Methods for solving word problems using an arithmetic method. Simple text arithmetic problems (their classification, examples and methods of solution) What is the arithmetic method of solution

Solve a math problem- this means finding such a sequence of general principles of mathematics, applying which to the conditions of the problem we obtain what we need to find - the answer.

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

Solve a problem arithmetic method - means finding the answer to the requirement of a problem by performing arithmetic operations on 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 a problem.

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

Solve using the algebraic method according to the following scheme:

1) identify the quantities discussed in the text of the problem and establish the relationship between them;

2) introduce variables (denote unknown quantities with letters);

3) using the entered variables and data, the problems create an equation or system of equations;

4) solve the resulting equation or system;

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

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

In primary school tasks are divided by the number of actions when solving for simple and compound ones. Problems in which only one action must be performed to answer a question are called simple. If to answer the question of a task you need to perform two or more actions, then such tasks are called compound.

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

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

Practical way.

Let's mark each fish with a circle. Let's draw 10 circles and designate the caught fish.

L L L O O O O O

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

Arithmetic method.

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

2) 10 - 7 = 3(p) - caught pikes.

Algebraic method.

Let x be the caught pikes. Then the number of all fish can be written as: 3 + 4 + x. According to the conditions 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 thereby answer the question of the problem.

Graphic method.

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.

The following is generally accepted in mathematics division of problem solving process :

1) analysis of the text of the problem, schematic recording of the problem, research 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 to the problem, verification.

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

1) Analysis: a) when reasoning moves from what is sought to the data of the problem; b) when the whole is divided into parts;

2) Synthesis: a) when moving from the task data to the required ones;
b) when elements are combined into a whole;

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

4) Inductive method of solving the problem: based on an accurate drawing, determine the properties of the figure, draw conclusions and prove them;

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

6) Forecasting - foreseeing the results that a search may lead to.

Let's take a closer look problem solving process:

Movement task. The boat traveled the distance along the river between two piers in 6 hours, and back in 8 hours. How long will it take a raft set along the river to travel the distance between the piers?

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 raft are floating have a certain flow speed. That is why the boat travels along 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 not these unknowns that need to be found, but the time during which the raft will travel this distance.

Schematic notation:

Boat 6 hours

raft boat

Finding a way to solve a problem. We need to find the time it takes the raft to travel the distance between the piers A and B. In order to find this time, you need to know the distance AB and the speed of the river flow. Both of them are unknown, so let’s denote the distance AB by the letter S (km), and the current speed and km/h. To relate these unknowns to the problem data, you need to know the boat's own speed. It is also unknown, let’s assume it is equal V km/h. Hence the solution plan arises, which consists in constructing a system of equations for the introduced unknowns.

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

Then the speed of the boat along the river is (V+a) km/h. Behind 6h the boat, moving at this speed, covered a distance of S (km). Therefore, 6( V + a) =S(1). This boat goes against the current at a speed of ( V - a)km/h and she passes this path for 8 hours, therefore 8( V - a) =S(2). Raft floating at the speed of the river and km/h, swam the distance S (km) behind x h, hence, Oh =S (3).

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

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

Checking the solution. We found that the raft will cover the distance between the piers in 48 hours. Therefore, its speed, equal to the speed of the river flow, is equal to . The speed of the boat along the river is equal to km/h, and against the current km/h In order to verify the correctness of the solution, it is enough to check whether the boat’s own speeds, found in two ways, are equal: + And
- . Having carried out the calculations, we obtain the correct equality: = . This means the problem was solved correctly.

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

Solution Analysis. We have reduced the solution to this problem to solving a system of three equations in four unknowns. However, one unknown had to be found. Therefore, the thought arises that this solution is not the most successful, although it is simple. We can offer another solution.

Knowing that the boat traveled the distance AB along the river in 6 hours, and against the current in 8 hours, we find that in 1 hour the boat, going with the river flow, covers part of this distance, and against the current. Then the difference between them - = is twice the distance AB covered by the raft in 1 hour. Means. The raft will cover part of the distance AB in 1 hour; therefore, it will travel the entire distance AB in 48 hours.

With this solution, we did not need to create a system of equations. However, this solution is more complicated than the one given above (not everyone can figure out the difference in speed of a boat downstream 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 whose speed in still water is 5 km/h, spending 10 hours on the entire journey. Find the speed of the river.

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

3. Two trains set off 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 speed of the trains.

4. In one warehouse there are 500 tons of coal, and in the other - 600 tons. The first warehouse supplies 9 tons daily, and the second - 11 tons of coal. In how many days will there be an equal amount of coal in the warehouses?

