Arithmetic method. Simple text arithmetic tasks (their classification, examples and solutions). Various approaches to classification of text tasks

In primary school teachers just need to know what types of tasks are. Today you will learn about simple textual arithmetic tasks. Simple text arithmetic tasks are tasks that are solved by one arithmetic action.. When we read the task, we automatically relate it to it with any kind, and here it is already easy to easily become clear what action it must be solved.

I will give you not only the classification of simple text tasks itself, but I will give their examples, and I will also tell you about solving text tasks with an arithmetic method. I took all the examples of mathematics textbooks for grade 2 (Part 1, part 2), for which they are trained in the schools of Belarus.

All simple arithmetic tasks are divided into two large groups:

- Hell I (+/-), that is, those that are solved by the arithmetic effects of the first order (addition or subtraction);

- Hell II (* / :), that is, those that are solved by the arithmetic actions of the second order (multiplication or division).

Consider the first group of simple text arithmetic tasks (Hell I):

1) Tasks revealing the specific meaning of addition (+)

In competitions, 4 girls and 5 boys took part in the run. How many students from the class participated in competitions?

After Sasha decided 9 examples, he remained to solve another 3 example. How many examples needed to solve Sasha?

Such tasks are solved by adding: a + b \u003d?

2) Tasks revealing the specific meaning of subtraction (-)

Mom baked 15 pies. How many pies remained after ate 10 pies?

There were 15 glasses of juice in the bank. For dinner drank 5 glasses. How many glasses of juice remains?

These tasks are solved by subtraction: A-B \u003d?

3) Tasks for the relationship between components and the result of adding or subtraction:

a) to find the unknown 1st terms (? + a \u003d b)

The boy put in a box of 4 pencil. There they became 13. How many pencils were in the box initially?

To solve this problem, it is necessary to take the known 2nd term from the result of the action: B-A \u003d?

b) to find the unknown 2nd terms (A +? \u003d B)

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

The tasks of this type are solved by subtraction, from the result of the action takes place the known 1st terms: B-A \u003d?

c) to find an unknown diminity (? -A \u003d b)

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

The arithmetic way to solve text objectives of this type is made by adding the result of the action and submitted: B + A \u003d?

d) to find an unknown subtractable (A -? \u003d B)

Bought 2 dozen eggs. After several eggs took for baking, it remained 15. How many eggs took?

These tasks are solved by subtraction: from a decrease of taking the result of the action: A-B \u003d?

4) Tasks for a decrease / increase by several units in a straight, indirect form

examples of tasks for a decrease of 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 was in the second box?

The first class gathered 19 apples boxes, and the second is less than 4 boxes. How many apple boxes ripped the second class?

These tasks are solved by subtraction (A-B \u003d?)

Examples of tasks for a decrease in indirect form, as well as an increase in a direct or indirect form in the textbook of the 2nd class in mathematics, I did not find. If there is a need to write in the comments - and I will add an article by my own examples.

5) Tasks for difference comparisons

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

In the first box 14 pencils, and in the second - 7. How many more pencils in the first box than in the second?

Solving text tasks for difference comparisons is made by subtracting from a larger number.

We have finished dealing with simple textual arithmetic objectives of 1 groups and proceed to tasks 2 groups. If you were not clear, ask in the comments.

The second group of simple text arithmetic tasks (blood pressure II):

1) Tasks revealing the specific meaning of multiplication

How many legs have two dogs? In three dogs?

Three cars are standing near the house. Each machine has 4 wheels. How many wheels in three cars?

These tasks are solved by multiplication: A * B \u003d?

2) Tasks revealing the specific meaning of division:

a) by content

10 Cakes distributed to children, two each. How many children got pastries?

In the packages of 2 kg there are 14 kg of flour. How many such packages?

In these tasks, we learn how many parts it turned out with an equal content.

b) on equal parts

10 cm long strip was cut into two equal parts. What length every part?

Nina laid 10 cupcakes on 2 plates equally. How many cupcakes on one plate?

And in these tasks we learn what is the content of one equal part.

Be that as it may, all these tasks are solved by division: A: B \u003d?

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

a) to find the unknown first factor :? * A \u003d B

Own example:

A few boxes of 6 pencils. Total in 24 pencil boxes. How many boxes?

Deciding by the division of the work on the famous second factor: B: A \u003d?

b) to find an unknown second multiplier: A *? \u003d B

In a cafe for one table, 3 people can be planted. How many such tables will be occupied if 15 people come there?

Deciding by the division of the work on the famous first factor: B: A \u003d?

c) to find an unknown divide:?: A \u003d B

Own example:

Kohl brought into the class of candy and shared them equally between all students. In the class of 16 children. Everyone received 3 candy. How many candies brought Kohl?

Is solved by multiplying the private on the divider: b * a \u003d?

d) on finding an unknown divider: A:? \u003d B

Own example:

Vitya brought 44 candy in class and divided them equally between all students. Everyone received 2 candy. How many students in the classroom?

Decides divided by private: A: B \u003d?

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

No examples of such text arithmetic tasks were found in the textbook 2 of the class of examples of such text arithmetic tasks.

5) Tasks on multiple comparison

Decided by dividing more to the smaller.

Friends, the entire above classification of simple text tasks is just a part of a large classification of all text tasks. In addition, there are still tasks to find interest, which I did not tell you about. You can learn about all this from this video:

And my gratitude will remain with you!

Training to solve text objectives plays an important role in the formation of mathematical knowledge. Text tasks give great space for the development of thinking of students. Learning to solve problems is not only training technique for the right answers in some typical situations, how much learning to the creative approach to the search for the solution, the accumulation of experience experience and demonstration of mathematics opportunities in solving a variety of tasks. However, when solving text problems in 5-6 classes, the equation is most often used. But thinking of the fifth graders is not yet ready for formal procedures performed in solving equations. The arithmetic method for solving problems have a number of advantages compared to algebraic because the result of each step in actions is visually and more specifically, it does not go beyond the framework of the experience of five-graders. Schoolchildren better and faster solve problems in actions than with equations. Child thinking specifically, and it is necessary to develop it on specific subjects and values, then gradually move to the operating abstract images.

Work on the task provides for a careful reading of the text condition, understanding in the meaning of each word. I will give examples of tasks that are easy and simply can be solved by an arithmetic way.

Task 1.For the preparation of jam into two parts of the raspberry take three parts of the sugar. How many kilograms of sugar should be taken by 2 kg of 600 g of raspberries?

When solving a task on "parts" it is necessary to accustom to visually represent the condition of the problem, i.e. It is better to rely on the drawing.

2600: 2 \u003d 1300 (g) - falls on one part of the jam;
1300 * 3 \u003d 3900 (D) - Sugar needs to be taken.

