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, 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.

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

Andrei -- ? 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

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 problem?
• 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.

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 greatest common divisor).

Let's show this with an example.

Task. For 4 kg of oranges and 5 kg of bananas they paid 620 rubles, and the next time 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.

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, found themselves 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 trial method (“false rule” and “rule of the ancient Babylonians”).

When using the sampling method, in particular the "false rule", one of the unknown quantities is given ("allowed") 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.

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 for 50 k. there will be 25 +15 = 40 (coins), and for 10 k. there will be 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.

Having offered students to independently choose different values ​​for 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.

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 on each page, 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.

The elementary school teacher simply needs to know what types of tasks are available. Today you will learn about simple text arithmetic problems. Simple text arithmetic problems are problems that are solved by one arithmetic operation.. When we read a task, we automatically correlate it with some kind, and here it immediately becomes easy to understand what action it needs to solve.

I will give you not only the very classification of simple word problems, but also give examples of them, and also talk about solving text problems in an arithmetic way. I took all the examples from mathematics textbooks for grade 2 (part 1, part 2), which are taught in Belarusian schools.

All simple arithmetic problems are divided into two large groups:

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

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

Consider the first group of simple text arithmetic problems (AD I):

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

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

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

Such tasks are solved by addition: a+b=?

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

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

There were 15 glasses of juice in the jar. At dinner we drank 5 glasses. How many glasses of juice are left?

Such problems are solved by subtraction: a-b=?

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

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

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

To solve this problem, it is necessary to subtract the known 2nd term from the result of the action: b-a=?

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

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

Problems of this type are solved by subtraction, the known 1st term is subtracted from the result of the action: b-a=?

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

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

Arithmetically, the solution of text problems of this type is carried out by adding the result of the action and the subtrahend: b+a=?

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

Bought 2 dozen eggs. After a few eggs were taken for baking, 15 remained. How many eggs were taken?

These tasks are solved by subtraction: subtract the result of the action from the reduced: a-b=?

4) Tasks for decreasing / increasing by several units in direct, indirect form

examples of tasks for reducing by several units in direct form:

In one box there were 20 kg of bananas, and in the second - 5 less. How many kilograms of bananas were in the second box?

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

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

I did not find examples of tasks for decreasing in an indirect form, as well as for increasing in a direct or indirect form in a 2nd grade textbook in mathematics. If necessary, write in the comments - and I will supplement the article with my own examples.

5) Tasks for difference comparisons

The mass of the goose is 7 kg, and the chicken is 3 kg. By how many kilograms is the weight of the chicken less than the weight of the goose?

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

The solution of text problems for difference comparisons is made by subtracting a smaller number from a larger number.

We have finished dealing with simple textual arithmetic problems of the 1st group and are moving on to the problems of the 2nd group. If you didn't understand something, ask in the comments.

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

1) Tasks that reveal the specific meaning of multiplication

How many legs do two dogs have? Three dogs?

There are three cars in front of the house. Each car has 4 wheels. How many wheels do three cars have?

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

2) Tasks that reveal the specific meaning of division:

a) content

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

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

In these tasks, we find out how many parts turned out with equal content.

b) in equal parts

A strip 10 cm long was cut into two equal parts. What is the length of each piece?

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

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

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

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

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

Own example:

Several boxes of 6 pencils. There are 24 pencils in the box. How many boxes?

It is solved by dividing the product by the known second factor: b:a=?

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

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

It is solved by dividing the product by the known first factor: b:a=?

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

Own example:

Kolya brought sweets to the class and divided them equally between all the students. There are 16 children in the class. Each received 3 candies. How many sweets did Kolya bring?

It is solved by multiplying the quotient by the divisor: b*a=?

d) finding an unknown divisor: a:?=b

Own example:

Vitya brought 44 sweets to the class and divided them equally between all the students. Each received 2 candies. How many students are in the class?

It is solved by dividing the dividend by the quotient: a:b=?

4) Tasks for increasing / decreasing several times in direct or indirect form

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

5) Tasks for multiple comparison

Solve by dividing the larger by the smaller.

Friends, the above classification of simple word problems is only a part of a large classification of all word problems. In addition, there are still tasks for finding percentages, which I did not tell you about. You can learn about all this from this video:

And my gratitude will remain with you!

Algebraic method for solving text problems to find an arithmetic way to solve them

Solving word problems by juniorsshkolniks can be considered as a means and as a teaching method, during the use of which the content of the initial course of mathematics is mastered: mathematical concepts, the meaning of arithmetic operations and their properties, the formation of computational skills and practical skills.

A teacher who directs the process of solving problems by schoolchildren must first of all be able to solve problems himself, as well as possess the necessary knowledge and skills to teach this to others.

The ability to solve problems is the basis of the teacher's mathematical preparation for teaching younger students to solve text problems.

Among the common methods for solving text problems (algebraic, arithmetic and geometric), the most widely used in elementary grades for most problems isarithmetic Method, including various ways to solve them. However, for the teacher, in many cases, this method of solving problems is more difficult than the algebraic one. This is due primarily to the, what fromhigh school mathematics course

the course of arithmetic, which provided for the formation of the ability of schoolchildren to solve problems by the arithmetic method, was practically excluded. Secondly, in the university course of mathematics, it is also not given due attention.

At the same time, the need to solve problems by the arithmetic method is dictated by the stock of mathematical knowledge of the younger student, which does not allow them to solve most problems using elements of algebra.

The teacher is able, as a rule, to solve any problem algebraically, but not everyone can solve any problem arithmetically.

At the same time, these methods are interconnected, and the teacher should not only notice this relationship, but also use it in his work. In this article, using the example of solving some problems, we will try to show the connection between algebraic and arithmetic methods for solving problems in order to help the teacher find an arithmetic way to solve a problem by solving it algebraically.