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

6. The product of a two-digit number and its sum of digits is 144. Find this number if its second digit is 2 more than the first.

7. Solve the following problems using the arithmetic method:

a) The motor boat spent 6 hours traveling down the river, and 10 hours on the way back. The speed of the boat in still water is 16 km/h. What is the speed of the river flow?

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. How much area will both tractor drivers plow while working for 5 days?

Solving problems using arithmetic methods

Math lesson in 5th grade.

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

Goals and objectives of the lesson:

developing the ability to solve problems using an arithmetic method;

development of creative abilities, cognitive interest;

development of logical thinking;

nurturing love for the subject;

fostering a 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 using an arithmetic method. Today we will solve problems of different types, but all of them will be solved without the help of equations.

II. Historical 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, someone who knew how to solve certain types of problems encountered in practice was considered trained.

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

(Students solve the problems correctly, and some not. All orally. They raise the cards and the teacher sees who solved the problem; the cards should contain the number 2.)

b) Problems in poetry and logical problems. (The teacher reads the problem aloud, the students raise the card with the correct answer.

The hedgehog gave the ducklings
Which one of the guys will answer?
Eight leather boots
How many ducklings were there?
(Four.)

Two nimble piglets
They were so cold, they were shaking.
Count and say:
How many boots should I buy them?
(Eight.)

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

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

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

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

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

8. When building the fence, the carpenters placed 5 pillars 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 minutes. Children solve problems at the board)
Tasks a) and b) are aimed at repeating the connection between the relations “by... more” and “by... less” with the operations of addition and subtraction.

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

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

Using problem c), students can be shown the solution to the problem “in reverse.”

c) There are 44 girls in three classes - this is 8 less than boys. How many boys are there in three classes?

In problem d) students can propose several solutions.

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

Task e) is intended to repeat the connection between “more in...” and “less in...”.

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

V. Physical education minute (2 minutes.)

Stand on one leg
It's like you're a steadfast soldier.
Raise your left leg.
Look, don't fall.
Now stand on the left,
If you are a brave soldier.

VI. Ancient, historical problems. Problems with fairy tale content (10 min.)

Problem e) 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 person have?

Movement task.

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

(It is easier to get to the answer using an equation. But students are encouraged to look for an arithmetic solution to the problem.)

1) 26 * 2 = 52 (versts) - the second train was so many miles behind the first;

2) 39 - 26 = 13 (versts) - by so many miles the second train was 1 hour behind the first;

3) 52: 13 = 4 (h) - that’s how long it took the first train to travel;

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.

The task is divided into parts.

h)Kikimora's task.The merman decided to marry the kikimore Ha-Ha. He planted several leeches on his kikimore 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 to be solved 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 and 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 (for example: the 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 problem 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 must come out whose number is the sum of all heard and seen numbers (for example: 3 claps, number 5 and number 1.);

what number is 4 greater than four?

I thought of a number, subtracted 3 from it, I got 7. What number did I think of?

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

We must try to select tasks so that the same numbers are not repeated in the answers, 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 using arithmetic?

- We must remember that the solution found to the problem must satisfy the conditions of the problem.

IX. Homework assignment. Grading (2 minutes.)

№ 387 (solve problems using arithmetic method), for weak students. For average and strong students, homework assignments are given on cards.

1. The bakery had 645 kg of black and white bread. After selling 215 kg of black and 287 kg of white bread, there was an equal amount of both types of bread left. How many kilograms of black and white bread were there in the bakery separately?

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

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

Literature

Vilenkin N., Zhokhov V., Chesnokov A.Mathematics. 5th grade. - M., “Mnemosyne”, 2002.

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

Volina V.Holiday of numbers. - M.: Knowledge, 1994.

Analyzing these problems, observing what the problems have in common from the point of view of mathematics, what are the differences, find an extraordinary way to solve problems, create a piggy bank of techniques for solving problems, learn to solve one problem in different ways. A simulator of problems grouped under the same theme “Arithmetic methods of solving problems” , tasks for group work and individual work.

“tasks for the simulator manual”

Trainer: “Arithmetic methods for solving problems”

“Comparing numbers by sum and difference.”

There are 80 boletus mushrooms in two baskets. The first basket contains 10 less boletus than the second. How many boletus mushrooms are there in each basket?

The sewing studio received 480 m of denim and drape. Denim fabric was supplied 140 m more than drapery. How many meters of denim did the studio receive?

