What is the algebraic method of solving. Video lesson “Arithmetic method for solving word problems

Learning to solve word problems plays an important role in the development of mathematical knowledge. Word problems provide a lot of scope for developing students' thinking. Learning to solve problems is not only about teaching the technique of obtaining the correct answers in some typical situations, but also about learning a creative approach to finding a solution, gaining experience in mental activity and demonstrating to students the capabilities of mathematics in solving a variety of problems. However, when solving word problems in grades 5-6, an equation is most often used. But the thinking of fifth-graders is not yet ready for the formal procedures involved in solving equations. The arithmetic method of solving problems has a number of advantages over the algebraic one because the result of each step of the actions is clearer and more specific, and does not go beyond the experience of fifth-graders. Students solve problems using actions better and faster than using equations. Children's thinking is concrete, and it must be developed on specific objects and quantities, then gradually move on to operating with abstract images.

Working on the task involves carefully reading the text of the condition, understanding the meaning of each word. I will give examples of problems that can be easily and simply solved using arithmetic.

Task 1. To make jam, take two parts of raspberries and three parts of sugar. How many kilograms of sugar do you need to take for 2 kg 600 g of raspberries?

When solving a problem into “parts,” you must learn to visualize the conditions of the problem, i.e. It's better to rely on the drawing.

1. 2600:2=1300 (g) - accounts for one part of the jam;
2. 1300*3= 3900 (g) - you need to take sugar.

Task 2. There were 3 times on the first shelf more books than the second. There were 120 books on the two shelves together. How many books were on each shelf?

1) 1+3=4 (parts) - accounts for all books;

2) 120:4=30 (books) - accounts for one part (books on the second shelf);

3) 30*3=90 (books) - stood on the first shelf.

Task 3. Pheasants and rabbits are sitting in a cage. In total it has 27 heads and 74 legs. Find out the number of pheasants and the number of rabbits in the cage.

Let's imagine that we put a carrot on the lid of the cage in which the pheasants and rabbits are sitting. Then all the rabbits will stand on their hind legs to reach it. Then:

1. 27*2=54 (legs) - will stand on the floor;
2. 74-54=20 (legs) - will be at the top;
3. 20:2=10 (rabbits);
4. 27-10=17 (pheasants).

Task 4. There are 30 students in our class. 23 people went on an excursion to the museum, and 21 went to the cinema, and 5 people did not go on either the excursion or the cinema. How many people went on both the excursion and the cinema?

“Eulerian circles” can be used to analyze the condition and select a solution plan.

1. 30-5=25 (persons) – went either to the cinema or on an excursion,
2. 25-23=2 (person) – went only to the cinema;
3. 21-2=19 (person) – went to the cinema and on an excursion.

Task 5. Three ducklings and four goslings weigh 2 kg 500 g, and four ducklings and three goslings weigh 2 kg 400 g. How much does one gosling weigh?

1. 2500+2400=2900 (g) – seven ducklings and seven goslings weigh;
2. 4900:7=700 (g) – the weight of one duckling and one gosling;
3. 700*3=2100 (g) – weight of 3 ducklings and 3 goslings;
4. 2500-2100=400 (g) – weight of the caterpillar.

Task 6. For kindergarten bought 20 pyramids: large and small - 7 and 5 rings each. All pyramids have 128 rings. How many large pyramids were there?

Let's imagine that we removed two rings from all the large pyramids. Then:

1) 20*5=100 (rings) – left;

2) 128-100-28 (rings) – we removed;

3) 28:2=14 (large pyramids).

Task 7. A watermelon weighing 20 kg contained 99% water. As it dried out a bit, its water content dropped to 98%. Determine the mass of the watermelon.

For convenience, the solution will be accompanied by an illustration of rectangles.

99% water	1% dry matter
98% water	2% dry matter

In this case, it is advisable to draw the rectangles of the “dry matter” equal, because the mass of the “dry matter” in the watermelon remains unchanged.

1) 20:100=0.2 (kg) – mass of “dry matter”;

2) 0.2:2=0.1 (kg) – accounts for 1% of dried watermelon;

3) 0.1*100=10 (kg) – mass of watermelon.

Task 8. The guests asked: how old was each of the three 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?

1. 38-28=10 (years) – Lyuba;
2. 23-10=13 (years) – Nadya;
3. 28-13=15 (years old) – Vera.

The arithmetic method of solving word problems teaches the child to act consciously, logically correctly, because when solving in this way, attention to the question “why” increases and there is great developmental potential. This contributes to the development of students, the formation of their interest in solving problems and in the science of mathematics itself.

To make learning feasible, exciting and instructive, you need to be very careful when choosing text problems, consider different ways to solve them, choosing the best ones, and develop logical thinking, which is necessary in the future when solving geometric problems.

Students can learn to solve problems only by solving them. “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,” writes D. Polya in the book “Mathematical Discovery.”

Arithmetic way to solve word problems

“...while we are trying to connect teaching mathematics with life, it will be difficult for us to do without word problems - a traditional means of teaching mathematics for the Russian methodology.”

A.V.Shevkin

The ability to solve word problems is one of the main indicators of students’ mathematical development and the depth of their learning educational material, clarity in reasoning, understanding of the logical aspects of various issues.

For most schoolchildren, word problems are difficult and therefore unfavorite educational material. However, in the school mathematics course he is given great value, since the tasks contribute to the development, first of all, of logical thinking, spatial imagination, and the practical application of mathematical knowledge in human activity.

In the process of solving problems, students gain experience working with quantities, understand the relationships between them, and gain experience in applying mathematics to solve real-life problems.Solving word problems develops logical culture, arousing interest first in the process of finding a solution to the problem, and then in the subject being studied.

The traditional Russian school has always paid special attentionteaching children to solve word problems. Historically, it was enough for a long time mathematical knowledge was passed down from generation to generation in the form of word problems with solutions. Their significance also lay in their applied significance, since in their content these were practical tasks (banking, trade, land calculations, etc.). In Russia, an educated person was considered to be someone who knew how to solve these typical problems, which are very important in everyday life.