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

1) 1 + 3 \u003d 4 (parts) - accounted for all books;

2) 120: 4 \u003d 30 (books) - falls on one part (books on the second shelf);

3) 30 * 3 \u003d 90 (books) - stood on the first shelf.

Task 3. Pheasants and rabbits are sitting in the cage. In total there are 27 heads and 74 legs. Learn the number of pheasants and the number of rabbits in the cage.

Imagine that on the cage lid in which the pheasants and rabbits are sitting, we put carrots. Then all rabbits will stand on the rear legs to reach it. Then:

27 * 2 \u003d 54 (legs) - will stand on the floor;
74-54 \u003d 20 (legs) - will be upstairs;
20: 2 \u003d 10 (rabbits);
27-10 \u003d 17 (Pheasants).

Task 4.In our class, 30 students. On an excursion to the museum there were 23 people, and in the cinema - 21, and 5 people did not go on a tour or movies. How many people went on an excursion and in the cinema?

To analyze the condition and selection of the Solution Plan, you can use "Euler circles".

30-5 \u003d 25 (man) - went or in the movies, or on a tour,
25-23 \u003d 2 (person) - went only into the movies;
21-2 \u003d 19 (man) - went to the cinema, and on a tour.

Task 5.Three duckling and four goes weigh 2 kg 500 g, and four duckling and three goes weigh 2kg 400g. How much does one goon weigh?

2500 + 2400 \u003d 2900 (d) - Weigh seven ducklings and seven geese;
4900: 7 \u003d 700 (g) - the weight of one duckling and one goover;
700 * 3 \u003d 2100 (g) - weight 3 ducklings and 3 gesyat;
2500-2100 \u003d 400 (g) - weight of the goer.

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

Imagine that from all big pyramids we shot two rings. Then:

1) 20 * 5 \u003d 100 (rings) - it remains;

2) 128-100-28 (rings) - we removed;

3) 28: 2 \u003d 14 (large pyramids).

Task 7.Watermelon weighing 20 kg contained 99% water. When he is a bit oral, the water content in it has decreased to 98%. Determine the mass of watermelon.

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

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

At the same time, it is desirable to draw the rectangles of the "dry matter" equal, because the mass of "dry matter" in the watermelon remains unchanged.

1) 20: 100 \u003d 0.2 (kg) - the mass of "dry matter";

2) 0.2: 2 \u003d 0.1 (kg) - accounted for 1% of the truncated watermelon;

3) 0.1 * 100 \u003d 10 (kg) - mass of watermelon.

Task 8.Guests asked: how old was each of the three sisters? Faith replied that she and Nada together 2 years old, Nae and anyone together together, and all three 38 years old. How many years each of the sisters?

38-28 \u003d 10 (years) - any;
23-10 \u003d 13 (years) - Nad;
28-13 \u003d 15 (years) - faith.

The arithmetic way to solve text objectives teaches a child to act consciously, logically correctly, because when solving this way, attention to the question "Why" is intensifying and there is a large developing potential. This contributes to the development of students, the formation of their interest in solving problems and to the science of mathematics.

To make learning to meet, fascinating and instructive, we must carefully consider the choice of text tasks, consider various ways to solve them, choosing optimal of them, develop logical thinking, which is further necessary when solving geometric tasks.

Learning to solve the tasks of schoolchildren will be able, only solving them. "If you want to learn to swim, then boldly enter the water, and if you want to learn to solve the tasks, then decide them," writes D.Poya in the "Mathematical Opening" book.

1. General comments to solving problems by algebraic method.

2. Move problems.

3. Work tasks.

4. Tasks for mixtures and interest.

The use of an algebraic method for finding an arithmetic solution to solve text tasks.

1. When solving problems with the algebraic method, the desired values \u200b\u200bor other values, knowing which you can define the desired are denoted by letters (usually x, y,z.). All independent relations between the data between the data and unknown values, which are either directly formulated in the condition (in verbal form), or flow out of the meaning of the problem (for example, physical laws, which are subject to the values \u200b\u200bunder consideration), or follow from the condition and certain reasoning, are recorded in The form of equality of inequalities. In general, these relations form some mixed system. In particular cases, this system may not contain inequalities or equations or it can consist only of one equation or inequality.

The solution of tasks by the algebraic method does not obey any single, quite universal scheme. Therefore, any indication relating to all tasks is the most common. The tasks that arise in solving practical and theoretical issues have their own individual characteristics. Therefore, their research and solution are most diverse.

Let us dwell on solving problems, the mathematical model of which is given by the equation with one unknown.

Recall that the task solution consists of four stages. Work at the first stage (analysis of the content of the problem) does not depend on the selected decision method and does not have fundamental differences. At the second stage (when searching for a solution to the problem and drawing up a plan for its solution), in the case of the use of an algebraic method of solution: the choice of the main relationship for the preparation of the equation; Choosing an unknown and introducing the designation for it; The expression of the values \u200b\u200bincluded in the main relationship through unknown and data. The third stage (implementation of the problem of solving the problem) implies the compilation of the equation and its decision. The fourth stage (verification of the problem of problem) is carried out standard.

Usually, when drawing up equations with one unknown h.adhere to the following two rules.

Rule I. . One of these values \u200b\u200bis expressed through an unknown h.and other data (that is, an equation is compiled in which one part contains a given value, and the other is the same value expressed by h.and other values).

Rule II. . For the same size, two algebraic expressions are drawn up, which are then equated to each other.

Externally, it seems that the first rule is easier than the second.

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

The process of solving text objectives by an algebraic method 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 basis for the preparation of the equation should be taken by a relation that sets some link between all unknowns.

Then choose an unknown, which is denoted by the corresponding letter.

All unknown values \u200b\u200bincluded in the ratio selected to compile the equation must be expressed through the selected unknown, relying on the remaining relations included in the task otherwise.

4. From the specified three operations directly implies the compilation of the equation as the design of verbal recording with the help of mathematical symbols.

The central place among the listed operations occupies the choice of the main relationship for the preparation of equations. The considered examples show that the choice of the main relationship is determined in the compilation of equations, it makes logical slightness in the threshold of the vague verbal text of the task, gives confidence in orientation and protects against disorderly actions to express all the values \u200b\u200bincluded in the task through data and the desired.

The algebraic method of solving problems is of great practical importance. With it, they solve a wide variety of tasks from the field of technology, agriculture, life. Already in high school, the equations are applied by students when studying physics, chemistry, astronomy. Where the arithmetic is powerless or, at best, requires extremely bulky reasoning, there is an algebraic method easily and quickly leads to the answer. And even in the so-called "typical" arithmetic tasks, relatively easily solved by an arithmetic pathway, an algebraic solution is usually also shorter and more natural.