Let's make some preliminary remarks:

1. Not always (and even far from always) a text problem solved by an algebraic method can be solved by an arithmetic one. It should be remembered that a problem can be solved using the arithmetic method when its algebraic model is reduced to a linear equation or a system of linear equations.

2. The form of a linear equation does not always “suggest” the arithmetic way of solving the problem, however, further transformations of the equation make it possible to find it. The solution of a system of linear equations, in our opinion, almost immediately makes it possible to outline the course of reasoning for solving the problem in an arithmetic way.

Consider examples.

Example 1 The problem is reduced to the equation

kind ah + b= s.

Task. At 8 o'clock in the morning a train left point A for point B at a speed of 60 km/h. At 11 o'clock another train left point B to meet him at a speed of 70 km/h. At what time will the trains meet if the distance between points is 440 km?

The algebraic method leads to the equation: (60 + 70) x + 60 3 \u003d 440 or 130x + 18 \u003d 440, where x hours is the time of the second train before the meeting. Then: 130x = 440- 180= 130

x= 260, x =2 (h).

The above reasoning and calculations "suggest" the following arithmetic way of solving the problem. Let's find: the sum of train speeds (60 + 70 = 130 (km / h), the time of the first train before the start of the second train (11-8 \u003d 3 (h), the distance traveled by the first train in 3 hours (60 3 \u003d 180 ( km), the distance left for the trains to meet before the meeting (440 - 180 = = 260 (km), the time of the second train before the meeting (260: 130-2 (h)).

In the future, the stages of solving each problem by the algebraic method and the corresponding stages of solving the problem by the arithmetic method will be written in parallel in a table that will allow you to visually trace how algebraic transformations in the course of solving equations that are a model of a text problem open an arithmetic method of solution. So, in this case we will have the following table (see table 1).

Table 1

Let x hours be the time of the second train before the meeting. According to the condition of the problem, we obtain the equation:

(60+70)-x+60*3=440 or 130x+180=440

Let's transform the equation:

130x=440-180 130x=260.

Let's find the known;

X=260:130; x=2

Let's find the sum of train speeds: 60+70=130(km/h).

Let's find the time of the first train before the start of the second train: 11-8=3(h). Find the distance traveled by the first train in 3 hours: 60*3=180(km)

Let's find the distance left for the trains to travel before meeting: 440-180=260(km).

Let's find the time of movement of the second train: 260:130=2(h).

Using the data in Table 1, we obtain an arithmetic solution.

1. 1. 3 (h)-the first train was on its way before the start of the second;

1. 3 = 180 (km) - the first train passed in 3 hours;

3) 440 - 180 \u003d 260 (km) - the distance traveled by trains while moving simultaneously;

1. 70 = 130 (km/h) - train approach speed;

1. 130 \u003d 2 (h) - the time of movement of the second train;

6) 11 + 2 = 13 (h) - at this time the trains will meet.

Answer: at 13:00.

Example 2 a 1 x + v 1 \u003d a x + b

Task. Schoolchildren bought 4 books, after which they had 40 rubles left. If they bought 7 of the same books, they would have 16 rubles left. How much does one book cost?

The algebraic method leads to the equation:4x + 40 = 7x + 16, where X - the cost of one book. In the course of solving this equation, we make the following calculations: 7 x - 4X \u003d 40-16 -\u003e Zx \u003d 24 -\u003e x \u003d 8, which, together with the reasoning used in compiling the equation, lead to an arithmetic way of solving the problem. Let's find: how many more books were bought: 7-4 = 3 (book); how much less money will be left, i.e. how much more money was spent: 40 - 16 = 24 (p); how much one book costs: 24:3 = 8 (p). The above considerations are summarized in Table 2.

Stages of problem solving

algebraic method

Stages of solving the problem by the arithmetic method

Let x be the cost of one book. According to the task

we get the equation: 4x+40=7x+16.

Let's transform the equation:

7x-4x=40-16 (7-4)x=24 3x=24

Let's find the known:

X=24:3; x=8

The cost of four books and another 40r. equal to the cost of 7 books and another 70r.

Let's find how many more books they would buy: 7-4=3(kn). Let's find how much more money they would pay: 40-16 = 24 (p.).

Let's find the cost of one book: 24:3=8(r.).

table 2

Using the data in Table 2, we obtain an arithmetic solution:

1) 7-4=3 (book) - so many more books would be bought;

1. 16 \u003d 24 (r.) - so many rubles they would pay more;

3) 24: 3 = 8 (p.) - there is one book.

Answer: 8 rubles.

Example 3 The problem is reduced to an equation of the form:Oh + b x + cx = d

Task. The tourist traveled 2,200 km, and on the ship he traveled twice as much as on the car, and on the train 4 times more than on the ship. How many kilometers did the tourist travel separately by boat, car and train?

Using the data in Table 3, we obtain an arithmetic solution.

Let's take the distance that the tourist traveled by car as one part:

1 2 \u003d 2 (h) - falls on the distance that the tourist covered on the boat;

2) 2 4 \u003d 8 (h) - falls on the distance that the tourist traveled by train;

3) 1+2+8=11(h) - falls on the whole journey

Table 3

Let x kilometers be the distance traveled on the boat.

According to the condition of the problem, we obtain the equation: x + 2x + 2 * 4x \u003d 2200.

Let's transform the equation:

(1+2+8)x=2200 11x=2200.

Let's find the known:

X=2200:11; x=200

Let's take the distance that the tourist traveled by car (at least) as 1 part. Then the distance that he traveled on the ship will correspond to two parts, and on the train - 2 - 4 parts. This means that the entire path of the tourist (2200 km) corresponds to 1+2+8=11 (hours).

Let's find how many parts make up the entire path of the tourist: 1 + 2 + 8 = 11 (hours).