It should be noted that learning to solve practical problems was not easy. Memorization of the solution method without conscious understanding of the condition was often observed. The main thing is to determine the type of problem and find a rule for solving it; understanding was not important.

Towards the middleXXcentury was developed good technique problem solving training. But, unfortunately, teachers were often observed coaching students to solve typical tasks, memorizing standard techniques. But it is impossible to learn to solve problems using a memorized pattern.

In the late 1960s, reform of school mathematics education involved the early introduction of equations in order to organize teaching problem solving in a new way. However, the role of the algebraic method of solving word problems in grades 5-6 was exaggerated precisely because school curriculum Arithmetic methods have been removed. And practice has proven that without sufficient preparation of students’ thinking, solving problems using equations is impractical. The student must be able to reason and imagine the actions that occur with objects.

In grades 5-6, it is necessary to pay enough attention to the arithmetic method of solving word problems and not rush to move on to the algebraic method - solving problems using an equation. Once a student has learned the algebraic method, it is almost impossible to return him to the “solution by actions.” Having drawn up an equation, the main thing is to solve it correctly and avoid a computational error. And you don’t need to think at all about what arithmetic operations are performed during the solution and what they lead to. And if we follow the solution of the equation step by step, we will see the same actions as in the arithmetic method. Only the student hardly thinks about this.

Very often we observe that a child is not ready to solve a problem algebraically when we introduce an abstract variable and the phrase “let x...” appears. Where this “X” came from and what words should be written next to it is not clear to the student at this stage. And this happens because it is necessary to take into account age characteristics children who at this moment have developed visual-figurative thinking. They are not yet capable of abstract models.

What do we mean by requirement - to solve a problem. This means finding a sequence of actions that, as a result of analyzing the condition, will lead to an answer to the question posed in the problem. To come to the answer, you need to go a long way, starting from the moment you understand the text, be able to highlight the main thing, “translate” the problem into the language of mathematics, replacing the words “faster”, “slower” with “less” or “more”, draw up a graphic model or table that makes it easier to understand the conditions of the problem, compare values, establishlogical relationships between the data according to the condition and the required ones. And this is very difficult for children.

It is important to note that the text of the tasks must be compiled in such a way that the child understands and imagines what the we're talking about. Often, before starting to solve a problem, a lot of time is spent analyzing the condition, when students have to explain what a cast iron blank is, how it differs from a part, as well as a reinforced concrete support, an automatic machine, living area etc. The text of the task must correspond to the level of his perception. Of course, the text of the problem must be brought closer to real life so you can see practical application this model.

When starting to solve a problem, it is necessary not only to imagine the situation in question, but also to depict it in a drawing, diagram, or table. It is impossible to solve a problem qualitatively without drawing up a brief recording of the condition. It is the schematic drawing up of the condition that allows, when discussing a solution, to identify all the actions that need to be performed to answer the question of the problem.

Let's look at some examples of solving word problems

Movement tasks

This type of problem is widespread in school mathematics courses. They address different types movements: towards, in opposite directions, in one direction (one catches up with the other).

To understand these tasks, it is convenient to draw a diagram. But, if a student is making a table, there is no need to convince him that this method The short description of the conditions is not very good. We perceive information differently. Maybe the child “sees” the task better in this display.

Example 1. Two cyclists simultaneously rode towards each other from two villages and met 3 hours later. The first cyclist was traveling at a speed of 12 km/h, and the second was traveling at 14 km/h. How far away are the villages?

Let’s draw up a diagram for the problem that sufficiently fully reflects the condition (directions of movement, speeds of cyclists, travel time to the meeting are indicated, the question is clear):

Let's consider two ways to solve this problem:

1 way:

Traditionally, we like to solve these problems by introducing the concept of “closing speed”, and finding it as the sum (or difference) of the speeds of the participants in the movement. When moving towards each other, we add up the speeds:

1)12 + 14 = 26 (km/h) – approach speed

Knowing that the movement time is the same, the second action allows using the path formula (S = vt) calculate the required distance and answer the question posed in the problem.

2) 26 3 = 78 (km)

Let's make an expression:

3(12 + 14) = 78(km)

Answer : 78 km.

But not all children understand what this abstract quantity is - the speed of approach. Why is it possible to add and in other cases subtract the speeds of two various participants movements, combining them common name. If your students solve this problem in a different way, do not try to win them over to your side. For some, the time has not yet come to understand this, and for others, the first method will never be available at all.

Method 2:

1)12 3 = 36 (km) – the path of the first cyclist to the meeting

2)14 3 = 42 (km) – the distance of the second cyclist to the meeting

3)36 + 42 = 78 (km) – distance between villages

Let's make an expression:

12 3 + 14 3 = 78 (km)

Answer : 78 km.

Gradually, when the child learns to understand such tasks, comparing numeric expressions, we can show that both methods are interconnected, and at the same time recall the distributive property of multiplication:

12 3 + 14 3 = 3(12 + 14) = 78

Example 2. There were 54 notebooks in two packs. When 10 notebooks were removed from the first pack, and 14 notebooks from the second, then there were an equal number of notebooks in both packs. How many notebooks were there in each pack initially?

How can I display a condition?

1.Make a table:

Was

Removed

It became

1 pack - ? 54 tet.

2 packs – ?

10 tet.

14 tet.

equally

2. Make a drawing

They took 14 pieces.

They took 10 pieces.

Equally

Total 54 pcs.

Let's analyze the solution to the problem, paying attention to what questions we answer when performing each arithmetic operation:

1) How many notebooks were removed from both packs?

10 + 14 = 24 (pieces);

2) How many notebooks are there in two packs?

– 24 = 30 (pieces);

3) How many are in each pack of notebooks?

30: 2 = 15 (pcs);

4) How many notebooks were there in the first pack initially?

10 = 25 (pieces);

5) How many notebooks were there in the second pack initially?

54 – 25 = 29 (pieces).