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

2. The problem of movement tasks include the tasks in which three values \u200b\u200bare referred: (s.), speeds ( v.) and time ( t.). As a rule, in them we are talking about a uniform rectilinear movement, when the speed is constant by module and direction. In this case, all three values \u200b\u200bare related to the following ratio: S. = vt.. For example, if a cyclist speed is 12 km / h, then in 1.5 hours. It will drive 12 km / h  1.5 h \u003d 18 km. There are tasks in which an equilibrium straight line movement is considered, that is, a constant acceleration movement (but).Distance traveled s. in this case, calculated by the formula: S. = v. 0 t. + aT. 2 /2, where v. 0 – Starting speed. So, for 10 from the fall at the initial speed of 5 m / s and the acceleration of the free fall of 9.8 m 2 / with the body, the distance equal to 5 m / s  10 ° C + 9.8 m 2 / s  10 2 С 2/2 \u003d 50 m + 490 m \u003d 540 m.

As already noted, during solving textual tasks and, first of all, in the tasks associated with the movement, it is very useful to make an illustrative drawing (build a supporting graphic model of the task). The drawing should be performed so that the dynamics of movement with all meetings, stops and turns is visible. A competent drawing drawing makes it possible not only to deeper the content of the problem, but also facilitates the compilation of equations and inequalities. Examples of such drawings will be shown below.

Typically, the following agreements are taken in the movement tasks.

If it is not specifically stipulated in the task, the movement in separate areas is considered uniform (be it movement in direct or around the circumference).

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

This group of tasks, in turn, can be divided into tasks in which the movements of the tel: 1) meet each other; 2) in one direction ("after"); 3) in opposite directions; 4) on a closed trajectory; 5) by the flow of the river.

If the distance between the bodies is S., and the speeds of the bodies are equal v. 1 and v. 2 (Fig. 16 but), then when moving bodies towards each other, through which they will meet, equal S./(v. 1 + v. 2).

2. If the distance between the bodies is equal S., and the speeds of the bodies are equal v. 1 I. v. 2 (Fig. 16 b.), then when moving bodies in one direction ( v. 1 > v. 2) the time through which the first body will catch up with the second, equal S./(v. 1 – v. 2).

3. If the distance between the bodies is S., and the speeds of the bodies are equal v. 1 I. v. 2 (Fig. 16 in), then going at the same time in opposite directions, the bodies will be through time t. be at a distance S. 1 = S. + (v. 1 + v. 2 ) t..

Fig. sixteen

4. If the bodies move in one direction on a closed trajectory length s. with speeds v. 1 I. v. 2, the time through which the bodies will again meet (one body will catch up with the other), going at the same time from one point, is on the formula t. = S./(v. 1 – v. 2) provided that v. 1 > v. 2 .

This follows from the fact that with a simultaneous start on a closed trajectory in one direction the body whose speed is greater, begins to catch up with the body whose speed is less. For the first time it caught up with him by passing the distance to S. more than another body. If it overtakes him in the second, for the third time, and so on, it means that it passes the distance to 2 S., 3. S. and so on more than another body.

If the bodies move in different directions on a closed trajectory length S. with speeds v. 1 I. v. 2, the time through which they will meet, going at the same time from one point, is on the formula t. = v.(v. 1 + v. 2). In this case, immediately after the start of the movement, the situation arises when the bodies begin to move towards each other.

5. If the body moves along the river flow, then its speed relative to the shore andcomplies from body velocity in standing water v. and river flow rates w.: and \u003d.v. + w.. If the body moves against the flow of the river, then its speed and \u003d.v. – w.. For example, if the speed of the boat v. \u003d 12 km / h, and the flow rate of the river w. \u003d 3 km / h, then for 3 hours. By the river, the boat saves (12 km / h + 3 km / h)  3 h. \u003d 45 km, and against current - (12 km / h - 3 km / h)  3 h. \u003d 27 km. It is believed that the speed of objects having a zero speed of movement in standing water (raft, log, etc.) is equal to the flow rate of the river.

Consider several examples.

Example. Is one point in one direction every 20 minutes. Cars leave. The second car is driving at a speed of 60 km / h, and the speed of the first 50% is higher than the speed of the second. Find the speed of the third car, if it is known that he 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% longer than the speed of the second, it means that it is equal

When driving in one direction, the meeting time is like the ratio between the objects to the difference of their speeds. The first car is 40 minutes. (2/3 h) erupt 90  (2/3) \u003d 60 km. Consequently, the third will catch it up (they will meet) after 60 / ( h. - 90) hours. The second in 20 minutes. (1/3 h) erupt 60  (1/3) \u003d 20 km. So, the third will catch it up (they will meet) after 20 / ( h. - 60) h. (Fig. 17).

P
about the condition of the task

Fig. 17.

After simple transformations, we get a square equation 11x 2 - 1730x + 63000 \u003d 0, solving which we find

Check shows that the second root does not satisfy the condition of the task, 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.

ExampleTreatment passed by the river 96 km, returned back and spent some time under loading, spending on all 32 hours. The river flow rate is 2 km / h. Determine the speed of the ship in standing water if the loading time is 37.5% of the time spent on the entire path and back.

Decision. Let x km / h be the speed of the ship in standing water. Then ( h.+ 2) km / h - its speed by flow; (x -2) km / h - against the flow; 96 / ( h. + 2) h. - time of movement by flow; 96 / ( h. - 2) h. - Time of movement against the flow. Since 37.5% of the total time of time, the ship was under loading, then a clean time of movement is 62.5%  32/100% \u003d 20 (h.). Consequently, under the condition of the problem, we have an equation:

Converted it, we get: 24 ( h. – 2 + h. + 2) = 5(h. + 2)(h. – 2) => 5h. 2 – 4h. - 20 \u003d 0. Deciding the square equation, we find: h. 1 = 10; h. 2 \u003d -0.4. The second root does not satisfy the condition of the problem.

Answer: 10 km / h - the speed of movement of the ship in standing water.

Example. The car drove the way from the city BUTin the city with through the city INwithout stops. Distance ABequal to 120 km, he drove at a constant speed of 1 h. faster than the distance Sun,equal 90 km. Determine the average vehicle speed from the city BUTto the city with, if it is known that the speed is on the plot AU30 km / h more speed on the plot Sun.

Decision. Let be h. km / h - car speed on the plot Sun.

Then ( h. + 30) km / h - speed on the plot AB120/(h. + 30) h, 90 / h. h - time, a knuckle car drives the way AU and Sunrespectively.