The TV tower model consists of two blocks. The lower block is 130 cm shorter than the upper one. What are the heights of the upper and lower blocks if the height of the tower is 4 m 70 cm?

Two boxes contain 16 kg of cookies. Find the mass of cookies in each box if one of them contains 4 kg more cookies.

Problem from “Arithmetic” by L. N. Tolstoy.

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

b) Two men have 40 sheep, and one has 6 sheep less than the other. How many sheep does each man have?

There were 23 cars and motorcycles with sidecars in the garage. Cars and motorcycles have 87 wheels. How many motorcycles are there in the garage if each sidecar has a spare wheel?

"Eulerian Circles".

The house has 120 residents, some of whom have dogs and cats. There is a circle in the picture WITH depicts residents with dogs, circle TO – residents with cats. How many tenants have both dogs and cats? How many tenants have only dogs? How many tenants have only cats? How many tenants have neither dogs nor cats?

Of the 52 schoolchildren, 23 play volleyball and 35 basketball, and 16 play both volleyball and basketball. The rest do not play any of these sports. How many schoolchildren do not play any of these sports?

There is a circle in the picture A depicts all university employees who know English, circle N – who know German and circle F - French. How many university employees know: a) 3 languages; b) English and German; c) French? How many university employees are there? How many of them don't speak French?

The international conference was attended by 120 people. Of these, 60 speak Russian, 48 speak English, 32 speak German, 21 speak Russian and German, 19 speak English and German, 15 speak Russian and English, and 10 people spoke all three languages. How many conference participants do not speak any of these languages?

82 students sing in the choir and do dance, 32 students do dance and rhythmic gymnastics, and 78 students sing in the choir and do rhythmic gymnastics. How many students sing in a choir, do dance and do rhythmic gymnastics separately, if it is known that each student does only one thing?

Every family living in our house subscribes to either a newspaper or a magazine, or both. 75 families subscribe to a newspaper, and 27 families subscribe to a magazine, and only 13 families subscribe to both a magazine and a newspaper. How many families live in our house?

"Method of data adjustment".

There are 29 flowers in 3 small and 4 large bouquets, and 35 flowers in 5 small and 4 large bouquets. How many flowers are in each bouquet individually?

The mass of 2 chocolate bars - large and small - is 120 g, and 3 large and 2 small - 320 g. What is the mass of each bar?

5 apples and 3 pears weigh 810 g, and 3 apples and 5 pears weigh 870 g. How much does one apple weigh? One pear?

Four ducklings and five goslings weigh 4 kg 100 g, five ducklings and four goslings weigh 4 kg. How much does one duckling weigh?

For one horse and two cows, 34 kg of hay is given daily, and for two horses and one cow - 35 kg of hay. How much hay is given to one horse and how much to one cow?

3 red cubes and 6 blue cubes cost 165 tenge rubles. Moreover, five red ones are 95 tenge more expensive than two blue ones. How much does each cube cost?

2 sketchbooks and 3 stamp albums together cost 160 rubles, and 3 sketchbooks cost 45 rubles. more expensive than two stamp albums.

"Counts".

Seryozha decided to give his mother a bouquet of flowers (roses, tulips or carnations) for her birthday and put them either in a vase or in a jug. In how many ways can he do this?

How many three-digit numbers can be made from the digits 0, 1, 3, 5 if the digits in the number are not repeated?

On Wednesday in 5th grade there are five lessons: mathematics, physical education, history, Russian and science. How many different schedules for Wednesday can you create?

“An ancient way to solve problems involving mixing substances.”

How to mix oils? A certain person had two types of oil for sale: one at a price of 10 hryvnia per bucket, the other at 6 hryvnia per bucket. He wanted to make oil from these two oils, mixing them, costing 7 hryvnia per bucket. What parts of these two oils do you need to take to get a bucket of oil worth 7 hryvnia?

How much caramel do you need to take at a price of 260 tenge per 1 kg and at a price of 190 tenge per 1 kg to make 21 kg of the mixture at a price of 210 tenge per kilogram?

Someone has three varieties of tea - Ceylon for 5 hryvnia per pound, Indian for 8 hryvnia per pound and Chinese for 12 hryvnia per pound. In what proportions should these three varieties be mixed to get tea worth 6 hryvnia per pound?

Someone has silver of different standards: one is 12th standard, another is 10th standard, the third is 6th standard. How much silver should you take to get 1 pound of 9th standard silver?

The merchant bought 138 arshins of black and blue cloth for 540 rubles. The question is, how many arshins did he buy for both, if the blue one cost 5 rubles? for an arshin, and black - 3 rubles?