In 5th grade, most likely, the student will choose exactly this method of solving the problem. Invite him to solve this problem in 6th or 7th grade. Perhaps the situation will change and the student will solve it using an equation. By performing the same actions, he will not think about numerous questions. By choosing an equation as a means of solving a problem, you will very quickly arrive at the same answer.

What then would the solution look like?

Let there be x notebooks in each pack after rearrangement,

then (x + 10) notebooks were initially in the first pack, and

(x + 14) notebooks were originally in the second pack.

Knowing that there were 54 notebooks in two packs, we can create the equation:

x + 10 + x + 14 = 54

The equation traces all the same actions that are performed in the arithmetic method of solving the problem.

x + x + (10 + 14) = 54; (1 operation of the arithmetic method)

2x = 54 – 24; (action 2)

x = 30:2; (action 3)

15 + 10 = 25 (pcs.) (4 action)

15 + 14 = 29 (pcs.) (5 action)

Answer: 25 notebooks, 29 notebooks.

But no one asks any questions about what we find as we complete each step.

I always show my students that the text of problems for 5th or 9th grades is often the same in meaning. And practice shows that fifth-graders are able to figure out the condition from the problem book for the 9th grade and even create an equation. Of course, there is still not enough knowledge to solve such an equation. But at the same time, not every ninth grader manages to solve a problem for 5th grade using an arithmetic method.

Schoolchildren usually choose the algebraic method of solving word problems; they almost never return to arithmetic. They simply stop seeing this method, getting carried away by introducing variables and composing equations.

Why do we value the arithmetic method of solving word problems? The first and most important thing is that when performing each arithmetic operation, the student thinks about the question: “What did I find as a result?” He imagines what the problem is about, since each action has a clear and specific interpretation. As a result, it is actively developing logical thinking. In the process of calculations, measurements, and searching for solutions to problems, the student develops cognitive universal learning activities, the formation of which isthe most important task modern system basic general education.

Word problems are studied throughout school course mathematics. But it is necessary to teach understanding problems, analyzing conditions, reasoning and finding rational solutions in grades 5-6, while their level of complexity is low, and the problem itself is one of the most important categories. The difficult can be comprehended with the easy.

The use of arithmetic methods for solving problems develops ingenuity and intelligence, the ability to pose questions and answer them, that is, it develops natural language, prepares schoolchildren for further education.

Arithmetic methods for solving word problems allow you to build a solution plan taking into account the relationships between known and unknown quantities (taking into account the type of problem), interpret the result of each action within the framework of the problem conditions, check the correctness of the solution by drawing up and solving the inverse problem, that is, form and develop important general educational skills.

If a student copes with word problems in mathematics lessons, that is, he can trace and explain the logical chain of his solution, give a description of all quantities, then he can also successfully solve problems in physics and chemistry, he can compare and analyze, transform information on all academic subjects school course.

The great D. Polya said: “If you want to learn to swim, then boldly enter the water, and if you want to learn to solve problems, then solve them.”If we teach children to solve problems, we will not only increase interest in the subject itself, we will have a significant impact on the formation of their mathematical thinking, which contributes to the successful development of new knowledge in other areas.

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 problem solving techniques, learn how to solve one problem in various ways.A simulator of problems grouped under the same theme “Arithmetic methods for solving problems”, tasks for working in a group and for 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 language, 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?

IN international conference 120 people participated. 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 practice dancing. rhythmic gymnastics There are 32 students, 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 you make up 0, 1, 3, 5 from the numbers if the numbers in the number are not repeated?

On Wednesday in 5th grade there are five lessons: mathematics, physical education, history, Russian and science. How many various options Can you make a schedule for Wednesday?

“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

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.

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

PROGRESS OF THE LESSON.

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, we will analyze ancient problems that have come down to us from different countries and times, tasks on equalization, comparison by sum and difference, and others.

The purpose of the lesson is to involve you in amazing world 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 executing arithmetic operations over the numbers.

With the algebraic method, the answer to the question of the problem is found as a result of composing and solving an 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 : Grandmother in autumn with summer cottage collected 51 kg of carrots and cabbage. 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. A total of 6 heads and 20 legs. 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 the 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 extra pair 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 are in this caravan dromedary camels? 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 languages(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 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.120 people took part in the international conference. 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 do you know how many ways Madina can use to choose her lunch? (3+3+3=9)

Yes, you're 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.

It is known that problem solving is practical art, similar to swimming or playing the piano. You can only learn it by imitating good examples, 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 resolve 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 math 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.

General notes on solving problems using the arithmetic method.

Problems of finding unknowns based on the results of actions.

Proportional division problems.

Problems involving percentages and parts.

Problems solved in reverse.

1. The arithmetic method is the main method for solving word problems in elementary school. It also finds its application in the middle management. secondary school. This method allows you to better understand and appreciate the importance and significance of each stage of working on a task.

In some cases, solving a problem using the arithmetic method is much simpler than using other methods.

While captivating with its simplicity and accessibility, the arithmetic method is at the same time quite complex, and mastering the techniques for solving problems using this method requires serious and painstaking work. The wide variety of types of problems does not allow us to form a universal approach to analyzing problems and finding ways to solve them: problems, even combined into one group, have completely different ways of solving them.

2 . To the tasks on finding unknowns by their difference and ratio These include problems in which, using the known difference and quotient of two values ​​of a certain quantity, it is required to find these values.

Algebraic model:

The answer is found using the formulas: X= ak/(k – 1), y = a/(k – 1).

Example. In the reserved seat carriages of the fast train there are 432 more passengers than in the compartment carriages. How many passengers are there in reserved seat and compartment carriages separately, if there are 4 times fewer passengers in compartment carriages than in reserved seat carriages?

Solution. A graphical model of the problem is presented in Fig. 4.

Rice. 4

We will take the number of passengers in compartment cars as 1 part. Then you can find how many parts there are per number of passengers in reserved seat cars, and then how many parts there are per 432 passengers. After this, you can determine the number of passengers making up 1 part (located in compartment cars). Knowing that there are 4 times more passengers in reserved seat carriages, we can find their number.