Let's find how many kilometers fall on one part: 2200:11=200 (km).

1. 200: 11= 200 (km) - the distance covered by the tourist by car;

1. 2 = 400 (km) - the distance covered by the tourist on the boat;

6) 200 -8 = 1 600 (km) - the distance traveled by the tourist by train.

Answer:200 km, 400 km, 1,600 km.

Example 4 The problem is reduced to the equationkind (X + a) in = cx + d.

Task. At the end of the performance, 174 spectators from the theater dispersed on foot, and the rest went on trams in 18 cars, and each car got 5 more people than there were seats in it. If the spectators leaving the theater on the tram got into it according to the number of seats, then 3 more cars would be needed, and in the last one there would be 6 empty seats. How many spectators were in the theatre?

Table 4

Let there be x seats in each tram. Then, according to the condition of the problem, we have the equation: (x+5)*18=x*(18+3)-6.

Let's transform the equation: 21x - 18x \u003d 90 + 6 or 3x \u003d 96.

Let's find the unknown:

X= 96: 3; x = 32.

Each carriage included 5 more people than there were seats in it. In 18 cars - 5 * 18 = 90 people more. 90 people entered 3 additional cars and there were 6 more empty seats. Therefore, there are 90 + 6 = 96 seats in three cars.

Find the number of seats in one car:

96: 3 = 32(m.)

Using the data in Table 4, we obtain an arithmetic solution:

1)5 18 \u003d 90 (persons) - so many people more than there were seats in 18 cars;

90 + 6 = 96 (m) - in three cars;

96: 3 = 32 (m) - in one car;

32 + 5 = 37 (people) - was in each of the 18 cars;

37 18 \u003d 666 (persons) - left on trams;

666 + 174 = 840 (people) - was in the theater.

Answer: 840 spectators.

Example 5 The problem is reduced to a system of equations of the form: x + y = a, x – y =b.

Task. A belt with a buckle costs 12 rubles, and the belt is 6 rubles more expensive than the buckle.

How much is the belt, how much is the buckle?

The algebraic method leads to a system of equations:

x+y=12,

x-y \u003d 6 where x: rubles - the price of the belt,atrubles - the price of the buckle.

This system can be solved by the substitution method: expressing one unknown in terms of another. From the first equation, substituting its value into the second equation, solve the resulting equation with one unknown, find the second unknown. However, in this case we will not be able to "feel" the arithmetic way of solving the problem.

Having added the equations of the system, we will immediately have the equation2x = 18.
Where do we find the cost of the beltx = 9 (R.). This method of solving the system allows us to obtain the following arithmetic line of reasoning. Suppose the buckle costs the same as the belt. Then a buckle with a belt (or 2 belts) will cost 12 + 6 = 18 (r.) (since in fact a buckle costs 6 rubles less). Therefore, one belt is worth 18:2=9 (p.).

If we subtract term by term from the first equation the second, then we get the equation 2at \u003d 6, whence y \u003d 3 (r.). In this case, solving the problem by the arithmetic method, one should argue as follows. Suppose the belt costs the same as the buckle. Then the buckle and belt (or two buckles) will cost 12-6=6 (p.) (because the belt actually costs 6 rubles more).
Therefore, one buckle costs 6:2=3 (p.)

Table 5

Let x rubles be the price of the belt, y rubles the price of the buckle. According to the condition of the problem, we obtain a system of equations:

X + y \u003d 12,

X - y \u003d 6.

Adding the equations of the system term by term, we get: 2x \u003d 12 + 6 2x \u003d 18.

Find unknown:

x = 18: 2; x = 9

Belt with buckle cost 12r. And the belt is 6r more expensive than the buckle.

Equalize the unknown:

Suppose the buckle costs the same as the belt, then two belts cost 12 + 6 = 18 (r.).

Find the price of the belt:

18: 2 = 9 (p.).

Using the data in Table 5, we obtain an arithmetic solution:

12 + 6 = 18 (r.) - two belts would cost if the buckle cost the same as the belt;

2) 18:2=9 (p.) - there is one belt;

3) 12-9=3 (p.) - there is one buckle.

Answer: 9 rubles, 3 rubles.

Example 6 The problem is reduced to a system of equations of the form:

ax + bu = c 1x+y=c2

Task. For the hike, 46 schoolchildren prepared four- and six-seater boats. How many of those and other boats were there if all the guys were accommodated in ten boats and there were no empty seats left ?

Table 6

Let x be the number of four-seater boats and y the number of six-seater boats. According to the condition of the problem, we have a system of equations:

x + y = 10,

4x + 6y = 46.

Multiply both sides of the first equation by 4.

We have:

4x + 4y = 40.

We subtract (term by term) the resulting equation from the second. We have:

(6 - 4) y \u003d 46 - 40 or 2y \u003d 6.

Let's find the unknown:

Y = 6: 2; y = 3.

There are 10 boats in all and 46 schoolchildren were accommodated in them.

Equalize the unknowns.

Let's assume that all boats were four-seater. Then m them would accommodate 40 people.

Let's find how many more people a six-seater boat can hold than a four-seater boat: 6 - 4 = 2 (persons). Let's find how many schoolchildren will not have enough places if all the boats are four-seater: 46 - 40 \u003d 6 (persons).

Let's find the number of six-seater boats: 6: 2 = 3 (pcs.).

Using the data in Table 6, we obtain an arithmetic solution:

1) 4- 10 \u003d 40 (persons) - would accommodate if all boats were four-seater;

2) 6 - 4 \u003d 2 (persons) - for so many people a six-seater boat holds more than a four-seater;

3) 46 - 40 - 6 (persons) - so many students will not have enough space if

all boats are quadruple;

4) 6: 2 = 3 (pcs.) - there were six-seater boats;

5) 10 - 3 = 7 (pcs.) - there were four-seater boats.

