Arithmetic solution method. Solving mathematical problems
The purpose of our lesson
The great mathematician Henri Poincaré said that "mathematics is the art of giving different things the same name." There is a deep meaning in this playful aphorism.
Work with the textbook.
When a problem is solved in an algebraic way, first of all, the condition of the problem is translated into the language of mathematics. The basis of such a translation, its first step, is the introduction of a letter to denote some unknown quantity.
As a result of the translation, an equality containing a letter is usually obtained. This equality, as you already know, is called equation .
Arithmetic solution of the problem:
There are four children. In 2000, the age of each of them is 2 years less, which means that their total age is 2 · 4 = 8 (years) less. Thus, in 2000, the twins were 50 - 8 = 42 (years) together.
If all of them were younger, then in 2000 they would have been
together 42 - 3 2 = 36 (years). So, the youngest in 2000 was
36: 4 \u003d 9 (years), and the oldest - 9 + 3 \u003d 12 (years).
Algebraic way of solving problems
There are two pairs of twins in the family, born three years apart. In 2012, everyone turned 50 together. How old were each of the twins in 2010?
Algebraic solution of the problem:
Denote by X the age of the younger twins in 2010. Then the older twins this year were x+ 3 years. In 2012, i.e. after 2 years, the younger twins were x+ 2 years, and older - by x+ 5 years.
According to the condition of the problem, the total age of twins in 2012 was
50 years. Means, ( X + 2) + ( X + 2) + ( X + 5) + ( X + 5) = 50.
Thus, the equation is made.
To find the unknown number x, this equation must be solved.
Workbook № 79
Workshop
Workbook No. 80
x op x op
12 op 12 op
(x - 12)op (x + 12)op
3(x - 12) = (x + 12)
Workbook No. 81
x + 8 = 3x
Workshop
Textbook No. 336
Denote by x people. - was in 1 wagon,
then there were (x + 14) people in the 2nd car.
According to the condition of the problem, the number of people in two cars was 86.
Write the equation: x + (x + 14) = 86
1 equation
2 equation
Denote by x people. - was in the 2 car,
Let's make an equation: x + (x - 14) \u003d 86
Textbook No. 337
Let x denote the number of sheets in the first bundle,
Then there were 4 sheets in 2 packs.
According to the condition of the problem, the number of sheets in two packs was 350.
Let's make the equation: x + 4x = 350
1 equation
2 equation
Let's denote by x the number of sheets in the second pack. Write the equation: x + x: 4 \u003d 350
Textbook No. 343
Let us denote Petya's age in x years,
then the age of the father is 3 years, and the age of the grandfather is 6 years.
According to the condition of the problem, the total age of Petya, father and grandfather is 110 years.
So 6x + 3x + x = 110
1 equation
2 equation
Let's make the equation: 110 - (6x + 3x) \u003d x
3 equation
Let's make the equation: 110 - 6x \u003d 3x + x
Textbook No. 345
the equation
Textbook No. 338
(x + 11): 2 = x + 2
right
(x + 3) + x = 21; 21 - (x + 3) = x;
x + 1.5x = 15; 15 - 1.5x \u003d x;
No. 336, 337, 343, 345 Orally: pp. 103-104
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Introduction
1.1 The concept of a text problem
1.2 Types of arithmetic problems
1.3 The role of the problem in mathematics
1.4 Stages of solving text problems and methods for their implementation
1.5 Some ways to solve word problems
2.4 Interest tasks
2.5 Tasks for collaboration
Conclusion
Literature
Introduction
You can teach students to solve quite a lot of types of problems, but true satisfaction will come only when we are able to transfer to our students not just knowledge, but flexibility of mind. W.U. sawyer
The ability to solve problems is one of the main indicators of the level mathematical development, development depth educational material. From the first days of school, the child is faced with a task. From the beginning to the end of schooling, a mathematical problem invariably helps the student to develop correct mathematical concepts, to better understand the various aspects of the relationships in the life around him, and makes it possible to apply the theoretical positions being studied. Word problems are an important means of teaching mathematics. With their help, students gain experience in working with quantities, comprehend the relationship between them, gain experience in applying mathematics to solving practical problems. The use of arithmetic methods for solving problems develops ingenuity and ingenuity, the ability to pose questions, answer them, that is, develops natural language. Arithmetic methods for solving text problems allow you to develop the ability to analyze problem situations, build a solution plan taking into account the relationship between known and unknown quantities (taking into account the type of problem), interpret the result of each action within the framework of the problem statement, check the correctness of the solution by compiling and solving an inverse problem, that is, to form and develop important general educational skills.
Arithmetic methods for solving text problems teach children to the first abstractions, allow them to cultivate a logical culture, and can contribute to the development of schoolchildren aesthetic sense in relation to problem solving and the study of mathematics, arousing interest first in the process of finding a solution to the problem, and then in the subject being studied.
Text tasks are traditionally difficult material for a significant part of schoolchildren. In practice, most teachers pay little attention to solving problems. Students often do not know how to identify the desired and data, to establish a connection between the quantities included in the problem; draw up a solution plan, check the result obtained.
my goal final work-- study of teaching methods for solving text problems in an arithmetic way, consider the structure of a text problem, the stages of solving problems using an arithmetic method, show difficulties in solving problems, the ability to overcome these difficulties, the use of an arithmetic method for solving text problems from personal practice.
The object of study is the educational process in mathematics lessons.
Work tasks:
- to analyze the psychological and pedagogical literature on this topic; to study the scientific and methodological literature aimed at teaching the solution of text problems;
- consider the characteristics of the text task and the methodology for working with it;
- show the use of the arithmetic method in solving word problems.
Work structure. My work consists of an introduction, chapters “Characteristics of a text problem and methods of working with it” and “Teaching schoolchildren how to solve text problems in an arithmetic way”, conclusion. In the first chapter, I examined the concept of a text problem, the types of problems, what it means to solve a problem, the stages of the process of solving a problem by arithmetic methods. fractions, tasks for percentage calculations, for joint work; tasks solved with the help of tables, arithmetic mean in tasks. I tried to show the methodology of teaching students to solve text problems, their place in the educational process in the classroom. In my work, I want to show a specific application of arithmetic methods for solving word problems, using my personal experience.
There is enough literature on this issue. Analyzing some of them, I would like to note the book by S. Lukyanova "The development of the "calculation of text problems in arithmetic ways." The book discusses various arithmetic methods for solving text problems and offers original methods for teaching this to students in grades 5-6. The author considers about 200 problems different levels of complexity, for most of which a solution is proposed (for some - several ways), each of which is implemented only with the help of arithmetic operations.In the book "Learning to solve text problems. A book for a teacher", author Shevkin A.V., offers are described in detail , bringing us back to the best traditions mathematics education, about the need to abandon the use of equations at an early stage of learning and return to a wider use of arithmetic methods for solving problems, making adjustments to the traditional teaching methodology and trying to avoid the characteristic shortcomings of its application. AT study guide Fridman L.M. “Subject problems in mathematics. History, theory, methodology” states that when solving problems various methods it is preferable to choose one that extends to a wider range of problems and there are a number of problems that are easier to solve arithmetically than algebraically, and there are those that are not at all inaccessible to algebra, although they do not present difficulties for arithmetic.
In the work I used the materials of the educational and methodical newspaper "Mathematics" No. 23 - 2005 (Publishing House "First of September"), "Non-traditional lessons. Mathematics 5-11 cells. " (M.E. Kozina, M.E. Fadeeva - Volgograd, 2008), Guidelines for grades 5-6, Didactic materials for grades 5-6 (M.K. Potapov, A.V. Shevkin) and others.
Chapter I. Characteristics of a text problem and methods of working with it
solution word problem arithmetic
Mathematics is a tool for reflection, in its arsenal there are a large number of tasks that for thousands of years have contributed to the formation of people's thinking, the ability to solve non-standard problems, and to get out of difficult situations with honor.
Working with text tasks a lot of time should be devoted, drawing the attention of children to the search and comparison of various ways to solve a problem, the construction of mathematical models, the literacy of presenting their own reasoning when solving problems.
1.1 The concept of a text problem
Solving text problems provides rich material for the development and education of students. These tasks are formulated in natural language, so they are called text tasks. They usually describe the quantitative side of some phenomena, events, so they are often called plot. Solving problems, students acquire new mathematical knowledge, prepare for practical activities. Tasks contribute to their development logical thinking. Great importance has a solution to problems in the education of the personality of students. Therefore, it is important that the teacher has a deep understanding of the text problem, its structure, and is able to solve such problems in various ways. “The task is a requirement or a question that needs to be answered, based on the conditions specified in the task and taking them into account,” L.M. Friedman in his work "Plot Problems in Mathematics".
A text task is a description of a certain situation in natural language with the requirement to give a quantitative description of any component of this situation, to establish the presence or absence of some relationship between its components, or to determine the type of this relationship. Text tasks can be of abstract content, when the text describes the relationship between numbers verbally (Find two numbers if one of them is 18 more than the other, and their sum is 80) or with a specific plot (A ticket to enter the stadium cost 160 rubles. After spectators increased by 50% and revenue increased by 25% after the entry fee was reduced (How much does a ticket cost after the entry fee is reduced?).
Each task is a unity of condition and goal. If one of these components is missing, then there is no task. It is very important to keep this in mind in order to analyze the text of the problem with such unity. This means that the analysis of the condition of the problem must be correlated with the question of the problem and, conversely, the question of the problem should be analyzed in a directed manner with the condition. They cannot be torn apart, as they are one whole.
A mathematical problem is a related laconic story in which the values of some quantities are entered and it is proposed to find others. unknown values quantities dependent on the data and associated with it by certain ratios specified in the condition.
Any text task consists of two parts: conditions and requirements (question), and conditions and requirements are interconnected.
The condition complies with information about the objects and some quantities that characterize the data of the object, about the known and unknown values of these quantities, about the relationships between them.
Task requirements are an indication of what needs to be found. It can be expressed in an imperative or interrogative sentence (“Find the speed of the cyclists or “How many kilometers did the tourist walk on each of the three days?”). There may be several requirements in a task.