1  4 = 4 (hours) – accounts for passengers in reserved seat carriages;

4 – 1 = 3 (h.) – accounts for the difference between the number of passengers in reserved seat and compartment carriages;

432: 3 = 144 (p.) – in compartment cars;

144  4 = 576 (p.) – in reserved seat carriages.

This problem can be verified by solving it in another way, namely:

1  4 = 4(h);

4 – 1 = 3 (h);

432: 3 = 144 (p.);

144 + 432 = 576 (p.).

Answer: there are 144 passengers in compartment carriages, and 576 in reserved seat carriages.

To the tasks on finding unknowns from two residues or two differences, include problems in which two directly or inversely proportional quantities are considered, such that two values ​​of one quantity and the difference of the corresponding values ​​of another quantity are known, and it is required to find the values ​​of this quantity themselves.

Algebraic model:

The answers are found using the formulas:

Example. Two trains traveled at the same speed - one 837 km, the other 248 km, and the first was on the road 19 hours longer than the second. How many hours did each train travel?

Solution. A graphical model of the problem is presented in Figure 5.

Rice. 5

To answer the question of the problem, how many hours was this or that train on the way, you need to know the distance it traveled and the speed. The distance is given in the condition. To find out the speed, you need to know the distance and the time during which this distance was covered. The condition says that the first train took 19 hours longer, and the distance it traveled during this time can be found. He walked for an extra 19 hours - obviously, during this time he also covered an extra distance.

837 – 248 = 589 (km) – the first train traveled so many kilometers more;

589: 19 = 31 (km/h) – speed of the first train;

837: 31 = 27 (hours) – the first train was on its way;

4) 248: 31 = 8 (hours) – the second train was on its way.

Let's check the solution to the problem by establishing a correspondence between the data and the numbers obtained when solving the problem.

Having found out how long each train was on the road, we will find how many hours more the first train was on the road than the second: 27 – 8 = 19 (hours). This number matches the one in the condition. Therefore, the problem was solved correctly.

This problem can be verified by solving it in another way. All four questions and the first three actions remain the same.

4) 27 –19 = 8 (hours).

Answer: the first train took 31 hours to travel, the second train took 8 hours.

Problems to find three unknowns from three sums of these unknowns, taken in pairs:

Algebraic model:

The answer is found using the formulas:

x =(A -b + c)/2, y = (a +b – c)/2, z = (b + With -a)/ 2.

Example. English and German languages 116 schoolchildren study German and spanish languages 46 students are studying, and 90 students are studying English and Spanish. How many students study English, German and Spanish separately if it is known that each student studies only one language?

Solution. A graphical model of the problem is presented in Figure 6.

How many students study each language?

The graphical model of the problem shows: if we add up the number of schoolchildren given in the condition (116 + 90 + 46), we get double the number of schoolchildren studying English, German and Spanish. Dividing it by two, we find total number schoolchildren. To find the number of schoolchildren studying English, it is enough to subtract from this number the number of schoolchildren studying German and Spanish. Similarly, we find the remaining required numbers.

Let's write down the decision on actions with explanations:

116 + 90 + 46 = 252 (schoolchildren) – double the number of schoolchildren studying languages;

252: 2 = 126 (school) – study languages;

126 – 46 = 80 (school) – study English;

126 – 90 = 36 (school) – study German;

126 – 116 = 10 (school) – learn Spanish.

This problem can be verified by solving it in another way.

116 – 46 = 70 (schoolchildren) – so many more schoolchildren study English than Spanish;

90 + 70 = 160 (schoolchildren) – double the number of schoolchildren studying English;

160: 2 = 80 (school) – learn English;

90 – 80 = 10 (school) – learn Spanish;

116 – 80 = 36 (school) – study German.

Answer: 80 schoolchildren study English, 36 schoolchildren study German, and 10 schoolchildren study Spanish.

3. Proportional division problems include problems in which a given value of a certain quantity needs to be divided into parts proportional to given numbers. In some of them, the parts are presented explicitly, while in others, these parts must be distinguished by taking one of the values ​​of this quantity as one part and determining how many such parts are accounted for by its other values.

There are five types of proportional division problems.

1) Problems involving dividing a number into parts, directlyproportional to a series of whole or fractional numbers

Problems of this type include tasks in which the number A X 1, X 2 , x 3 , ..., X n directly proportional to the numbers A 1 , A 2 , A 3 , ..., A n .

Algebraic model:

The answer is found using the formulas:

Example. The travel company has four recreation centers, which have buildings of the same capacity. On the territory of the 1st recreation center there are 6 buildings, the 2nd - 4 buildings, the 3rd - 5 buildings, the 4th - 7 buildings. How many campers can each base accommodate if all 4 bases can accommodate 2,112 people?

Solution. A summary of the task is shown in Figure 7.

Rice. 7

To answer the question of the problem, how many vacationers can be accommodated at each base, you need to know how many vacationers can be accommodated in one building and how many buildings are located on the territory of each base. The number of buildings on each base is given in the condition. To find out how many vacationers can be accommodated in one building, you need to know how many vacationers can be accommodated at all 4 bases (this is given in the condition) and how many buildings are located on the territory of all 4 bases. The latter can be determined by knowing from the condition how many buildings are located on the territory of each base.

Let's write down the decision on actions with explanations:

6 + 4 + 5 + 7 = 22 (k.) – located on the territory of 4 bases;

2112: 22 = 96 (hours) – can be placed in one building;

96  6 = 576 (h) – can be placed on first base;

96  4 = 384 (h) – can be placed on second base;

96  5 = 480 (h) – can be placed on third base;

96  7 = 672 (h) – can be placed on fourth base.

Examination. We calculate how many vacationers can be accommodated at 4 bases: 576 + 384 + 480 + 672 = 2,112 (hours). There is no discrepancy with the task conditions. The problem was solved correctly.

Answer: the first base can accommodate 576 vacationers, the second - 384 vacationers, the third - 480 vacationers, and the fourth - 672 vacationers.

