black and white bread in the bakery. After they sold 215 kg of black and 287 kg of white bread, both types of bread remained equal. How many kilograms of black and white bread separately were there in the bakery?

Brother and sister found 25 white mushrooms in the forest. The brother found 7 more mushrooms than the sister. How many white mushrooms did the brother find?

For compote, we took 6 parts of apples, 5 parts of pears and 3 parts of words. It turned out that pears and plums together took 2 kg 400 g. Determine the mass of apples taken; mass of all fruits.

Literature

Vilenkin N., Zhokhov V., Chesnokov A.Mathematics. Grade 5 - M., "Mnemosyne", 2002.

Shevkin A.V.Text problems in school course mathematics. - M.: Pedagogical University "First of September", 2006.

Wolina V.Number holiday. - M.: Knowledge, 1994.

Page 1

The arithmetic solution is rather confusing, but the problem is solved simply if you turn to the services of algebra and write an equation.

At arithmetic solution all the questions of the plan and the arithmetic operations that serve as answers to them must be written out, and in the case of algebraic - the motives for choosing unknowns, the equations drawn up and their solution.

Schultz gave an arithmetic solution to this equation, using arbitrary values of the constants, and concluded that the efficiency of fractionation should greatly increase when working with dilute solutions.

The problem admits a purely arithmetic solution, and even operations on fractions can be dispensed with.

And now we give an arithmetic solution to this problem - a solution in which it is possible to do without writing equations at all.

Other arithmetic solutions are also possible.

In this section, some problems admit both an algebraic and an arithmetic solution; they can be used when reviewing the course of arithmetic.

They involve the use of arithmetic operations according to the plan for solving the problem. The arithmetic solution is often used in calculations for chemical formulas and equations, according to the concentrations of solutions, etc.

But here we present only arithmetic solutions of the problems.

We do not divide problems into algebraic and arithmetic, since problems that can be solved arithmetically can always be solved algebraically. On the contrary, problems solved with the help of equations often allow a simpler arithmetic solution. In the decision department we give sometimes arithmetic, sometimes algebraic solution, but this should not in any way hamper the student's initiative in choosing a solution.

We do not divide problems into algebraic and arithmetic, since problems that can be solved arithmetically can always be solved algebraically. On the contrary, problems solved with the help of equations often allow a simpler arithmetic solution. In the department of solutions, we sometimes give an arithmetic, sometimes an algebraic solution, but this should not in any way hinder the student's initiative in choosing the method of solution.

Here is an example of an indirect problem: a piece of an alloy of copper and zinc with a volume of 1 dm3 has a mass of 8 14 kg. Here, from the condition of the problem, it is not clear what actions lead to its solution. With the so-called arithmetic solution, sometimes great ingenuity must be shown in order to outline a plan for solving an indirect problem. Each new task requires the creation of a new plan. The work of the calculator is spent irrationally.

To confirm his thought, Petrov invented problems that, due to their non-shabda-like nature, made experienced skillful teachers very difficult, but were easily solved by more capable students who had not yet been spoiled by their studies. Among such problems (Petrov composed several of them) is the problem of the artel of mowers. Experienced teachers, of course, could easily solve it using an equation, but a simple arithmetic solution eluded them. Meanwhile, the problem is so simple that it is not at all worth using the algebraic apparatus to solve it.

Here is an example of an indirect problem: a piece of an alloy of copper and zinc with a volume of dm3 weighs 8 14 kg. Here, from the condition of the problem, it is not clear what actions lead to its solution. With the so-called arithmetic solution, sometimes great ingenuity must be shown in order to outline the plan for solving an indirect problem. Each new task requires the creation of a new plan. The work of the calculator is spent irrationally.

Despite the fact that computational activity is of interest to children, and the problem itself is given a significant place in the curriculum in kindergarten, many older preschoolers and even junior schoolchildren(students of grades 1-3) experience significant difficulties precisely in solving arithmetic problems. About 20% of children of the seventh year of life experience difficulties in choosing an arithmetic operation and arguing it. These children, when solving arithmetic problems, in choosing an arithmetic operation, are guided mainly by external non-essential "pseudo-mathematical" connections and relationships between numerical data in the problem statement, as well as between the condition and the question of the problem. This is manifested primarily in their misunderstanding of the generalized content of the concepts: “condition”, “question”, “action”, as well as signs (+, -, =), in their inability to correctly select the necessary sign, an arithmetic operation in the case when given in condition, a specific display does not correspond to an arithmetic operation (arrived, added, more expensive - addition; departed, taken, cheaper - subtraction). Moreover, sometimes individual educators orient children precisely towards these pseudo-mathematical connections. In such situations, computational activity is formed insufficiently consciously (M. A. Bantova, N. I. Moro, A. M. Pyshkalo, E. A. Tarkhanova, etc.).