Consequently, under the condition of the problem, we have an equation:

We transform it:

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

Deciding the square equation, we find: h. 1 = 30, h. 2 \u003d -90. The second root does not satisfy the condition of the problem. It means speed on the plot Sunequal to 30 km / h, on the plot AB 60 km / h. It follows that distance AUthe car drove for 2 hours (120 km: 60 km / h \u003d 2 h.), and the distance Sun - For 3 hours (90 km: 30 km / h \u003d 3 h.), So all distance AChe drove in 5 hours (3 hours. + 2 h. \u003d 5 h.). Then the average speed of movement on the plot Acthe length of which is 210 km, is 210 km: 5 h. \u003d 42 km / h.

Answer: 42 km / h - average vehicle speed on the site AU.

The task group includes tasks in which three quantities refer to: work BUT, time t.during which work is performed, performance R -work produced per unit of time. These three values \u200b\u200bare associated with the equation BUT = Rt.. Tasks are related to the tasks associated with filling and emptying of tanks (vessels, tanks, pools, etc.) with pipes, pumps and other devices. As a work done in this case, the volume of pumping water is considered.

The tasks of work, generally speaking, can be attributed to the group of tasks in motion, since in the tasks of this type we can assume that all the work or the full volume of the tank play the role of the distance, and the performance of work facilities, similar to the speeds of movement. However, by Fabule, these tasks differ in a natural way, and part of the tasks to work have its own specific decisions of the solution. So, in those tasks in which the amount of work performed is not specified, all work is taken per unit.

Example.Two brigades had to fulfill the order for 12 days. After 8 days of collaboration, the first brigade received another task, so the second brigade finished the order for another 7 days. How many days could each of the brigades be fulfilled, working separately?

Decision. Let the first brigade performs the task for h.days, second brigade - for y. days. We will take all the work per unit. Then 1 / x - Performance of the first brigade, A 1 / y. – second. Since two brigades must fulfill the order for 12 days, we obtain the first equation 12 (1 / h. + 1/w.) = 1.

From the second condition it follows that the second brigade worked 15 days, and the first is only 8 days. It means that the second equation has the form:

8/h.+ 15/w.= 1.

Thus, we have a system:

The first will be subtracted from the second equation, we get:

21/y. = 1 \u003d\u003e y \u003d21.

Then 12 / h. + 12/21 = 1 => 12/ H. – = 3/7 => x \u003d28.

Answer: For 28 days I will execute the first brigade order, for 21 days - the second.

Example. Working BUT and workers IN can perform work for 12 days, working BUTand workers FROM - For 9 days, worker INand working C - for 12 days. For how many days they do work, working in the threesome?

Decision. Let the worker BUTcan perform work for h.days, worker IN - per w.days, worker FROM - per z. days. We will take all the work per unit. Then 1 / x, 1 /y. and 1 / z. – Performance workers A, B.and FROM respectively. Using the condition of the problem, we arrive at the next system of the equations presented in the table.

Table 1

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

After folding the system equation, we get:

The amount is the joint performance of the workers, so the time for which they will perform all the work will be equal

Answer: 7.2 days.

Example. Two pipes were held in the pool - the feeding and discharge, and through the first pipe, the pool is filled 2 hours longer than through the second water from the pool is poured. When filled with one third, both pipes were opened, and the pool turned out to be empty after 8 hours. For how many hours through one first pipe can be filled with a pool and for how many hours through one second pipe can a full swimming pool can be drunk?

Decision. Let be V. m 3 - the volume of the pool, h.m 3 / h - productivity of the feed pipe, w.m 3 / h - discharge. Then V./ x. h. - the time required by the feed pipe to fill the pool, V./ y. h. - the time required by the discharge pipe for the drainage of the pool. Under the condition of the task V./ x. – V./ y. = 2.

Since the performance of the discharge pipe is more productivity of the filling, then when both pipes are turned on, the basin will occur and one third of the pool will dry out during the time (V./3)/(y. – x.), which, by the condition of the problem, is 8 hours. So, the condition of the task can be recorded as a system of two equations with three unknowns:

In the task you need to find V./ x. and V./ y.. Highlight in the equations a combination of unknown V./ x. and V./ y., recovering the system:

Introducing new unknowns V./ x. \u003d A.and V./ y. = b., we get the following system:

Substituting in the second equation but= b. + 2, have an equation regarding b.:

solving which we find b. 1 = 6, b. 2 = -eight. The task condition satisfies the first root 6, \u003d 6 (h.). From the first equation of the last system we find but\u003d 8 (h), that is, the first pipe fills the pool for 8 hours.

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

Example. One tractor brigade should plow 240 hectares, and another is 35% more than the first. The first brigade, plowing daily by 3 hectares less than the second, finished work for 2 days earlier than the second brigade. How many hectares plowed every brigade daily?

Decision. We find 35% of 240 hectares: 240 hectares  35% / 100% \u003d 84 hectares.

Consequently, the second brigade had to plow 240 hectares + 84 hectares \u003d 324 hectares. Let the first brigade plowed daily h.ha. Then the second brigade plowed daily ( h. + 3) ha; 240 / h. - the time of work of the first brigade; 324 / ( h. + 3) - the time of operation of the second brigade. By the condition of the task, the first brigade finished work 2 days earlier than the second, so we have an equation

which after transformations can be written as follows:

324h. – 240x -720 \u003d 2x 2 + 6x\u003d\u003e 2x 2 - 78x + 720 \u003d 0 \u003d\u003e x 2 - 39x + 360 \u003d 0.

Deciding the square equation, we find x 1 \u003d 24, x 2 \u003d 15. This is the norm of the first brigade.

Consequently, the second brigade plowed on day 27 hectares and 18 hectares, respectively. Both solutions satisfy the condition of the task.

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

Example. In May, two workshops produced 1080 details. In June, the first workshop increased the production of details by 15%, and the second increased the production of parts by 12%, so both workshops made 1224 parts. How many parts made every workshop in June?

Decision. Let be h. details made in May the first workshop, w.details - second. Since 1080 parts made in May, then by the condition of the task we have an equation x. + y. = 1080.

We find 15% of h.:

So, 0.15 h. details increased production of products The first shop, therefore, in June he released x +.0,15 h. = 1,15 x. details. Similarly, we find that the second workshop in June made 1.12 y. details. So, the second equation will look: 1.15 x. + 1,12 w. \u003d 1224. Thus, we have a system:

from which we find x \u003d480, y \u003d.600. Consequently, in June, 552 parts and 672 parts were made, respectively.

Answer: The first workshop made 552 details, the second - 672 parts.

4. A group of tasks on the mixture and interest relate to the tasks in which it is about mixing various substances in certain proportions, as well as interest tasks.