2) Problems involving dividing a number into parts inversely proportional to a series of integers or fractions

These include tasks in which the number A(the value of a certain quantity) needs to be divided into parts x 1 i , x 2 , x 3 i , ..., X" inversely proportional to numbers A 1b A 2 , A 3 ,..., A n .

Algebraic model:

or

x 1 : x 2 :X 3 :...:х„ = a 2 a 3 ...A n :A 1 A 3 ...A n :A 1 A 2 A 4 ...A n :...:A 1 A 2 ...A n -1

The answer is found using the formulas:

Where S = A 2 A 3 ...a„ +a l a i ... a n + a ] A 2 A 4 ...A n + ... + a 1 A 2 ...A n -1.

Example. For four months, the fur farm’s income from the sale of furs amounted to 1,925,000 rubles, and by month the money received was distributed in inverse proportion to the numbers 2, 3, 5, 4. What is the farm’s income in each month separately?

Solution. To determine the income mentioned in the condition, the total income for four months is given, that is, the sum of the four required numbers, as well as the relationship between the required numbers. The required income is inversely proportional to the numbers 2, 3, 5, 4.

Let's denote the required incomes respectively through x, X 2 , X 3 , X 4 . Then the problem can be briefly written as shown in Figure 8.

Rice. 8

Knowing the number of parts per each of the required numbers, we will find the number of parts contained in their sum. Based on the given total income for four months, that is, based on the sum of the required numbers and the number of parts contained in this amount, we find out the value of one part, and then the required income.

Let's write down the decision on actions with explanations:

1. The required incomes are inversely proportional to the numbers 2, 3, 5, 4, which means they are directly proportional to the numbers inverse to the data, that is, there are relations . Let us replace these ratios in fractional numbers with ratios of integers:

2. Knowing that X contains 30 equal parts, X 2 – 20, X 3 – 12, X 4 – 15, let’s find how many parts are contained in their sum:

30 + 20 + 12+ 15 = 77 (hours).

3. How many rubles are there for one part?

1,925,000: 77 = 25,000 (r.).

4. What is the farm's income in the first month?

25,000 30 = 750,000 (r.).

5. What is the farm income in the second month?

25,000 20 = 500,000 (r.).

6. What is the farm income in the third month?

25,000–12 = 300,000 (r.).

7. What is the farm income in the fourth month?

25,000–15 = 375,000 (r.).

Answer: in the first month, the farm’s income was 750,000 rubles, in the second – 500,000 rubles, in the third – 300,000 rubles, in the fourth – 375,000 rubles.

3) Problems involving dividing a number into parts, when separate ratios are given for each pair of required numbers

Problems of this type include those tasks in which the number A(the value of a certain quantity) must be divided into parts x 1, X 2 , x 3, ..., X", when a series of relations is given for the required numbers, taken in pairs. Algebraic model:

x 1: X 2 = a 1 : b 1, X 2 : X 3 = a 2 : b 2, x 3 : X 4 = a 3 : b 3 , ..., X n-1 : X n = a n -1 : b n-1 .

n = 4. Algebraic model:

X X :X 2 = a 1 : b 1, X 2 :X 3= A 2 : b 2, X 3 : X 4 = a 3: b 3 .

So, X 1: X 2 : x 3: X 4 = A 1 A 2 A 3 : b 1 A 2 A 3 : b 1 b 2 A 3 : b 1 b 2 b 3 .

Where S = A 1 A 2 A 3 + b 1 A G A 3 + b 1 b 2 A 3 + b 1 b 2 b 3

Example. The three cities have 168,000 inhabitants. The numbers of residents of the first and second cities are in the ratio , and the second and third cities – in relation to . How many inhabitants are there in each city?

Solution. Let us denote the required numbers of inhabitants, respectively, by X 1 , X 2 , X 3 . Then the problem can be briefly written as shown in Figure 9.

Rice. 9

To determine the number of inhabitants, the numbers of inhabitants in three cities are given, that is, the sum of the three required numbers, as well as individual relationships between the required numbers. Replacing these relations with a series of relations, we express the number of inhabitants of the three cities in equal parts. Knowing the number of parts per each of the required numbers, we will find the number of parts contained in their sum. From the given total number of inhabitants in three cities, that is, from the sum of the required numbers and from the number of parts contained in this sum, we find out the size of one part, and then the required numbers of inhabitants.

Let's write down the decision on the actions with explanations.

1. Replace the ratio of fractional numbers with the ratio of integers:

We match the number of residents of the second city with the number 15 (the least common multiple of the numbers 3 and 5).

We change the resulting relationships accordingly:

X 1: X 2 = 4: 3 = (4-5): (3-5) = 20: 15, x 2: x 3 = 5: 7 = (5-3): (7-3) = 15: 21.

From individual relations we create a series of relations:

X 1: X 2 : X 3 = 20: 15: 21.

2. 20 + 15 + 21 = 56 (h) – the number 168,000 corresponds to so many equal parts;

3. 168,000: 56 = 3,000 (f.) – per part;

4. 3,000 20 = 60,000 (f.) – in the first city;

5. 3,000 15 = 45,000 (f.) – in the second city;

3,000 21 = 63,000 (f.) - in the third city.

Answer: 60,000 inhabitants; 45,000 inhabitants; 63,000 inhabitants.

4) Problems involving dividing a number into parts proportional to two, three, and so on rows of numbers

Problems of this type include problems in which the number A(the value of a certain quantity) needs to be divided into parts X 1, X 2 , X 3 ,..., X n proportional to two, three, ..., N rows of numbers.

Due to the cumbersomeness of the formulas for solving the problem in general view Let's consider a special case when n = 3 and N = 2. Let X 1 X 2 , X 3 directly proportional to numbers A 1 , A 2 , A 3 and inversely proportional to the numbers b 1 , b 2 , b 3 .

Algebraic model:

(see paragraph 1 of this paragraph),

Example. Two workers received 1,800 rubles. One worked 3 days for 8 hours, the other 6 days for 6 hours. How much did each earn if for 1 hour of work they received equally?

