Demonstration versions of the OGE (GIA) in mathematics - Archive of files. Demonstration versions of the OGE (GIA) in mathematics - Archive of OGE files demonstration material

The demo version is intended to enable the exam participant and the general public to get an idea of ​​the structure of the future examination paper, the number and form of tasks, as well as their level of difficulty. This information makes it possible to develop a strategy for preparing for the mathematics exam.

Demo version of the OGE 2018 in mathematics, grade 9

Changes in CMM 2018 compared to 2017

Compared to the 2017 structure, the “Real Mathematics” module was excluded from the work. The tasks in this module are distributed across the Algebra and Geometry modules.

Characteristics of the structure and content of KIM OGE 2018 in mathematics

The work consists of two modules: “Algebra” and “Geometry”. Each module has two parts, corresponding to testing at basic and advanced levels. When testing basic mathematical competence, students must demonstrate mastery of basic algorithms, knowledge and understanding of key content elements (mathematical concepts, their properties, methods of solving problems, etc.), the ability to use mathematical notation, and apply knowledge to solving mathematical problems that are not reducible to direct application algorithm, as well as apply mathematical knowledge in the simplest practical situations.

Parts 2 of the modules “Algebra” and “Geometry” are aimed at testing mastery of the material at an advanced level. Their purpose is to differentiate well-performing schoolchildren by level of training, to identify the most prepared part of graduates, who make up the potential contingent of specialized classes. These parts contain tasks of an increased level of complexity from various sections of the mathematics course. All tasks require recording solutions and answers. The tasks are arranged in order of increasing difficulty - from relatively simple to complex, assuming fluency in the material and a good level of mathematical culture.

The Algebra module contains 17 tasks: in part 1 - 14 tasks; in part 2 - 3 tasks.

The "Geometry" module contains 9 tasks: in part 1 - 6 tasks; in part 2 - 3 tasks. There are 26 tasks in total, of which 20 are basic level tasks, 4 advanced level tasks and 2 high level tasks.

Duration of the OGE 2018 in mathematics- 235 minutes.

Requirements Codifier to the level of preparation of students for the main state exam in mathematics is one of the documents defining the structure and content of control measuring materials - KIM. The codifier is a systematized list of requirements for the level of training of graduates and tested content elements, in which each object corresponds to a specific code.

Content Element Codifier for the main state exam in mathematics is one of the documents defining the structure and content of control measuring materials - CMM. The codifier is a systematized list of requirements for the level of training of graduates and tested content elements, in which each object corresponds to a specific code.

Algebra module

1 . Find the meaning of the expression

2. The table shows the standards for running 30 meters for 9th grade students.
What mark will a girl get if she runs this distance in 5.62 seconds?
1) mark “5” 2) mark “4”
3) mark “3” 4) standard not met

3 . A point is marked on the coordinate line A. It is known to correspond to one of the four numbers below.
Which number does the point correspond to? A?
1) 2) 3) 4)

4 . Find the meaning of the expression

5 . The graph shows the dependence of atmospheric pressure on altitude above sea level. The horizontal axis shows the altitude above sea level in kilometers, and the vertical axis shows the pressure in millimeters of mercury. Determine from the graph at what altitude the atmospheric pressure is 620 millimeters of mercury. Give your answer in kilometers.

6. Solve the equation. If an equation has more than one root, write down the larger root as your answer.

7. The cost of travel on an electric train is 198 rubles. Schoolchildren receive a 50% discount. How many rubles will it cost for 4 adults and 12 schoolchildren?

8. The chart shows the nutrient content of dried porcini mushrooms.
Which of the following statements are true?
1) 1000 grams of mushrooms contain approximately 360 g of fat.
2) 1000 grams of mushrooms contain approximately 240 g of carbohydrates.
3) 1000 grams of mushrooms contain approximately 140 g of proteins.
4) 1000 grams of mushrooms contain approximately 500 g of fats, proteins and carbohydrates.
In response, write down the numbers of the selected statements without spaces, commas and other additional characters

9. On the plate are pies that look identical: 4 with meat, 8 with cabbage and 3 with apples. Petya chooses one pie at random. Find the probability that the pie will contain apples.

10. Establish a correspondence between the graphs of functions and the formulas that define them.

11. In a sequence of numbers, the first number is 6, and each next number is greater than the previous one by 4. Find the fifteenth number.

12. Find the value of the expression at .

13. To convert the temperature value on the Celsius scale to the Fahrenheit scale, use the formula, where - temperature in degrees Celsius, - temperature in degrees Fahrenheit. How many degrees on the Fahrenheit scale does -25 degrees Celsius correspond to?