Answer: 3 six-seater boats, 7 four-seater boats.

Example 7 The problem is reduced to a system of equations of the form: a x+b y=c1; a x + b y \u003d c2

Task. 3 pens and 4 notebooks cost 26 rubles, and 7 pens and 6 similar notebooks cost 44 rubles. How much does a notepad cost?

Table 7

Let x rubles be the price of a pen, and y rubles be the price of a notebook. According to the condition of the problem, we obtain a system of equations:

3 x + 4 y \u003d 26,

7 x + 6 y = 44.

Multiply both sides of the first equation by 7. We get:

21 x + 28 y \u003d 182,

21 x + 18 y = 132.

Subtract (term by term) from the first equation the second.

We have:

(28 - 18) y \u003d 182 - 132 or 10 y \u003d 50.

Let's find the unknown:

Y \u003d 50: 10, y \u003d 5.

3 pens and 4 notepads cost 26 rubles. 7 pens and 6 notebooks cost 44 rubles.

Equalize the number of pens in two purchases. To do this, we find the smallest multiple of the numbers 3 and 7 (21). Then, as a result of the first purchase, 21 pens and 28 notebooks were purchased, and the second - 21 pens and 18 notebooks. Let's find the cost of each purchase in this case:

26 * 7 \u003d 182 (r.), 44 * 3 \u003d 132 (r.).

Let's find how many more notebooks were bought for the first time:

28 - 18 \u003d 10 (pcs.).

Find how much more you would pay for the first purchase:

182 - 132 \u003d 50 (p.).

Find out how much Notepad costs:

50: 10 = 5 (p.).

Using the data in Table 7, we obtain an arithmetic solution:

1) 26 7 \u003d 182 (p.) - there are 21 pens and 28 notebooks;

2) 44 3 \u003d 132 (p.) - there are 21 pens and 18 notebooks;

3) 28 - 18 \u003d 10 (pcs.) - so many notebooks in the first purchase would be more than in the second;

4) 182 - 132 = 50 (p.) - there are 10 notebooks;

5) 50: 10=5 (p.) - there is a notebook.

Answer: 5 rubles.

We have considered some types of text problems found in various mathematics textbooks for elementary grades. Despite the apparent simplicity of establishing a connection between algebraic and arithmetic methods, this technique still requires careful practice with students in practical classes and painstaking work of the teacher in the course of self-preparation for the lesson.

1. General remarks on the solution of problems by the algebraic method.

2. Tasks for movement.

3. Tasks for work.

4. Tasks for mixtures and percentages.

Using the algebraic method to find an arithmetic way to solve text problems.

1. When solving problems by the algebraic method, the desired quantities or other quantities, knowing which it is possible to determine the desired ones, are denoted by letters (usually x, y,z). All independent relations between data and unknown quantities, which are either directly formulated in the condition (in verbal form), or follow from the meaning of the problem (for example, the physical laws that the quantities under consideration obey), or follow from the condition and some reasoning, are written in form of equality of inequalities. In the general case, these relations form a certain mixed system. In special cases, this system may not contain inequalities or equations, or it may consist of only one equation or inequality.

The solution of problems by the algebraic method is not subject to any single, sufficiently universal scheme. Therefore, any indication relating to all tasks is of the most general nature. The tasks that arise in solving practical and theoretical issues have their own individual characteristics. Therefore, their study and solution are of the most diverse nature.

Let us dwell on solving problems whose mathematical model is given by an equation with one unknown.

Recall that the activity for solving the problem consists of four stages. Work at the first stage (analysis of the content of the problem) does not depend on the chosen solution method and has no fundamental differences. At the second stage (when searching for a way to solve the problem and drawing up a plan for solving it), in the case of using the algebraic method of solving, the following are carried out: the choice of the main ratio for compiling the equation; the choice of the unknown and the introduction of a designation for it; expression of the quantities included in the main ratio, through the unknown and the data. The third stage (implementation of the plan for solving the problem) involves the compilation of an equation and its solution. The fourth stage (checking the solution of the problem) is carried out in the standard way.

Usually when writing equations with one unknown X adhere to the following two rules.

rule I . One of these quantities is expressed in terms of the unknown X and other data (that is, an equation is drawn up in which one part contains a given value, and the other contains the same value, expressed by X and other given quantities).

rule II . For the same quantity, two algebraic expressions are compiled, which are then equated to each other.

Outwardly, it seems that the first rule is simpler than the second.

In the first case, it is always required to compose one algebraic expression, and in the second, two. However, there are often problems in which it is more convenient to make two algebraic expressions for the same quantity than to choose an already known one and make one expression for it.

The process of solving text problems in an algebraic way is performed according to the following algorithm:

1. First, choose the ratio on the basis of which the equation will be drawn up. If the problem contains more than two ratios, then the ratio that establishes some connection between all unknowns should be taken as the basis for compiling the equation.

Then the unknown is chosen, which is denoted by the corresponding letter.

All unknown quantities included in the ratio chosen for compiling the equation must be expressed in terms of the chosen unknown, based on the rest of the ratios included in the problem, except for the main one.

4. From these three operations, the compilation of an equation directly follows as the design of a verbal record with the help of mathematical symbols.

The central place among the listed operations is occupied by the choice of the main relation for compiling equations. The considered examples show that the choice of the main ratio is decisive in the formulation of equations, introduces logical harmony into the sometimes vague verbal text of the problem, gives confidence in the orientation and protects against chaotic actions for expressing all the quantities included in the problem through the data and the desired ones.

The algebraic method of solving problems is of great practical importance. With its help, they solve a wide variety of tasks from the field of technology, agriculture, and everyday life. Already in secondary school, equations are used by students in the study of physics, chemistry, and astronomy. Where arithmetic turns out to be powerless or, at best, requires extremely cumbersome reasoning, there the algebraic method leads easily and quickly to the answer. And even in the so-called "typical" arithmetic problems, relatively easily solved by arithmetic, the algebraic solution, as a rule, is both shorter and more natural.