Consider the problem: A sweater, a hat and a scarf were knitted from 1 kg 200 g of wool. The scarf required 100 g more wool than the hat and 400 g less than the sweater. How much wool was used for each item?
Task objects: scarf, hat, sweater. Regarding these objects, there are certain statements and requirements.
Statements: Sweater, hat, scarf are knitted from 1200 g of wool.
We spent 100 g more on a scarf than on a hat.
400 g less was spent on a hat than on a sweater.
Requirements: How much wool did you use for the sweater?
How much wool did you use for the hat?
How much wool did you use for the scarf?
There are three unknown values of quantities in the problem, one of which is contained in the requirement of the problem. This value is called the desired value.
Sometimes tasks are formed in such a way that part of the condition or the entire condition is included in one sentence with the task requirement.
In real life, a wide variety of problem situations often arise. Tasks formulated on their basis may contain redundant information, that is, information that is not needed to fulfill the requirements of the task.
On the basis of problem situations that arise in life, tasks can also be formulated in which there is not enough information to fulfill the requirements. So in the problem: “How many liters of water are in each barrel, if the first one has 48 liters more than the other?” - not enough data to answer her question. To solve this problem, it is necessary to supplement it with the missing data.
The same problem can be considered as a problem with a sufficient number of data, depending on the available and decisive values.
Considering the problem in the narrow sense of this concept, the following constituent elements can be distinguished in it:
1. A verbal presentation of the plot, in which, explicitly or in a veiled form, a functional relationship is indicated between the quantities whose numerical values are included in the problem.
2. Numerical values of quantities or numerical data referred to in the text of the problem.
A task, usually formulated as a question, in which it is proposed to find out the unknown values of one or more quantities. These values are called desired.
Understanding the role of the task and its place in the education and upbringing of the student, the teacher must approach the selection of the problem and the choice of methods for solving it reasonably and clearly know what work should give the student when solving the problem given to him.
1.2 Types of arithmetic problems
All arithmetic problems, according to the number of actions performed to solve them, are divided into simple and compound. The problem, for the solution of which it is necessary to perform an arithmetic operation once, is called a simple one. A task that requires several actions to solve is called a composite task.
Simple tasks in the system of teaching mathematics play extremely important role. With the help of solving simple problems, one of the central concepts of the initial course of mathematics is formed - the concept of arithmetic operations and a number of other concepts. Ability to decide simple tasks is a preparatory stage for students to master the ability to solve compound problems, since the solution of a compound problem is reduced to solving a number of simple problems. When solving simple problems, the first acquaintance with the problem and its components occurs. In connection with solving simple problems, children master the basic methods of working on a problem.
A composite problem includes a number of simple problems interconnected in such a way that the desired ones of some simple problems serve as data for others. The solution of a composite problem is reduced to dividing it into a number of simple problems and their sequential solution. Thus, to solve a composite problem, it is necessary to establish a system of relationships between the data and the desired, in accordance with which to choose, and then perform arithmetic operations.
Recording the solution of a compound problem by compiling an expression on it allows students to focus on the logical side of the work on the problem, to see the progress of solving it as a whole. At the same time, children learn to write down a plan for solving a problem and save time.
In the solution of a compound problem, something essentially new has appeared in comparison with the solution of a simple problem: here not one connection is established, but several, in accordance with which arithmetic operations are developed. Therefore, special work is being carried out to familiarize children with a compound problem, as well as to develop their skills to solve compound problems.
1.3 The role of the problem in mathematics
A significant place in mathematics is occupied by word problems. When considering the meaning of arithmetic operations, the connection existing between the actions and the relationship between the components and the results of actions, the corresponding simple tasks (tasks solved by one arithmetic operation) are certainly used. Text tasks are one of the essential funds familiarization of children with mathematical relationships, are used to understand the share, and help in the formation of a number of geometric concepts, as well as in the consideration of elements of algebra.
Acting as a specific material for the formation of knowledge, tasks provide an opportunity to connect theory with practice, learning with life. Solving problems forms in children the practical skills necessary for every person in Everyday life. For example, calculate the cost of a purchase, calculate what time you need to leave so as not to miss the train, etc.
The use of tasks as a concrete basis for introducing new knowledge and for applying the knowledge that children already have plays an extremely important role in shaping the elements of a materialistic worldview in children. Solving problems, the student is convinced that many mathematical concepts have roots in real life, in the practice of people. Through problem solving, children get acquainted with important cognitive and educational attitude facts. The content of many tasks reflects the work of children and adults, the achievements of our country in the field of the national economy, technology, science and culture.
The very process of solving problems with a certain methodology has a very positive effect on the mental development of schoolchildren, since it requires the performance of mental operations: analysis and synthesis, concretization and abstraction, comparison, generalization. So, when solving any problem, the student performs analysis: separates the question from the condition, highlights the data and the desired numbers; outlining a plan for the solution, he performs a synthesis, using concretization (mentally draws the condition of the problem), and then abstraction (distracting from the specific situation, chooses arithmetic operations); as a result of multiple solving problems of a certain type, the student generalizes knowledge of the relationships between data and what is sought in problems of this type, as a result of which a method for solving problems of this type is generalized.
Tasks are a useful means of developing logical thinking in children, the ability to analyze and synthesize, generalize, abstract and concretize, and reveal the connections that exist between the phenomena under consideration. Problem solving is an exercise that develops thinking. Moreover, problem solving contributes to the development of patience, perseverance, will, contributes to the awakening of interest in the very process of finding a solution, makes it possible to experience deep satisfaction associated with a successful solution.
Mastering the basics of mathematics is unthinkable without solving and analyzing the problem, which is one of the important links in the chain of knowledge of mathematics, this type of activity not only activates the study of mathematics, but also paves the way for a deep understanding of it. Work on understanding the course of solving a particular mathematical problem gives impetus to the development of the child's thinking. Solving problems cannot be considered an end in itself, they should be seen as a means to in-depth study theoretical positions and at the same time a means of developing thinking, a way of understanding the surrounding reality, a path to understanding the world. In addition, we must not forget that solving problems brings up positive qualities of character in children and develops them aesthetically.
1.4 Solution steps test tasks and how to do them
Problems and their solution occupy a very significant place in the education of schoolchildren both in terms of time and in terms of their influence on the mental development of the child. The solution of the problem is the result, that is, the answer to the requirement of the problem, the process of finding the result. Moreover, this process is considered in two ways: the method of finding the result and the sequence of those actions that the decisive one performs, using one method or another. That is, in this case, the solution of the problem is understood as all human activity, problem solving. The main methods for solving text problems are arithmetic and algebraic. To solve a problem in an arithmetic way means to find the answer to the requirement of the problem by performing arithmetic operations on numbers.
Problem solving is a somewhat unusual job, namely mental work. And in order to learn any work, you must first thoroughly study the material on which you will have to work, the tools with which this work is performed.
So, in order to learn how to solve problems, you need to understand what they are, how they are arranged, from which constituent parts they consist of what tools are used to solve problems.
Consider an example: “A certain person hired a worker for a year, promised to give him 12 rubles and a caftan. But he, having worked for 7 months, wanted to leave and asked for a decent pay with a caftan. The owner gave him a worthy settlement of 5 rubles and a caftan. The question is, what was the price of that caftan?
Solution of the problem: the employee did not receive 12 - 5 = 7 (rubles) for 12 - 7 = 5 (months),
therefore, for one month he was paid 7: 5 = 1.4 (rubles),
and in 7 months he received 7 * 1.4 = 9.8 (rubles),
then the caftan cost 9.8 - 5 = 4.8 (rubles).
Answer: the cost of the caftan is 4.8 rubles.
The same problem can be solved in different arithmetic ways. They differ from each other in the logic of reasoning performed in the process of solving the problem.
In an extended form, the solution of a text problem can be represented as a sequence of the following steps:
1) task analysis;
2) building a model;
3) search for a solution (drawing up a solution plan);
4) record of the decision;
5) verification of the solution;
6) study of the problem and its solution;
7) formulation of the answer;
8) educational and cognitive analysis of the problem and its solution.
Most often, only four stages are implemented: problem analysis, drawing up a solution plan, writing a solution, formulating an answer, and at all stages they stop only when solving complex, problematic tasks or tasks that have a certain generalizing - theoretical value.
The analysis of a task is always directed to its requirement.
Stage objectives: - to understand the situation described in the problem;
Highlight conditions and requirements;
Name known and sought objects;
Select all relationships (dependencies) between them.
To understand the content of the task, to isolate the conditions and requirements, you need to ask special questions:
1. What is the task about?
2. What is required to find in the problem?
3. What do certain words in the text of the problem mean?
4. What is unknown in the problem?
5. What is being sought?
Consider an example: “Two boys are walking along the road in the same direction. At first, the distance between them was 2 km, but since the speed of the boy walking in front is 4 km/h, and the speed of the second is 5 km/h, the second overtakes the first. From the beginning of the movement until the second boy catches up with the first, a dog runs between them at a speed of 8 km / h. From the boy walking behind, she runs to the one walking in front, having run, she comes back and runs like that until the boys are nearby. How far will the dog run in all this time?
Problem analysis: 1) What is this problem about?
The problem of the movement of two boys and a dog. It is characterized for each of the participants in the movement by speed, time and distance traveled.
2) What is required to find in the problem?
The task is to find the distance that the dog will run for the entire time from the start of the movement until the boys are nearby, i.e. the second one does not catch up with the first one.
3) What is known in the problem about the movement of each of its participants?
In the problem it is known: a) the boys go in the same direction;
b) before the start of the movement, the distance between the boys was 2 km;
c) the speed of the first boy walking in front is 4 km/h;
d) the speed of the second boy walking behind is 5 km/h;
e) the speed at which the dog runs, 8 km/h;
f) the time of movement, when the distance between the boys was 2 km before the meeting.
4) What is unknown in the problem?
In the problem, it is not known: a) the time in which the second boy will catch up with the first one (the time of movement of all its participants);
b) how fast the boys are approaching;
c) the distance that the dog ran (this is required to be found in the problem).
5) What is the desired one: a number, a value of a quantity, a kind of some relation?