Various tasks.

For New Year's gifts, we bought 87 kg of fruit, and there were 17 kg more apples than oranges. How many apples and how many oranges did you buy?

At the New Year's tree, there were 3 times more snowflakes for children in carnival costumes than in Parsley costumes. How many children were there in Parsley costumes if there were 12 fewer of them?

Masha received 2 times less New Year's greetings than Kolya. How many congratulations did each person receive if there were 27 in total? (9 and 18).

28 kg of sweets were purchased for New Year's prizes. Candies “Swallow” made up 2 parts, “Muse” - 3 parts, “Romashka” - 2 parts. How many sweets of each type did you buy? (8, 8, 12).

There are 2004 kg of flour in the warehouse. Can it be put into bags weighing 9 kg and weighing 18 kg?

There are 5 different cups and 3 different saucers in the "Everything for Tea" store. In how many ways can you buy a cup and saucer?

A horse eats a haystack in 2 days, a cow in 3, a sheep in 6. How many days will it take them to eat the haystack if they eat it together?

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"lesson summary arif sp"

"Arithmetic methods for solving word problems."

For a student of mathematics, it is often more useful to solve the same problem in three different ways than to solve three or four different problems. By solving one problem in different ways, you can find out by comparison which one is shorter and more efficient. This is how experience is developed.

W.W. Sawyer

The purpose of the lesson: use the knowledge acquired in previous lessons, show imagination, intuition, imagination, and ingenuity to solve test problems in various ways.

Lesson objectives: educational: by analyzing these problems, observing what the problems have in common from the point of view of a mathematician, what are the differences, finding an extraordinary way to solve problems, creating a piggy bank of techniques for solving problems, learning to solve one problem in different ways.

Developmental: feel the need for self-realization when finding yourself in a certain role situation.

Educational: develop personal qualities, form a communicative culture.

Means of education: a simulator of problems grouped under the same theme “Arithmetic methods for solving problems”, tasks for working in a group and for individual work.

DURING THE CLASSES.

I. Organizational moment

Hello guys. Sit down. Today we have a lesson on the topic “Arithmetic methods for solving word problems.”

II. Updating knowledge.

Mathematics is one of the ancient and important sciences. People used a lot of mathematical knowledge in ancient times - thousands of years ago. They were necessary for merchants and builders, warriors and land surveyors, priests and travelers.

And nowadays, not a single person can get by in life without a good knowledge of mathematics. The basis of a good understanding of mathematics is the ability to count, think, reason, and find successful solutions to problems.

Today we will look at arithmetic methods for solving word problems, analyze ancient problems that have come down to us from different countries and times, problems on equalization, comparison by sum and difference, and others.

The purpose of the lesson is to involve you in the wonderful world of beauty, richness and diversity - a world of interesting challenges. And, therefore, introduce you to some arithmetic methods that lead to very elegant and instructive solutions.

A task is almost always a search, the discovery of some properties and relationships, and the means of solving it are intuition and conjecture, erudition and mastery of mathematical methods.

The main ones in mathematics are arithmetic and algebraic methods of solving problems.

Solving a problem using the arithmetic method means finding the answer to the requirement of the problem by performing arithmetic operations on numbers.

With the algebraic method, the answer to the question of the problem is found as a result of composing and solving the equation.

It is no secret that a person who owns different tools and uses them depending on the nature of the work being performed achieves significantly better results than a person who owns only one universal tool.

There are many arithmetic methods and non-standard techniques for solving problems. Today I want to introduce you to some of them.

1.Method of solving word problems “Comparing numbers by sum and difference.”

Task : Grandma collected 51 kg of carrots and cabbage from her summer cottage in the fall. There was 15 kg more cabbage than carrots. How many kilograms of carrots and how many kilograms of cabbage did grandma collect?

Questions that correspond to the points of the algorithm for solving problems of this class.

1. Find out what quantities are being discussed in the problem

About the number of carrots and cabbage that grandma collected, together and separately.

2. Indicate the values of which quantities need to be found in the problem.

How many kilograms of carrots and how many kilograms of cabbage did grandma collect?

3. Name the relationship between the quantities in the problem.

The problem talks about the sum and difference of quantities.

4. Name the sum and difference of the values of quantities.

Sum – 51 kg, difference – 15 kg.

5. By equalizing the quantities, find the double value of the smaller quantity (subtract the difference of the quantities from the sum of the quantities).

51 – 15 = 36 (kg) – double the amount of carrots.