Solution. A summary of the task is shown in Figure 10.

Rice.10

To find out how much each worker received, you need to know how many rubles they paid for 1 hour of work and how many hours each worker worked. To find out how many rubles were paid for 1 hour of work, you need to know how much they paid for the entire work (given in the condition) and how many hours both workers worked together. To find out the total number of hours worked, you need to know how many hours each person worked, and for this you need to know how many days each person worked and how many hours a day. This data is included in the condition.

Let's write down the decision on actions with explanations:

8  3 = 24 (hours) – the first worker worked;

6  6 = 36 (hours) – the second worker worked;

24 + 36 = 60 (hours) – both workers worked together;

1800: 60 = 30 (r.) – workers received for 1 hour of work;

30  24 = 720 (r.) – earned by the first worker;

30  36 = 1080 (r.) - earned by the second worker. Answer: 720 rub.; 1080 rub.

5) Problems on finding several numbersaccording to their relations and the sum or difference (the sum or difference of some of them)

Example. The school administration spent 49,000 rubles on equipment for the playground, greenhouse and gym. Equipping a playground cost half as much as greenhouses, and greenhouses cost 3 times less than a gym and playground together. How much money was spent on equipment for each of these facilities?

Solution. A summary of the task is shown in Figure 11.

Rice. 11

To find out the amount of money spent on the equipment of each object, you need to know how many parts of all the money spent were on the equipment of each object and how many rubles were for each part. The number of parts of money spent on the equipment of each object is determined from the conditions of the problem. Having determined the number of parts for the equipment of each object separately, and then finding their sum, we calculate the value of one part (in rubles).

Let's write down the decision on the actions with explanations.

We take as 1 part the amount of money spent on equipment for the playground. According to the condition, 2 times more was spent on greenhouse equipment, that is, 1  2 = 2 (h); 3 times more was spent on the equipment of the playground and sports hall than on the greenhouse, that is, 2  3 = 6 (hours), therefore, 6 – 1 = 5 (hours) were spent on the equipment of the sports hall.

1 part was spent on equipment for the playground, 2 parts for greenhouses, and 5 parts for the gym. The entire flow rate was 1 + 2 + + 5 = 8 (h).

8 parts amount to 49,000 rubles, one part is 8 times less than this amount: 49,000: 8 = 6,125 (rub.). Consequently, 6,125 rubles were spent on equipment for the playground.

Twice as much was spent on greenhouse equipment: 6,125  2 = 12,250 (r.).

5 parts were spent on equipment for the gym: 6,125  5 = 30,625 (r.).

Answer: 6,125 rubles; RUB 12,250; RUR 30,625

6) Problems to exclude one of the unknowns

Problems in this group include problems in which the sums of two products that have two repeating factors are given, and it is required to find the values ​​of these factors. Algebraic model

The answer is found using the formulas:

These problems are solved by the method of equalizing data, the method of equalizing data and the required ones, the method of replacing data, as well as the so-called “assumption” method.

Example. At a garment factory, 24 coats and 45 suits used 204 m of fabric, and 24 coats and 30 suits used 162 m. How much fabric is used for one suit and how much for one coat?

Solution. Let's solve the problem using data adjustment method. Brief description of the task.

Based on the similarity in mathematical meaning and the interchangeability of different solution methods, all arithmetic methods can be combined into the following groups:

• 1) method of reduction to unity, reduction to a general measure, inverse reduction to unity, method of relations;
• 2) a way to solve problems from the “end”;
• 3) a method for eliminating unknowns (replacing one unknown with another, comparing unknowns, comparing data, comparing two conditions by subtraction, combining two conditions into one); way of guessing;
• 4) proportional division, similarity or finding of parts;
• 5) a method of transforming one problem into another (decomposing a complex problem into simple, preparatory ones; bringing unknowns to such values ​​for which their relationship becomes known; the method of determining an arbitrary number for one of the unknown quantities).

In addition to the above methods, it is advisable to consider also the arithmetic mean method, the surplus method, the method of rearranging the known and the unknown, and the method of “false” rules.

Since it is usually impossible to determine in advance which of the methods is rational, to foresee which of them will lead to the simplest and most understandable solution for the student, then students should be introduced to in different ways and give them the opportunity to choose which one to use when solving a specific problem.

Method for excluding unknowns

This method is used when there are several unknowns in the problem. This problem can be solved using one of five techniques: 1) replacing one unknown with another; 2) comparison of unknowns; 3) comparison of two conditions by subtraction; 4) comparison of data; 5) combining several conditions into one.

As a result of using one of the listed techniques instead of several unknowns, there is only one that can be found. Having calculated it, they use the data in the dependence condition to find other unknowns.

Let's take a closer look at some of the techniques.

1. Replacing one unknown with another

The name of the technique reveals its idea: based on the dependencies (multiple or difference) that are given according to the conditions of the problem, it is necessary to express all the unknowns through one of them.

Task. Sergei and Andrey have only 126 stamps. Sergei has 14 marks more than Andrey. How many stamps did each boy have?

Brief description of the condition:

Sergey -- ? marks, 14 marks more

Andrey -- ? stamps

Total -- 126 stamps

Solution 1.

• (replacing a larger unknown with a smaller one)
• 1) Let Sergei have as many stamps as Andrey. Then total quantity there would be 126 marks - 14 = 112 (marks).
• 2) Since the boys now have the same number of marks, we will find how many marks Andrei had at the beginning: 112: 2 = 56 (stamps).
• 3) Considering that Sergei has 14 marks more than Andrey, we get: 56 + 14 = 70 (marks).

Solution 2.

• (replacing a smaller unknown with a larger one)
• 1) Let Andrei have the same number of stamps as Sergei. Then the total number of stamps would be 126 + 14 = 140 (stamps).
• 2) Since the boys now have the same number of marks, let’s find how many marks Sergei had at first: 140: 2 = 70 (marks).
• 3) Considering that Andrey had 14 marks less than Sergei, we get: 70 - 14 = 56 (marks).