The desired value is the value of the quantity - the distance that the dog ran during the time from the start of the boys' movement to the moment of the meeting.
A great help in understanding the task is provided by the technique - paraphrasing the text of the task. That is, everything superfluous (not essential) is discarded from the text of the problem, and the descriptions of some concepts are replaced by the corresponding terms, and vice versa, some terms are replaced by a description of the content of the corresponding concepts.
Paraphrasing the text of the task - transforming the text of the task into a form convenient for finding a solution plan. The result of the paraphrase should be the highlighting of the main situations. For ease of understanding the problem, you can write it down in the form of a table or a schematic drawing. Both the table and the schematic drawing are auxiliary models of the task. They serve as a form of fixing the analysis of a text problem and are the main means of finding a plan for its solution. After building the auxiliary model, you need to check:
1) whether all objects of the task are shown on the model;
2) whether all relations between objects are reflected;
3) whether all numerical data are given;
4) whether there is a question (requirement) and whether it correctly indicates what is being sought.
Finding a plan for solving a problem
Stage goals: to establish a connection between data and source objects;
outline a sequence of actions.
A plan for solving a problem is just an idea for a solution, its idea. It may happen that the found idea is wrong. Then it is necessary to return to the analysis of the problem again and start all over again.
One of the most well-known methods for finding a plan for solving a problem in an arithmetic way is to parse the problem from the text or from its auxiliary model. The analysis of the problem is carried out in the form of a chain of reasoning, which can begin both from the data of the problem and from its questions. When analyzing the problem from data to question, the solver singles out two data in the text of the problem and, based on knowledge of the relationship between them (such knowledge should be obtained when analyzing the problem), determine which unknown can be found from these data and with the help of which arithmetic operation. Then, considering this unknown as data, the solver again singles out two interconnected data, determines the unknown that can be found from them and with the help of what action, etc., until it is clarified which action leads to obtaining the object sought in the problem. When parsing the problem from the question to the data, you need to pay attention to the question of the problem and establish (based on the information obtained in the analysis of the problem) what is enough to know to answer this question. Why you need to refer to the conditions and find out if there is the necessary data for this. If there is no such data or there is only one data, then establish what you need to know in order to find the missing data (missing data), etc. Then a plan is drawn up for solving the problem. The reasoning is carried out in reverse order. Analysis according to the text of the problem: “A tourist traveled 6 hours on a train that was moving at a speed of 56 km / h. After that, he had to drive 4 times more than he drove. What is the whole path of a tourist?
Reasoning from the data to the question: it is known: the tourist traveled by train for 6 hours;
the speed of the train is 56 km/h.
From these data, you can find out the distance traveled by a tourist in 6 hours (multiply the speed by the time). Knowing the part of the distance traveled and the fact that the remaining distance is 4 times greater, you can find what it is equal to (the distance traveled must be multiplied by 4 (increase 4 times)). Knowing how many kilometers the tourist has traveled and how much he has left to go, you can find the whole path by adding the found segments of the path.
So actions: 1) the distance that the tourist traveled by train;
2) the distance that he has left to travel; . 3) all the way.
Reasoning from the question to the data: In the problem it is required to know the entire path of the tourist. We have established that the path consists of two parts. This means that to fulfill the requirement of the task, it is enough to know how many kilometers the tourist has traveled and how many kilometers he has left to travel. Both are unknown. To find the distance traveled, it is enough to know the time and speed with which the tourist was traveling. This is known in the problem. Multiplying the speed by the time, we find out the path that the tourist traveled. The remaining path can be found by increasing the distance traveled by 4 times (multiplying by 4). So, first you can find out the path traveled, then the remaining one, after which, by addition, find the whole path.
Implementation of the plan for solving the problem:
The purpose of the stage: to find the answer to the requirement of the task by completing all the actions in accordance with the plan.
For text problems that solve in an arithmetic way, the following techniques are used:
Record of actions (with explanation, without explanation, with questions);
Recording as an expression.
a) Recording a decision on actions with an explanation for each action performed: 1) 56 * 6 \u003d 336 (km) - the tourist traveled in 6 hours.
2) 336 * 4 = 1344 (km) - it remains for the tourist to pass;
3) 336 + 1344 = 1680 (km) - a tourist had to pass.
If explanations are given orally (or not given at all), then the entry will be as follows: 1) 56 * 6 = 336 (km);
2) 336 * 4 = 1344(km);
3) 336 + 1344 = 1680(km)
b) Recording a decision on actions with questions:
1) How many kilometers did the tourist travel by train?
56 * 6 = 336(km)
2) How many kilometers is left for the tourist to drive?
336 * 4 = 1344(km)
3) How many kilometers did the tourist have to travel?
336 + 1344 = 1680(km)
Checking the solution of the problem:
The purpose of the stage: to establish the correctness or erroneousness of the solution.
Several techniques are known to help determine whether the problem is solved correctly. Consider the main ones:
1. Establishing a correspondence between the result and the conditions of the problem. To do this, the found result is entered into the text of the problem and, on the basis of reasoning, it is established whether a contradiction arises in this case.
2. Solving the problem in a different way.
Let some result be obtained by solving the problem in some way. If its solution in another way leads to the same result, then the problem is solved correctly.
1.5 Some ways of solving word problems.
Based on the similarity in mathematical meaning and interchangeability of different methods of solution, all arithmetic methods can be combined into the following groups:
1) the method of reduction to a unit, reduction to a common measure, inverse reduction to a unit, the method of relations;
2) a way to solve problems from the "end";
3) a method of 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 parts;
5) a method for transforming one problem into another (decomposing a complex problem into simple, preparatory ones; bringing the unknowns to such values for which their ratio becomes known; the method of determining an arbitrary number for one of the unknown quantities).
In addition to these methods, it is advisable to consider the arithmetic mean method, the surplus method, the method of permuting the known and the unknown, the method of "false" rules.
Since it is usually impossible to determine in advance which of the methods is nairational, to foresee which of them will lead to the simplest and most understandable solution for the student, then students should be introduced to different methods and given the opportunity to choose which one to use when solving a specific problem.
Unknown Exclusion Method
This method is used when there are several unknowns in the problem. Such a problem can be solved using one of five methods: 1) replacing one unknown with another; 2) comparison of unknowns; 3) comparison of two conditions by subtraction; 4) data comparison; 5) combining several conditions into one.
As a result of applying one of the above methods, instead of several unknowns, one remains that can be found. Having calculated it, use the data in the dependency condition to find other unknowns.
Let's take a closer look at some of the methods.
1. Replacing one unknown with another
The name of the technique reveals its idea: based on the dependencies (multiple or difference), which are given according to the condition of the problem, it is necessary to express all the unknowns through one of them.
A task. Sergey and Andrey have 126 stamps in total. Sergey has 14 marks more than Andrey. How many stamps did each boy have?
Brief statement of the condition:
Sergey -- ? stamps, 14 stamps more
Andrew -- ? stamps
Total -- 126 stamps
Solution 1
(replacing a larger unknown with a smaller one)
1) Let Sergey have as many stamps as Andrey. Then the total number of stamps would be 126 -- 14 = 112 (marks).
2) Since the boys now have the same number of stamps, we will find how many stamps Andrey had at first: 112: 2 = 56 (marks).
3) Considering that Sergey has 14 marks more than Andrey, we get: 56 + 14 = 70 (marks).
Solution 2
(replacing the 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 stamps, we will find how many stamps Sergey had at first: 140: 2 = 70 (marks).
3) Considering that Andrei 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 the following fact with students: if the number A is greater than the number B by C units, then in order to compare the numbers A and B it is necessary:
a) subtract the number C from the number A (then both numbers are equal to the number B);
b) add the number C to the number B (then both numbers are equal to the 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 the unknown and express other quantities through it when drawing up an equation.
2. Comparison of unknowns
A 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 less than on the third shelf. How many books are on each shelf?
Task Analysis
For a better understanding of the dependencies between four unknown quantities (the number of books on each shelf), we use the scheme:
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, 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, we will remove 16 books from the first shelf, 8 books from the third, and put 4 books on the fourth shelf. Then on all shelves there will be the same number of books, namely, as on the second one it was at first.
1) How many books are on all the shelves after the operations described in the analysis of the task?
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 offer students to solve this problem in other ways, if we compare the unknown number of books that were on the first, or on the second, or on the fourth shelves.
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 numerical values of another quantity proportional to them.
A task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time they paid 500 rubles for 4 kg of oranges and 3 kg of bananas bought at the same prices. How much does 1kg of oranges and 1kg of bananas cost?
Brief statement of the condition:
4kg app. and 5kg ban. - 620 rubles,
4kg app. and 3kg ban. - 500 rubles.
1) Compare the cost of two purchases. Both the first time and the second time they bought the same number of oranges at the same price. The first time they paid more because they bought more bananas. Let's find how many kilograms of bananas were bought more for the first time: 5 - 3 = 2 (kg).
2) Let's find how much more they paid the first time than the second (that is, we find out how much 2 kg of bananas cost): 620 - 500 = 120 (rubles).
3) Find the price of 1 kg of bananas: 120: 2 = 60 (rubles).
4) Knowing the cost of the first and second purchases, we can find the price of 1 kg of oranges. To do this, first we find the cost of purchased bananas, then the cost of oranges, and then the price of 1 kg. We have: (620 - 60 * 5): 4 \u003d 80 (rubles).
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.
A task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time they paid 660 rubles for 6 kg of oranges and 3 kg of bananas bought at the same prices. How much does 1kg of oranges and 1kg of bananas cost?
Brief statement of the condition:
4kg app. and 5kg ban. - 620 rubles,
6kg app. and 3kg ban. - 660 rubles.
Let's equalize the number of oranges and bananas by comparing them with the least common multiple: LCM(4;6) = 12.
Solution 1.
1) Let's increase the number of purchased fruits and their cost in the first case by 3 times, and in the second - by 2 times. We get the following shorthand for the condition:
12kg app. and 15kg ban. - 1860 rubles,
12kg app. and 6kg ban. - 1320 rubles.