6. Knowing the doubled value, find the smaller value (divide the doubled value by two).

36: 2 = 18 (kg) – carrots.

7. Using the difference between the quantities and the value of the smaller quantity, find the value of the larger quantity.

18 + 15 = 33 (kg) – cabbage. Answer: 18 kg, 33 kg. Task.There are pheasants and rabbits in the cage. There are 6 heads and 20 legs in total. How many rabbits and how many pheasants are in a cage ?
Method 1. Selection method:
2 pheasants, 4 rabbits.
Check: 2 + 4 = 6 (goals); 4 4 + 2 2 = 20 (feet).
This is a selection method (from the word “to select”). Advantages and disadvantages of this solution method (difficult to select if the numbers are large) Thus, there is an incentive to search for more convenient solution methods.
Discussion results: the selection method is convenient when working with small numbers; when the values increase, it becomes irrational and labor-intensive.
Method 2. Complete search of options.

A table is compiled:

Answer: 4 rabbits, 2 pheasants.
The name of this method is “full”. Discussion results: the exhaustive search method is convenient, but for large values it is quite labor-intensive.
Method 3. Guessing method.

Let's take an old Chinese problem:

The cage contains an unknown number of pheasants and rabbits. It is known that the entire cell contains 35 heads and 94 legs. Find out the number of pheasants and the number of rabbits.(Problem from the Chinese mathematical book “Kiu-Chang”, compiled 2600 BC).

Here is a dialogue found in the old masters of mathematics. - Let’s imagine that we put a carrot on the cage in which the pheasants and rabbits are sitting. All rabbits will stand on their hind legs to reach the carrot. How many feet will be on the ground at this moment?

But in the problem statement, 94 legs are given, where are the rest?

The remaining legs are not counted - these are the front legs of the rabbits.

How many are there?

24 (94 – 70 = 24)

How many rabbits are there?

12 (24: 2 = 12)

What about pheasants?

23 (35- 12 = 23)

The name of this method is “deficiency guessing method.” Try to explain this name yourself (those sitting in a cage have 2 or 4 legs, and we assumed that everyone has the smallest of these numbers - 2 legs).

Another way to solve the same problem. - Let’s try to solve this problem using the “surplus assumption method”: Let’s imagine that pheasants now have two more legs, then there will be all legs 35 × 4 =140.

But according to the conditions of the problem, there are only 94 legs, i.e. 140 – 94= 46 extra legs, whose are they? These are the legs of pheasants, they have an extra pair of legs. Means, pheasants will 46: 2 = 23, then rabbits 35 -23 = 12.
Discussion results: the assumption method has two options- By deficiency and excess; Compared to previous methods, it is more convenient because it is less labor-intensive.
Task. A caravan of camels is slowly walking through the desert, there are 40 of them in total. If you count all the humps on these camels, you will get 57 humps. How many dromedary camels are there in this caravan?1 way. Solve using equation.

Number of humps per person Number of camels Total humps

2 x 2 x

1 40 - X 40 - X 57

2 x + 40 - X = 57

x + 40 = 57

X = 57 -40

X = 17

Method 2.

- How many humps can camels have?

(there may be two or one)

Let's attach a flower to each camel's hump.

- How many flowers will you need? (40 camels – 40 flowers)

- How many humps will be left without flowers?

(There will be such 57-40=17 . This second humps Bactrian camels).

How many Bactrian camels? (17)

How many dromedary camels? (40-17=23)

What is the answer to the problem? ( 17 and 23 camels).

Task.In the garage there were cars and motorcycles with sidecars, 18 of them all together. The cars and motorcycles had 65 wheels. How many motorcycles with sidecars were in the garage, if cars have 4 wheels and motorcycles have 3 wheels?

1 way. Using the equation:

Number of wheels for 1 Number of total wheels

Mash. 4x 4 x

Mot. 3 18 -X 3(18 - X ) 65

4 x + 3(18 - X ) = 65

4 x + 5 4 -3 X =65

X = 65 - 54

X = 11, 18 – 11 = 7.

Let's reformulate the problem : The robbers, who came to the garage where 18 cars and motorcycles with sidecars were parked, removed three wheels from each car and motorcycle and took them away. How many wheels are left in the garage if there were 65 of them? Do they belong to a car or a motorcycle?

3×18=54 – that’s how many wheels the robbers took away,

65- 54 = 11 – so many wheels left (cars in the garage),

18 - 11 = 7 motorcycles.

Answer: 7 motorcycles.