The algebraic method of solving problems makes it easy to show that some problems that differ from each other only in the plot have not only the same relationships between the data and the desired values, but also lead to typical reasoning through which these relationships are established. Such problems give only different specific interpretations of the same mathematical reasoning, the same relationships, that is, they have the same mathematical model.

2. The group of tasks for movement includes tasks that talk about three quantities: paths (s), speed ( v) and time ( t). As a rule, they are talking about uniform rectilinear motion, when the speed is constant in magnitude and direction. In this case, all three quantities are related by the following relation: S = vt. For example, if the speed of a cyclist is 12 km/h, then in 1.5 hours he will travel 12 km/h  1.5 h = 18 km. There are problems in which uniformly accelerated rectilinear motion is considered, that is, motion with constant acceleration (a). Distance traveled s in this case is calculated by the formula: S = v 0 t + at 2 /2, where v 0 – initial speed. So, in 10 s of falling with an initial speed of 5 m/s and a free fall acceleration of 9.8 m 2 /s, the body will fly a distance equal to 5 m/s  10s + 9.8 m 2 /s  10 2 s 2/2 = 50 m + 490 m = 540 m.

As already noted, in the course of solving text problems and, first of all, in problems related to movement, it is very useful to make an illustrative drawing (to build an auxiliary graphical model of the problem). The drawing should be done in such a way that it shows the dynamics of movement with all meetings, stops and turns. A well-designed drawing allows not only a deeper understanding of the content of the problem, but also facilitates the compilation of equations and inequalities. Examples of such drawings will be given below.

The following conventions are usually adopted in motion problems.

Unless specifically stated in the task, the movement in individual sections is considered uniform (whether it is movement in a straight line or in a circle).

Turns of moving bodies are considered instantaneous, that is, they occur without spending time; the speed also changes instantly.

This group of tasks, in turn, can be divided into tasks in which the movements of bodies are considered: 1) towards each other; 2) in one direction ("after"); 3) in opposite directions; 4) along a closed trajectory; 5) along the river.

If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 a), then when the bodies move towards each other, the time after which they will meet is equal to S/(v 1 + v 2).

2. If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 b), then when the bodies move in one direction ( v 1 > v 2) the time after which the first body overtakes the second is S/(v 1 – v 2).

3. If the distance between the bodies is S, and the velocities of the bodies are equal v 1 and v 2 (Fig. 16 in), then, having set off simultaneously in opposite directions, the bodies will be in time t be at a distance S 1 = S + (v 1 + v 2 ) t.

Rice. sixteen

4. If the bodies move in one direction along a closed trajectory of length s with speeds v 1 and v 2 , the time after which the bodies will meet again (one body will overtake the other), leaving simultaneously from one point, is found by the formula t = S/(v 1 – v 2) provided that v 1 > v 2 .

This follows from the fact that with a simultaneous start along a closed trajectory in one direction, a body with a higher speed begins to catch up with a body with a lower speed. The first time it catches up with him, having traveled a distance of S more than another body. If it overtakes him for the second, third time, and so on, this means that it travels a distance of 2 S, by 3 S and so on more than another body.

If the bodies move in different directions along a closed path of length S with speeds v 1 and v 2 , the time after which they will meet, having departed simultaneously from one point, is found by the formula t = v(v 1 + v 2). In this case, immediately after the start of motion, a situation arises when the bodies begin to move towards each other.

5. If the body moves along the river, then its speed relative to the shore and is the sum of the speed of the body in still water v and the speed of the river w: and =v + w. If a body moves against the current of a river, then its speed is and =v – w. For example, if the speed of the boat v\u003d 12 km / h, and the speed of the river w \u003d 3 km / h, then in 3 hours the boat will sail along the river (12 km / h + 3 km / h)  3 hours = 45 km, and against the current - (12 km / h - 3 km / h)  3 hours = 27 km. It is believed that the speed of objects with zero speed in still water (raft, log, etc.) is equal to the speed of the river.

Let's look at a few examples.

Example.From one point in one direction every 20 min. cars are leaving. The second car travels at a speed of 60 km/h, and the speed of the first is 50% more than the speed of the second. Find the speed of the third car if it is known that it overtook the first car 5.5 hours later than the second.

Decision. Let x km/h be the speed of the third car. The speed of the first car is 50% greater than the speed of the second, so it is equal to

When moving in one direction, the meeting time is found as the ratio of the distance between objects to the difference in their speeds. First car in 40 min. (2/3 h) travels 90  (2/3) = 60 km. Therefore, the third one will overtake him (they will meet) in 60/( X– 90) hours. Second in 20 min. (1/3 h) travels 60  (1/3) = 20 km. This means that the third one will catch up with him (they will meet) in 20/( X- 60) hours (Fig. 17).

P
about the condition of the problem

Rice. 17

After simple transformations, we obtain a quadratic equation 11x 2 - 1730x + 63000 = 0, solving which we find

The check shows that the second root does not satisfy the condition of the problem, since in this case the third car will not catch up with other cars. Answer: The speed of the third car is 100 km/h.

Example The motor ship passed 96 km along the river, returned back and stood for some time under loading, spending 32 hours for all. The speed of the river is 2 km / h. Determine the speed of the ship in still water if the loading time is 37.5% of the time spent on the whole round trip.

Decision. Let x km/h be the speed of the ship in still water. Then ( X+ 2) km/h - its speed downstream; (X - 2) km/h - against the current; 96/( X+ 2) hours - the time of movement with the flow; 96/( X- 2) hours - the time of movement against the current. Since 37.5% of the total time the ship was under loading, the net time of movement is 62.5%  32/100% = 20 (hours). Therefore, according to the condition of the problem, we have the equation:

Transforming it, we get: 24( X – 2 + X + 2) = 5(X + 2)(X – 2) => 5X 2 – 4X– 20 = 0. Having solved the quadratic equation, we find: X 1 = 10; X 2 = -0.4. The second root does not satisfy the condition of the problem.