2) Find out how many more bananas were bought for the first time: 15-6 = 9 (kg).
3) How much does 9kg of bananas cost? 1860 - 1320 = 540 (rubles).
4) Find the price of 1 kg of bananas: 540: 9 = 60 (rubles).
5) Find the cost of 3 kg of bananas: 60 * 3 = 180 (rubles).
6) Find the cost of 6 kg of oranges: 660 - 180 = 480 (rubles).
7) Find the price of 1 kg of oranges: 480: 6 = 80 (rubles).
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 bought the first time and the second time, we reduce the quantity of the purchased goods and its cost in the first case by 2 times, in the second - by 3 times. Let's get a task that has such a short condition record
2kg app. and 2.5 kg ban. - 310 rubles,
2kg app. and 1kg ban. - 220 rubles.
2) How many more bananas are now bought: 2.5 - 1 = 1.5 (kg).
3) Find how much 1.5 kg of bananas costs: 310 - 220 = 90 (rubles).
4) Find the price of 1 kg of bananas: 90: 1.5 = 60 (rubles).
5) Find the price of 1 kg of oranges: (660 - 60 * 3): 6 = 80 (rubles).
Answer: the price of 1 kg of oranges is 80 rubles, 1 kg of bananas is 60 rubles.
When solving problems using the method of comparing data, you can not do such a detailed analysis and records, but only record the changes that were made for comparison, and write them down in the form of a table.
5. Combining multiple conditions into one
Sometimes you can get rid of unnecessary unknowns by combining several conditions into one.
A task. The tourists left the camp and at first walked for 4 hours, and then for another 4 hours they rode bicycles 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, moving on foot for 4 hours, ended up at a distance of 50 km from the camp. How fast were the tourists cycling?
There are two unknowns in the problem: the speed with which the tourists rode bicycles, and the speed with which they walked. In order to exclude one of them, you can combine two conditions into one. Then the distance traveled by tourists in 4 hours, moving forward the first time on foot, is equal to the distance they traveled in 4 hours, moving backward 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 50 + 60 = 110 (km).
Then the speed of tourists on bicycles: 110: 11 = 10 (km/h).
Answer: Bicycles travel at a speed of 10 km/h.
6. Method of admission
The use of the assumption method in solving problems does not cause difficulties for most students. Therefore, in order to avoid mechanical memorization by students of the scheme of steps of this method and misunderstanding of the essence of the actions performed on each of them, students should first be shown the method of trials (“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") some value. Then, using all the conditions, they find the value of another quantity. The resulting value is compared with the one specified in the condition. If the value obtained is different from the given one 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 is found. So it is necessary to do until we get the value of another quantity such as in the condition of the problem.
A task. The cashier has 50 coins of 50 kopecks and 10 kopecks, totaling 21 rubles. Find how many 50k coins the cashier had separately. and 10k.
Solution 1. (sampling method)
Let's use the rule of the "ancient" Babylonians. Suppose that the cashier has equal coins of each denomination, that is, 25 pieces. Then the amount of money will be 50 * 25 + 10 * 25 \u003d 1250 + 250 \u003d 1500 (k.), Or 15 rubles. But in the condition of 21 rubles, that is, more than received, by 21 UAH - 15 rubles = 6 rubles. This means that it is necessary to increase the number of coins of 50 kopecks and reduce the number of coins of 10 kopecks, until we get a total of 21 rubles. We write 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 than or greater than condition |
|
Less than 6 rubles. |
||||||
Less than 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 equal coins of 50k. and 10k each, then in total he had money 15 rubles. It is easy to see that each replacement of a coin is 10k. for a 50k coin. increases the total amount by 40k. This means that it is necessary to find how many such replacements need to be made. To do this, we first find by how much money it is necessary to increase the total amount:
21 rub - 15 rub. = 6 rubles. = 600 k.
Let's find how many times such a replacement needs to be done: 600 k. : 40 k. = 15.
Then 50 k. each will be 25 +15 =40 (coins), and 10 k. coins will remain
25 -- 15 = 10.
Verification 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.
Inviting students to choose different meanings the 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 the same denomination (for example, all 50 coins of 50 kopecks or all 50 coins of 10k 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 by trial and error. We use the method of residuals when solving problems for movement in one direction, namely, when it is necessary to find the time during which the first object, which moves behind with 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 that the first object needs to overcome the distance that was between it and the second at the beginning of the movement, it is necessary to 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 cover) or their difference. Unknown multipliers (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 by the method of residuals are called problems for finding numbers by two differences.
A task. The students decided to paste photos from the holiday into the album. If they stick 4 photos on each page, then there will not be enough space for 20 photos in the album. If you stick 6 photos on each page, then 5 pages will remain free. How many photos are the 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 condition of the problem, it is unknown, but it can be found if the number of photos that are placed on one page and the number of pages in the album are known.
The number of photos that are pasted on 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 photos could be pasted into the album: 6 * 5 = 30 (photos).
So, 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 two more photos were pasted on each page and a total of 50 more photos were pasted, we find the number of pages in the album: 50: 2 = 25 (p.).
Therefore, there were 4 * 25 + 20 = 120 photos in total.
Answer: There were 25 pages in the album and 120 photos were pasted.
Chapter II. Teaching schoolchildren how to solve text arithmetic problems
I conduct training in methods for solving text problems systematically, when studying each topic of the school course.
2.1 Solving joint motion problems
Beginning in 5th grade, students often come across these problems. Also in primary school students are given the concept of “general speed”. As a result, they form not quite correct ideas about the speed of approach and the speed of removal (there is no such terminology in elementary school). Most often, when solving a problem, students find the sum. It is best to start solving these problems with the introduction of the concepts: “rapprochement rate”, “removal rate”. For clarity, you can use the movement of the hands, explaining that bodies can move in one direction and in different directions. In both cases, there may be an approach speed and a removal speed, but in different cases they are found in different ways. After that, students write down the following table:
Table 1.
Methods for finding the speed of approach and speed of removal
When analyzing the problem, the following questions are given
1. Using the movement of the hands, we find out how the bodies move relative to each other (in one direction, in different ones).
2. We find out what action is the speed (addition, subtraction).
3. Determine what speed it is (approach, removal).
Write down the solution to the problem.
Example No. 1. From cities A and B, the distance between which is 600 km, at the same time, a truck and a car left towards each other. The speed of a car is 100 km/h, and that of a truck is 50 km/h. In how many hours will they meet?
Students use their hands to show how cars move and draw the following conclusions:
a. cars move in different directions;
b. the speed will be found by addition;
in. since they are moving towards each other, this is the speed of approach.
1. 100 + 50 = 150 (km/h) - closing speed.
2. 600: 150 = 4 (h) - the time of movement before the meeting.
Answer: after 4 hours.
Example #2. The man and the boy left the house for the dacha at the same time and go the same way. The man's speed is 5 km/h and the boy's speed is 3 km/h. How far apart will they be after 3 hours?
With the help of hand movements, we find out:
a. the boy and the man are moving in the same direction;
b. speed is the difference;
in. the man walks faster, i.e., moves away from the boy (removal speed).
1. 5 - 3 \u003d 2 (km / h) - removal speed.
2. 2 * 2 \u003d 4 (km / h) - the distance between a man and a boy after 2 hours
Answer: 4 km.
2.2 Tasks solved using tables
When preparing for solving such problems, you can successfully use signal maps.
Oral counting should be carried out using card data that each student should have, which allows the whole class to be involved in the work.
Example number 1. The first boy has 5 marks more than the second. How to find how many stamps the second one has?
Students raise card number 1 and explain that 5 should be added to the number of the first, since it has 5 more, emphasizing with the intonation “by ... more”
Example #2. The second boy has 30 marks, and the first has 3 times less. How many stamps does the first boy have?
Students should hold up card number 4 and answer: 10 marks, since 30:3 \u003d 10. Supporting words - "in ... less."
The selection of tasks for oral counting should be varied, but each time the student must give an explanation, naming the key words. It is better to underline key words in the table.
Example #3. The rider traveled 80 km in 5 hours. How much time will the cyclist spend on this path if his speed is 24 km / h more than the speed of the rider?
When filling out the table, the student must underline the key words and explain that the speed of the cyclist is found by adding 16 km / h and 24 km / h. Then, establishing a functional relationship between the values, students fill in all the rows and columns of the table. After that, depending on the task, the student either answers the question or draws up a solution. When working with a table, the student must understand that when solving a problem, all rows and columns must be filled with task data, and data that are obtained as a result of using a functional relationship between quantities.
2.3 Solving problems on finding a part of a number and a number by part
To prepare for solving these problems, work is underway to master the concept of a fraction. With oral counting, it is necessary to ensure that each student knows: a. what action does the fractional bar indicate;
b. which means fraction.
The fractional line indicates the action of division, and the fraction 3/4 indicates that the given was divided into 4 equal parts and 3 parts were taken. To do this, it is good to use the envelopes that all students prepare with the help of their parents. Circles are enclosed in envelopes: whole, cut in half, into 3 equal parts, into 4; 6; 8 parts. Each share of one circle has the same color. Using this material, students visually see how fractions are obtained.
For example. Lay out a figure depicting a fraction 5/6. Knowing the colors of the shares, the teacher sees the mistakes made by the students and analyzes the task. When answering, the student says that the circle was divided by 6 equal parts and took 5 such parts.
The presence of such envelopes makes it possible to visualize the addition of fractions with the same denominators and the subtraction of a fraction from a unit. Since all students are involved in the work and the addition is clearly visible, after two examples, the students themselves formulate the rule for adding fractions with the same denominators.
Consider subtraction. Subtract 1/4 from 1. Students put a circle on the table, but notice that nothing can be removed from it yet. Then they offer to cut the circle into 4 equal parts and remove one. We conclude that 1 must be replaced by a fraction 4/4. After 2-3 examples, students make their own conclusion.
With the use of this material, the concept of the main property of a fraction is given, when they impose 2/6 on the fraction 1/3, etc. Having worked out this material, we proceed to solve problems.
Example #1. There are 120 trees in the garden. Birches make up 2/3 of all trees, and the rest are pines. How many pines were there?
