In this article, you will learn how to solve math problems if you don't know where to start.

Often, when solving problems, schoolchildren "go into a stupor" - there is a fog in their heads, thoughts have scattered somewhere, and it seems that it is no longer possible to collect them.

I want an example of solving a problem from open bank tasks to show which simple steps need to do to gather your thoughts and how to solve problems correctly.

How to solve problems. Task B13 (No. 26582)

The cyclist left constant speed from city A to city B, the distance between them is 98 km. The next day he went back at a speed of 7 km / h more than before. On the way he made a stop for 7 hours. As a result, he spent as much time on the way back as he did on the way from A to B. Find the speed of the cyclist on the way from A to B. Give the answer in km/h.

1. Read the problem carefully. Perhaps several times.

2. We determine what process the problem is about, and what formulas describe this process. We write out these formulas. AT this case this is a task for movement, and the formula that describes this process is S=vt.

3. We write out the dimension of each variable that is part of the equation:

• S - distance - km
• v - speed - km/h
• t - time - h

Knowing the dimension will help us in checking the resulting formulas.

4. We write out all the numbers that are found in the condition of the problem, we write what they mean and their dimension:

98 km - distance between cities,

7 km / h - as much as the speed of a cyclist way back more than the speed on the way from city A to city B,

7 hours - the time the cyclist stopped (this time he did not ride)

5. Read the problem question again.

6. We decide what value we will take for the unknown. It is convenient to take for the unknown the value that needs to be known in the problem. In this case, this is the speed of the cyclist on the way from A to B.

So: let the speed of the cyclist on the way from A to B be x. Then, since the speed of the cyclist on the way back is 7 km/h greater than the speed on the way from city A to city B, then it is equal to x+7.

7. We make an equation. To do this, we express the third value of the equation of motion (time) through the first two. Then:

• the time it took the cyclist to travel from A to B is 98/x,

• and on the road from B to A - 98 / (x + 7) + 7 - remember that on the way back the cyclist made a stop for 7 hours, that is, his travel time is the sum of the travel time and the parking time.

The equation is for time. Once again we read in the condition of the problem that it says about time: As a result, he spent as much time on the way back as on the way from A to B. That is, the time "there" is equal to the time "back". We equate the time "there" and the time "back" We get the equation:

98/x=98/(x+7)+7.

Once again, we check the dimensions of the quantities that are included in the equation - you need to make sure that, for example, do not add hours to kilometers.

8. We solve the equation. Now we need to focus on solving the equation. To do this, we determine what type this equation is. Since the unknown is in the denominator of the fractions, this is rational equation. To solve it, you need to move all the terms to the left and bring the fractions to common denominator. Note that the numbers 98 and 7 are multiples of 7.

To simplify the solution, we divide both sides of the equation by 7. We get the equation: 14/x=14/(x+7)+1

After that, we transfer all the terms to the left, reduce to a common denominator, and equate the numerator to zero.

We get in the numerator: 14(x+7)-14x-x(x+7)=0 like terms and solve the quadratic equation.

Its roots are -14 and 7.

The number -14 does not fit the condition of the problem: the speed must be positive.

Once again we read the question of the problem and correlate it with the value that we found: for the unknown we took the speed of the cyclist on the way from A to B, and we need to find the same value.

Answer: 7 km/h.

How to solve problems. Outcome

Note that we divided the entire path of solving the problem into small pieces, and in each section we focused precisely on thinking specific action. And only after this action was performed, did the next step.

When it is not clear what to do, you need to decide which small step you can do it right now, do it, and then think about the next one.

To learn how to solve typical logical tasks, simple and non-standard math problems, it is important to know the basic techniques and methods for solving them. After all, in many cases it is possible to solve the same problem and come to the correct answer in different ways.

Knowing and understanding the different solution methods will help you determine which method is best for each case, so that you can choose the quickest and easiest way to get an answer.

The "classic" logical tasks include text tasks, the purpose of which is to recognize objects or arrange them in a certain order in accordance with given conditions.

More complex and exciting types of tasks are tasks in which certain statements are true and others are false. The tasks of moving, shifting, weighing, pouring are the most bright examples wide range non-standard tasks to logic.

Basic methods for solving logical problems

• method of reasoning;
• using truth tables;
• block diagram method;
• means of logic algebra (propositional algebra);
• graphic (including "tree logical conditions”, the Euler circle method);
• method of mathematical billiards.

Let's take a closer look with examples of three popular ways to solve logical problems that we recommend using in elementary school (children 6-12 years old):

• method of sequential reasoning;
• a kind of method of reasoning - "from the end";
• tabular way.

Sequential reasoning method

The easiest way to solve simple problems is to reason sequentially using all known conditions. Conclusions from the statements that are the conditions of the problem gradually lead to the answer to the question posed.

On the table are Blue , Green , Brown and Orange

The third is the pencil with the most letters in its name. Blue the pencil lies between Brown and orange .

Lay out the pencils in the order described.

Decision:

We argue. We consistently use the conditions of the problem to formulate conclusions about the position on which each next pencil should lie.

• Most of the letters in the word "brown", so it lies third.
• It is known that a blue pencil lies between brown and orange. There is only one position to the right of brown, which means that it is possible to place blue between brown and another pencil only to the left of brown.
• The next conclusion is based on the previous one: the blue pencil is in the second position, and the orange one is in the first one.
• For the green pencil left last position- He is fourth.

End method

This way of solving is a kind of reasoning method and is great for problems in which we know the result of certain actions, and the question is to restore the original picture.

Grandma baked bagels for her three grandchildren and left them on the table. Kolya ran to have a bite to eat first. I counted all the bagels, took my share and ran away.
Anya came into the house later. She did not know that Kolya had already taken the bagels, counted them and, dividing them into three, took her share.
The third came Gena, who also divided the rest of the pastry into three and took his share.
There are 8 bagels left on the table.

Out of the eight remaining bagels, how many should each person eat so that they all eat equally?

Decision:

Let's start the discussion from the end.
Gena left 8 bagels for Anya and Kolya (4 for each). It turns out that he himself ate 4 bagels: 8 + 4 = 12.
Anya left 12 bagels for the brothers (each 6). So she herself ate 6 pieces: 12 + 6 = 18.
Kolya left 18 bagels for the guys. So he ate 9 himself: 18 + 9 = 27.

Grandmother put 27 bagels on the table, hoping that everyone would get 9 pieces. Since Kolya has already eaten his share, Anya must eat 3 and Gena must eat 5 bagels.

Solving Logic Problems Using Truth Tables

The essence of the method is to fix the conditions of the problem and the results of the reasoning in tables specially compiled for the problem. Depending on whether the statement is true or false, the corresponding cells of the table are filled with the signs "+" and "-" or "1" and "0".

Three athletes ( red , blue and green) played basketball.
When the ball was in the basket, the red one exclaimed: "The ball was scored by the blue one."
Blue objected: "Green scored the ball."
Zeleny said, "I didn't score."

Who scored the ball if only one of the three told a lie?

Decision:

First, a table is compiled: on the left, they write down all the statements that are contained in the condition, and on top - possible options response.

Then the table is filled in sequentially: true statements mark with a “+” sign, and false statements with a “-” sign.

Consider the first answer option (“the ball was thrown red"), analyze the statements written on the left, and fill in first column.
Based on our assumption (“the ball was thrown red"), the statement "the ball was thrown by blue" is a lie. We put in the cell "-".
The statement "ball scored green" is also a lie. We fill the cell with the sign "-".
The green statement "I didn't score" is true. We put in the cell "+".

Consider the second answer (assume that ball thrown green) and fill second column.
The statement "Blue has thrown the ball" is a lie. We put in the cell "-".
The statement "ball scored green « - truth. Fill in the cell with a "+" sign.
The green statement "I didn't score" is a lie. We put in the cell "-".