Obviously, the main reason for the low level of children's knowledge lies in the very essence of what distinguishes computational activity from counting. While counting, the child deals with specific sets (objects, sounds, movements). He sees, hears, feels these multitudes, has the opportunity to practically act with them (impose, apply, directly compare). As for computational activity, it is connected with numbers. And the numbers are abstract concepts. Computational activity is based on various arithmetic operations, which are also generalized, abstracted operations with sets.

Understanding the simplest arithmetic problem requires analyzing its content, extracting its numerical data, understanding the relationships between them, and, of course, the very actions that the child must perform.

It is especially difficult for preschoolers to understand the question of the problem, which reflects the mathematical essence of the actions, although it is the question of the problem that directs the child's attention to the relationship between numerical data.

Teaching preschoolers to solve arithmetic problems leads them to understand the content of arithmetic operations (added - added, reduced - subtracted). This is also possible at a certain level of development of the child's analytical-synthetic activity. In order for children to learn elementary methods of computational activity, preliminary work is necessary, aimed at mastering knowledge about the relationship between adjacent numbers of the natural series, about the composition of a number, counting in groups, etc.

Of particular importance in the formation of computational activity is a clear systematic and phased work.

Solve by addition (add one to three). Children conclude: "Four birds flew to the feeder."

“There were five TVs in the store, one of them was sold. How many TVs are left in the store? Solving this problem, the educator teaches to argue his actions as follows: there were five televisions, one was sold, therefore, there are one less of them left. To find out how many TVs are left, you need to subtract one from five and you get four.

The teacher forms children's ideas about the actions of addition and subtraction, at the same time introduces them to the signs "+" (add, add), "-" (subtract, subtract) and "=" (equal, it will turn out).

Thus, the child gradually moves from actions with concrete sets to actions with numbers, that is, he solves an arithmetic problem.

Already in the second or third lesson, along with dramatization and illustration tasks, children can be offered to solve oral (text) tasks. This stage of work is closely related to the use of cards with numbers and signs. Especially useful are the exercises of children in independently compiling similar tasks by them. At the same time, the educator must remember that the main thing is to find not so much the answer (the name of the number), as the path to it. So, the children solve the problem: “Four trees were planted on the kindergarten site on the first day, and on the next day, another tree. How many trees were planted in two days? The teacher teaches the child to think while solving the problem. He asks the children: “What is the problem about?” - "The fact that trees were planted on the site of the kindergarten." “How many trees were planted on the first day?” -- "Four". “How many trees were planted on the second day?” - One tree. - "What is asked in the problem?" - "How many trees were planted on the site in two days?" - "How can I find out how many trees have been planted on the site?" “Add one to four.”

The teacher leads the children to such a generalization: in order to add one (one) to a number, you do not need to count all the items, you just need to name next number. When we add one to four, we simply call the next number "four" the number "five". And when you need to subtract, subtract one, you should name previous number standing in front of him. Thus, relying on the knowledge that the children have, the educator equips them with methods of counting (adding) to the number of one and subtracting one. Below are several tasks of the first type.

1. Five sparrows sat on a branch. Another sparrow flew towards them. How many birds are on the branch?
2. Tanya and Vova helped their mother. Tanya peeled three potatoes, and Vova peeled one carrot. How many vegetables did the children clean?
3. Five tulips bloomed on one flower bed, one peony on the other. How many flowers bloomed in both flower beds together?

If from the first steps of learning children realize the need, the value of analysis simple tasks, then later it will help them in solving complex math problems. The activity of the mental activity of the child largely depends on the ability of the educator to raise questions, to encourage him to think. So, the teacher asks the children: “What should be learned in the task? How can you answer the question? Why do you think it should be folded? How do you add one to four?”