Question: What does the fraction 2/3 mean?
Answer: The entire number of trees was divided into 3 equal parts and birches make up 2 parts.
40 * 2 \u003d 80 (village) - there were birches.
120 - 80 \u003d 40 (village) - there were pines.
II way:
120: 3 = 40 (der.) - make up one part.
3 - 2 \u003d 1 (part) - make up pines.
40 * 1 \u003d 40 (village) - make up pines.
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1. The arithmetic method is the main method for solving text problems in elementary school. It finds its application in the middle link secondary school. This method allows you to better understand and appreciate the importance and significance of each stage of the work on the task.
In some cases, the solution of the problem by the arithmetic method is much simpler than by other methods.
Bribing with its simplicity and accessibility, the arithmetic method is at the same time quite complicated, and mastering the methods of solving problems by this method requires serious and painstaking work. A wide variety of types of problems does not allow the formation of a universal approach to the analysis of problems, the search for a way to solve them: problems, even combined into one group, have completely different ways of solving.
2 . For tasks on finding unknowns by their difference and ratio include problems in which it is required to find these values from the known difference and the quotient of two values of a certain quantity.
Algebraic model:
The answer is given by the formulas: X= ak / (k - 1), y \u003d a / (k - 1).
Example. There are 432 more passengers in the second-class carriages of the fast train than in compartments. How many passengers are in reserved seat and compartment cars separately, if there are 4 times fewer passengers in compartment cars than in reserved seats?
Solution. The graphical model of the problem is shown in fig. four.
Rice. four
The number of passengers in compartment cars will be taken as 1 part. Then you can find how many parts are for the number of passengers in second-class cars, and then how many parts are for 432 passengers. After that, you can determine the number of passengers that make up 1 part (located in compartment cars). Knowing that there are 4 times more passengers in second-class carriages, we find their number.
1 4 \u003d 4 (h) - accounts for passengers in reserved seat cars;
4 - 1 \u003d 3 (hours) - falls on the difference between the number of passengers in reserved seat and compartment cars;
432: 3 = 144 (p.) - in compartment cars;
144 4 \u003d 576 (p.) - in reserved seat cars.
This problem can be checked by solving it in another way, namely:
1 4 \u003d 4 (h);
4 - 1 = 3 (h);
432: 3 = 144 (p.);
144 + 432 = 576 (p.).
Answer: there are 144 passengers in compartment cars, 576 in reserved seats.
For tasks on finding unknowns by two remainders 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 \u200b\u200bof the other quantity are known, and it is required to find the values of this quantity themselves.
Algebraic model:
The answers are found according to the formulas:
Example. Two trains traveled at the same speed - one 837 km, the other 248 km, and the first was on the way 19 hours more than the second. How many hours did each train travel?
Solution. The graphical model of the task is shown in Figure 5.
Rice. 5
To answer the question of the problem, how many hours this or that train was on the way, you need to know the distance traveled by it and the speed. The distance is given in the condition. To know the speed, you need to know the distance and the time it took to cover that distance. The condition says that the first train went 19 hours longer, and the distance traveled by it during this time can be found. He walked an extra 19 hours - obviously, during this time he walked an extra distance.
837 - 248 \u003d 589 (km) - the first train traveled so many kilometers more;
589: 19 = 31 (km/h) – speed of the first train;
837: 31 = 27 (h) - the first train was on the way;
4) 248: 31 = 8 (h) - the second train was on the way.
Let's check the solution of the problem by establishing a correspondence between the data and the numbers obtained in solving the problem.
Having learned how long each train was on the way, we find how many hours the first train was on the way more than the second: 27 - 8 \u003d 19 (h). This number matches the given in the condition. Therefore, the problem is solved correctly.
This problem can be verified by solving it in a different way. All four questions and the first three actions remain the same.
4) 27 –19 = 8 (hours).
Answer: the first train was on the way for 31 hours, the second train - 8 hours.
Tasks for finding three unknowns for three sums of these unknowns, taken in pairs:
Algebraic model:
The answer is given by the formulas:
x =(a -b + c)/2, y = (a +b – c)/2, z = (b + With -a)/ 2.
Example. English and German languages 116 students study German and Spanish, 46 students study English and Spanish, and 90 students study English and Spanish. How many students study English, German and Spanish separately if each student is known to study only one language?
Solution. The graphical model of the task is shown in Figure 6.
How many students are learning 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 will get twice the number of schoolchildren studying English, German and Spanish. Dividing it by two, we find the total number of students. To find the number of students studying English language, it is enough to subtract from this number the number of students studying German and Spanish. Similarly, we find the remaining desired numbers.
Let's write down the decision on actions with explanations:
116 + 90 + 46 = 252 (school) - twice the number of students 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 a different way.
116 - 46 = 70 (school) - so many more students learn English than Spanish;
90 + 70 = 160 (school) - twice the number of students studying English;
160: 2 = 80 (school) - study English;
90 - 80 = 10 (school) - learn Spanish;
116 - 80 = 36 (school) - learn German.
Answer: English is studied by 80 students, German by 36 students, Spanish by 10 students.
3. Proportional division problems include problems in which a given value of a certain quantity needs to be divided into parts in proportion to given numbers. In some of them, the parts are explicitly presented, while in others these parts must be distinguished by taking one of the values of this quantity for one part and determining how many such parts fall into its other values.
There are five types of tasks for proportional division.
1) Tasks for dividing a number into parts, directlyproportional to a series of whole or fractional numbers
To tasks of this type include tasks in which the number BUT 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 given by 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 vacationers can be accommodated at each base if all 4 bases can accommodate 2112 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 hulls 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 in 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 (c.) - located on the territory of 4 bases;
2112: 22 = 96 (hours) - can be accommodated in one building;
96 6 \u003d 576 (h) - can be placed on first base;
96 4 \u003d 384 (h) - can be placed on the second base;
96 5 \u003d 480 (h) - can be placed on third base;
96 7 \u003d 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 condition of the problem. Problem solved correctly.
Answer: the first base can accommodate 576 vacationers, the second - 384 vacationers, the third - 480 vacationers, the fourth - 672 vacationers.
2) Problems for dividing a number into parts inversely proportional to a series of integer or fractional numbers
These include tasks in which the number BUT(the value of some quantity) must 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 P :a 1 a 2 a 4 ...a n :...:a 1 a 2 ...a n -1
The answer is given by 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 income of the fur farm from the sale of furs amounted to 1,925,000 rubles, and by months the money received was distributed in inverse proportion to the numbers 2, 3, 5, 4. What is the income of the farm in each month separately?
Solution. To determine the income named 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 desired income is inversely proportional to the numbers 2, 3, 5, 4.
Denote the desired income, respectively, through x, X 2 , X 3 , X 4 . Then the problem can be briefly written as shown in Figure 8.
Rice. eight
Knowing the number of parts that fall on each of the desired numbers, we find the number of parts that are in their sum. From the given total income for four months, that is, from the sum of the desired numbers and the number of parts contained in this sum, we find out the value of one part, and then the desired income.
Let's write down the decision on actions with explanations:
1. The desired income is inversely proportional to the numbers 2, 3, 5, 4, which means that they are directly proportional to the numbers inverse to the data, that is, there are relations . We replace these relations in fractional numbers with relations of integers:
2. Knowing that X contains 30 equal parts, X 2 – 20, X 3 – 12, X 4 – 15, find how many parts are contained in their sum:
30 + 20 + 12+ 15 = 77 (hours).
3. How many rubles per part?
1,925,000: 77 = 25,000 (r.).
4. What is the farm 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 (p.).
7. What is the farm income in the fourth month?
25,000–15 = 375,000 (p.).
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) Tasks for dividing a number into parts, when separate ratios are given for each pair of desired numbers
Tasks of this type include those tasks in which the number BUT(the value of a certain quantity) must be divided into parts x 1, X 2 , x 3, ..., X", when given a series of ratios for the desired 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 inhabitants of the first and second cities are in the ratio , and the second and third cities - in relation to . How many inhabitants in each city?
Solution. Let us denote the desired number of inhabitants, respectively, through 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 desired numbers, as well as individual relationships between the desired numbers. Replacing these ratios with a number of ratios, we express the number of inhabitants of three cities in equal parts. Knowing the number of parts that fall on each of the desired numbers, we find the number of parts that are in their sum. From the given total number of inhabitants in three cities, that is, from the sum of the desired numbers and the number of parts contained in this sum, we find out the value of one part, and then the desired population.
Let's write down the decision on actions with explanations.
1. We replace the ratio of fractional numbers with the ratio of integers:
The number of inhabitants of the second city is assigned the number 15 (the least common multiple of the numbers 3 and 5).
Change the resulting relationship accordingly:
X 1: X 2 \u003d 4: 3 \u003d (4-5): (3-5) \u003d 20: 15, x 2: x 3 \u003d 5: 7 \u003d (5-3): (7-3) \u003d 15: 21.
From individual relations we make a series of relations:
X 1: X 2 : X 3 = 20: 15: 21.
2. 20 + 15 + 21 \u003d 56 (h) - so many equal parts correspond to the number 168,000;
3. 168,000: 56 \u003d 3,000 (f.) - accounts for one part;
4. 3,000 20 = 60,000 (female) - in the first city;
5. 3,000 15 = 45,000 (female) - in the second city;
3,000 21 = 63,000 (female) - in the third city.
Answer: 60,000 inhabitants; 45,000 inhabitants; 63,000 inhabitants.
4) Tasks for dividing a number into parts in proportion to two, three, and so on rows of numbers
Problems of this type include problems in which the number BUT(the value of some quantity) must be divided into parts X 1, X 2 , X 3 ,..., X n in proportion to two, three, ..., N rows of numbers.
Due to the cumbersome 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 the numbers a 1 , a 2 , a 3 and inversely proportional to the numbers b 1 , b 2 , b 3 .
Algebraic model:
(see point 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, one must know how many rubles were 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 worked, and for this you need to know how many days each worked and how many hours a day. This information is included in the condition.