And finally, the third option: suppose that "the ball is thrown blue«.
Then the statement "ball thrown blue « - truth. We put in the cell "+".
The statement "the ball scored green" is a lie. We fill the cell with the sign "-". The green statement "I didn't score" is true. We put in the cell "+".

Since, according to the condition, only one of the three guys told a lie, in the completed table we select such an answer option, where it will be only one false statement (in the column one sign "-"). The third column fits.

So, the correct answer is that the ball was thrown by the blue one.

Flowchart method

The flowchart method is considered the best option for solving problems of weighing and pouring liquids. Alternative way solving this type of problem - the method of enumeration of options - is not always optimal, and it is rather difficult to call it systemic.

The procedure for solving problems using the flowchart method is as follows:

• graphically (flowchart) describe the sequence of operations;
• determine the order of their implementation;
• in the table we fix the current states.

More about this and other ways to solve logical problems with examples and a description of the solution, we tell in full course LogicLike on the development of logical thinking.

Guess the most collected especially for regular readers of our blog and students of LogicLike, solve logical problems online together with thousands of children and adults!

The average general education

UMK line G. K. Muravina. Algebra and beginnings mathematical analysis(10-11) (deep)

Line UMK Merzlyak. Algebra and the Beginnings of Analysis (10-11) (U)

Mathematics

Preparation for the exam in mathematics (profile level): tasks, solutions and explanations

We analyze tasks and solve examples with the teacher

Examination paper profile level lasts 3 hours 55 minutes (235 minutes).

Minimum Threshold- 27 points.

The examination paper consists of two parts, which differ in content, complexity and number of tasks.

The defining feature of each part of the work is the form of tasks:

• part 1 contains 8 tasks (tasks 1-8) with a short answer in the form of an integer or a final decimal fraction;
• part 2 contains 4 tasks (tasks 9-12) with a short answer in the form of an integer or a final decimal fraction and 7 tasks (tasks 13-19) with a detailed answer (full record of the decision with the rationale for the actions performed).

Panova Svetlana Anatolievna, mathematic teacher the highest category schools, 20 years of work experience:

“In order to receive a school certificate, a graduate must pass two compulsory exam in USE form, one of which is mathematics. In accordance with the Development Concept mathematics education in Russian Federation The USE in mathematics is divided into two levels: basic and specialized. Today we will consider options for the profile level.

Task number 1- checks with participants USE skill apply the skills acquired in the course of 5-9 grades in elementary mathematics, in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals be able to convert one unit of measurement to another.

Example 1 An expense meter was installed in the apartment where Petr lives cold water(counter). On the first of May, the meter showed an consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water for May, if the price of 1 cu. m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.

Decision:

1) Find the amount of water spent per month:

177 - 172 = 5 (cu m)

2) Find how much money will be paid for the spent water:

34.17 5 = 170.85 (rub)

Answer: 170,85.

Task number 2- is one of the simplest tasks of the exam. The majority of graduates successfully cope with it, which indicates the possession of the definition of the concept of function. Task type No. 2 according to the requirements codifier is a task for using acquired knowledge and skills in practical activities and Everyday life. Task number 2 consists of a description using the functions of various real dependencies between quantities and interpretation of their graphs. Task number 2 tests the ability to extract information presented in tables, diagrams, graphs. Graduates need to be able to determine the value of a function by the value of the argument when various ways defining a function and describing the behavior and properties of the function according to its graph. It is also necessary to be able to find the maximum or smallest value and build graphs of the studied functions. The mistakes made are of a random nature in reading the conditions of the problem, reading the diagram.

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Example 2 The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the purchased shares, and on April 13 he sold all the remaining ones. How much did the businessman lose as a result of these operations?

Decision:

2) 1000 3/4 = 750 (shares) - make up 3/4 of all purchased shares.

6) 247500 + 77500 = 325000 (rubles) - the businessman received after the sale of 1000 shares.

7) 340,000 - 325,000 = 15,000 (rubles) - the businessman lost as a result of all operations.

Answer: 15000.

Task number 3- is a task basic level the first part, tests the ability to perform actions with geometric shapes on the content of the course "Planimetry". In task 3, the ability to calculate the area of ​​\u200b\u200ba figure on checkered paper, ability to calculate degree measures corners, calculate perimeters, etc.

Example 3 Find the area of ​​a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.

Decision: To calculate the area of ​​this figure, you can use the Peak formula:

To calculate the area given rectangle Let's use Pick's formula:

S= B +	G
	2

where V = 10, G = 6, therefore

S = 18 +	6
	2

Answer: 20.

See also: Unified State Examination in Physics: solving vibration problems

Task number 4- the task of the course "Probability Theory and Statistics". The ability to calculate the probability of an event in the simplest situation is tested.

Example 4 There are 5 red and 1 blue dots on the circle. Determine which polygons are larger: those with all red vertices, or those with one of the blue vertices. In your answer, indicate how many more of one than the other.

Decision: 1) We use the formula for the number of combinations from n elements by k:

all of whose vertices are red.

3) One pentagon with all red vertices.

4) 10 + 5 + 1 = 16 polygons with all red vertices.

whose vertices are red or with one blue vertex.

whose vertices are red or with one blue vertex.

8) One hexagon whose vertices are red with one blue vertex.

9) 20 + 15 + 6 + 1 = 42 polygons that have all red vertices or one blue vertex.

10) 42 - 16 = 26 polygons that use the blue dot.

11) 26 - 16 = 10 polygons - how many polygons, in which one of the vertices is a blue dot, are more than polygons, in which all vertices are only red.

Answer: 10.

Task number 5- the basic level of the first part tests the ability to solve the simplest equations (irrational, exponential, trigonometric, logarithmic).

Example 5 Solve Equation 2 3 + x= 0.4 5 3 + x .

Decision. Divide both sides of this equation by 5 3 + X≠ 0, we get

2 3 + x	= 0.4 or		2		3 + X	=	2	,
5 3 + X			5				5

whence it follows that 3 + x = 1, x = –2.

Answer: –2.

Task number 6 by planimetry to find geometric quantities(lengths, angles, areas), modeling real situations in the language of geometry. Study of built models using geometric concepts and theorems. The source of difficulties is usually ignorance or misapplication necessary theorems of planimetry.

Area of ​​a triangle ABC equals 129. DE- middle line, side parallel AB. Find the area of ​​the trapezoid ABED.

Decision. Triangle CDE similar to a triangle CAB at two corners, since the corner at the vertex C general, angle CDE equal to the angle CAB as corresponding angles at DE || AB secant AC. As DE is the midline of the triangle by condition, then by property middle line | DE = (1/2)AB. So the similarity coefficient is 0.5. squares similar figures are related as the square of the similarity coefficient, so

Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.

Task number 7- checks the application of the derivative to the study of the function. For successful implementation a meaningful, non-formal possession of the concept of a derivative is necessary.

Example 7 To the graph of the function y = f(x) at the point with the abscissa x 0 a tangent is drawn, which is perpendicular to the straight line passing through the points (4; 3) and (3; -1) of this graph. Find f′( x 0).

Decision. 1) We use the equation of a straight line passing through two given points and find the equation of a straight line passing through the points (4; 3) and (3; -1).

(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)

(y – 3)(3 – 4) = (x – 4)(–1 – 3)

(y – 3)(–1) = (x – 4)(–4)

–y + 3 = –4x+ 16| · (-one)

y – 3 = 4x – 16

y = 4x– 13, where k 1 = 4.

2) Find the slope of the tangent k 2 which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:

3) Slope tangent - the derivative of the function at the point of contact. Means, f′( x 0) = k 2 = –0,25.

Answer: –0,25.

Task number 8- checks the knowledge of elementary stereometry among the participants of the exam, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric shapes, coordinates and vectors, etc.

The volume of a cube circumscribed around a sphere is 216. Find the radius of the sphere.

Decision. 1) V cube = a 3 (where a is the length of the edge of the cube), so

a 3 = 216

a = 3 √216

2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.