The next stage in the work is connected with familiarizing children with new tasks (tasks of the second type) on the relationship "more - less by several units." In these tasks, arithmetic operations are prompted in the very condition of the problem. The relation "more than one" requires the child to increase, count, add. The expression "more (less) by one" children have already learned in groups of the fifth and sixth years of life, comparing adjacent numbers. At the same time, it is not recommended to focus the attention of children on the individual words “more”, “less”, and even more so to orient them to the choice of an arithmetic operation only depending on these words. Later, when solving "indirect, indirect" problems, there is a need to retrain children, and this is much more difficult than teaching them to make the right choice of an arithmetic operation. Example tasks of the second type are given below.

1. Mom put two spoons of sugar in the Car, and one spoon more in dad's big cup. How much sugar did mom put in dad's cup?
2. There were four passenger trains at the station, and one less freight train. How many freight trains were at the station?
3. The children gathered three boxes of tomatoes in the garden, and one less of cucumbers. How many boxes of cucumbers did the children collect?

At the beginning of education, preschoolers are offered only. direct tasks, in which both the condition and the question seem to suggest what action should be performed: addition or subtraction.

Six-year-olds should be encouraged to compare tasks different types, although this is difficult for them, since children do not see the text, and both tasks must be kept in memory. The main criterion for comparison is the question. The question emphasizes that only the number of the second set that is more (less) by one needs to be determined, or total(remainder, difference). Arithmetic operations are the same, but the goal is different. This is what contributes to the development of children's thinking. The teacher gradually brings them to this understanding.

An even more important and responsible stage in teaching children to solve arithmetic problems is to familiarize them with the third type of problems - the difference comparison of numbers. Problems of this type are solved only by subtraction. When introducing children to this type of task, their attention is drawn to the main thing - the question in the task. The question begins with the words “by how much?”, i.e. it is always necessary to determine the difference, the difference relationship between numerical data. The teacher teaches children to understand the relationship of dependence between numerical data. The task analysis should be more detailed. During the analysis, children should go from the question to the condition of the problem. It should be explained that in the choice of an arithmetic operation, the main question is always the problem, the solution depends on its content and formulation. Therefore, you should start with an analysis of the issue. First, the children are presented with a task without a question. For example: “For a walk, the children took four large balls and one small one. What it is? Can this be called an arithmetic problem?” the teacher addresses the children. “No, this is only a condition of the problem,” the children answer. "Now put your own question to this problem."

Children should be led to the fact that two questions can be posed to this condition of the problem:

1. How many balls were taken for a walk?
2. How many more big balls were taken than small ones?

In accordance with the first question, addition should be performed, and in accordance with the second, subtraction. This convinces the children that the analysis of the problem should begin with a question. The line of reasoning can be as follows: to find out how many balls the children took for a walk, you need to know how many large and small ones they took separately and find their total number. In the second case, you need to find how many more balls there are than others, that is, determine the difference. The difference is always found by subtraction: subtract the smaller from the larger number.

So, tasks of the third type help the educator to consolidate knowledge about the structure of the task and contribute to the development in children of the ability to distinguish and find the appropriate arithmetic operation.

In these classes, not mechanically, but more or less consciously, children perform actions, argue the choice of an arithmetic operation. Tasks of this type should also be compared with tasks of the first and second types.

Computing activity at preschool age involves mastering by children the arithmetic operations of addition and subtraction related to operating system mathematics and subject to special laws of operational actions.

In order for children to better remember numerical data, cards with numbers are used, and a little later, signs.

At first, it is better to limit numerical data in tasks to the first five numbers of the natural series. Children in such cases, as a rule, easily find the answer. The main goal of these classes is to teach to analyze the problem, its structure, to understand the mathematical essence. Children learn to identify structural components tasks, numerical data, argue arithmetic operations, etc.

Particular attention during this period should be given to teaching children how to compose and solve problems using illustrations and numerical examples.

So, the teacher addresses the children: "Now we will compose and solve problems in the picture." At the same time, the attention of children is drawn to the picture, which depicts a river, five children play on the shore, and two children swim to the shore in boats. It is proposed to consider the picture and answer the question: “What is shown in the picture? What did the artist want to talk about? Where kids play? How many children are on the beach? What are these kids doing? (Points to the children in the boat.) How many are there? When they come ashore, will there be more or less of them on the shore? Make up a problem with this picture.

The teacher calls two or three children and listens to the tasks they have compiled. Then he chooses the most successful problem, and all together solve it. “What is the issue about? How many children were playing on the beach? How many children came in the boat? What needs to be done to solve the problem? How can the number "five" be added to the number "two"? -- 5+1 + 1=7.

The teacher makes sure that the children correctly formulate the arithmetic operation and explain the method of counting by one.