Let's write down the decision on actions with explanations:
8 3 \u003d 24 (h) - the first worker worked;
6 6 \u003d 36 (h) - the second worker worked;
24 + 36 = 60 (hours) - both workers worked together;
1800: 60 = 30 (p.) - workers received for 1 hour of work;
30 24 \u003d 720 (p.) - the first worker earned;
30 36 \u003d 1080 (p.) - the second worker earned. Answer: 720 rubles; 1080 rubles
5) Tasks for finding several numbersaccording to their ratios and sum or difference (sum or difference of some of them)
Example. The school administration spent 49,000 rubles on equipping the playground, greenhouse and sports hall. The equipment of the playground cost half as much as greenhouses, and greenhouses cost 3 times cheaper than sports hall and playground together. How much money was spent on equipping each of these facilities?
Solution. A summary of the task is shown in Figure 11.
Rice. elevenIn order to find out the amount of money spent on the equipment of each object, it is necessary to know how many parts of all the money spent were on the equipment of each object and how many rubles were accounted for by each part. The number of parts of the money spent on the equipment of each object is determined from the condition of the problem. Having determined the number of parts for the equipment of each facility separately, and then finding their sum, we calculate the value of one part (in rubles).
Let's write down the decision on actions with explanations.
We accept as 1 part the amount of money spent on the equipment of the playground. According to the condition, 2 times more was spent on the equipment of the greenhouse, that is, 1 2 = 2 (hours); 3 times more was spent on the equipment of the playground and the gym than on the greenhouse, that is, 2 3 = 6 (hours), therefore, 6 - 1 = 5 (hours) were spent on the equipment of the gym.
1 part was spent on the equipment of the playground, 2 parts for greenhouses, 5 parts for the gym. The entire consumption was 1 + 2 + + 5 = 8 (hours).
8 parts are 49,000 rubles, one part is 8 times less than this amount: 49,000: 8 \u003d 6,125 (r.). Consequently, 6,125 rubles were spent on the equipment of the playground.
2 times more was spent on the equipment of the greenhouse: 6 125 2 = 12 250 (r.).
5 parts were spent on the equipment of the sports hall: 6 125 5 = 30 625 (r.).
Answer: 6 125 rubles; 12 250 rubles; 30 625 rubles
6) Tasks to eliminate one of the unknowns
The problems of this group include problems in which the sums of two products with two repeating factors are given, and it is required to find the values of these factors. Algebraic model
The answer is given by the formulas:
These tasks are solved by the method of data adjustment, the method of adjustment of data and the desired ones, the method of replacing data, as well as the so-called “guessing” method.
Example. At a garment factory, 204 m of fabric was used for 24 coats and 45 suits, and 162 m for 24 coats and 30 suits. How much fabric is used for one suit and how much for one coat?
Solution. Let's solve the problem by way of data equalization. A brief record of the task.
Learning to solve text problems plays an important role in the formation of mathematical knowledge. Text tasks give a lot of scope for the development of students' thinking. Learning to solve problems is not only learning the technique of getting the right answers in some typical situations, but learning a creative approach to finding solutions, gaining experience in mental activity and demonstrating by students the possibilities of mathematics in solving various problems. However, when solving word problems in grades 5-6, the equation is most often used. But the thinking of the fifth graders is not yet ready for the formal procedures performed when solving equations. The arithmetic method of solving problems has a number of advantages over the algebraic one, because the result of each step in the actions is clearer and more specific, and does not go beyond the experience of fifth-graders. Students solve problems better and faster by actions than by using equations. Children's thinking is concrete, and it is necessary to develop it on specific objects and quantities, then gradually move on to operating with abstract images.
Work on the task involves a careful reading of the text of the condition, understanding the meaning of each word. I will give examples of problems that can be easily and simply solved in an arithmetic way.
Task 1. To make jam, three parts of sugar are taken for two parts of raspberries. How many kilograms of sugar should be taken for 2 kg of 600 g of raspberries?
When solving a problem in “parts”, one must learn to visualize the condition of the problem, i.e. it is better to rely on the drawing.
- 2600:2=1300 (g) - accounts for one part of the jam;
- 1300 * 3 = 3900 (g) - you need to take sugar.
Task 2. On the first shelf stood 3 times more books than on the second. There were 120 books on two shelves together. How many books were on each shelf?
1) 1+3=4 (parts) - falls on 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. There are 27 heads and 74 legs in total. Find out the number of pheasants and the number of rabbits in a cage.
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 her. Then:
- 27*2=54 (legs) - will stand on the floor;
- 74-54=20 (feet) - will be at the top;
- 20:2=10 (rabbits);
- 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 either to the excursion or to the cinema. How many people went on both the tour and the cinema?
To analyze the condition and choose a solution plan, you can use “Euler circles”.
- 30-5=25 (persons) - went to the cinema or on a tour,
- 25-23=2 (people) - went only to the cinema;
- 21-2=19 (people) - went to the cinema and on an excursion.
Task 5. Three ducklings and four caterpillars weigh 2 kg 500 g, and four ducklings and three caterpillars weigh 2 kg 400 g. How much does one gosling weigh?
- 2500+2400=2900 (g) - seven ducklings and seven goslings weigh;
- 4900:7=700 (g) - the weight of one duckling and one caterpillar;
- 700 * 3 \u003d 2100 (g) - weight 3 ducklings and 3 goslings;
- 2500-2100 \u003d 400 (g) - the 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 great pyramids were there?
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 (great pyramids).
Task 7. A watermelon weighing 20 kg contained 99% water. When it shrunk 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 desirable to draw the “dry matter” rectangles equal, because the mass of “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 \u003d 10 (kg) - the mass of watermelon.
Task 8. The guests asked: how old was each of the three sisters? Vera replied that she and Nadia were together for 28 years, Nadia and Lyuba were together for 23 years, and all three were 38 years old. How old are each sisters?
- 38-28=10 (years) - Luba;
- 23-10=13 (years) - Nadia;
- 28-13=15 (years) - Vera.
The arithmetic method of solving text problems teaches the child to act consciously, logically correctly, because when solving in this way, attention to the question “why” increases and there is a 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, one must be very careful in choosing text problems, consider various ways to solve them, choosing the best ones, develop logical thinking, which is necessary in the future when solving geometric problems.
Students can learn how 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. Poya in the book “Mathematical Discovery”.
Generalization of experience.
Text tasks in the school course of mathematics.
Arithmetic ways of solving problems.
Soldatova Svetlana Anatolievna
mathematics teacher of the first category
MOU Uglich Physics and Mathematics Lyceum
2017
"... while we are trying to link the teaching of mathematics with life, it will be difficult for us to do without word problems - a traditional means of teaching mathematics for the domestic methodology."
A.V. Shevkin
We encounter the term "task" all the time in our daily lives. Each of us solves certain problems, which we call tasks. In the broad sense of the wordA task is a situation that requires research and decision by a person. .
Tasks in which objects are mathematical (proving theorems, computational exercises, properties and signs of the studied mathematical concept, geometric figure) are often calledmath problems . Mathematical problems in which there is at least one object that is a real object are usually calledtext. The role of word problems is great in elementary education in mathematics.
Solving text problems, students acquire new mathematical knowledge, prepare for practical activities. Tasks contribute to the development of their logical thinking.
There are various methods for solving text problems: arithmetic, algebraic, geometric, logical, practical, etc. Each method is based on various types of mathematical models. For example, whenalgebraic method solving the problem, equations or inequalities are compiled, withgeometric - charts or graphs are built. The solution of the problemlogical method begins with the compilation of the algorithm.
It should be borne in mind that almost every problem within the framework of the chosen method can be solved using various models. So, using the algebraic method, the answer to the requirement of the same problem can be obtained by compiling and solving completely different equations, using the logical method - by building different algorithms. It is clear that in these cases we are also dealing with various methods for solving a specific problem, which I callsolutions.
To solve the task arithmetic method - means to find the answer to the requirement of the problem by performing arithmetic operations on numbers. One and the same problem in many cases can be solved by different arithmetic methods. A problem is considered to be solved in different ways if its solutions differ in the connections between the data and the desired ones underlying the solutions, or in the sequence of these connections.
In traditional Russian school teaching of mathematics, word problems have always occupied special place. On the one hand, the practice of using text tasks in the learning process in all civilized states comes from the clay tablets of Ancient Babylon and other ancient written sources, that is, it has related roots. On the other hand, the close attention of teachers to text tasks, which was typical for Russia, is an almost exclusively Russian phenomenon.
One of the reasons for the great attention to problems is that historically for a long time the goal of teaching children arithmetic was to master a certain range of computational skills associated with practical calculations. At the same time, the main line of arithmetic - the line of numbers - has not yet been developed, and calculations were taught through tasks.
The second reason for the increased attention to the use of word problems in Russia is that in Russia not only have they adopted and developed the old fashioned way transfer mathematical knowledge and reasoning techniques with the help of text problems, but also learned to form important general educational skills related to text analysis, highlighting the conditions of a problem and a question, drawing up a solution plan, posing a question and searching for conditions from which you can get an answer to it checking the result.
By the mid 50sXXin. text tasks were well systematized,a developed typology of tasks has developed, including tasks for parts, for finding two numbers by their sum and difference, by their ratio and sum (difference), for fractions, for percentages, for joint work, for solutions and alloys, for direct and inverse proportionality and etc.
By this time, the methodology for their application in the educational process was well developed, but during the reform of mathematical education in the late 60s, the attitude towards them changed. Revising the role and place of arithmetic in the system of school subjects, striving to increase the scientific character of the presentation of mathematics through the earlier introduction of equations and functions, mathematicians and methodologists-mathematicians considered that too much time was spent on teaching arithmetic methods for solving problems.
But it is word problems and arithmetic methods for solving them that prepare the child for mastering algebra. And when this happens, then algebra will teach simpler than arithmetic ways to solve some (but not all!) Problems. Other arithmetic solutions will remain in the active baggage of the student. For example, if a student was taught to divide a number in this ratio, then even in high school he will not divide the number 15 in a ratio of 2: 3 using an equation, he will perform arithmetic operations:
1) ,
2) ,
3) 15 – 6 = 9.