Task number 9- requires the graduate to transform and simplify algebraic expressions. Task number 9 advanced level Difficulty with short answers. Tasks from the section "Calculations and transformations" in the USE are divided into several types:

numeric conversions rational expressions;

transformations of algebraic expressions and fractions;

numeric/alphabetic conversions irrational expressions;

actions with degrees;

transformation logarithmic expressions;

1. conversion of numeric/letter trigonometric expressions.

Example 9 Calculate tgα if it is known that cos2α = 0.6 and

3π	< α < π.
4

Decision. 1) Let's use the formula double argument: cos2α = 2 cos 2 α – 1 and find

tan 2 α =	1	– 1 =	1	– 1 =	10	– 1 =	5	– 1 = 1	1	– 1 =	1	= 0,25.
	cos 2 α		0,8		8		4		4		4

Hence, tan 2 α = ± 0.5.

3) By condition

3π	< α < π,
4

hence α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.

Answer: –0,5.

#ADVERTISING_INSERT# Task number 10- checks students' ability to use the acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all necessary formulas and values ​​are given in the condition. The problems are reduced to solving a linear or quadratic equation, either linear or square inequality. Therefore, it is necessary to be able to solve such equations and inequalities, and determine the answer. The answer must be in the form of a whole number or a final decimal fraction.

Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Decision. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).

mv 2 sin 2 α ≥ 50

2 10 2 sin 2 α ≥ 50

200 sin2α ≥ 50

Since α ∈ (0°; 90°), we will only solve

We represent the solution of the inequality graphically:

Since by assumption α ∈ (0°; 90°), it means that 30° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.

Task number 11- is typical, but it turns out to be difficult for students. The main source of difficulties is the construction of a mathematical model (drawing up an equation). Task number 11 tests the ability to solve word problems.

Example 11. On the spring break 11-grader Vasya had to solve 560 training problems to prepare for the exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2 on the last day of vacation.

Decision: Denote a 1 = 5 - the number of tasks that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 - the number of days from March 18 to April 2 inclusive, S 16 = 560 – total tasks, a 16 - the number of tasks that Vasya solved on April 2. Knowing that every day Vasya solved the same number of tasks more than the previous day, then you can use the formulas for finding the sum arithmetic progression:

560 = (5 + a 16) 8,

5 + a 16 = 560: 8,

5 + a 16 = 70,

a 16 = 70 – 5

a 16 = 65.

Answer: 65.

Task number 12- check students' ability to perform actions with functions, be able to apply the derivative to the study of the function.

Find the maximum point of a function y= 10ln( x + 9) – 10x + 1.

Decision: 1) Find the domain of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).

2) Find the derivative of the function:

4) The found point belongs to the interval (–9; ∞). We define the signs of the derivative of the function and depict the behavior of the function in the figure:

The desired maximum point x = –8.

Download for free the work program in mathematics to the line of UMK G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free algebra manuals

Task number 13- an increased level of complexity with a detailed answer, which tests the ability to solve equations, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

a) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0

b) Find all the roots of this equation, belonging to the segment.

Decision: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,

	log3(2cos x) =	2	⇔		2cos x = 9	⇔		cos x =	4,5	⇔ because |cos x| ≤ 1,
	log3(2cos x) =	1			2cos x = √3			cos x =	√3
		2							2

then cos x =	√3
	2

	x =	π	+ 2π k
		6
	x = –	π	+ 2π k, k ∈ Z
		6

b) Find the roots lying on the segment .

It can be seen from the figure that given segment belong to the roots

11π	and	13π	.
6		6

Answer: a)	π	+ 2π k; –	π	+ 2π k, k ∈ Z; b)	11π	;	13π	.
	6		6		6		6

Task number 14- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

The diameter of the circle of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its bases along chords of length 12 and 16. The distance between the chords is 2√197.

a) Prove that the centers of the bases of the cylinder lie on the same side of this plane.

b) Find the angle between this plane and the plane of the base of the cylinder.

Decision: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections on a plane parallel to the bases of the cylinders is either 8 + 6 = 14, or 8 − 6 = 2.

Then the distance between chords is either

= = √980 = = 2√245

= = √788 = = 2√197.

According to the condition, the second case was realized, in which the projections of the chords lie on one side of the axis of the cylinder. So the axis does not intersect given plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.

b) Let's denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 the perpendicular bisector to this chord (it has a length of 8, as already noted) and from the center of the other base to another chord. They lie in the same plane β perpendicular to these chords. Let's call the midpoint of the smaller chord B, greater than A, and the projection of A onto the second base H (H ∈ β). Then AB,AH ∈ β and, therefore, AB,AH are perpendicular to the chord, that is, the line of intersection of the base with the given plane.

So the required angle is

∠ABH = arctan	AH	= arctg	28	= arctg14.
	BH		8 – 6

Task number 15- an increased level of complexity with a detailed answer, checks the ability to solve inequalities, the most successfully solved among tasks with a detailed answer of an increased level of complexity.

Example 15 Solve the inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .

Decision: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:

1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values ​​are included in the solution.

2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). In this case, this inequality can be rewritten in the form ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 -1 or x≤ -0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].

3) Finally, consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by a positive expression 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the area, we have x ∈ (0; 1].

Combining the obtained solutions, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.

Answer: (–1; –0.5] ∪ ∪ {3}.

Task number 16- advanced level refers to the tasks of the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two items. In the first paragraph, the task must be proved, and in the second paragraph, it must be calculated.

AT isosceles triangle ABC with an angle of 120° at vertex A, a bisector BD is drawn. AT triangle ABC rectangle DEFH is inscribed so that side FH lies on segment BC and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of ​​the rectangle DEFH if AB = 4.

Decision: a)

1) ΔBEF - rectangular, EF⊥BC, ∠B = (180° - 120°) : 2 = 30°, then EF = BE due to the property of the leg opposite the angle of 30°.

2) Let EF = DH = x, then BE = 2 x, BF = x√3 by the Pythagorean theorem.

3) Since Δ ABC isosceles, so ∠B = ∠C = 30˚.

BD is the bisector of ∠B, so ∠ABD = ∠DBC = 15˚.

4) Consider ΔDBH - rectangular, because DH⊥BC.

2x	=	4 – 2x
2x(√3 + 1)		4

1	=	2 – x
√3 + 1		2

√3 – 1 = 2 – x

x = 3 – √3

EF = 3 - √3

2) S DEFH = ED EF = (3 - √3 ) 2(3 - √3 )

S DEFH = 24 - 12√3.

Answer: 24 – 12√3.

Task number 17- a task with a detailed answer, this task tests the application of knowledge and skills in practical activities and everyday life, the ability to build and explore mathematical models. This task - text task with economic content.

Example 17. The deposit in the amount of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the depositor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, at which the bank will add less than 17 million rubles to the deposit in four years.

Decision: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X) 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year, the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality

(29,282 + 2,31x) – 20 – 2x < 17

29,282 + 2,31x – 20 – 2x < 17

0,31x < 17 + 20 – 29,282

0,31x < 7,718

x <	7718
	310

x <	3859
	155

x < 24	139
	155

The largest integer solution to this inequality is the number 24.

Answer: 24.

Task number 18- a task of an increased level of complexity with a detailed answer. This task is for competitive selection universities with higher requirements mathematical training applicants. Exercise high level complexity is not a task for applying one solution method, but for a combination of different methods. For the successful completion of task 18, in addition to solid mathematical knowledge, a high level of mathematical culture is also required.

At what a system of inequalities

	x 2 + y 2 ≤ 2ay – a 2 + 1
	y + a ≤ |x| – a

has exactly two solutions?

Decision: This system can be rewritten as

	x 2 + (y– a) 2 ≤ 1
	y ≤ |x| – a

If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 centered at the point (0, a). The set of solutions of the second inequality is the part of the plane that lies under the graph of the function y = | x| – a, and the latter is the graph of the function
y = | x| , shifted down by a. The solution of this system is the intersection of the solution sets of each of the inequalities.