Tasks for concentration and percentage

We clarify some concepts. Let there be a mixture from pvarious substances (components) BUT 1 BUT 2 , ..., BUT n. accordingly, the volumes of which are equal V. 1 , V. 2 , ..., V. n. . Volume of mix V. 0 it consists of pure components: V. 0 = V. 1 + V. 2 + ... + V. n. .

Bulk concentrationsubstances BUT i. (i. = 1, 2, ..., p)in the mixture is called the value with i. calculated by the formula:

Volume percentage of substance A i. (i. = 1, 2, ..., p)in the mixture is called the magnitude p. i. , Calculated by formula r i. = from i. ,  100%. Concentration from 1, from 2 , ..., from n. which are dimensionless values \u200b\u200bare associated with equality. from 1 + S. 2 + ... + with n. \u003d 1, and ratios

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

If a percentage is known i.-to component, its concentration is located by the formula:

i.e PI – this is a concentration i.-Ho substances in a mixture expressed as a percentage. For example, if the percentage of the substance is 70%, then its corresponding concentration is 0.7. Conversely, if the concentration is equal to 0.33, then the percentage is 33%. Thus, the amount r 1 + R. 2 + ... + p n. \u003d 100%. If concentration is known from 1 , from 2 , ..., from n. components that make up this volume mixture V. 0 , then the corresponding volume components are in the formulas:

The concepts are similar in the same way. weight (mass) concenterscomponents of the mixture and the corresponding percentages. They are defined as the ratio of weight (mass) of pure substance BUT i. , in alloy to weight (mass) of the whole alloy. What concentration, volume or weight, is in question in a specific task, it is always clear from its condition.

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

wherein V. 1 and V. 2 – The volume components of the mixture of components. Weight concentrations of components are from equalities:

which determine the connection of these values \u200b\u200bwith volume concentrations.

As a rule, in the texts of such tasks, the same repeated condition is found: from two or several mixtures containing components A. 1 , A. 2 , BUT 3 , ..., BUT n. , a new mixture is drawn up by mixing the initial mixtures taken in a certain proportion. At the same time, it is required to find in what the components BUT 1, BUT 2 , BUT 3 , ..., BUT n. enter the resulting mixture. To solve this problem, it is convenient to introduce a volume or weight number of each mixture into consideration, as well as the concentration of component components. BUT 1, BUT 2 , BUT 3 , ..., BUT n. . With the help of concentrations, it is necessary to "split" each mixture into individual components, and then the method specified in the condition method to make a new mixture. It is easy to calculate how much of each component enters the resulting mixture, as well as the total amount of this mixture. After that, the concentrations of components are determined. BUT 1, BUT 2 , BUT 3 , ..., BUT n. in a new mixture.

Example. There are two pieces of copper and zinc alloy with percentage of copper 80% and 30%, respectively. What is the matter of these alloys to, remember the pieces taken together, get an alloy containing 60% of copper?

Decision. Let the first alloy taken h. kg, and the second - w.kg. Under the condition, the concentration of copper in the first alloy is 80/100 \u003d 0.8, in the second - 30/100 \u003d 0.3 (it is clear that we are talking about weight concentrations), it means that in the first alloy 0.8 h. CG Copper and (1 - 0.8) h. = 0,2h. kg zinc, in the second - 0.3 w.cG Copper and (1 - 0.3) y. = 0,7w. kg zinc. The amount of copper in the resulting alloy is equal to (0.8  h. + 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 definition, is equal to

Under the problem of problem, this concentration should be 0.6. Therefore, we obtain the equation:

This equation contains two unknown h.and yHowever, by the condition of the task, it is required to determine the values \u200b\u200bthemselves h.and y,but only their attitude. After easy transformations we get

Answer: Alloys must be taken in respect of 3: 2.

Example. There are two sulfuric acid solutions 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 80% solution was added, then 70% solution would be obtained. How many 40% and 60% solutions were?

Decision. Let be h.kg - the mass of the first solution, w.kg - second. Then the mass of a 20% solution ( h. + w.+ 5) kg. As B. h.kg 40% solution contains 0.4 h. kg acid, in w.kg 60% solution contains 0.6 y. kg acid, and in (x + y +5) kg of a 20% solution contains 0.2 ( h. + in +.5) kg of acid, then under the condition we have the first equation 0.4 h. + 0,6y. = 0,2(h. + U +.5).

If instead of 5 kg of water add 5 kg of 80% solution, then the solution will be solved (x + u+ 5) kg in which there will be (0.4 h. + 0,6w. + 0.8  5) kg of acid, which will be 70% of (x + y+ 5) kg.

Analyzing the data of the task, observing that in common in the tasks from the point of view of mathematics, what is the difference, find an extraordinary way to solve problems, create a piggy bank of the tasks solving, learn to solve one problem in various ways. Taxes of tasks grouped by a single subject "Arithmetic methods solving problems" , tasks for working in a group and for individual work.

"Tasks for the simulator technique"

Simulator: "Arithmetic ways to solve problems"

"Comparison of numbers in sum and difference."

In two baskets 80 borovikov. In the first basket on 10 boroviks less than in the second. How many Boroviki in each basket?

The sewing studio received 480 m denim and drape. Denim tissue was 140 m more than Drapa. How many meters of denim entered the studio?

The television model consists of two blocks. The lower unit is 130 cm shorter than the top. What is the height of the upper and lower blocks, if the height of the tower is 4 m 70 cm?

Two boxes of 16 kg of cookies. Find a lot of cookies in each box if in one of them biscuits per 4 kg more.

The task of "arithmetic" L. N. Tolstoy.

a) Two men have 35 sheep. One for 9 sheep is greater than that of another. How many sheep have everyone?

b) two men have 40 sheep, and one is less against another 6 sheep. How many sheep have every man?

There were 23 passenger cars and motorcycles with a carriage in the garage. Machine and motorcycles 87 wheels. How many motorcycle garages, if a spare wheel put a spare wheel in each carriage?

"Euler circles."

There are 120 residents in the house, some of them have dogs and cats. In the picture circle FROM pictures tenants with dogs, circle TO – residents with cats. How many tenants have dogs and cats? How many tenants have only dogs? How many tenants have only cats? How many tenants do not have dogs or cats?

Of the 52 schoolchildren 23 engage in volleyball and 35 basketball, and 16 - and volleyball, and basketball. The rest do not engage in any of these sports. How many schoolchildren do not deal with any of these sports?

In the picture circle BUT depicts all university employees who know English, circle N. - knowledgeable German and circle F. - French. How many university employees know: a) 3 languages; b) English and German; c) French? How many university staff? How many of them do not speak French?

120 people participated in the international conference. Of these, 60 are owned by the Russian language, 48 - English, 32 - German, 21 - Russian and German, 19 - English and German, 15 - Russian and English, and 10 people owned all three languages. How many conference participants do not own any of these languages?