On one's own:

There were 23 cars and motorcycles with sidecars in the garage. Cars and motorcycles have 87 wheels. How many motorcycles are there in the garage if each sidecar has a spare wheel?

- How many wheels do cars and motorcycles have together? (4×23=92)

- How many spare wheels did you put in each stroller? (92 - 87= 5)

- How many cars are in the garage? (23 - 5=18).

Task.In our class you can study English or French (optional). It is known that 20 schoolchildren study English, and 17 study French. In total, there are 32 students in the class. How many students study both English and French?

Let's draw two circles. In one we will record the number of schoolchildren studying English, in the other - schoolchildren studying French. Since according to the conditions of the problem there are students studyingboth languages: English and French, then the circles will have a common part. The conditions of this problem are not so easy to understand. If you add 20 and 17, you get more than 32. This is explained by the fact that we counted some schoolchildren here twice - namely those who study both languages: English and French. So, (20 + 17) – 32 = 5 Students learn both languages: English and French.

English Fran.

20 lessons 17 school

(20 + 17) – 32 = 5 (students).

Schemes similar to the one we used to solve the problem are called in mathematics Euler circles (or diagrams). Leonhard Euler (1736) born in Switzerland. But for many years he lived and worked in Russia.

Task.Every family living in our house subscribes to either a newspaper or a magazine, or both. 75 families subscribe to a newspaper, and 27 families subscribe to a magazine, and only 13 families subscribe to both a magazine and a newspaper. How many families live in our house?

Newspapers magazines

The picture shows that 89 families live in the house.

Task.The international conference was attended by 120 people. Of these, 60 speak Russian, 48 speak English, 32 speak German, 21 speak Russian and German, 19 speak English and German, 15 speak Russian and English, and 10 people spoke all three languages. How many conference participants do not speak any of these languages?

Russian 15 English

21 10 19

German

Solution: 120 – (60 + 48 + 32 -21 – 19 – 15 + 10) = 25 (persons).

Task. Three kittens and two puppies weigh 2 kg 600 g, and two kittens and three puppies weigh 2 kg 900 g. How much does the puppy weigh?

3 kittens and 2 puppies – 2kg 600 g

2 kittens and 3 puppies – 2 kg 900 g.

It follows from the condition that 5 kittens and 5 puppies weigh 5 kg 500 g. This means that 1 kitten and 1 puppy weigh 1 kg 100 g

2 cats and 2 puppies. weigh 2 kg 200 g

Let's compare the conditions -

2 kittens + 3 puppies = 2kg 900 g

2 kittens + 2 puppies = 2 kg 200 g, we see that the puppy weighs 700 g.

Task.For one horse and two cows, 34 kg of hay is given daily, and for two horses and one cow - 35 kg of hay. How much hay is given to one horse and how much to one cow?

Let's write down a brief statement of the problem:

1 horse and 2 cows -34kg.

2 horses and 1 cow -35kg.

Is it possible to know how much hay is needed for 3 horses and 3 cows?

(for 3 horses and 3 cows – 34+35=69 kg)

Is it possible to find out how much hay is needed for one horse and one cow? (69: 3 – 23kg)

How much hay does one horse need? (35-23=12kg)

How much hay does one cow need? (23 -13 =11kg)

Answer: 12kg and 11kg.

Task.Madina decided to have breakfast at the school cafeteria. Study the menu and answer, in how many ways can she choose a drink and a confectionery item?

	Confectionery
	Cheesecake

Let's assume that Madina chooses tea as a drink. What confectionery product can she choose for tea? (tea - cheesecake, tea - cookies, tea - bun)

How many ways? (3)

What if it's compote? (also 3)

How can you find out how many ways Madina can use to choose her lunch? (3+3+3=9)

Yes you are right. But to make it easier for us to solve this problem, we will use graphs. The word "graph" in mathematics means a picture with several points drawn, some of which are connected by lines. Let's denote drinks and confectionery products with dots and connect the pairs of those dishes that Madina chooses.

tea milk compote

cheesecake cookies bun

Now let's count the number of lines. There are 9 of them. This means that there are 9 ways to choose dishes.

Task.Seryozha decided to give his mother a bouquet of flowers (roses, tulips or carnations) for her birthday and put them either in a vase or in a jug. In how many ways can he do this?

How many ways do you think? (3)

Why? (3 colors)

Yes. But there are also different types of dishes: either a vase or a jug. Let's try to complete the task graphically.

vase jug

roses tulips carnations

Count the lines. How many are there? (6)

So, how many ways does Seryozha have to choose? (6)

Lesson summary.