Answer: 10 km/h is the speed of the ship in still water.

Example. Car drove way out of town BUT to city C through city AT Without stops. Distance AB, equal to 120 km, he traveled at a constant speed 1 hour faster than the distance sun, equal to 90 km. Determine the average speed of the car from the city BUT to city C, if it is known that the speed on the section AB 30 km/h more speed on the site Sun.

Decision. Let be X km / h - the speed of the car on the site Sun.

Then ( X+ 30) km/h – speed on the section AB, 120/(X+ 30) h, 90/ X h is the time the car travels AB and sun respectively.

Therefore, according to the condition of the problem, we have the equation:

.

Let's transform it:

120X+ 1(X + 30)X = 90(X + 30) => X 2 + 60X – 2700 = 0.

Solving the quadratic equation, we find: X 1 = 30, X 2 = -90. The second root does not satisfy the condition of the problem. So the speed in the section sun equal to 30 km/h, on the section AB - 60 km/h It follows that the distance AB the car traveled in 2 hours (120 km: 60 km/h = 2 hours), and the distance Sun - in 3 hours (90 km: 30 km/h = 3 hours), so the whole distance AC he traveled in 5 hours (3 hours + 2 hours = 5 hours). Then the average speed of movement on the site AU, the length of which is 210 km, is equal to 210 km: 5 hours \u003d 42 km / h.

Answer: 42 km / h - the average speed of the car on the site AS.

The group of tasks for work includes tasks that talk about three quantities: work BUT, time t, during which work is performed, productivity R - work done per unit of time. These three quantities are related by the equation BUT = Rt. Tasks for work also include tasks related to filling and emptying tanks (vessels, tanks, pools, etc.) using pipes, pumps and other devices. In this case, the volume of pumped water is considered as the work done.

Tasks for work, generally speaking, can be attributed to the group of tasks for movement, since in tasks of this type it can be considered that all work or the total volume of the reservoir plays the role of distance, and the productivity of objects that do work is similar to the speed of movement. However, according to the plot, these tasks naturally differ, and some tasks for work have their own specific methods of solving. So, in those tasks in which the amount of work performed is not specified, all work is taken as a unit.

Example. Two teams had to complete the order in 12 days. After 8 days of joint work, the first team received another task, so the second team finished the order for another 7 days. In how many days could each of the teams complete the order, working separately?

Decision. Let the first brigade complete the task for X days, the second brigade - for y days. Let's take all the work as a unit. Then 1/ X - productivity of the first brigade, a 1/ y – second. Since two teams must complete the order in 12 days, we get the first equation 12(1/ X + 1/at) = 1.

From the second condition it follows that the second team worked 15 days, and the first - only 8 days. So the second equation is:

8/X+ 15/at= 1.

Thus, we have a system:

Subtracting the first equation from the second equation, we get:

21/y = 1 => y= 21.

Then 12/ X + 12/21 = 1 => 12/X – = 3/7 => x = 28.

Answer: the first brigade will complete the order in 28 days, the second in 21 days.

Example. Worker BUT and working AT can complete the job in 12 days BUT and working With– in 9 days, working AT and working C - in 12 days. How many days will it take them to complete the job, working together?

Decision. Let the worker BUT can do the job for X days, working AT- behind at days, working With- behind z days. Let's take all the work as a unit. Then 1/ x, 1/y and 1/ z – worker productivity A, B and With respectively. Using the condition of the problem, we arrive at the following system of equations presented in the table.

Table 1

Having transformed the equations, we have a system of three equations with three unknowns:

Adding the equations of the system term by term, we get:

or

The sum is the joint productivity of the workers, so the time in which they complete all the work will be equal to

Answer: 7.2 days.

Example. Two pipes are laid in the pool - supply and discharge, and through the first pipe the pool is filled for 2 hours longer than through the second pipe the water is poured out of the pool. When the pool was one-third full, both pipes were opened, and the pool turned out to be empty after 8 hours. How many hours can the pool fill through one first pipe and how many hours can a full pool drain through one second pipe?

Decision. Let be V m 3 - the volume of the pool, X m 3 / h - the performance of the supply pipe, at m 3 / h - outlet. Then V/ x hours - the time required for the supply pipe to fill the pool, V/ y hours - the time required by the outlet pipe to drain the pool. According to the task V/ x – V/ y = 2.

Since the productivity of the outlet pipe is greater than the productivity of the filling pipe, when both pipes are turned on, the pool will dry out and one third of the pool will dry out in time (V/3)/(y – x), which, according to the condition of the problem, is equal to 8 hours. So, the condition of the problem can be written as a system of two equations with three unknowns:

The task is to find V/ x and V/ y. Let us single out a combination of unknowns in the equations V/ x and V/ y, writing the system as:

Introducing new unknowns V/ x= a and V/ y = b, we get the following system:

Substituting into the second equation the expression a= b + 2, we have an equation for b:

deciding which we find b 1 = 6, b 2 = -eight. The condition of the problem is satisfied by the first root 6, = 6 (p.). From the first equation of the last system we find a= 8 (h), that is, the first pipe fills the pool in 8 hours.

Answer: through the first pipe the pool will be filled in 8 hours, through the second pipe the pool will be drained after 6 hours.

Example. One tractor team has to plow 240 hectares, and the other 35% more than the first. The first brigade, plowing 3 ha less than the second brigade every day, finished work 2 days earlier than the second brigade. How many hectares did each brigade plow daily?