They sing in the choir and are engaged in dancing 82 students, are engaged in dancing and rhythmic gymnastics 32 student, and sing in the choir and are engaged in rhythmic gymnastics of 78 students. How many students sing in the choir are engaged in dancing and rhythmic gymnastics separately, if it is known that every student is only doing something alone?

Each family living in our house discharges or newspaper, or magazine, or both. 75 families discharge the newspaper, and 27 families discharge the magazine, and only 13 families discharge the magazine and the newspaper. How many families live in our house?

"Data equalization method".

In 3 small and 4 large bouquets of 29 flowers, and in 5 small and 4 large bouquets of 35 flowers. How many flowers in each bouquet separately?

The mass of 2 chocolate tiles is large and small - 120 g, and 3 large and 2 small - 320 What is the mass of each tile?

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

Four duckling and five geussy weigh 4kg 100g, five ducklings and four goes weigh 4 kg. How much is one duckling weigh?

For one horse and two cows produce 34 kg of hay daily, and for two horses and one cow - 35 kg of hay. How many hay give one horse and how much one cow?

3 red cubes and 6 blue cubes stand 165tg rub. And, five red is more expensive than two blue at 95 tg. How much is every cube?

2 Albums for drawing and 3 albums for stamps are worth 160 rubles together, and 3 drawing albums are 45 rubles. More expensive two albums for brands.

"Graphs".

Seryozha decided to give mom for a birthday bouquet of flowers (roses, tulips or carnations) and put them or in a vase, or in a jar. How many ways can he do it?

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

On Wednesday in grade 5, five lessons: mathematics, physical education, history, Russian language and natural science. How many different variants of the schedule on Wednesday can be made up?

"An old way to solve problems for mixing substances."

How to mix oils? Some person had for sale of the oil of two varieties: one price is 10 hryvnia per bucket, the other than 6 hryvnia per bucket. I wanted to make it from these two oils, mixing them, the oil at the price of 7 hryvnia per bucket. What parts of these two oils need to take to get a bucket of oil worth 7 hryvnia?

How much do you need to take caramels at a price of 260 tg per 1 kg and at a price of 190 tg per 1 kg to make up 21 kg of mixture at a price of 210 tg per kilogram?

Someone has three varieties tea - Ceylon 5 hryvnia per pound, Indian 8 hryvnia for Pound and Chinese 12 hryvnia per pound. In what fractions you need to mix these three varieties to get tea worth 6 hryvnia per pound?

Someone has silver samples: one - 12 - oh sample, another - 10 sample, the third - 6 - oh sample. How many silver should be taken to get 1 pound of silver 9 - oh sample?

The merchant bought 138 Arshin of Black and Blue Sukna for 540 rubles. It is asked how many Arshin bought it both, if there was a blue 5 rubles. For Arshin, and black - 3 rubles.?

Different tasks.

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

On the Christmas tree of children in the carnival suits of snowflakes 3 times more than in the costumes of parsley. How many children were in parsley costumes, if they were 12 less?

Masha received 2 times less than New Year's congratulations than Kohl. How many congratulations did everyone, if all of them were 27? (9 and 18).

For New Year's prizes, 28 kg of candy was bought. Candy "Swallow" amounted to 2 parts, "Muse" - 3 parts, "chamomile" - 2 parts. How many candies of each grade bought? (8, 8, 12).

The warehouse has 2004 kg of flour. Is it possible to decompose it in bags weighing 9 kg and weighing 18 kg?

In the store "Everything for tea", there are 5 different cups and 3 different saucers. How many ways can I buy a cup with a saucer?

The horse eats a hay stack for 2 days, cow - for 3, sheep - for 6. For how many days they will eat a stack if there are it together?

View the contents of the document
"Abstract lesson Arif SP"

"Arithmetic ways to solve text tasks."

A person who studies mathematics is often more useful to solve the same task in three different ways than to solve three - four different tasks. Solving one task in various ways, it is possible by comparison to find out which one is shorter and more efficient. So the experience is produced.

U.U.Soyer

The purpose of the lesson: Use knowledge obtained in previous lessons, show fantasy, intuition, imagination, mixtalk to solve test problems in various ways.

Tasks lesson: educational: Analyzing these tasks, observing that in common in the tasks in terms of mathematics, what is the difference, find an extraordinary way to solve problems, create a piggyback of task solutions, learn to solve one problem in various ways.

Developing: Feel the need for self-realization, being in a certain role situation.

Educational:develop personal qualities, form a communicative culture.

Means of education: Simulator of tasks grouped by a single theme "Arithmetic ways to solve problems", tasks for working in a group and for individual work.

DURING THE CLASSES.

I. Organizational moment

Hello guys. Sit down. Today we have a lesson on the topic "Arithmetic methods for solving textual tasks."

II. Actualization of knowledge.

Mathematics is one of the ancient and important sciences. Many mathematical knowledge people used in ancient times - thousands of years ago. They were needed merchants and builders, soldiers and farmers, priests and travelers.

And nowadays, no person can do in life without good knowledge of mathematics. The basis of a good understanding of mathematics is the ability to count, think, reason, find successful solutions to tasks.

Today we consider arithmetic ways to solve text objectives, we will analyze the tasks of the old, which have come down from different countries and times, tasks on the equalization, on the comparison of the amount and difference and others.

The purpose of the lesson is to involve you in the amazing world of beauty, wealth and diversity - the world of interesting tasks. And, it means to introduce some arithmetic methods, leading to very elegant and instructive solutions.

The task is almost always the search, the disclosure of some properties and relationships, and the means of solving it is intuition and guess, erudition and possession of mathematics methods.

As the main in mathematics, arithmetic and algebraic methods of solving problems are distinguished.

Solve the task arithmetic method - it means to find an answer to the requirement of the problem by performing arithmetic action on numbers.

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

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

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

1. Method of solving textual tasks "Comparison of numbers in sum and difference".

A task : Grandmother in the fall from the country area collected 51 kg of carrots and cabbage. The cabbage was 15 kg more than carrots. How many kilograms of carrots and how many kilograms of cabbage gathered her grandmother?

Questions that correspond to the items of the algorithm for solving the tasks of this class.

1. Find out what values \u200b\u200bis in question

On the number of carrots and cabbage, which gathered grandmother, together and separately.

2. Specify, what values \u200b\u200bmust be found in the task.

How many kilograms of carrots and how many kilograms of cabbage gathered her grandmother?

3. Call the relationship between values \u200b\u200bin the task.

The task refers to the amount and difference of quantities.

4. Name the amount and difference of values \u200b\u200bof values.

The amount is 51 kg, the difference is 15 kg.