Today we solved a number of problems. But the work is not completed, there is a desire to continue it, and I hope that this will help you successfully solve word problems.

We know that problem solving is a practical art, like swimming or playing the piano. You can learn it only by imitating good examples and constantly practicing.

These are only the simplest of problems; complex ones remain a subject for future study. But there are still many more of them than we could solve. And if at the end of the lesson you can solve problems “behind the pages of the educational material,” then we can consider that I have completed my task.

Knowledge of mathematics helps to solve a certain life problem. In life, you will have to regularly resolve certain issues, for this you need to develop intellectual abilities, thanks to which internal potential develops, the ability to foresee the situation, make predictions, and make non-standard decisions develops.

I want to end the lesson with the words: “Every well-solved mathematical problem gives mental pleasure.” (G. Hesse).

Do you agree with this?

Homework.

The following assignment will be given at home: using the texts of solved problems as a sample, solve problems No. 8, 17, 26 using the methods that we have studied.

A primary school teacher simply needs to know what types of problems are available. Today you will learn about simple text arithmetic problems. Simple text arithmetic problems are problems that can be solved with one arithmetic operation. When we read a problem, we automatically associate it with some type, and then it immediately becomes easy to understand what action should be used to solve it.

I will provide you not only with the classification of simple word problems, but also give examples of them, and also tell you about solving word problems using an arithmetic method. I took all the examples from mathematics textbooks for grade 2 (part 1, part 2), which are used in schools in Belarus.

All simple arithmetic problems are divided into two large groups:

— AD I (+/-), that is, those that are solved by first-order arithmetic operations (addition or subtraction);

— AD II (*/:), that is, those that are solved by second-order arithmetic operations (multiplication or division).

Let's consider the first group of simple text arithmetic problems (AD I):

1) Problems that reveal the specific meaning of addition (+)

4 girls and 5 boys took part in the running competition. How many students from the class took part in the competition?

After Sasha solved 9 examples, he still had 3 more examples to solve. How many examples did Sasha need to solve?

The following problems are solved by addition: a+b=?

2) Problems that reveal the specific meaning of subtraction (-)

Mom baked 15 pies. How many pies are left after eating 10 pies?

There were 15 glasses of juice in the jar. We drank 5 glasses at lunch. How many glasses of juice are left?

The following problems are solved by subtraction: a-b=?

3) Tasks on the relationship between components and the result of addition or subtraction:

a) to find the unknown 1st term (?+a=b)

The boy put 4 pencils in the box. There were 13 of them there. How many pencils were in the box initially?

To solve this problem, you need to subtract the well-known 2nd term from the result of the action: b-a=?

b) to find the unknown 2nd term (a+?=b)

13 glasses of water were poured into a saucepan and a kettle. How many glasses of water were poured into the kettle if 5 glasses were poured into the pan?

Problems of this type are solved by subtraction; the known first term is subtracted from the result of the action: b-a=?

c) to find the unknown minuend (?-a=b)

Olga collected a bouquet. She put 3 colors in the vase, and she had 7 flowers left. How many flowers were in the bouquet?

In an arithmetic way, word problems of this type are solved by adding the result of the action and the subtrahend: b+a=?

d) to find the unknown subtrahend (a-?=b)

We bought 2 dozen eggs. After taking several eggs for baking, there are 15 left. How many eggs did you take?

These problems are solved by subtraction: from the minuend we subtract the result of the action: a-b=?

4) Tasks to decrease / increase by several units in direct, indirect form

examples of problems involving reduction by several units in direct form:

One box contained 20 kg of bananas, and the second had 5 less. How many kilograms of bananas were in the second box?

The first class collected 19 boxes of apples, and the second class collected 4 boxes less. How many boxes of apples did the second grade pick?

These problems are solved by subtraction (a-b=?)

I did not find any examples of problems involving reduction in indirect form, as well as increasing in direct or indirect form, in the 2nd grade textbook on mathematics. If necessary, write in the comments and I will supplement the article with my own examples.

5) Difference comparison problems

The weight of a goose is 7 kg, and that of a chicken is 3 kg. How many kilograms does a chicken weigh less than a goose?

The first box contains 14 pencils, and the second box contains 7. How many more pencils are in the first box than in the second?

Solving word problems involving difference comparisons is done by subtracting a smaller number from a larger number.

We have finished dealing with simple text arithmetic problems of group 1 and are moving on to problems of group 2. If anything was unclear to you, ask in the comments.

The second group of simple text arithmetic problems (AD II):

1) Problems that reveal the specific meaning of multiplication

How many legs do two dogs have? Three dogs?