Therefore, two solutions this system will have only in the case shown in Fig. one.

The points of contact between the circle and the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So the triangle PQR- rectangular isosceles. Dot Q has coordinates (0, a), and the point R– coordinates (0, – a). In addition, cuts PR and PQ are equal to the circle radius equal to 1. Hence,

QR= 2a = √2, a =	√2	.
	2

Answer: a =	√2	.
	2

Task number 19- a task of an increased level of complexity with a detailed answer. This task is intended for competitive selection to universities with increased requirements for the mathematical preparation of applicants. A task of a high level of complexity is not a task for applying one solution method, but for a combination of different methods. To successfully complete task 19, you must be able to search for a solution by choosing different approaches from among the known, modifying the studied methods.

Let be sn sum P members of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.

a) Give the formula P th member of this progression.

b) Find the smallest modulo sum S n.

c) Find the smallest P, at which S n will be the square of an integer.

Decision: a) Obviously, a n = S n – S n- one . Using this formula, we get:

S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,

S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27

means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.

B) because S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Her graph can be seen in the figure.

It is obvious that the smallest value is reached at the integer points located closest to the zeros of the function. Obviously these are points. X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 144 – 25 12| = 12, S(13) = |S 13 | = |2 169 – 25 13| = 13, then the smallest value is 12.

c) It follows from the previous paragraph that sn positive since n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case when given expression is a perfect square, is realized when n = 2n- 25, that is, with P= 25.

It remains to check the values ​​​​from 13 to 25:

S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13 S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.

It turns out that for smaller values P full square is not achieved.

Answer: a) a n = 4n- 27; b) 12; c) 25.

________________

*Since May 2017, the joint publishing group "DROFA-VENTANA" has been part of the corporation " Russian textbook". The corporation also included Astrel publishing house and digital educational platform"lecta". CEO appointed Alexander Brychkin, graduate Financial Academy under the Government of the Russian Federation, candidate economic sciences, supervisor innovative projects DROFA publishing house in the field of digital education ( electronic forms textbooks, "Russian Electronic School", digital educational platform LECTA). Prior to joining the DROFA publishing house, he held the position of Vice President for Strategic Development and Investments of the EKSMO-AST publishing holding. Today, the Russian Textbook Publishing Corporation has the largest portfolio of textbooks included in the federal list- 485 titles (approximately 40%, excluding textbooks for remedial school). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed to develop the country's production potential. The corporation's portfolio includes textbooks and study guides for elementary school awarded the Presidential Prize in Education. These are textbooks and manuals on subject areas that are necessary for the development of the scientific, technical and industrial potential of Russia.

The solution to the problem usually comes down to logical reasoning and calculations to find the value of some quantity. For example, find the speed, time, distance, mass of an object or the amount of something.

This problem can be solved using an equation. To do this, the desired value is denoted through a variable, then, by logical reasoning, they compose and solve an equation. Having solved the equation, they check whether the solution of the equation satisfies the conditions of the problem.

Lesson content

Writing Expressions Containing the Unknown

The solution of the problem is accompanied by the compilation of an equation for this problem. On the initial stage studying problems, it is desirable to learn how to compose literal expressions describing one or another life situation. This stage is not difficult and can be studied in the process of solving the problem itself.

Consider several situations that can be written using a mathematical expression.

Task 1. Father's age x years. Mom is two years younger. Son younger than father 3 times. Record the age of each using expressions.

Decision:

Task 2. Father's age x years, mother is 2 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. Record the age of each using expressions.

Decision:

Task 3. Father's age x years, mother is 3 years younger than father. The son is 3 times younger than the father, the daughter is 3 times younger than the mother. How old is each general age father, mother, son and daughter is 92 years old?

Decision:

In this problem, in addition to writing expressions, it is necessary to calculate the age of each family member.

First, we write down the age of each family member using expressions. Per variable x let's take the age of the father, and then using this variable we will compose the remaining expressions:

Now let's determine the age of each family member. To do this, we need to write and solve an equation. We have all the components of the equation ready. It remains only to collect them together.

The total age of 92 years was obtained by adding the ages of dad, mom, son and daughter:

For each age, we have mathematical expression. These expressions will be the components of our equation. Let's assemble our equation according to this scheme and the table that was given above. That is, the words dad, mom, son, daughter will be replaced by the expression corresponding to them in the table:

Expression for mother's age x − 3 for clarity, it was taken in brackets.

Now let's solve the resulting equation. To begin with, you can open the brackets where possible:

To free the equation from fractions, multiply both sides by 3

We solve the resulting equation using the known identical transformations:

We found the value of the variable x. This variable was responsible for the age of the father. So the age of the father is 36 years.

Knowing the age of the father, you can calculate the ages of the rest of the family. To do this, you need to substitute the value of the variable x in those expressions that are responsible for the age of a particular family member.

In the problem, it was said that the mother is 3 years younger than the father. We denoted her age through the expression x−3. Variable value x is now known, and in order to calculate the age of the mother, it is necessary in the expression x − 3 instead of x substitute the found value 36

x - 3 \u003d 36 - 3 \u003d 33 years old mom.

Similarly, the age of the remaining family members is determined:

Examination:

Task 4. A kilogram of apples is worth x rubles. Write down an expression that calculates how many kilograms of apples you can buy for 300 rubles.

Decision

If a kilogram of apples costs x rubles, then for 300 rubles you can buy a kilogram of apples.

Example. A kilogram of apples costs 50 rubles. Then for 300 rubles you can buy, that is, 6 kilograms of apples.

Task 5. On the x rubles, 5 kg of apples were bought. Write down an expression that calculates how many rubles one kilogram of apples costs.

Decision

If for 5 kg of apples was paid x rubles, then one kilogram will cost rubles

Example. For 300 rubles, 5 kg of apples were bought. Then one kilogram of apples will cost, that is, 60 rubles.

Task 6. Tom, John and Leo went to the cafeteria during recess and bought a sandwich and a mug of coffee. The sandwich is worth x rubles, and a mug of coffee - 15 rubles. Determine the cost of a sandwich if it is known that 120 rubles were paid for everything?

Decision

Of course, this problem is as simple as three pennies and can be solved without resorting to an equation. To do this, subtract the cost of three cups of coffee (15 × 3) from 120 rubles, and divide the result by 3

But our goal is to write an equation for the problem and solve this equation. So the cost of a sandwich x rubles. Bought only three. So, having tripled the cost, we get an expression describing how many rubles were paid for three sandwiches

3x - cost of three sandwiches

And the cost of three cups of coffee can be written as 15 × 3. 15 is the cost of one mug of coffee, and 3 is a multiplier (Tom, John and Leo) that triples this cost.

According to the condition of the problem, 120 rubles were paid for everything. We already have exemplary scheme, What do we have to do:

We already have expressions describing the cost of three sandwiches and three cups of coffee. These are expressions 3 x and 15×3. Using the scheme, we will write an equation and solve it:

So, the cost of one sandwich is 25 rubles.

The problem is solved correctly only if the equation for it is compiled correctly. Unlike ordinary equations, by which we learn to find roots, equations for solving problems have their own specific application. Each component of such an equation can be described in verbal form. When compiling an equation, it is imperative to understand why we include one or another component in its composition and why it is needed.

It is also necessary to remember that the equation is an equality, after solving which the left side will have to be equal to the right side. The resulting equation should not contradict this idea.

Imagine that the equation is a balance with two bowls and a screen showing the state of the balance.

AT this moment the screen shows an equal sign. It is clear why the left bowl is equal to the right bowl - there is nothing on the bowls. We write the state of the scales and the absence of something on the bowls using the following equality:

0 = 0

Let's put a watermelon on the left scale:

The left bowl outweighed the right bowl and the screen sounded the alarm, showing the not equal sign (≠). This sign indicates that the left bowl is not equal to the right bowl.