Similarly, they compose and solve other problems. At the end of the lesson, the teacher asks what the children were doing, clarifies their answers: “That's right, we learned to compose and solve problems, choose the appropriate action, add and subtract the number 2 by counting and counting by one.”

In approximately the same way, children compose and solve problems according to a numerical example. Drawing up and solving arithmetic problems based on a numerical example requires even more complex mental activity, since the content of the problem cannot be arbitrary, but is based on numerical example as per diagram. At the beginning, the attention of children is drawn to the action itself. In accordance with the action (addition or subtraction), a condition and a question in the task are compiled. It is possible to complicate the goal - not for each numerical example a new problem is compiled, and sometimes several problems of different types are compiled for the same example. This, of course, is much more difficult, but it is most effective for mental development child.

So, according to the numerical example 4 + 2, children compose and solve two problems: the first - to find the sum (how much in total), the second - to the ratio "more by several units" (by 2). At the same time, the child must be aware of the relationships and dependencies between numerical data.

Based on example 4 - 2, children should make three tasks: the first, second and third types. First, the teacher helps the children with questions, suggestions: “Now we will make up a task where there will be the words “2 less”, and then, using this very example, we will make up a task where there will be no such words, and it will be necessary to determine the difference in quantity (how much is left)” . And then the teacher asks: “Is it possible to create a new, completely different task based on this example?” If the children themselves cannot orient themselves, then the teacher prompts them: "Make a problem where the question would begin with the words" how much more (less) "".

Such activities with children help them understand the main thing: arithmetic problems can be different in content, but a mathematical expression (solution) can be the same. During this period of study great importance has a "detailed" method of calculation, activating mental activity child. The day before, the teacher repeats with the children the quantitative composition of the number of units and suggests adding the number 2 not immediately, but first counting 1, then another 1. The inclusion of a detailed method in computational activity ensures the development of logical thinking, while contributing to the assimilation of the essence of this activity.

After the children have formed ideas and some concepts about the arithmetic problem, the relationship between numerical data, between the condition and the question of the problem, you can proceed to the next stage in learning - familiarizing them with the transformation of direct problems into inverse ones. This will enable you to get deeper mathematical formula tasks, the specifics of each type of tasks. The teacher explains to the children that each simple arithmetic problem can be converted into a new one if the desired task is taken as one of the data new task, and consider one of the data of the transformed problem to be the required one in the new problem.

Such problems, where one of the data of the first is the desired in the second, and the desired of the second problem is included in the data of the first, are called mutually inverse problems.

So, from each direct arithmetic problem, 2 inverse problems can be made by transformation.

If children, when solving problems from the first steps, will be guided by significant connections and relationships, then the words “became”, remained “and others do not disorient them. Regardless of these words, children choose the correct arithmetic operation. Moreover, it is at this stage that the teacher should draw the attention of children to the independence of the choice of solving the problem from individual words and expressions.

Familiarization with direct and inverse problems increases cognitive activity children, develops their ability to think logically. When solving any problems, children should proceed from the question of the problem. An adult teaches a child to justify his actions, in this case justify the choice of arithmetic operation. At the same time, the train of thought can go according to the scheme: “To find out ... we need ... because ...”, etc.

In the group of the seventh year of life, children can be introduced to new methods of calculation - based on counting in groups. Children, having learned to count in pairs, triples, can immediately add the number 2, and then 3. However, this should not be rushed. It is important that children develop strong, fairly conscious skills and abilities of counting and counting by one.

AT modern research according to the methodology of mathematical development, there are some recommendations for the formation of generalized methods for solving arithmetic problems in children. One of these methods is the solution of problems according to the scheme-formula. This position is substantiated and experimentally verified in the studies of N. I. Nepomnyashchaya, L. P. Klyueva, E. A. Tarkhanova, R. L. Nepomnyashchaya. The formula proposed by the authors is a schematic representation of the relationship between the part and the whole. The work preceding this stage is the practical division of an object (circle, square, strip of paper) into parts. What the children do in practice, the educator then depicts in a formula scheme (Fig. 29). At the same time, he argues as follows: “If the circle is divided in half, then two halves will be obtained. If these halves are added together, then a whole circle is formed again. If we subtract one part from the whole circle, we get another part of this circle. And now let's try, before solving some problems (the word “some” is underlined), to determine what the question orients us to in the problem: to find a part or a whole. An unknown whole is always found by adding parts, and a part of a whole by subtracting.