I would like to note that I am a representative of exactly that generation of schoolchildren who were participants in the above reform. I went to school in 1968 and my first grade textbook was called Arithmetic. It turns out that we were the last ones to learn from it. In the second grade, it was surprising and unusual for me that the subject, and, accordingly, the textbook, of my first-grader girlfriends was called “mathematics”. In the third grade, we already studied "mathematics". In the middle link, and accordingly in the senior classes, the main way to solve text problems was algebraic. I feel the influence of the reform of the late 60s to this day, because. parents who participate in educational process children, due to the fact that they developed a certain stereotype, an opinion was formed that problems should be solved with the help of equations. Moms and dads, not knowing other methods, persistently try to explain at home in their own way, which is not always beneficial, even sometimes it only complicates the teacher's work.
In no case should one belittle the value of the algebraic method of solving problems, which is universal and sometimes the only one in solving more complex problems. In addition, quite often it is the equation that gives a hint for finding a way to solve by actions. But practice has shown that the early application of this promising, from the point of view of further use in training, method of solving problems without sufficient preparation is ineffective.
In grades 5-6, the arithmetic method of solving text problems should be given maximum attention and not in a hurry to move on to solving problems using an equation. Once a student has learned the algebraic way, it is almost impossible to bring him back to "decision by action." Having compiled an equation, the main thing is to solve it correctly, to prevent a computational error. And there is absolutely no need to think about what arithmetic operations are performed in the course of solving, what is the result of each action. And if we trace the solution of the equation step by step, we will see the same actions as in the arithmetic method.
Very often you can see that the child is not ready to solve the problem in an algebraic way, when an abstract variable is introduced and the phrase "let x ..." appears. Where did this "X" come from, what words should be written next to it - at this stage it is not clear to the student. And this happens because children of this age have developed visual-figurative thinking. And the equation is an abstract model. Yes, and there are no tools for solving equations in children of the fifth, beginning of the sixth grade. Historically, people came to use equations by generalizing solutions to problems in which they had to operate with such concepts as “part”, “heap”, etc. The child must go the same way!
For successful work, it is important that the teacher has a deep understanding of the text problem, its structure, and is able to solve such problems in various ways.
Many years ago, I had in my hands a long-released manual for teachers in grades 5-8 (in modern school- 5-9 grades) "Collection of Moscow Mathematical Olympiads (with solutions)" 1967, the author of which is Galina Ivanovna Zubelevich. The vast majority of problems in it are solved arithmetically, which interested me very much. Later, my attention was attracted by two textbooks "Arithmetic, 6", and "Arithmetic, 6" by A.V. Shevkin, and a teacher's guide "Teaching text problem solving in grades 5-6" by the same author. These sources were the beginning of my work on this topic. The proposed ideas seemed to me very relevant and consonant with my understanding of the stated topic, namely:
1) abandoning the use of equations at an early stage of learning and returning to a wider use of arithmetic methods for solving problems;
2) wider use of "historical" problems and ancient ways of solving them;
3) refusal of a chaotic offer to students of tasks on different topics and consideration of a chain of tasks from the simplest, accessible to all students, to complex and very complex.
Types of word problems according to the method of solution.
Text tasks can be conditionally divided into arithmetic and algebraic. This separation is due to the choice of a solution method that is more characteristic (rational) for a particular problem.
Arithmetic problems conceal great opportunities for teaching schoolchildren to think independently, analyzing non-obvious life situations. Arithmetic is the shortest way to understanding nature, as it deals with the simplest, most fundamental, experimental facts (for example, that recalculation
stones "by rows" and "by columns" always leads to one
result):
5+5+5 = 3+3+3+3+3.
Let's consider some types of tasks.
“Two grades of goods were bought for the same amount, the first grade is half as much as the second. They were mixed and sold half of the mixture at the price of the highest, the rest at the price of the lowest grade. What percentage of the profit or loss was made on the sale?
This is, in essence, a typical problem solved by introducing arbitrary units of measure. However, even under this condition, the operation of unknown quantities necessary for the solution is here clearly expressed.algebraic character. Along with this, there are often problems in which, on the contrary, the arithmetic way of solving is much simpler than the algebraic one. This may depend on two reasons. In some cases, the transition from the known to the unknown is so simple that the formulation of equations (the transition from the unknown to the known) would introduce unnecessary cumbersomeness that slows down the solution process. Such, for example, is the following task:
“Once the Devil offered to earn money for the idler. “As soon as you cross this bridge,” he said, the money will double. You can cross it as many times as you want, but after each transition, give me 24 kopecks for it. The loafer agreed and ... after the third transition he was left penniless. How much money did he have at first?
The second is a classical problem, interesting because of the paradoxical formulation of the condition. The stages of the "synthetic" solution unfold in it, as in the previous problem, in the order opposite to the course of the described events.
“The egg merchant sold the first buyer half of the total number of eggs in her basket and another half an egg; the second buyer - half the remainder and another half an egg, the third - half the remainder and another half an egg, after which she had nothing left. How many eggs were in the basket at the beginning?
In other cases, the formulation of an equation requires a kind of reasoning, which in itself is sufficient to achieve the goal. These are arithmetic problems in the full sense of the word: their algebraic solution is not easier, but more difficult, and is usually associated with the introduction of extra unknowns, which then have to be excluded, and so on.
So, if, for example, in the problem“Tanya said: I have 3 more brothers than sisters. How many more brothers are there in Tanya's family than sisters? denote the number of brothers through x, the number of sisters through y, then the equation will be x − (y − 1) = 3, but if we have already guessed that we need to write y−1 (the sister did not consider herself), then it is already clear that not 3 brothers, but only 2 more than sisters.
Let's take a few more examples.
“I was paddling upstream and, passing under a bridge, I lost my hat. After 10 minutes, I noticed this and, turning and rowing with the same force, caught up with the hat 1 km below the bridge. What is the speed of the river flow?
Solution: 1 (60:(10+10))=3(km/h)
“When I arrived at the station, they usually sent a car for me. Arriving one hour earlier, I went on foot and, meeting the car sent for me, arrived with it at the place 10 minutes earlier than usual. How many times is the car going faster than I am walking?
Consider the solution to this problem by actions:
1) 10:2=5 (min) - the time left for the car to arrive at the station on time from the meeting point.
2) 60-5=55 (min) - the time spent by the pedestrian for the same distance.
3) 55:5=11(times) the car is going faster.
“To swim a certain distance downstream on a boat, it takes three times less time than against the current. How many times the speed of the boat is greater than the speed of the current?
In this problem, you have to guess to go from time to distance.
These are very good arithmetic problems: they require a clear understanding of the relevant specific situation, and not actions according to memorized formal patterns.
Here is another example of an arithmetic problem, for the solution of which it is not necessary to perform any “actions”:
« Some mischievous person from a bottle of tar poured a spoonful of tar into a jar of honey. I mixed it thoroughly, and then poured the same spoonful of the mixture from a jar into a bottle with tar. Then he did it again. What turned out more: honey in a bottle with tar or tar in a jar of honey? »
To solve the problem, it is enough to ask yourself the question: where did the tar go from the bottle, which was displaced by honey?
This is not algebra, not reduction of like terms, and not "transfer from one part to another with the opposite sign." This is exactly the kind of logic associated with imaginary, but having quite real significance in the field of the quantities under study, the development and improvement of which is included in the direct tasks of arithmetic.
The distinctions between arithmetic and algebraic in nature problems are somewhat blurred, as they depend on quantitative signs, in the assessment of which one can disagree, just as one cannot draw a line between “a few grains” and “a bunch of grains”.
Let us dwell in more detail on the types of text problems and how to solve them. Consider those problems that many tend to solve with the help of equations, and at the same time they have simple and sometimes very beautiful solutions for actions.
1. Finding tasks by their multiple ratio and sum or difference (into "parts").
Acquaintance with such problems should begin with those where we are talking about parts in their pure form. When solving them, a basis is created for solving problems of finding two numbers by their ratio and sum (difference). Students must learn to take a suitable value for 1 part, determine how many such parts fall on another value, their sum (difference).
a) For jam, 3 parts of sugar are taken for 2 parts of strawberries. How much sugar should be taken for 3 kg of strawberries?
b) Bought 2700 g of dried fruits. Apples make 4 parts, pears - 3 parts, plums - 2 parts. How many grams of apples, pears and plums separately?
c) The girl read 3 times less pages than she had left. How many pages are in the book if she read 42 pages less?
It is advisable to start the solution of this problem with a drawing:
1) - account for 42 pages.
2) - 1 part, or so many pages the girl read.
3) - in the book.
In the future, students will be able to solve more complex problems.
c) The task of S.A. Rachinsky. I spent a year in Moscow, in the countryside and on the road - and, moreover, in Moscow 8 times more time than on the road, and in the countryside 8 times more than in Moscow. How many days did I spend on the road, in Moscow and in the countryside?
d) When harvesting at the state farm, the students harvested 2 times more tomatoes than cucumbers, and 3 times less than potatoes. How many vegetables did the students pick separately if potatoes were picked 200 kg more than tomatoes?
e) The grandfather says to his grandchildren: “Here are 130 nuts for you. Divide them into 2 parts so that the smaller part, increased by 4 times, would be equal to the larger part, reduced by 3 times.
f) The sum of two numbers is 37.75. If the first term is increased by 5 times, and the second term by 3 times, then the new sum will be equal to 154.25. Find these numbers.
Tasks on the division of a number in this respect belong to this type.
2. Finding two numbers by their sum and difference.
a) There are 50 notebooks in two packs, and in the first pack there are 8 more notebooks. How many notebooks are in each pack?
Solving problems of this kind, I always start with a drawing. Then I propose to equalize the values. The guys offer two ways: remove from the first pack or add to the second. So the main two ways are determined: through a doubled smaller number or a doubled larger number.
When these methods are worked out, it is appropriate to show the "old" way of solving problems of this kind. After the question "How can the stacks of notebooks be equalized without changing the total number of notebooks?" students guess how to do this, and conclude: to find a smaller number, you need to subtract the half-difference from the half-sum, and to find a larger number, you need to add the half-difference to the half-sum. Strong learners can justify this by converting literal expressions:
,
Using this method The following task is solved in one step:
b) The arithmetic mean of two numbers is 3, and their half-difference is 1. What is the value of the smaller number?