Now let's try to solve the problem. Let it be required to find out how much the watermelon weighs, which lies on the left bowl. But how do you know? After all, our scales are designed only to check whether the left bowl is equal to the right one.

Equations come to the rescue. Recall that by definition the equation is equality A that contains the variable whose value you want to find. The scales in this case play the role of this very equation, and the mass of the watermelon is a variable whose value must be found. Our goal is to get this equation right. Understand, align the scales so that you can calculate the mass of the watermelon.

To level the scales, you can put some weight on the right bowl. heavy object. For example, let's put a weight of 7 kg there.

Now, on the contrary, the right bowl outweighed the left. The screen still shows that the bowls are not equal.

Let's try to put a weight of 4 kg on the left bowl

Now the scales have leveled out. The figure shows that the left bowl is at the level of the right bowl. And the screen shows an equal sign. This sign indicates that the left bowl is equal to the right bowl.

Thus, we have obtained an equation - an equality containing an unknown. The left pan is the left side of the equation, consisting of the 4 components and the variable x(masses of watermelon), and the right bowl is right part equation, consisting of component 7.

Well, it is not difficult to guess that the root of the equation 4 + x\u003d 7 is 3. So the mass of the watermelon is 3 kg.

The same is true for other tasks. To find some unknown value, add to the left or right side of the equation various elements: terms, factors, expressions. In school problems, these elements are already given. It remains only to correctly structure them and build an equation. We are in this example engaged in selection, trying weights of different masses in order to calculate the mass of a watermelon.

Naturally, the data that is given in the problem must first be brought to a form in which they can be included in the equation. Therefore, as they say "Whether you like it or not, you have to think".

Consider the following problem. Father's age equal to age son and daughter together. Son halved older than daughter and twenty years younger than his father. How old is each?

The daughter's age can be expressed as x. If the son is twice as old as the daughter, then his age will be indicated as 2 x. The condition of the problem says that together the age of the daughter and son is equal to the age of the father. So the father's age will be denoted by the sum x + 2x

You can add like terms in an expression. Then the father's age will be denoted as 3 x

Now let's make an equation. We need to get an equality in which we can find the unknown x. Let's use weights. On the left bowl we put the age of the father (3 x), and on the right bowl the age of the son (2 x)

It is clear why the left bowl outweighed the right one and why the screen shows the sign (≠) . After all, it is logical that the age of the father is greater than the age of the son.

But we need to balance the scales so that we can calculate the unknown x. To do this, you need to add some number to the right bowl. What number is indicated in the problem. The condition stated that the son was 20 years younger than the father. So 20 years is the same number that needs to be put on the scales.

The scales will even out if we add these 20 years to the right side of the scales. In other words, let's raise the son to the age of the father

Now the scales have leveled out. It turned out the equation , which is easily solved:

x we marked the age of the daughter. Now we have found the value of this variable. Daughter 20 years old.

And finally, we calculate the age of the father. The assignment said that he is equal to the sum the ages of the son and daughter, i.e. (20 + 40) years.

Let's return to the middle of the task and pay attention to one point. When we put the age of the father and the age of the son on the scale, the left bowl outweighed the right

But we solved this problem by adding another 20 years to the right bowl. As a result, the scales leveled out and we got the equality

But it was possible not to add these 20 years to the right bowl, but subtract them from the left. We would get equality in this case

This time the equation is . The root of the equation is still 20

That is, the equations and are equivalent. And we remember that equivalent equations roots match. If you look closely at these two equations, you can see that the second equation is obtained by transferring the number 20 from the right side to the left side with opposite sign. And this action, as indicated in the previous lesson, does not change the roots of the equation.

You also need to pay attention to the fact that at the beginning of solving the problem, the ages of each family member could be denoted through other expressions.

Let's say the son's age is denoted by x and since he is two older than his daughter, then designate the age of the daughter through (understand to make her younger than son twice). And the age of the father, since it is the sum of the ages of the son and daughter, is denoted through the expression . And finally, to build a logically correct equation, you need to add the number 20 to the age of the son, because the father is twenty years older. The result is a completely different equation. . Let's solve this equation

As you can see, the answers to the problem have not changed. My son is still 40 years old. The daughters are still years old, and the father is 40 + 20 years old.

In other words, the problem can be solved various methods. Therefore, one should not despair that it is not possible to solve this or that problem. But it must be borne in mind that there is a simple ways problem solving. You can get to the city center various routes, but there is always the most convenient, fastest and safest route.

Examples of problem solving

Task 1. There are 30 notebooks in two packs. If 2 notebooks were transferred from the first bundle to the second, then there would be twice as many notebooks in the first bundle as in the second. How many notebooks were in each pack?

Decision

Denote by x the number of notebooks that were in the first pack. If there were 30 notebooks in total, and the variable x this is the number of notebooks from the first pack, then the number of notebooks in the second pack will be denoted by the expression 30 − x. That is, from the total number of notebooks, we subtract the number of notebooks from the first pack and thereby obtain the number of notebooks from the second pack.

and add these two notebooks to the second pack

Let's try to make an equation from the existing expressions. We put both packs of notebooks on the scales

The left bowl is heavier than the right. This is because the condition of the problem says that after two notebooks were taken from the first bundle and placed in the second, the number of notebooks in the first bundle became twice as large as in the second.

To equalize the scales and get the equation, double the right side. To do this, multiply it by 2

It turns out an equation. We will decide given equation:

We denoted the first pack by the variable x. Now we have found its meaning. Variable x equal to 22. So there were 22 notebooks in the first pack.

And we denoted the second pack through the expression 30 − x and since the value of the variable x Now we know, we can calculate the number of notebooks in the second pack. It is equal to 30 − 22, that is, 8 pieces.

Task 2. Two people were peeling potatoes. One peeled two potatoes a minute, and the other three potatoes. Together they cleared 400 pieces. How long did each work if the second worked 25 minutes more than the first?

Decision

Denote by x time of the first person. Since the second person worked 25 minutes more than the first, his time will be denoted by the expression

The first worker peeled 2 potatoes per minute, and since he worked x minutes, then in total he cleared 2 x potatoes.

The second person peeled three potatoes per minute, and since he worked for minutes, he peeled potatoes in total.

Together they peeled 400 potatoes

From the available components, we will compose and solve the equation. On the left side of the equation there will be potatoes peeled by each person, and on the right side of their sum:

At the beginning of the solution of this problem through the variable x we marked the time of work of the first person. Now we have found the value of this variable. The first person worked 65 minutes.

And the second person worked for minutes, and since the value of the variable x now it is known, then you can calculate the time of the second person - it is equal to 65 + 25, that is, 90 minutes.

Problem from Andrey Petrovich Kiselev's Algebra Textbook. A mixture of 32 kg was made from the varieties of tea. A kilogram of the first grade costs 8 rubles, and of the second grade 6 rubles. 50 kop. How many kilograms are taken of both varieties, if a kilogram of the mixture costs (without profit or loss) 7 rubles. 10 kopecks?

Decision

Denote by x a lot of tea of ​​the first grade. Then the mass of tea of ​​the second grade will be denoted through the expression 32 − x

A kilogram of tea of ​​the first grade costs 8 rubles. If these eight rubles are multiplied by the number of kilograms of tea of ​​the first grade, then it will be possible to find out how much the rubles cost x kg of tea of ​​the first grade.

A kilogram of second-class tea costs 6 rubles. 50 kop. If these 6 rubles. 50 kop. multiply by 32 − x, then you can find out how many rubles cost 32 − x kg of tea of ​​the second grade.

The condition says that a kilogram of the mixture costs 7 rubles. 10 kop. In total, 32 kg of the mixture was prepared. Multiply 7 rubles. 10 kop. at 32 we can find out how much 32 kg of the mixture costs.