For example: “To make a pattern, the girl took 4 blue and 3 red circles. From how many circles did the girl make a pattern? Children reason like this: “According to the condition of the problem, the drawing is made up of blue and red circles. These are parts. You need to find out how many circles the pattern is made up of. This is the whole. The whole is always found by adding the parts (4 + 3 =)."

For high level children intellectual development problematic (indirect) tasks can be proposed. Familiarization of children of the seventh year of life with tasks of this type is possible and is of great importance for their mental development. On this basis, in the future, skills will be formed to analyze an arithmetic problem, explain the course of a solution, and choose an arithmetic operation. Indirect tasks differ in that both numbers characterize the same object in them, and the question is aimed at determining the amount of another object. Difficulties in solving such problems are determined by the very structure and content of the problem. As a rule, in these tasks there are words that disorient the child when choosing an arithmetic operation. Despite the fact that the condition of the problem contains the words “more”, “arrived”, “older”, etc., you should perform the opposite action - subtraction. In order for the child to correctly orient himself, the teacher teaches him to analyze the problem more carefully. To choose an arithmetic operation, the child must be able to reason, think logically. An example of an indirect problem: “There were 5 mushrooms in the basket, which is 2 mushrooms more than they are on the table. How many mushrooms are on the table? Often children, focusing on non-essential signs, namely individual words(in this case, the word "more"), rush to perform the addition operation, making a gross mathematical error.

The teacher emphasizes the features of such tasks, suggesting that they reason together like this: “In the condition of the problem, both numbers characterize one object - the number of mushrooms in the basket. There are 5 mushrooms in it and 2 more in it than on the table. You need to find out how many mushrooms are on the table. If there are 2 more mushrooms in the basket, then there are 2 less mushrooms on the table. To find out how many of them are on the table, subtract 2 from 5 (5-2 = ?).”

When compiling tasks, the educator should remember that it is important to diversify the wording in the condition and question of the task: how much higher, harder, more expensive, etc.

Along with solving arithmetic problems, children are offered arithmetic examples that help to consolidate their computational skills. At the same time, children are introduced to some laws of addition.

It is known that it is always easier to perform addition if the second term is less than the first. However, this is not always the case in the example; it may be the other way around - the first term is less and the second is greater (for example, 2 + 1 = 1). In this case, there is a need to introduce the children to the commutative law of addition: 2 + 7 = 7 + 2. First, the teacher shows this on concrete examples, for example on bars. At the same time, he updates the knowledge of children about the composition of the number of two smaller ones. Children learned well that the number 9 can be formed (made up) from two smaller numbers: 2 and 7 or, which is the same, 7 and 2. Based on numerous examples with visual material, children make a conclusion-generalization: the action of addition is easier to perform if more add less, and the result will not change if you rearrange these numbers, swap them.

For school year it is enough to conduct 10-12 lessons on teaching children to solve arithmetic problems and examples (Table 1).

Below is the program content of these classes.

1. Familiarize yourself with the concept of "task". Condition and question in the problem. Dramatization tasks, illustration tasks of the first type. Numbers within 5, one of the numbers is 1.
2. To consolidate the concept of the structure of the task. Solving problems with pictures. Tasks of the second type. Signs "+", "--", "=". oral tasks. Numbers within 5, one of the numbers is 1. Learning how to calculate based on understanding the relationship between adjacent numbers.
3. Comparison of problems of the first and second types. Self-compilation of tasks according to the picture, according to numerical data and according to the condition.
4. Tasks for adding and subtracting numbers over 1 (2 = 1 + 1; 3=1 + 1 + 1). Tasks of the third type - on the relationship between numbers. Comparison of tasks of all three types.
5. Mutually inverse problems. Transformation of arithmetic problems. Drawing up tasks according to the numerical example 4 + 2; 4 - 2 of all three types.
6. Acquaintance with arithmetic examples. Formation of computing skills. Compilation of tasks on a numerical example.
7. Solving problems within 10 based on the composition of the number of two smaller numbers. Ability to justify your actions. The reasoning algorithm for solving a problem is from a question to a condition.
8. Solving problems by formula. The logic of reasoning from the question to the condition of the problem.
9. Indirect tasks. Problem tasks. Solving arithmetic examples.
10. Non-standard tasks(in poetic form, jokes, etc.). Connection with measurement and temporal relations.
11. Solving addition problems based on the commutative law of addition. Solving problems by formula.
12. Solving problems of the first, second and third types. The logic of reasoning in solving problems. Graphic image task content. pseudomath arithmetic numeric child

So, the program of education in kindergarten and the methodology of mathematical development great attention focus on the problem of teaching computational activity. However, only as a result of targeted systematic work, children develop sufficiently strong and conscious knowledge and skills in computational activity, and this is an important prerequisite for mastering mathematics at school.