Decision. Let's find 35% of 240 ha: 240 ha  35% / 100% = 84 ha.

Consequently, the second team had to plow 240 ha + 84 ha = 324 ha. Let the first brigade plow daily X ha. Then the second brigade plowed daily ( X+ 3) ha; 240/ X– working hours of the first brigade; 324/( X+ 3) - the time of the second brigade. According to the condition of the problem, the first team finished work 2 days earlier than the second, so we have the equation

which after transformations can be written as follows:

324X – 240X - 720 = 2x 2 + 6x=> 2x 2 - 78x + 720 = 0 => x 2 - 39x + 360 = 0.

Having solved the quadratic equation, we find x 1 \u003d 24, x 2 \u003d 15. This is the norm of the first brigade.

Consequently, the second brigade plowed 27 ha and 18 ha, respectively, per day. Both solutions satisfy the condition of the problem.

Answer: 24 hectares per day were plowed by the first brigade, 27 hectares by the second; 15 hectares per day were plowed by the first brigade, 18 hectares by the second.

Example. In May, two workshops produced 1080 parts. In June, the first shop increased the output of parts by 15%, and the second increased the output of parts by 12%, so both shops produced 1224 parts. How many parts did each shop produce in June?

Decision. Let be X parts were made in May by the first workshop, at details - the second. Since 1080 parts were manufactured in May, according to the condition of the problem, we have the equation x + y = 1080.

Find 15% off X:

So, at 0.15 X parts increased the output of the first workshop, therefore, in June it produced x + 0,15 X = 1,15 x details. Similarly, we find that the second shop in June produced 1.12 y details. So the second equation will look like: 1.15 x + 1,12 at= 1224. Thus, we have the system:

from which we find x = 480, y= 600. Consequently, in June the workshops produced 552 parts and 672 parts, respectively.

Answer: the first workshop produced 552 parts, the second - 672 parts.

4. The group of tasks on mixtures and percentages includes tasks in which we are talking about mixing various substances in certain proportions, as well as tasks on percentages.

Tasks for concentration and percentage

Let's clarify some concepts. Let there be a mixture of P various substances (components) BUT 1 BUT 2 , ..., BUT n respectively, the volumes of which are equal V 1 , V 2 , ..., V n . Mix volume V 0 consists of the volumes of pure components: V 0 = V 1 + V 2 + ... + V n .

Volume concentration substances BUT i (i = 1, 2, ..., P) in the mixture is called the quantity c i, calculated by the formula:

Volume percentage of substance A i (i = 1, 2, ..., P) in the mixture is called the quantity p i , calculated by formula R i = with i ,  100%. Concentrations with 1, with 2 , ..., with n, which are dimensionless quantities, are related by the equality with 1 + with 2 + ... + with n = 1, and the relations

show what part of the total volume of the mixture is the volume of the individual components.

If the percentage is known i-th component, then its concentration is found by the formula:

i.e Pi – is the concentration i th substance in the mixture, expressed as a percentage. For example, if the percentage of a substance is 70%, then its corresponding concentration is 0.7. Conversely, if the concentration is 0.33, then the percentage is 33%. So the sum R 1 + p 2 + …+ p n = 100%. If concentrations are known with 1 , with 2 , ..., with n components that make up this mixture of volume V 0 , then the corresponding volumes of the components are found by the formulas:

The concepts weight (mass) concentralization components of the mixture and the corresponding percentages. They are defined as the ratio of the weight (mass) of a pure substance BUT i , in the alloy to the weight (mass) of the entire alloy. What concentration, volume or weight, is involved in a particular problem is always clear from its conditions.

There are tasks in which it is necessary to recalculate the volume concentration to the weight concentration or vice versa. In order to do this, it is necessary to know the density (specific gravity) of the components that make up the solution or alloy. Consider, for example, a two-component mixture with volume concentrations of the components with 1 and with 2 (with 1 + with 2 = 1) and the specific gravity of the components d 1 and d 2 . The mass of the mixture can be found by the formula:

wherein V 1 and V 2 – the volumes of the components that make up the mixture. The weight concentrations of the components are found from the equalities:

which determine the relationship of these quantities with volumetric concentrations.

As a rule, in the texts of such problems one and the same repeated condition occurs: from two or more mixtures containing components A 1 , A 2 , BUT 3 , ..., BUT n , a new mixture is compiled by mixing the original mixtures, taken in a certain proportion. In this case, it is required to find in what ratio the components BUT 1, BUT 2 , BUT 3 , ..., BUT n enter the resulting mixture. To solve this problem, it is convenient to introduce into consideration the volume or weight amount of each mixture, as well as the concentrations of its constituent components BUT 1, BUT 2 , BUT 3 , ..., BUT n . With the help of concentrations, it is necessary to “split” each mixture into separate components, and then, in the manner indicated in the condition of the problem, compose a new mixture. In this case, it is easy to calculate how much of each component is included in the resulting mixture, as well as the total amount of this mixture. After that, the concentrations of the components are determined BUT 1, BUT 2 , BUT 3 , ..., BUT n in the new mix.

Example.There are two pieces of copper-zinc alloy with copper percentage of 80% and 30% respectively. In what ratio should these alloys be taken in order, by melting the pieces taken together, to obtain an alloy containing 60% copper?

Decision. Let the first alloy be taken X kg, and the second - at kg. By condition, the concentration of copper in the first alloy is 80/100 = 0.8, in the second - 30/100 = 0.3 (it is clear that we are talking about weight concentrations), which means that in the first alloy 0.8 X kg of copper and (1 - 0.8) X = 0,2X kg of zinc, in the second - 0.3 at kg of copper and (1 - 0.3) y = 0,7at kg of zinc. The amount of copper in the resulting alloy is (0.8  X + 0,3  y) kg, and the mass of this alloy will be (x + y) kg. Therefore, the new concentration of copper in the alloy, according to the definition, is equal to

According to the condition of the problem, this concentration should be equal to 0.6. Therefore, we get the equation:

This equation contains two unknowns X and y. However, according to the condition of the problem, it is not required to determine the quantities themselves X and y, but only their attitude. After simple transformations, we get

Answer: alloys must be taken in a ratio of 3: 2.