14. Specify the solution to the system of inequalities

15. The sloping roof is mounted on three vertical supports, the bases of which are located on the same straight line. The middle support stands in the middle between the small and large supports (see figure). The height of the small support is 1.7 m, the height of the middle support is 2.1 m. Find the height of the large support. Give your answer in meters.

16 . In an isosceles triangle ABC with base A.C. external vertex angle C equals 123°. Find the angle YOU. Give your answer in degrees.

17 . Find the length of the chord of a circle with a radius of 13 cm if the distance from the center of the circle to the chord is 5 cm. Give your answer in cm.

18. Find the area of ​​the trapezoid shown in the figure.

19 . Find the tangent of the acute angle shown in the figure.

20 . Which of the following statements are true?
1) Through a point not lying on a given line, you can draw a line parallel to this line.
2) A triangle with sides 1, 2, 4 exists.
3) Any parallelogram has two equal angles.
In response, write down the numbers of the selected statements without spaces, commas or other additional characters.

Algebra module

21 . Solve the equation

22 . At 5 o'clock in the morning, the fisherman set off from the pier against the flow of the river in a motor boat, after some time dropped anchor, fished for 2 hours and returned back at 10 o'clock in the morning of the same day. How far did he sail from the pier if the speed of the river current is 2 km/h and the boat’s own speed is 6 km/h?

23 . Graph the Function and determine at what values the straight line has exactly one common point with the graph.

Module "Geometry"

24 . In a right triangle ABC with right angle C known legs: A.C.= 6, B.C.= 8. Find the median CK this triangle.

25 . In a parallelogram ABCD dot E- middle of the side AB. It is known that EC = ED. Prove that this parallelogram is a rectangle.

26 . Base A.C. isosceles triangle ABC is equal to 12. A circle of radius 8 with a center outside this triangle touches the extensions of the sides of the triangle and touches the base A.C.. Find the radius of a circle inscribed in a triangle ABC.

Answers

The OGE in Russian language in 2019 will be held in two stages.

The final interview (oral part) is one of the conditions for students’ admission to the written part of the OGE in the Russian language, held at the end of the school year.

The final interview in the Russian language is conducted for students and external students on the second Wednesday of February according to texts, topics and tasks generated by time zones by the Federal Service for Supervision of Education and Science

Oral part in Russian language OGE 2019 (final interview) - demo version from FIPI

Demo version of the OGE in Russian language 2019 (State Examination Grade 9)

Final interview in Russian language consists of two parts, including four tasks.

Part 1 consists of two tasks. Tasks 1 and 2 are completed using the same text.

Task 1 – reading a short text aloud. Preparation time: 2 minutes.

In task 2 it is proposed to retell the text read, supplementing it with a statement. Preparation time: 2 minutes. Part 2 consists of two tasks.

Tasks 3 and 4 are not related to the text that you read and retold while completing tasks 1 and 2. You have to choose one topic for monologue and dialogue.

In task 3, you are asked to choose one of three proposed conversation options: a description of a photograph, a narration based on life experience, a reasoning on one of the formulated problems. Preparation time – 1 minute.

In task 4 you will have to participate in a conversation on the topic of the previous task. Your total response time (including preparation time) is 15 minutes.

An audio recording is made throughout the entire response time.

Try to fully complete the assigned tasks, speak clearly and clearly, and stay on topic. This way you can score the most points.

The final interview is assessed according to the system pass - fail

Examination work of the OGE in the Russian language (written part) consists of three parts, including 15 tasks.

3 hours 55 minutes (235 minutes) are allotted to complete the examination work in the Russian language.

Part 1 includes one task and is a short written work based on the listened text (condensed presentation). The source text for the condensed presentation is listened to 2 times. This task is completed on answer sheet No. 2.

Part 2 consists of 13 tasks (2–14). Part 2 tasks are completed based on the text read. Write down the answer to tasks 2 and 3 in answer form No. 1 in the form of one number, which corresponds to the number of the correct answer.

The answers to tasks 4–14 are a word (phrase), number or sequence of numbers. write in the answer field in the text of the work, and then transfer it to answer form No. 1.

The task of part 3 is performed on the basis of the same text that you read while working on the tasks of part 2. When starting part 3 of the work, choose one of the three proposed tasks (15.1, 15.2 or 15.3) and give a written, detailed, reasoned answer.

Basic general education

Line UMK A. G. Merzlyak. Algebra (7-9) (basic)

Mathematics

Demo version of OGE-2020 in mathematics

Download the demo version of the OGE 2020 along with the codifier and specification from the link below:

Major changes in the new demo

The CMM includes a new block of practice-oriented tasks 1-5.