Questions and tasks

1. Expand the specifics of counting and computing activities, justify the connection between counting and computing.
2. Analyze several alternative programs (or programs different years publications) in terms of their orientation to the level of intellectual development of each child.
3. Compose perspective plan for one quarter to familiarize older preschoolers with computing activities. On his example, prove the developmental nature of learning.
4. What is your attitude to the method of gradual development of computational activity in children preschool age?

§ 1 Ways to solve text problems

There are several ways to solve word problems:

arithmetic method - this is a way to solve a text problem using numbers and signs of arithmetic operations of addition, subtraction, multiplication and division, that is, using several operations on numbers that are interconnected;

The algebraic method is a way of solving a text problem by introducing variables and compiling corresponding equation or inequalities, or systems of equations or inequalities;

The geometric method is a way of solving a text problem using geometric knowledge;

Schematic method - this is a way to solve a text problem using diagrams;

graphic method is a way to solve a text problem using graphs in rectangular system coordinates.

Each of these methods involves translating the conditions of the problem into the language of mathematics. This action of mathematics is called mathematical modeling. The result of this action is called mathematical model. When applied various ways solutions are obtained by various mathematical models. In the arithmetic method, the mathematical model is a numerical expression, that is, a numerical example with several actions, and the final result of the calculations will be the solution to the problem. In the algebraic way, the mathematical model is most often an equation, and solving the equation gives a solution to the problem. In a geometric way, a mathematical model can be geometric figure, and the solution to the problem - for example, one of the found elements of this figure. In the schematic method, a mathematical model is a diagram with the help of which a solution to a problem is found. AT graphical way a mathematical model is a graph built according to the condition of the problem. With this method, the solution of the problem can be the coordinates certain points graphs.

§ 2 An example of solving a text problem in an arithmetic way

In this lesson, we will consider the arithmetic method of solving the problem in more detail.

To solve a problem in an arithmetic way means to find the answer to main question task by performing arithmetic operations on numeric data from the task condition. The same problem can be solved in different arithmetic ways. They differ from each other in the number of actions and the sequence of these actions in the process of solving the problem.

For example. Consider the following problem. Three friends Sasha, Kolya and Vitya were picking mushrooms in the forest. Kolya collected 2 times less mushrooms than Sasha, Vitya - 6 more mushrooms than Kolya. How many mushrooms did the three friends pick together if Sasha picked 22 mushrooms?

Helps to determine the correct course of logical reasoning short entry task conditions in the form of a table.

Let's solve this problem by actions or the so-called method of solving problems by questions. To begin with, let's answer the first question "How many mushrooms did Kolya collect?".

According to the condition of the problem, “Kolya collected 2 times less mushrooms than Sasha,” which means that in order to answer the question, 22 must be divided by 2. As a result, it turned out that Kolya collected 11 mushrooms. (22:2=11(mushrooms) - collected Kolya).

The next step is to answer the second question of the problem "How many mushrooms did Vitya collect?". According to the condition of the problem, “Vitya collected 6 more mushrooms than Kolya,” which means that to answer the question, you need to add 6 to 11. As a result, it turned out that Vitya collected 17 mushrooms.

22+22:2+(22:2+6)=50 mushrooms collected by three friends together.

Ability to solve problems using arithmetic numeric expressions talking about more high level mathematical preparation in comparison with the ability to solve word problems by actions.

List of used literature:

G.N. Timofeev Mathematics for applicants to universities. Tutorial. Text problems. - Yoshkar-Ola: Mar. state university, 2006
V. Bulynin Application graphic methods when solving word problems. - Weekly educational and methodical newspaper "Mathematics", No. 14, 2005.
N.I. Popov, A.N. Marasanov Tasks for drawing up equations. Tutorial. Yoshkar-Ola: Mar. state university, 2003
ON THE. Zaripova The program of the elective course "Text Problems". http://festival.1september.ru/articles/310281/
ON THE. Zaripova Methodology for solving problems of the vts group. Materials for the elective course "Solving text problems" http://festival.1september.ru/articles/415044/

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