The expressions from which we will compose the equation now take the following form:

Let's try to make an equation from the existing expressions. Let's put the cost of mixtures of teas of the first and second grade on the left pan of the scales, and put the cost of 32 kg of the mixture on the right pan, that is total cost mixture, which includes both varieties of tea:

At the beginning of the solution of this problem through the variable x we designated the mass of tea of ​​the first grade. Now we have found the value of this variable. Variable x equals 12.8. This means that 12.8 kg of tea of ​​the first grade was taken to prepare the mixture.

And through expression 32 − x we denoted the mass of tea of ​​the second grade, and since the value of the change x now known, we can calculate the mass of tea of ​​the second grade. It is equal to 32 − 12.8, that is, 19.2. This means that 19.2 kg of second grade tea was taken to prepare the mixture.

Task 3. A cyclist traveled a distance at a speed of 8 km/h. He had to return by another road, which was 3 km longer than the first, and although returning, he was traveling at a speed of 9 km / h, he used time for more than minutes. How long were the roads?

Decision

Some tasks may cover topics that the person may not have studied. This task belongs to this range of tasks. It deals with the concepts of distance, speed and time. Accordingly, in order to solve such a problem, you need to have an idea about the things that are said in the problem. In our case, we need to know what is the distance, speed and time.

The task is to find the distances of two roads. We must write an equation that will allow us to calculate these distances.

Consider the relationship between distance, speed and time. Each of these quantities can be described using a literal equation:

We will use the right side of one of these equations to draw up our equation. To find out which one, you need to return to the text of the task and look for what you can catch on

You can catch on to the moment where the cyclist on the way back used time for more than a minute. This hint tells us that we can use the equation , namely its right side. This will allow us to write an equation that contains the variable S .

So let's denote the length of the first road as S. The cyclist traveled this path at a speed of 8 km/h. The time for which he covered this path will be denoted by the expression, since time is the ratio of the distance traveled to the speed

The way back for the cyclist was 3 km longer. Therefore, its distance will be denoted by the expression S+ 3 . A cyclist traveled this road at a speed of 9 km/h. So the time for which he overcame this path will be denoted by the expression .

Now let's make an equation from the existing expressions

The right bowl is heavier than the left. This is because the problem says that the cyclist spent more time on the way back.

To equalize the scales, add these same minutes to the left side. But first, let's convert minutes to hours, since in the problem the speed is measured in kilometers per hour, and not in meters per minute.

To convert minutes to hours, you need to divide them by 60

Minutes make hours. Add these hours to the left side of the equation:

It turns out the equation . Let's solve this equation. To get rid of fractions, both parts of the part can be multiplied by 72. Further, using the known identical transformations, find the value variable S

Through a variable S we marked the distance of the first road. Now we have found the value of this variable. Variable S is 15. So the distance of the first road is 15 km.

And we denoted the distance of the second road through the expression S+ 3 , and since the value of the variable S Now we know, we can calculate the distance of the second road. This distance is equal to the sum of 15 + 3, that is, 18 km.

Task 4. Two cars are driving down the highway at the same speed. If the first one increases the speed by 10 km / h, and the second one reduces the speed by 10 km / h, then the first one will cover the same distance in 2 hours as the second one in 3 hours. At what speed do cars go?

Decision

Denote by v the speed of each car. Further in the problem, hints are given: increase the speed of the first car by 10 km/h, and decrease the speed of the second car by 10 km/h. Let's use this hint

It is further stated that at such speeds (increased and decreased by 10 km / h), the first car will cover the same distance in 2 hours as the second in 3 hours. Phrase "as many" can be understood as "the distance traveled by the first car will be equals distance traveled by the second car.

The distance, as we remember, is determined by the formula. We are interested in the right side of this literal equation - it will allow us to write an equation containing a variable v .

So, at speed v + 10 km/h the first car will pass 2(v+10) km, and the second will pass 3(v − 10) km. Under this condition, the cars will cover the same distances, therefore, to obtain an equation, it is enough to connect these two expressions with an equal sign. Then we get the equation. Let's solve it:

In the condition of the problem, it was said that the cars go at the same speed. We denoted this speed by the variable v. Now we have found the value of this variable. Variable v equals 50. So the speed of both cars was 50 km/h.

Task 5. In 9 hours downstream the ship travels the same distance as in 11 hours upstream. Find the speed of the boat if the speed of the river is 2 km/h.

Decision

Denote by v own speed of the ship. The speed of the river flow is 2 km/h. In the course of the river, the speed of the ship will be v + 2 km/h, and against the current - (v − 2) km/h.

The condition of the problem states that in 9 hours the ship travels the same distance along the river as in 11 hours against the current. Phrase "same way" can be understood as the distance traveled by the boat along the river in 9 hours, equals distance traveled by the ship against the current of the river in 11 hours. That is, the distances will be the same.

The distance is determined by the formula . Let's use the right side of this literal equation to write our own equation.

So, in 9 hours, the ship will pass along the river 9(v + 2) km, and in 11 hours upstream - 11(v − 2) km. Since both expressions describe the same distance, we equate the first expression to the second. As a result, we get the equation . Let's solve it:

Means own speed the ship is 20 km / h.

When solving problems good habit is to determine in advance on which solution is sought for it.

Suppose that the task was to find the time it takes for a pedestrian to overcome specified path. We denoted the time through the variable t, then we made an equation containing this variable and found its value.

From practice, we know that the time of movement of an object can take both integer values ​​and fractional values, for example, 2 hours, 1.5 hours, 0.5 hours. Then we can say that the solution to this problem is sought on the set rational numbers Q, since each of the values ​​2 h, 1.5 h, 0.5 h can be represented as a fraction.

Therefore, after unknown quantity denoted by a variable, it is useful to indicate to which set this value belongs. In our example, the time t belongs to the set of rational numbers Q

t ∈ Q

You can also introduce a constraint on the variable t, indicating that it can only accept positive values. Indeed, if the object spent on the path certain time, then this time cannot be negative. Therefore, next to the expression t∈ Q specify that its value must be greater than zero:

t∈ R, t > 0

If we solve the equation, we get negative meaning for a variable t, then it will be possible to conclude that the problem was solved incorrectly, since this solution will not satisfy the condition t∈ Q , t> 0 .

Another example. If we were solving a problem in which it was required to find the number of people to perform a particular job, then we would denote this number through a variable x. In such a problem, the solution would be sought on the set natural numbers

x ∈ N

Indeed, the number of people is an integer, such as 2 people, 3 people, 5 people. But not 1.5 (one whole person and half a person) or 2.3 (two whole people and another three tenths of a person).

Here one could indicate that the number of people must be greater than zero, but the numbers included in the set of natural numbers N are themselves positive and greater than zero. This set does not negative numbers and the number 0. Therefore, the expression x > 0 can be omitted.

Task 6. To repair the school, a team arrived in which there were 2.5 times more painters than carpenters. Soon the foreman included four more painters in the team, and transferred two carpenters to another object. As a result, there were 4 times more painters in the brigade than carpenters. How many painters and how many carpenters were in the brigade initially

Decision

Denote by x carpenters who arrived for repairs initially.

The number of carpenters is an integer greater than zero. Therefore, we point out that x belongs to the set of natural numbers

x∈N

There were 2.5 times more painters than carpenters. Therefore, the number of painters will be denoted as 2.5x.

And the number of painters will increase by 4

Now the number of carpenters and painters will be denoted by the following expressions:

Let's try to make an equation from the existing expressions:

The right bowl is larger, because after adding four more painters to the team, and moving two carpenters to another object, the number of painters in the team turned out to be 4 times more than carpenters. To equalize the scales, you need to increase the left bowl by 4 times:

Got an equation. Let's solve it:

Through a variable x the initial number of carpenters was designated. Now we have found the value of this variable. Variable x equals 8. So 8 carpenters were in the brigade initially.

And the number of painters was indicated through the expression 2.5 x and since the value of the variable x now it is known, then you can calculate the number of painters - it is equal to 2.5 × 8, that is, 20.