– smaller number.
The adjustment method is also applicable in the problem:
c) 8 calves and 5 sheep ate 835 kg of feed. During this time, each calf was given 28 kg more feed than a sheep. How much feed did each calf and each sheep eat?
3. Tasks on the "assumption".
Tasks of this type are associated with the intended actions with objects and quantities. In the traditional methodology, problems of this type also had other names for the most famous problems: for “blue and red cloth”, for “mixing the ΙΙ kind”. I think that the most famous among the "guess" problems is the old Chinese problem.
a) Pheasants and rabbits are sitting in a cage. They are known to have 35 heads and 94 legs. Find out the number of pheasants and the number of rabbits.
Imagine that only pheasants are sitting in a cage. How many legs do they have?
Why are there fewer legs? (Not all pheasants, there are rabbits among them). How many more legs?
If one pheasant is replaced by a rabbit, by how much will the number of legs increase? (On 2)
You can choose another way, imagining that all the rabbits.
Very interesting is another reasoning given by the old masters of the methodology of mathematics and arousing great interest in children.
- Imagine that we put a carrot on top of the cage in which the pheasants and rabbits are sitting. All rabbits will stand on their hind legs to reach for the carrot. How many feet will be on the ground at this moment?
2 35= 70(n.)
- But in the condition of the problem, 94 legs are given, where are the rest?
- The rest are not counted - these are the front paws of rabbits.
- How many of them?
94 - 70 \u003d 24 (n.)
- How many rabbits?
24:2
= 12
And the pheasants?
35
– 12 = 23
Having mastered the reasoning algorithm, the guys easily solve the following problems:
b) Mixed 135 pounds of tea of two varieties with a total cost of 540 rubles. How many pounds of both grades were taken separately, if a pound of the first grade cost 5 rubles, and a pound of the second grade cost 3 rubles?
c) For 94 rubles. bought 35 arshins of blue and red cloth. For an arshin of blue cloth they paid 2 rubles, and for an arshin of red cloth they paid 4 rubles. How many arshins of both cloths did you buy separately?
d) The owner bought 112 sheep, old and young, and paid 49 rubles. 20 Altyn. For an old ram, he paid 15 altyns and 4 polushkas, and for a young ram, 10 altyns. How many and which rams were bought? Altyn - 3 kopecks, half - a quarter of a kopeck.
The problem from the article by I.V. Arnold "Principles for the selection and compilation of arithmetic problems" (1946) about cars:
e)“Passing past the station, I noticed a freight train of 31 cars standing at the station and heard a conversation between a greaser and a coupler. The first one said: "105 axles in total had to be checked." The second noticed that there were many four-axle cars in the composition - three times more than two-axle ones, the rest were three-axle ones. On the next stage, I wanted to, out of nothing to do, calculate how many cars were in this train. How to do it?"
An arithmetic solution is simpler than an algebraic one and requires a clear idea that two-axle and four-axle cars are included (in quantitative terms) in certain groups (4 cars each). The imaginary “replacement” of all wagons by three-axle ones is a common and already well-known technique for students.
An aid can begraphic linear display of task conditions.
4. Tasks for movement.
These tasks are traditionally difficult. Students should have well-formed concepts such as the speed of approach and the speed of removal. When students learn how to solve such problems using an equation, it will be much easier for them to get to the answer. But easier doesn't mean better. Many years ago, one of my students, quite strong in mathematics, enthusiastically searched for arithmetic way solving the problem, at a time when the whole class solved it with the help of an equation. I well remembered his words, very understandable to me: "I'm not interested in the equation."
I will give the conditions and solution of several problems.
a) An old problem. Two trains left Moscow for Tver at the same time. The first passed at 39 versts and arrived in Tver two hours earlier than the second, which passed at 26 versts. How many miles from Moscow to Tver?
Solution:
1) the second train was so far behind.
2) - removal rate.
3) the first train was on the way.
4) distance from Moscow to Tver.
b) Two planes took off simultaneously from Moscow in the same direction: one at a speed of 350 km/h, the other at a speed of 280 km/h. Two hours later, the first one reduced the speed to 230 km/h. At what distance from Moscow will the second plane overtake the first?
Solution:
1) removal speed.
2) - the second plane is so far behind.
3) approach speed.
4) how long will it take for the second plane to catch up with the first.
5) (km) - at this distance to Moscow, the second plane will catch up with the first.
c) From two cities, the distance between which is 560 km, two cars left towards each other and met after 4 hours. If the speed of the first car is reduced by 15%, and the speed of the second car is increased by 20%, then the meeting will also take place in 4 hours. Find the speed of each car.
Solution:
Let's take it as 100% or 1 speed of the first car.
1) approach speed.
2) - is the speed of the second from the speed of the first.
3) is related to the speed of approach.
4) the speed of the first car.
5) second car speed.
d) The train passes a telegraph pole in a quarter of a minute, and a bridge 0.7 km long in 50 seconds. Calculate the average speed of the train and its length.
Solution: When solving this problem, students should understand that, to pass the bridge - to pass the path, equal to the length bridge and the length of the train, go past the telegraph pole - go the way equal to the length of the train.
1) the train travels a distance equal to the length of the bridge.
2) is the speed of the train.
3) train length.
e) The passage of the way between two piers requires a steamer 40 minutes more than a boat. The speed of the boat is 40 km/h and the speed of the steamer is 30 km/h. Find the distance between the marinas.
Solution: 40 min h
1) ship delay.
2) - removal rate
2) - There was a boat on the way.
3) distance between piers.
These are just a few of the vast variety of movement tasks. Using their example, I wanted to show how you can do without equations until the ability to solve them in students is not formed. Naturally, such tasks are within the power of strong students, but this is a great opportunity for their mathematical development.
5. Tasks for "pools".
This is another type of task that causes both interest and difficulty in children. It can also be called tasks for joint work, and some of the tasks for movement also apply to them.
The name of this type is given by a well-known old problem:
a) In the city of Athens there was a body of water into which 3 pipes were laid. One of the pipes can fill the pool at 1 o'clock, the other, thinner, at 2 o'clock, the third, even thinner, at 3 o'clock. So, find out, in what fraction of an hour will all three pipes together fill the pool?
Solution:
1) (v./h) - filling speed through the ΙΙ pipe pipe.
2) (v./h) - filling speed through the ΙΙΙ pipe.
3) (v./h) - total speed.
4) (h) - 3 pipes will fill the reservoir.
You can offer children another interesting solution:
In 6 hours, 6 reservoirs are filled through the Ι pipe, 3 reservoirs through the ΙΙ pipe, 2 reservoirs through the ΙΙΙ pipe. All pipes in 6 hours will fill 11 reservoirs, respectively, to fill one reservoir will require h.
The following problem has a similar solution:
b) The lion ate the sheep in one hour, and the wolf ate the sheep in two hours, and the dog ate the sheep in three hours. No matter how soon, all three - a lion, a wolf and a dog - ate that sheep, count. (Mathematical manuscripts of the 17th century).
c) One man will drink a cup of drink in 14 days, and with his wife he will drink the same cup of drink in 10 days, and knowingly there is, in how many days his wife will especially drink the same cup. (from Arithmetic by Magnitsky)
Solution:
1) (h) - drink a day together.
) (h) - the husband drinks a day.
3) (h) - the wife drinks a day.
4) (d.) - the wife will need to drink the cup of drink.
d) An old problem. A wild duck flies from the South Sea to the North Sea for 7 days. The wild goose flies from the northern sea to the southern sea for 9 days. Now the wild duck and wild goose fly out at the same time. In how many days will they meet? (similar solution)
e) Two pedestrians left points A and B at the same time towards each other. They met 40 minutes after leaving, and 32 minutes after the meeting, the first one arrived at B. How many hours after leaving B did the second one arrive at A?
Solution:
1) (way / min) - the speed of approach.
) (paths / min) - the speed of the first pedestrian.
3) (paths / min) - the speed of the second pedestrian.
4) (min) – there was a second pedestrian on the way.
90 min1.5 h
f) The ship from Nizhny Novgorod to Astrakhan sails for 5 days, and back 7 days. How many days will rafts sail from Nizhny Novgorod to Astrakhan?
Solution:
1) (way / day) - speed downstream.
) (way / day) - speed against the current.
3) (way / day) - twice the speed of the current. The problem was first published in General Arithmetic.I. Newton, but since then it has not lost its relevance and is oneone of the beautiful arithmetic problems, which, although it can be solved by drawing up an equation, is much more beautiful - to do it with the help of sequential reasoning. I had to watch how high school students puzzled over it, introducing several variables, and at the same time, fifth graders easily figured out the solution if they were prompted with the idea of a solution.
The grass in the meadow grows equally thick and fast. It is known that 70 cows would eat all the grass in 24 days, and 30 cows in 60 days. How many cows will eat all the grass in the meadow in 96 days?
In this paper, examples are given and only some of the huge number of word problems are analyzed.
In conclusion, I would like to note that it is necessary to welcome various ways of solving problems. ExactlySolving problems in different ways is an extremely exciting activity for students of different age groups. Interest, curiosity, creativity, the desire to succeed - these are the attractive aspects of the activity.If a student copes with text problems in mathematics lessons, that is, he can trace and explain the logical chain of his decision, give a description of all quantities, then he can also successfully solve problems in physics and chemistry, he can compare and analyze, transform information in all academic subjects school course.
Literature.
1. Arnold I.V. Principles of selection and compilation of arithmetic problems // Izvestiya APN RSFSR. 1946. - Issue. 6 - S. 8-28.
2. Zubelevich G. I. Collection of problems of the Moscow Mathematical Olympiads. – M.: Enlightenment, 1971.
3. Shevkin A. V. Learning to solve text problems in grades 5-6. – M.: Gals plus, 1998.
4 . Shevkin A.V. Materials of the course "Text tasks in school course Mathematics": Lectures 1-4. - M .: Pedagogical University "First of September", 2006. 88 p.