5. By equalizing the magnitudes to find a double value of a smaller value (from the amount of values \u200b\u200bto take away the difference in quantities).

51 - 15 \u003d 36 (kg) - twice the number of carrots.

6. Knowing doubled, finding a smaller value (doubled to divide into two).

36: 2 \u003d 18 (kg) - carrots.

7. Using the difference between the values \u200b\u200band the value of a smaller value, find the value of a greater value.

18 + 15 \u003d 33 (kg) - cabbage. Answer: 18 kg, 33 kg. A task.There are pheasants and rabbits in the cage. Total 6 goals and 20 legs. How many rabbits and how many pheasans in the cage ?
Method 1. Method of selection:
2 Pheasant, 4 rabbits.
Check: 2 + 4 \u003d 6 (heads); 4 4 + 2 2 \u003d 20 (legs).
This is the selection method (from the word "pick up"). The advantages and disadvantages of this method of solution (it is difficult to select, if the numbers are large) in this way, an incentive appears to search for more convenient solutions.
Results of the discussion: The selection method is convenient when actions with small numbers, with an increase in the values \u200b\u200bit becomes irrational and time consuming.
Method 2. Full bust of options.

Compiled table:

Answer: 4 rabbits, 2 pheasant.
The name of this method is "full". Results of the discussion: the method of complete exemption is convenient, but at large quantities enough time-consuming.
Method 3. Method of assumption.

Take an old Chinese task:

The cell contains an unknown number of pheasants and rabbits. It is known that the entire cell contains 35 heads and 94 legs. Learn the number of pheasants and the number of rabbits. (The challenge from the Chinese mathematical book "Kiu-Chang", compiled in 2600 years BC. E.).

We give a dialogue found from old mathematics masters. - Imagine that the cage in which the pheasants and rabbits are sitting, we put carrots. All rabbits will stand on the rear legs to reach the carrot. How many legs at this moment will stand on earth?

But in the condition of the task, 94 legs are given, where are the rest?

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

How many of them?

24 (94 – 70 = 24)

How many rabbits?

12 (24: 2 = 12)

And pheasants?

23 (35- 12 = 23)

The name of this method is "Method of Assumption for Lack." Try to explain this name (in a cell sitting 2 or 4 legs, and we suggested that everyone is the smallest of these numbers - 2 legs).

Another way to solve the same task. - Let's try to solve this task - "by the method of excess to excess": we will imagine that the Pheasans appeared two more legs, then all the legs will 35 × 4 \u003d 140.

But under the condition of the problem, only 94 legs, i.e. 140 - 94 \u003d 46 Feet extra, whose? These are the feet of pheasants, they have an extra couple of feet. It means pheasanov will be 46: 2 = 23, then rabbits 35 -23 = 12.
Results of the discussion: the assumption method has two options - by disadvantage and excess; Compared with previous methods, it is more convenient, as less time consuming.
A task. In the desert, a caravan of camels, all of them, are slowly walking. If you recalculate all the humps from these camels, then 57 horseship will be. How many alogy camels in this caravan? 1 way. Solve using the equation.

Number of humps from one number of camels of all humps

2 x 2

1 40 - h. 40 - h. 57

2 x +. 40 - h. = 57

x +. 40 = 57

h. = 57 -40

h. = 17

2 way.

- How many humps can have camels?

(there may be two or one)

Let's make each camel on one hump. I will attach a flower.

- How many flowers will need? (40 camels - 40 colors)

- How many humps will remain without flowers?

(Such will be 57-40=17 . it second gorge dugorble camels).

how many dugorby camels? (17)

how many single-burnt camels? (40-17 \u003d 23)

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

A task.In the garage there were passenger cars and motorcycles with strollers, all together 18. Machine and motorcycles - 65 wheels. How many motorcycles with wheelchairs stood in the garage, if cars have 4 wheels, and at a motorcycle - 3 wheels?

1 way. With the help of the equation:

Kol- wheels in 1 coat

Mash. fourx 4 x.

ILO. 3 18 -h. 3(18 - h. ) 65

4 x +. 3(18 - h. ) = 65

4 x + 5. 4 -3 h. =65

h. = 65 - 54

h. = 11, 18 – 11 = 7.

We reformulate the task : Robbers who came to the garage, where 18 cars and motorcycles were stood with wheelchairs, removed from each machine and each motorcycle three wheels and took. How many wheels remain in the garage if there were 65? Do they belong to the car or motorcycle?

3 × 18 \u003d 54 - how many wheels were taken by robbers,

65- 54 \u003d 11 - so many wheels left (cars in the garage),

18 - 11 \u003d 7-Motorocycles.

Answer: 7 motorcycles.

Alone:

There were 23 passenger cars and motorcycles with a carriage in the garage. Machine and motorcycles 87 wheels. How many motorcycle garages, if a spare wheel put a spare wheel in each carriage?

- How many wheels have machines and motorcycles together? (4 × 23 \u003d 92)

- How many spare wheels put in each stroller? (92 - 87 \u003d 5)

- How many cars in the garage? (23 - 5 \u003d 18).

A task.In our class you can learn English or French (optional). It is known that English is studying 20 schoolchildren, and French - 17. Total in class 32 student. How many students are learning both languages: and English and french?

Show two circles. In one we will fix the number of schoolchildren studying English, in the other-clerk studies studying French. As under the condition of the problem there are students learningboth languages: English and French, Circles will have a common part. In the condition of this task, it is not easy to figure out. If you fold 20 and 17, it will turn out more than 32. This is explained by the fact that some schoolchildren we took into account twice - namely those who study both languages: both English and French. So, (20 + 17) - 32 \u003d 5 pupils are learning both languages: both English and French.

English Fran.

20 uch. 17 uch.

(20 + 17) - 32 \u003d 5 (students).

Schemes like the one we took advantage of the task in solving problems in mathematics circles (or diagrams) Euler. Leonard Euler (1736) Born in Switzerland. But for many years I lived in Russia.

A task. Each family living in our house discharges or newspaper, or magazine, or both. 75 families discharge the newspaper, and 27 families discharge the magazine, and only 13 families discharge the magazine and the newspaper. How many families live in our house?

Newspapers magazines

Figure shows that 89 families live in the house.

A task.120 people participated in the international conference. Of these, 60 are owned by the Russian language, 48 - English, 32 - German, 21 - Russian and German, 19 - English and German, 15 - Russian and English, and 10 people owned all three languages. How many conference participants do not own any of these languages?

English 15 English

21 10 19

German

Solution: 120 - (60 + 48 + 32 -21 - 19 - 15 + 10) \u003d 25 (people).

A task. Three kitten and two puppies weigh 2 kg 600 g, and two kittens and three puppies weigh 2 kg 900 g. How much is the puppy weigh?