Answer: Sergei had 70 marks, and Andrey had 56 marks.

For the best assimilation by students of the method of replacing a smaller unknown with a larger one, before considering it, it is necessary to clarify with students the following fact: if the number A more number B by C units, then to compare numbers A and B you need to:

• a) subtract number C from number A (then both numbers are equal to number B);
• b) add number C to number B (then both numbers are equal to number A).

The ability of students to replace a larger unknown with a smaller one, and vice versa, further contributes to the development of the ability to choose an unknown and express other quantities through it when composing an equation.

2. Comparison of unknowns

Task. There were 188 books on four shelves. On the second shelf there were 16 fewer books than on the first, on the third - 8 more than on the second, and on the fourth - 12 fewer than on the third shelf. How many books are on each shelf?

Task analysis

To better understand the dependencies between four unknown quantities (the number of books on each shelf), we use the following diagram:

I_________________________________

II___________________________

III______________________________

IV_______________________ _ _ _ _ _

Comparing the segments that schematically depict the number of books on each shelf, we come to the following conclusions: there are 16 more books on the first shelf than on the second; on the third there are 8 more than on the second; on the fourth - 12 - 8 = 4 (books) less than on the second. Therefore, the problem can be solved by comparing the number of books on each shelf. To do this, remove 16 books from the first shelf, 8 books from the third, and put 4 books on the fourth shelf. Then there will be the same number of books on all shelves, namely, as there were on the second one at first.

• 1) How many books are there on all shelves after the operations described in the problem analysis?
• 188 -- 16 -- 8 + 4 = 168 (books)
• 2) How many books were on the second shelf?
• 168: 4 = 42 (books)
• 3) How many books were on the first shelf?
• 42 + 16 = 58 (books)
• 4) How many books were on the third shelf?
• 42 + 8 = 50 (books)
• 5) How many books were on the fourth shelf?
• 50 -- 12 = 38 (books)

Answer: There were 58, 42, 50 and 38 books on each of the four shelves.

Comment. You can invite students to solve this problem in other ways by comparing the unknown number of books that were on the first, or on the second, or on the fourth shelf.

3. Comparison of two conditions by subtraction

The plot of the problem that is solved by this technique often includes two proportional quantities (the quantity of goods and its cost, the number of workers and the work they performed, etc.). The condition gives two values ​​of one quantity and the difference of two proportional to them numerical values of a different size.

Task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time for 4 kg of oranges and 3 kg of bananas bought at the same prices they paid 500 rubles. How much does 1kg of oranges and 1kg of bananas cost?

Brief description of the condition:

• 4kg app. and 5kg ban. - 620 rubles,
• 4kg app. and 3kg ban. - 500 rub.

• 1) Let's compare the cost of two purchases. Both the first and second time they bought the same number of oranges at the same price. The first time we paid more because we bought more bananas. Let's find how many more kilograms of bananas were purchased the first time: 5 -- 3 = 2 (kg).
• 2) Let's find out how much more we paid the first time than the second time (that is, we find out how much 2 kg of bananas cost): 620 - 500 = 120 (rub.).
• 3) Find the price of 1 kg of bananas: 120: 2 = 60 (rub.).
• 4) Knowing the cost of the first and second purchases, we can find the price of 1 kg of oranges. To do this, first find the cost of purchased bananas, then the cost of oranges, and then the price of 1 kg. We have: (620 -- 60*5) : 4 = 80 (rub).

Answer: the price of 1 kg of oranges is 80 rubles, and the price of 1 kg of bananas is 60 rubles.

4. Data comparison

The use of this technique makes it possible to compare data and apply the subtraction method. You can compare data values:

• 1) using multiplication (comparing them with the least common multiple);
• 2) using division (comparing them with the largest common divisor).

Let's show this with an example.

Task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time for 6 kg of oranges and 3 kg of bananas bought at the same prices they paid 660 rubles. How much does 1kg of oranges and 1kg of bananas cost?

Brief description of the condition:

• 4kg app. and 5kg ban. - 620 rubles,
• 6kg app. and 3kg ban. - 660 rub.

Let's equalize the number of oranges and bananas by comparing them with the least common multiple: LCM(4;6) = 12.

Solution1.

• 1) Let us increase the number of fruits purchased and their cost in the first case by 3 times, and in the second - by 2 times. We get the following short statement of the condition:
• 12kg app. and 15kg ban. - 1860 rubles,
• 12kg app. and 6kg ban. - 1320 rub.
• 2) Find out how many more bananas you bought the first time: 15-6 = 9 (kg).
• 3) How much does 9kg of bananas cost? 1860 -- 1320 = 540 (rub).
• 4) Find the price of 1 kg of bananas: 540: 9 = 60 (rub).
• 5) Find the cost of 3 kg of bananas: 60 * 3 = 180 (rub).
• 6) Find the cost of 6 kg of oranges: 660 -- 180 = 480 (rub).
• 7) Find the price of 1 kg of oranges: 480: 6 = 80 (rub).

Solution2.

Let's equalize the number of oranges and bananas by comparing them with the greatest common divisor: GCD (4; 6) = 2.

• 1) To equalize the number of oranges purchased the first time and the second time, we reduce the quantity of the purchased product and its cost in the first case by 2 times, in the second - by 3 times. Let us obtain a problem that has the following short form of condition:
• 2kg app. and 2.5 kg ban. - 310 rubles,
• 2kg app. and 1kg ban. - 220 rub.
• 2) How many more bananas are they buying now: 2.5 -- 1 = 1.5 (kg).
• 3) Let's find how much 1.5 kg of bananas costs: 310 -- 220 = 90 (rub).
• 4) Find the price of 1 kg of bananas: 90: 1.5 = 60 (rub).
• 5) Find the price of 1 kg of oranges: (660 -- 60*3) : 6 = 80 (rub).

Answer: the price of 1 kg of oranges is 80 rubles, 1 kg of bananas is 60 rubles.