We return to the beginning of the task and make sure that the condition is met x∈ N. Variable x equals 8, and the elements of the set of natural numbers N these are all numbers starting with 1, 2, 3 and so on ad infinitum. The same set includes the number 8, which we found.

8 ∈N

The same can be said about the number of painters. The number 20 belongs to the set of natural numbers:

20 ∈N

To understand the essence of the problem and correct compilation equation, it is not necessary to use the scale model with bowls. You can use other models: segments, tables, diagrams. You can come up with your own model that would describe the essence of the problem well.

Task 9. 30% of milk was poured from the can. As a result, 14 liters remained in it. How many liters of milk were in the can originally?

Decision

The desired value is the initial number of liters in the can. Draw the number of liters as a line and label this line as X

It is said that 30% of the milk was poured out of the can. We select in the figure approximately 30%

A percentage, by definition, is one hundredth of something. If 30% of the milk was poured out, then the remaining 70% remained in the can. These 70% account for 14 liters indicated in the problem. Select the remaining 70% in the figure

Now you can make an equation. Let's remember how to find the percentage of a number. To do this, the total amount of something is divided by 100 and the result is multiplied by the desired percentage. Note that 14 liters, which is 70%, can be obtained in the same way: the initial number of liters X divide by 100 and multiply the result by 70. Equate all this to the number 14

Or get a simpler equation: write 70% as 0.70, then multiply by X and equate this expression to 14

This means that initially there were 20 liters of milk in the can.

Task 9. They took two alloys of gold and silver. In one, the ratio of these metals is 1: 9, and in the other 2: 3. How much of each alloy should be taken to get 15 kg of a new alloy in which gold and silver would be related as 1: 4?

Decision

Let's first try to find out how much gold and silver will be contained in 15 kg of the new alloy. The task says that the content of these metals should be in a ratio of 1: 4, that is, gold should be one part of the alloy, and silver should be four parts. Then the total number of parts in the alloy will be 1 + 4 = 5, and the mass of one part will be 15: 5 = 3 kg.

Let's determine how much gold will be contained in 15 kg of alloy. To do this, multiply 3 kg by the number of parts of gold:

3 kg × 1 = 3 kg

Let's determine how much silver will be contained in 15 kg of alloy:

3 kg × 4 = 12 kg

This means that an alloy weighing 15 kg will contain 3 kg of gold and 12 kg of silver. Now back to the original alloys. You need to use each of them. Denote by x the mass of the first alloy, and the mass of the second alloy can be denoted by 15 − x

Let's express as a percentage all the relationships that are given in the problem and fill in the following table with them:

In the first alloy, gold and silver are in a ratio of 1: 9. Then the total parts will be 1 + 9 = 10. Of these, there will be gold , and silver .

Let's transfer this data to the table. 10% will be entered in the first line in the column "percentage of gold in the alloy", 90% will also be entered in the first line of the column "percentage of silver in the alloy", and in the last column "weight of alloy" enter a variable x, since this is how we denoted the mass of the first alloy:

We do the same with the second alloy. Gold and silver in it are in a ratio of 2: 3. Then there will be 2 + 3 = 5 parts in total. Of these, gold will be , and silver .

Let's transfer this data to the table. 40% will be entered in the second line in the column "percentage of gold in the alloy", 60% will also be entered in the second line of the column "percentage of silver in the alloy", and in the last column "weight of alloy" enter the expression 15 − x, because this is how we denoted the mass of the second alloy:

Let's fill in the last line. The resulting alloy weighing 15 kg will contain 3 kg of gold, which is alloy, and silver will be alloy. In the last column we write down the mass of the resulting alloy 15

You can now write equations using this table. We remember. If we separately add up the gold of both alloys and equate this amount to the mass of gold of the resulting alloy, we can find out what the value is x.

The first gold alloy had 0.10 x, and in the second gold alloy it was 0.40(15 − x) . Then in the resulting alloy, the mass of gold will be the sum of the masses of gold of the first and second alloys, and this mass is 20% of the new alloy. And 20% of the new alloy is 3 kg of gold, calculated by us earlier. As a result, we obtain the equation 0,10x+ 0.40(15 − x) = 3 . Let's solve this equation:

Initially through x we have designated the mass of the first alloy. Now we have found the value of this variable. Variable x is equal to 10. And we denoted the mass of the second alloy through 15 − x, and since the value of the variable x now it is known, then we can calculate the mass of the second alloy, it is equal to 15 − 10 = 5 kg.

This means that to obtain a new alloy weighing 15 kg in which gold and silver would be treated as 1: 4, you need to take 10 kg of the first and 5 kg of the second alloy.

The equation could be made using the second column of the resulting table. Then we would get the equation 0,90x+ 0.60(15 − x) = 12. The root of this equation is also 10

Task 10. There is ore from two layers with a copper content of 6% and 11%. How much low-grade ore should be taken to get it when mixed with rich 20 tons with a copper content of 8%?

Decision

Denote by x mass of poor ore. Since you need to get 20 tons of ore, then 20 rich ore will be taken − x. Since the copper content in poor ore is 6%, then in x tons of ore will contain 0.06 x tons of copper. In rich ore, the copper content is 11%, and in 20 - x tons of rich ore will contain 0.11(20 − x) tons of copper.

In the resulting 20 tons of ore, the copper content should be 8%. This means that 20 tons of copper ore will contain 20 × 0.08 = 1.6 tons.

Add expressions 0.06 x and 0.11(20 − x) and equate this sum to 1.6. We get the equation 0,06x + 0,11(20 − x) = 1,6

Let's solve this equation:

This means that to obtain 20 tons of ore with a copper content of 8%, you need to take 12 tons of poor ore. The rich will take 20 − 12 = 8 tons.

Task 11. Increasing average speed from 250 to 300 m/min, the athlete began to run the distance 1 minute faster. What is the length of the distance?

Decision

The length of the distance (or the distance of the distance) can be described by the following letter equation:

Let's use the right side of this equation to write our own equation. Initially, the athlete ran the distance at a speed of 250 meters per minute. At this speed, the length of the distance will be described by the expression 250 t

Then the athlete increased her speed to 300 meters per minute. At this speed, the length of the distance will be described by the expression 300t

Note that the length of the distance is a constant value. From the fact that the athlete increases the speed or reduces it, the length of the distance will remain unchanged.

This allows us to equate the expression 250 t to expression 300 t, since both expressions describe the length of the same distance

250t = 300t

But the task says that at a speed of 300 meters per minute, the athlete began to run the distance 1 minute faster. In other words, at a speed of 300 meters per minute, the travel time will decrease by one. Therefore, in equation 250 t= 300t on the right side, the time must be reduced by one:

At a speed of 250 meters per minute, the athlete runs the distance in 6 minutes. Knowing the speed and time, you can determine the length of the distance:

S= 250 × 6 = 1500 m

And at a speed of 300 meters per minute, the athlete runs the distance for t− 1 , that is, in 5 minutes. As mentioned earlier, the length of the distance does not change:

S= 300 × 5 = 1500 m

Task 12. A rider overtakes a pedestrian who is 15 km ahead of him. In how many hours will the rider catch up with the pedestrian if every hour the first rider travels 10 km, and the second travels only 4 km?

Decision

This task is . It can be solved by determining the approach speed and dividing the initial distance between the rider and the pedestrian by this speed.

The closing speed is determined by subtracting the lower speed from the larger one:

10 km/h − 4 km/h = 6 km/h (speed of approach)

Every hour the distance of 15 kilometers will be reduced by 6 kilometers. To find out when it will decrease completely (when the rider catches up with the pedestrian), you need to divide 15 by 6

15:6 = 2.5h

2,5 h it's two whole hours and half an hour. And half an hour is 30 minutes. So the rider will overtake the pedestrian in 2 hours and 30 minutes.

Let's solve this problem using the equation.

After that, after him, a rider set out on the road at a speed of 10 km / h. And the walking speed is only 4 km/h. This means that the rider will overtake the pedestrian after some time. We need to find this time.