There are three cars parked near the house. Each car has 4 wheels. How many wheels do three cars have?

These problems are solved by multiplication: a*b=?

2) Tasks that reveal the specific meaning of division:

a) by content

10 cakes were distributed to the children, two each. How many children received cakes?

2 kg bags contain 14 kg of flour. How many such packages?

In these problems we will find out how many parts were obtained with equal content.

b) into equal parts

A strip 10 cm long was cut into two equal parts. How long is each part?

Nina divided 10 cakes equally onto 2 plates. How many cakes are there on one plate?

And in these problems we find out what the content of one equal part is.

Be that as it may, all these problems are solved by division: a:b=?

3) Problems on the relationship between the component and the result of multiplication and division:

a) to find the unknown first factor: ?*a=b

Own example:

Several boxes contain 6 pencils. There are 24 pencils in total in the boxes. How many boxes?

Solved by dividing the product by the known second factor: b:a=?

b) to find the unknown second factor: a*?=b

In the cafe you can seat 3 people at one table. How many of these tables will be occupied if 15 people come there?

Solved by dividing the product by the known first factor: b:a=?

c) to find the unknown dividend: ?:a=b

Own example:

Kolya brought candy to the class and divided it equally among all the students. There are 16 children in the class. Everyone received 3 candies. How many sweets did Kolya bring?

Solved by multiplying the quotient by the divisor: b*a=?

d) to find an unknown divisor: a:?=b

Own example:

Vitya brought 44 candies to the class and divided them equally among all the students. Everyone received 2 candies. How many students are there in the class?

Solved by dividing the dividend by the quotient: a:b=?

4) Tasks to increase / decrease several times in direct or indirect form

No examples of such text arithmetic problems were found in the 2nd grade textbook.

5) Multiple comparison problems

Solved by dividing the larger by the smaller.

Friends, the entire above classification of simple word problems is only part of a larger classification of all word problems. In addition, there are also problems for finding percentages that I didn’t tell you about. You can learn about all this from this video:

And my gratitude will remain with you!

introduce different ways of solving problems;
give ideas about the algebraic method of solving,
teach children to choose different solutions and create inverse problems.

develop logical thinking,
development of mental operations such as analysis, synthesis.

During the classes

1. Warm up

(Students stand at their seats, the teacher asks a question, if the student answered correctly, then sits down).

What is an equation?
What does it mean to find the root of an equation?
How to find an unknown multiplier? Divider? Minuend?
Continue with the definitions: Speed is...
To find the distance you need...
To find time, you need...

2. Checking homework

(At home, the children looked for definitions in reference books: algebra , arithmetic, geometry).

What does algebra study? arithmetic? geometry?

Algebra – the science that studies questions of equations and inequalities.
Geometry- one of the oldest parts of mathematics, studying spatial relationships and shapes of bodies.
Arithmetic– the science of numbers and operations on them.

(We will need these terms later in the lesson.)

3. Listen to the problem

Each of the four cells contains 1 animal. There are inscriptions on each cell, but none of them correspond to reality. Indicate who is in each cell. Place the animals in their cells (each child has a set of canvas and cards with pictures of animals).

Show what you got. How did you reason? (check on the board).
How did you solve this problem? (Reasoning, thinking logically).
What is this task? (Logical).

But mostly in mathematics lessons we solve problems in which it is necessary to perform mathematical transformations.

4. Read the problems

12 kg of wool was sheared from two camels. The second one cut 3 times more than the first one. How many kilograms of wool were sheared from each camel?
A leopard weighs 340 kg, a giraffe is 3 times heavier than a leopard, and a lion is 790 kg lighter than a giraffe. How many kilograms is a leopard heavier than a lion?
Two giraffes ran towards each other. One ran at a speed of 12 m/s, the speed of the other was 15 m/s. After how many seconds will they meet if the distance between them was 135 meters?

Compare tasks. What common? What are their differences?

Read the problem to be solved by writing an equation.
Read the problem that needs to be solved by action?
Which problem can be solved in two ways?
Formulate the topic of our lesson.

Different ways to solve problems

5. Solve any problem by making a short note (in the form of a table, drawing)

Two people are working at the board.

Examination

How did you solve the first problem? (Equation).
What is the name of the branch of mathematics that studies equations? (Algebra).
(Algebraic).
How were the second and third problems solved? (By actions).
What branch of mathematics studies this? (Arithmetic).
What will this solution be called? (Arithmetic).

(Hang it on the board):