Example.There are two solutions of sulfuric acid in water: the first is 40%, the second is 60%. These two solutions were mixed, after which 5 kg of pure water was added and a 20% solution was obtained. If, instead of 5 kg of pure water, 5 kg of an 80% solution were added, then a 70% solution would be obtained. How many were 40% and 60% solutions?

Decision. Let be X kg is the mass of the first solution, at kg - the second. Then the mass of a 20% solution ( X + at+ 5) kg. Since in X kg 40% solution contains 0.4 X kg of acid at kg of 60% solution contains 0.6 y kg of acid, and (x + y + 5) kg of 20% solution contains 0.2( X + y + 5) kg of acid, then by condition we have the first equation 0.4 X + 0,6y = 0,2(X +y + 5).

If, instead of 5 kg of water, add 5 kg of an 80% solution, you get a solution with a mass (x + y+ 5) kg, in which there will be (0.4 X + 0,6at+ 0.8  5) kg of acid, which will be 70% of (x + y+ 5) kg.

Solving problems in an algebraic way (using equations) According to the textbook by I.I. Zubareva, A.G. Mordkovich

teacher of mathematics, MOU "LSOSh No. 2"

Likhoslavl, Tver region

Goals:- show the rule for solving problems in an algebraic way; - to form the ability to solve problems in arithmetic and algebraic ways.

Ways

problem solving

Arithmetic (problem solving by actions)

Algebraic (solving a problem using an equation)

Problem #509

Read the task.

Try to find different solutions.

There are 16 kg of cookies in two boxes. Find the mass of biscuits in each box if one of them contains 4 kg more biscuits than the other.

1 solution

(look)

3 way to solve

(look)

2 way to solve

4 way to solve

1 way (arithmetic)

• 16 - 4 \u003d 12 (kg) - cookies will remain in two boxes if 4 kg of cookies are taken from the first box.

• 12: 2 = 6 (kg) - cookies were in the second box.

• 6 + 4 = 10 (kg) - cookies were in the first box.

Answer

The solution used adjustment method .

Question: Why did it get such a name?

Back)

2 way (arithmetic)

• 16 + 4 \u003d 20 (kg) - cookies will be in two boxes if 4 kg of cookies are added to the second box.

• 20: 2 = 10 (kg) - cookies were in the first box.

• 10 - 4 = 6 (kg) - cookies were in the second box.

Answer: the mass of cookies in the first box is 10 kg, and in the second 6 kg.

The solution used adjustment method .

Back)

3 way (algebraic)

Denote the mass of cookies in the second box letter X kg. Then the mass of cookies in the first box will be ( X+4) kg, and the mass of cookies in two boxes is (( X +4)+ X) kg.

(X +4)+ X =16

X +4+ X =16

2 X +4=16

2 X =16-4

2 X =12

X =12:2

The second box contained 6 kg of cookies.

6+4=10 (kg) - cookies were in the first box.

The solution used algebraic way.

Exercise: Explain the difference between the arithmetic method and the algebraic one?

Back)

4 way (algebraic)

Denote the mass of cookies in the first box letter X kg. Then the mass of cookies in the second box will be equal to ( X-4) kg, and the mass of cookies in two boxes is ( X +(X-4)) kg.

According to the condition of the problem, two boxes contained 16 kg of cookies. We get the equation:

X +(X -4)=16

X + X -4=16

2 X -4=16

2 X =16+4

2 X =20

X =20:2

The first box contained 10 kg of cookies.

10-4=6 (kg) - cookies were in the second box.

The solution used algebraic way.

Back)

• What two methods were used to solve the problem?
• What is an equalization method?
• How does the first alignment method differ from the second one?
• One pocket has 10 rubles more than the other. How can you equalize the amount of money in both pockets?
• What is the algebraic way to solve a problem?
• What is the difference between the 3rd method of solving the problem and the 4th?
• One pocket has 10 rubles more than the other. It is known that a smaller amount of money was designated as a variable X. How will it be expressed through X
• If for X denote more money in your pocket, while it will be expressed through X the amount of money in the other pocket?
• In the store, shampoo costs 25 rubles more than in the supermarket. Label one variable with a letter at and express another cost in terms of this variable.

Problem #510

Solve the problem in arithmetic and algebraic ways.

156 centners of potatoes were harvested from three plots of land. Potatoes were harvested equally from the first and second plots, and 12 q more from the third plot than from each of the first two. How many potatoes were harvested from each plot.

Algebraic way

(look)

Arithmetic way

(look)

output)

Arithmetic way

• 156 - 12 \u003d 144 (q) - potatoes would be harvested from three plots if the yield of all plots were the same.

• 144: 3 = 48 (c) - potatoes were harvested from the first and harvested from the second plots.
• 48 + 12 = 60 (c) - potatoes were collected from the third plot.

Answer

Back)

Algebraic way

Let them collect from the first site X c potatoes. Then from the second site they also collected X q potatoes, and harvested from the third site ( X+12) c potatoes.

According to the condition, 156 centners of potatoes were collected from all three plots.

We get the equation:

x + x + (x +12) =156

x + x + x + 12 = 156

3 X +12 = 156

3 X = 156 – 12

3 X = 144

X = 144: 3

From the first and second plots, 48 ​​centners of potatoes were collected.

48 +12 \u003d 60 (c) - potatoes were collected from the third site.

Answer: 48 quintals of potatoes were collected from the first and second sections, and 60 quintals of potatoes were collected from the third section.

Back