When the rider catches up with the pedestrian, it will mean that they have traveled the same distance together. The distance traveled by the rider and the pedestrian is described by the following equation:

Let's use the right side of this equation to write our own equation.

The distance traveled by the rider will be described by the expression 10 t. Since the pedestrian set off before the rider and managed to overcome 15 km, the distance traveled by him will be described by the expression 4 t + 15 .

By the time the rider catches up with the pedestrian, both of them will have covered the same distance. This allows us to equate the distances traveled by the rider and the walker:

The result is a simple equation. Let's solve it:

Tasks for independent solution

Problem 1. A passenger train arrives from one city to another 45 minutes faster than a freight train. Calculate the distance between cities if the speed of the passenger train is 48 km/h and the speed of the freight train is 36 km/h.

Decision

The train speeds in this problem are measured in kilometers per hour. Therefore, we will convert the 45 minutes indicated in the task into hours. 45 minutes is 0.75 hours

Let us denote the time during which a freight train arrives in the city through the variable t. Since the passenger train arrives in this city 0.75 hours faster, the time of its movement will be denoted by the expression t - 0,75

Passenger train overcame 48( t - 0.75) km, and commodity 36 t km. Insofar as we are talking about the same distance, we equate the first expression to the second. As a result, we obtain the equation 48(t - 0.75) = 36t . Let's solve it:

Now let's calculate the distance between cities. To do this, the speed of a freight train (36 km / h) is multiplied by the time of its movement t. Variable value t now known - it is equal to three hours

36 × 3 = 108 km

To calculate the distance, you can also use the speed of the passenger train. But in this case the value of the variable

Variable value t equals 1.2. So the cars met after 1.2 hours.

Answer: the cars met after 1.2 hours.

Task 3. There are a total of 685 workers in three workshops of the plant. In the second shop there are three times more workers than in the first, and in the third - 15 workers less than in the second shop. How many workers are in each shop?

Decision

Let be x workers were in the first shop. In the second workshop there were three times more than in the first, so the number of workers in the second workshop can be denoted by the expression 3 x. The third shop had 15 fewer workers than the second. Therefore, the number of workers in the third workshop can be denoted by the expression 3 x - 15 .

The problem says that there were 685 workers in total. Therefore, we can add the expressions x, 3x, 3x - 15 and equate this sum to the number 685. As a result, we obtain the equation x + 3x + ( 3x - 15) = 685

Through a variable x the number of workers in the first workshop was indicated. Now we have found the value of this variable, it is equal to 100. So there were 100 workers in the first shop.

In the second workshop there were 3 x workers, i.e. 3 × 100 = 300. And in the third workshop there were 3 x - 15, i.e. 3 × 100 − 15 = 285

Answer: in the first workshop there were 100 workers, in the second - 300, in the third - 285.

Task 4. Two repair shops within a week should repair 18 motors according to the plan. The first workshop completed the plan by 120%, and the second by 125%, so 22 motors were repaired within a week. What weekly engine repair plan did each workshop have?

Decision

Let be x the motors were to be repaired by the first workshop. Then the second workshop had to renovate 18 − x motors.

Since the first workshop completed its plan by 120%, this means that it has repaired 1.2 x motors. And the second workshop fulfilled its plan by 125%, which means that it repaired 1.25 (18 − x) motors.

The task says that 22 motors were repaired. Therefore, we can add the expressions 1,2x and 1.25(18 − x) , then equate this sum to the number 22. As a result, we obtain the equation 1,2x + 1,25(18− x) = 22

Through a variable x the number of motors that the first workshop was supposed to repair was indicated. Now we have found the value of this variable, it is equal to 10. So the first workshop had to repair 10 motors.

And through the expression 18 − x the number of motors that the second workshop was supposed to repair was indicated. So the second workshop had to repair 18 − 10 = 8 motors.

Answer: the first workshop was to repair 10 motors, and the second 8 motors.

Problem 5. The price of the goods has increased by 30% and is now 91 rubles. How much was the product before the price increase?

Decision

Let be x rubles worth of goods before the price increase. If the price has increased by 30% it means that it has increased by 0.30 x rubles. After the price increase, the goods began to cost 91 rubles. Add x with 0.30 x and equate this sum to 91. As a result, we obtain the equation Decreasing the number by 10% resulted in 45. Find the original value of the number. x -

Answer: to get a 12% salt solution, you need to add 0.25 kg of a 20% solution to 1 kg of a 10% solution.

Problem 12. Two solutions of salt in water are given, the concentrations of which are 20% and 30%. How many kilograms of each solution must be mixed in one vessel to obtain 25 kg of a 25.2% solution?

Decision

Let be x kg of the first solution must be taken. Since it is required to prepare 25 kg of solution, the mass of the second solution can be denoted by the expression 25 − x.

The first solution will contain 0.20x kg of salt, and the second will contain 0.30(25 − x) kg of salt. In the resulting solution, the salt content will be 25 × 0.252 = 6.3 kg. Add the expressions 0.20x and 0.30(25 − x), then equate this sum to 6.3. As a result, we obtain the equation

So the first solution needs to be taken 12 kg, and the second 25 - 12 = 13 kg.

Answer: the first solution you need to take 12 kg, and the second 13 kg.

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The concept of percentage occurs in our lives too often, so it is very important to know how to solve problems with percentages. In principle, this is not a difficult matter, the main thing is to understand the principle of working with interest.

What is a percentage

We operate with the concept of 100 percent, and accordingly, one percent is a hundredth certain number. And all the calculations are already based on this ratio.

For example, 1% of 50 is 0.5, 15 of 700 is 7.

How to decide

1. Knowing that one percent is one hundredth of the number presented, you can find any number of required percentages. In order to make it clearer, let's try to find 6 percent of the number 800. This is done simply.
   • First we find one percent. To do this, divide 800 by 100. It turns out 8.
   • Now we multiply this very one percent, that is, 8, by the number of percent we need, that is, by 6. It turns out 48.
   • Fix the result by repetition.

   15% of 150. Solution: 150/100*15=22.

   28% of 1582. Solution: 1582/100*28=442.

2. There are other problems when you are given values, and you need to find percentages. For example, you know that the store has 5 Red roses out of 75 whites, and you need to find out what percentage is scarlet. If we do not know this percentage, then we will denote it as x.

   There is a formula for this: 75 - 100%

   In this formula, the numbers are multiplied cross by cross, that is, x \u003d 5 * 100/75. It turns out that x \u003d 6% So the percentage of scarlet roses is 6%.

3. There is another type of problem for percentages, when you need to find by what percentage one number is greater or less than another. How to solve problems with percentages in this case?

   There are 30 students in the class, 16 of them are boys. The question is how many percent of boys are more than girls. First you need to calculate what percentage of students are boys, then you need to find out what percentage of girls. And finally find the difference.

   So let's get started. We make a proportion of 30 accounts. - 100%

   16 accounts -X %

   Now we count. X=16*100/30, x=53.4% ​​of all students in the class are boys.

   Now find the percentage of girls in the same class. 100-53.4=46.6%

It remains now only to find the difference. 53.4-46.6=6.8%. Answer: there are more boys than girls by 6.8%.

Key Points in Interest Solving

So, so that you do not have problems with how to solve problems for percentages, remember a few basic rules:

1. In order not to get confused in problems with percentages, always be vigilant: go from specific values ​​to percentages and vice versa if necessary. The main thing is never to confuse one with the other.
2. Be careful when calculating percentages. It is important to know from what specific value you need to count. For successive changes in values, the percentage is calculated from the last value.
3. Before writing down the answer, read the entire problem again, because it may be that you have found only an intermediate answer, and you need to perform one or two more actions.

Thus, solving problems with percentages is not such a difficult matter, the main thing in it is attentiveness and accuracy, as, indeed, in all mathematics. And don't forget that practice is required to improve any skill. So decide more, and everything will be fine or even excellent for you.