3 Kitten and 2 puppies - 2kg 600 g

2 Kitten and 3chenchenka - 2kg 900 g

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

2 cat. And 2 pueps. Weigh 2 kg 200 g

Compare the conditions -

2 Kitten + 3 Schedule \u003d 2kg 900 g

2 kittens + 2 puppies \u003d 2 kg 200 g, we see that the puppy weighs 700 g.

A task.For one horse and two cows produce 34 kg of hay daily, and for two horses and one cow - 35 kg of hay. How many hay give one horse and how much one cow?

We write a brief condition of the task:

1 horses and 2 cows -34kg.

2 horses and 1 cows -35kg.

Is it possible to find out how much hay will need for 3 horses and 3 cows?

(for 3 horses and 3 cows - 34 + 35 \u003d 69 kg)

Is it possible to find out how much hay will need for one horse and one cow? (69: 3 - 23 kg)

How many hay will need for one horse? (35-23 \u003d 12kg)

How much hay will need for one cow? (23 -13 \u003d 11kg)

Answer: 12kg and 11 kg.

A task.Madina decided to have breakfast in the school buffet. Learn the menu and answer, how many ways can it choose a drink and confectionery?

	Confectionery
	Cheesecake

Let's assume that Madina's drinks will choose tea. What confectionery it can pick up for tea? (tea - cheese, tea - cookies, tea - bun)

How many ways? (3)

And if compote? (also 3)

How to find out how many ways can Madina use to choose a lunch? (3 + 3 + 3 \u003d 9)

Yes you are right. But to us easier to solve such a task, we will use graphs. The word "graph" in mathematics means a picture where several points are drawn, some of which are connected by lines. Denote drinks and confectionery dots and connect pairs of those dishes that Madina will choose.

tea milk compote

vatrushka biscuit bun

Now count the number of lines. There are 9. Therefore, there are 9 ways to choose dishes.

A task.Seryozha decided to give mom for a birthday bouquet of flowers (roses, tulips or carnations) and put them or in a vase, or in a jar. How many ways can he do it?

What do you think, how many ways? (3)

Why? (colors 3)

Yes. But there is still different dishes: or a vase, or a jar. Let's try to perform the task graphically.

vase Kuvshin

roses Tulips Carnations

Count lines. How many of them? (6)

So how many ways to choose from Serge? (6)

The outcome of the lesson.

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

It is known that the solution of tasks is practical art, similar to swimming or a game for piano. You can learn only by imitating good samples, constantly practicing.

This is just the simplest of tasks, complex yet remain the subject for the future study. But they are still much more than we could solve them. And if at the end of the lesson you can solve the tasks "behind the pages of the educational material", then we can assume that I performed my task.

Knowledge of mathematics helps solve a certain vital problem. In life, you will have to regularly resolve certain issues, for this it is necessary to develop intellectual abilities, thanks to which the internal potential develops, develop the ability to foresee the situation, predict, adopt a non-standard solution.

I want to finish the lesson with the words: "Any well-solved mathematical task delivers mental pleasure." (Gesse).

Do you agree with this?

Homework.

There will be such a task to the house: using the texts of solved problems, as a sample, solve Tasks No. 8, 17, 26 by the methods that we studied.

Problem solving by algebraic (using equations) According to the textbook I.I. Zubareva, A.G. Mordkovich

mathematics teacher MOU "LSOS №2"

likhoslavl Tver region

Objectives: - show the problem of solving problems by an algebraic method; - to form the ability to solve problems with arithmetic and algebraic methods.

Methods

task solutions

Arithmetic (solution of the task of action)

Algebraic (solving the problem with the equation)

Task number 509.

Read the task.

Try to find different ways to solve.

Two boxes of 16 kg of cookies. Find a lot of cookies in each box if in one of them cookies per 4 kg more than in another.

1 way solution

(look)

3 way solution

(look)

2 way solution

4 way solution

1 method (arithmetic)

16 - 4 \u003d 12 (kg) - cookies will remain in two boxes, if you get 4 kg of cookies from the first box.

12: 2 \u003d 6 (kg) - cookies were in the second box.

6 + 4 \u003d 10 (kg) - cookies were in the first box.

Answer

The solution is used the method of equalization .

Question : Why did he get such a name?

Back)

2 method (arithmetic)

16 + 4 \u003d 20 (kg) - cookies will be in two boxes, if you add 4 kg of cookies to the second box.

20: 2 \u003d 10 (kg) - cookies were in the first box.

10 - 4 \u003d 6 (kg) - the 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 is used the method of equalization .

Back)

3 method (algebraic)

Denote a lot of cookies in the second Box letter h. kg. Then the mass of cookies in the first box will be equal ( h. +4) kg, and the mass of cookies in two boxes - (( h. +4)+ h.) kg.

(h. +4)+ h. =16

h. +4+ h. =16

2 h. +4=16

2 h. =16-4

2 h. =12

h. =12:2

In the second box there were 6 kg of cookies.

6 + 4 \u003d 10 (kg) - cookies were in the first box.

The solution is used algebraic method.

The task : Explain what the difference between the arithmetic method from algebraic?

Back)

4 method (algebraic)

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

By the condition of the task, there were 16 kg of cookies in two boxes. We get the equation:

h. +(h. -4)=16

h. + h. -4=16

2 h. -4=16

2 h. =16+4

2 h. =20

h. =20:2

The first box had 10 kg of cookies.

10-4 \u003d 6 (kg) - cookies were in the second box.

The solution is used algebraic method.

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Task number 510

Decide the task of arithmetic and algebraic methods.

From three plots of land collected 156 C potatoes. From the first and second sections of potatoes, they collected robusts, and from the third - on 12 c more than from each of the first two. How many potatoes collected from each site.

Algebraic method

(look)

Arithmetic method

(look)

output)

Arithmetic method

156 - 12 \u003d 144 (C) - potatoes would collect from three sites if the yields of all sites would be the same.

144: 3 \u003d 48 (C) - potatoes collected from the first and collected from the second sections.
48 + 12 \u003d 60 (C) - potatoes collected from the third site.

Answer

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Algebraic method

Let from the first plot collected h. C potatoes. Then from the second site collected too h. C potatoes, and from the third plot collected ( h. +12) C potatoes.

By condition from all three sites, 156 s potatoes were collected.

We get the equation:

x + x + (x +12) =156

x + x + x + 12 = 156

3 h. +12 = 156

3 h. = 156 – 12

3 h. = 144

h. = 144: 3

From the first and second sections, they collected 48 C potatoes.

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

Answer : From the first and second sections, they collected 48 C potatoes, and from the third site collected 60 C potatoes.

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