When solving problems using the technique of comparing data, you can not do such detailed analysis and recordings, but only record the changes that were made for comparison and write them down in the form of a table.

5. Combining several conditions into one

Sometimes you can get rid of unnecessary unknowns by combining several conditions into one.

Task. The tourists left the camp and first walked for 4 hours, and then rode bicycles for another 4 hours at a certain constant speed and moved 60 km away from the camp. The second time they left the camp and first rode bicycles at the same speed for 7 hours, and then turned in the opposite direction and, walking for 4 hours, found themselves at a distance of 50 km from the camp. How fast did the tourists ride their bikes?

There are two unknowns in the problem: the speed at which the tourists rode their bicycles, and the speed at which they walked. In order to exclude one of them, you can combine two conditions into one. Then the distance that tourists will cover in 4 hours, moving forward on foot the first time, is equal to the distance they covered in 4 hours, moving back the second time. Therefore, we do not pay attention to these distances. This means that the distance that tourists will cover in 4 + 7 = 11 (hours) on bicycles will be equal to 50 + 60 = 110 (km).

Then the speed of tourists on bicycles is: 110: 11 = 10 (km/h).

Answer: The speed of bicycles is 10 km/h.

6. Method of assumption

Using the assumption method when solving problems does not cause difficulties for most students. Therefore, to avoid students mechanically memorizing the diagram of steps of this method and misunderstanding the essence of the actions performed on each of them, students should first be shown the trial method (“false rule” and “rule of the ancient Babylonians”).

When using the sampling method, in particular the “false rule,” one of the unknown quantities is given (“allowed”) a certain value. Then, using all the conditions, they find the value of another quantity. The resulting value is checked against the one specified in the condition. If the resulting value is different from the one given in the condition, then the first value specified is not correct and it must be increased or decreased by 1, and again the value of another value must be found. This must be done until we obtain the value of another quantity such as in the problem statement.

Task. The cashier has 50 coins of 50 kopecks and 10 kopecks, totaling 21 rubles. Find how many separate 50k coins the cashier had. and 10k each.

Solution1. (sampling method)

Let's use the rule of the “ancient” Babylonians. Let's assume that the cashier has an equal number of coins of each denomination, that is, 25 pieces each. Then the amount of money will be 50*25 + 10*25 = 1250+250=1500 (k.), or 15 rubles. But in the condition 21 rubles, that is, 21 UAH more than received - 15 rubles = 6 rubles. This means that it is necessary to increase the number of 50-kopeck coins and decrease the number of 10-kopeck coins until we get a total of 21 rubles. We will record the change in the number of coins and the total amount in the table.

Number of coins	Number of coins	Amount of money	Amount of money	Total amount	Less or more than in the condition
					Less by 6 rubles.
					Less by 5rub60k
					As in condition

As can be seen from the table, the cashier had 40 coins of 50 kopecks and 10 coins of 10 kopecks.

As it turned out in solution 1, if the cashier had an equal number of 50k coins. and 10k each, then in total he had 15 rubles of money. It is easy to see that each coin replacement is 10k. per coin 50k. increases the total amount by 40k. This means we need to find how many such replacements need to be made. To do this, let’s first find how much money we need to increase the total amount by:

21 rubles -- 15 rubles. = 6 rub. = 600 k.

Let's find how many times such a replacement needs to be made: 600 k. : 40 k. = 15.

Then 50 kopecks will be 25 +15 = 40 (coins), and 10 kopeck coins will remain 25 -- 15 = 10.

The check confirms that the total amount of money in this case is 21 rubles.

Answer: The cashier had 40 coins of 50 kopecks and 10 coins of 10 kopecks.

By asking students to choose for themselves different meanings number of coins of 50 kopecks, it is necessary to bring them to the idea that the best from the point of view of rationality is the assumption that the cashier had only coins of one denomination (for example, all 50 coins of 50 kopecks or all 50 coins of 10 kopecks each). Due to this, one of the unknowns is excluded and replaced by another unknown.

7. Residue method

This method has some similarities with thinking when solving problems using trial and guesswork methods. We use the method of remainders when solving problems involving movement in one direction, namely, when it is necessary to find the time during which the first object, which is moving behind at a higher speed, will catch up with the second object, which has a lower speed. In 1 hour, the first object approaches the second at a distance that is equal to the difference in their speeds, that is, equal to the “remainder” of the speed that it has in comparison with the speed of the second. To find the time it takes for the first object to cover the distance that was between it and the second at the beginning of the movement, you should determine how many times the “remainder” is placed in this distance.

If we abstract from the plot and consider only the mathematical structure of the problem, then it talks about two factors (the speed of movement of both objects) or the difference between these factors and two products (the distances they travel) or their difference. The unknown factors (time) are the same and need to be found. From a mathematical point of view, the unknown factor shows how many times the difference of the known factors is contained in the difference of the products. Therefore, problems that are solved using the method of remainders are called problems of finding numbers by two differences.

Task. The students decided to paste photos from the holiday into an album. If they stick 4 photos on each page, there won't be enough space in the album for 20 photos. If you paste 6 photos onto each page, then 5 pages will remain free. How many photos are students going to put in the album?

Task analysis

The number of photos remains the same for the first and second gluing options. According to the conditions of the problem, it is unknown, but it can be found if the number of photographs that are placed on one page and the number of pages in the album are known.

The number of photographs that are pasted onto one page is known (the first multiplier). The number of pages in the album is unknown and remains unchanged (second multiplier). Since it is known that 5 pages of the album remain free for the second time, you can find how many more photographs could be pasted into the album: 6 * 5 = 30 (photos).

This means that by increasing the number of photos on one page by 6 - 4 = 2, the number of pasted photos increases by 20 + 30 = 50.

Since the second time they pasted two more photographs onto each page and in total they pasted 50 more photographs, we will find the number of pages in the album: 50: 2 = 25 (pages).

Therefore, there were 4*25 + 20 = 120 (photos) in total.

Answer: The album had 25 pages and 120 photographs.