OGE schedule in mathematics in 2020

At the moment, it is known that the Ministry of Education and Rosobrnadzor have published draft OGE schedules for public discussion. Estimated dates for the main wave of mathematics exams: June 9, reserve days June 24, 25, 30.

Soon we will talk about the upcoming Unified State Exam on and on air our YouTube channel.

A new manual for preparing for the main state exam in mathematics is offered to 9th grade graduates. The collection includes tasks for all sections and topics tested in the main state exam: “Numbers and calculations”, “Practice-oriented problems”, “Equations and inequalities”, “Algebraic expressions”, “Geometry”, “Sequences, functions and graphs” " Tasks of different difficulty levels are presented. At the end of the book, answers are given that will help in monitoring and assessing knowledge, skills and abilities. The materials in the manual can be used for systematic repetition of the studied material and training in performing tasks of various types in preparation for the OGE. They will help the teacher organize preparation for the main state exam, and students will independently test their knowledge and readiness to take the exam.

The examination paper (OGE) consists of two modules: “Algebra” and “Geometry”, included in two parts: basic level (part 1), advanced and high level (part 2). There are 26 tasks in total, of which 20 are basic level tasks, 4 advanced level tasks and 2 high level tasks. The Algebra module contains 17 tasks: in part 1 - 14 tasks; in part 2 - 3 tasks. The “Geometry” module contains 9 tasks: in part 1 – 6 tasks; in part 2 - 3 tasks. 3 hours 55 minutes (235 minutes) are allotted to complete the examination paper in mathematics.

Part 1

Exercise 1

Find the meaning of the expression

Solution

Answer: 0,32.

Solution

Since the time is 5.62 seconds, the girl’s standard for a grade of “4” was not met, however, this time does not exceed 5.9 seconds. – standard for rating “3”. Therefore, its mark is “3”.

Answer: 3.

Solution

The first number is greater than 11, therefore it cannot be the number A. Note that point A is located on the second half of the segment, which means it is certainly greater than 5 (for reasons of the scale of the coordinate line). Therefore, this is not the number 3) and not the number 4). We note that the number satisfies the inequality:

Answer: 2.

Task 4

Find the meaning of the expression

Solution

By the property of the arithmetic square root (at a ≥ 0, b≥ 0), we have:

Answer: 165.

Solution

To answer the question posed, it is enough to determine the price of division along the horizontal and vertical axes. One notch along the horizontal axis is 0.5 km, and one notch along the vertical axis is 20 mm. r.s. Therefore the pressure is 620 mm. r.s. is reached at an altitude of 1.5 km.

Answer: 1,5.

Task 6

Solve the equation x 2 + x – 12 = 0.

If an equation has more than one root, write down the larger root as your answer.

Solution

Let's use the formula for the roots of a quadratic equation

Where x 1 = –4, x 2 = 3.

Answer: 3.

Task 7

The cost of travel on an electric train is 198 rubles. Schoolchildren receive a 50% discount. How many rubles will it cost for 4 adults and 12 schoolchildren?

Solution

A student ticket will cost 0.5 · 198 = 99 rubles. This means that travel for 4 adults and 12 schoolchildren will cost

4 198 + 12 99 = 792 + 1188 = 1980.

Answer: 1980.

Solution

Statements 1) and 2) can be considered correct, since the areas corresponding to proteins and carbohydrates occupy approximately 36% and 24% of the total part of the pie chart. At the same time, it is clear from the diagram that fats occupy less than 16% of the entire diagram, and therefore statement 3) is incorrect, as is statement 4), since fats, proteins and carbohydrates together make up the majority of the diagram.

Answer: 12 or 21.

Task 9

On the plate are pies that look identical: 4 with meat, 8 with cabbage and 3 with apples. Petya chooses one pie at random. Find the probability that the pie will contain apples.

Solution

The probability of an event in the classical definition is the ratio of the number of favorable outcomes to the total number of possible outcomes:

In this case, the number of all possible outcomes is 4 + 8 + 3 = 15. The number of favorable outcomes is 3. Therefore

Answer: 0,2.

Establish a correspondence between the graphs of functions and the formulas that define them.

Solution

The first graph obviously corresponds to a parabola, the general equation of which is:

y = ax 2 + bx + c.

Therefore, this is formula 1). The second graph corresponds to a hyperbola, the general equation of which is:

Therefore, this is formula 3). The third graph remains, which is a direct proportionality graph:

y = kx.

This is formula 2).

Answer: 132.

Task 11

In a sequence of numbers, the first number is 6, and each next number is greater than the previous one by 4. Find the fifteenth number.

Solution

The problem deals with an arithmetic progression with the first term a 1 = 6 and difference d= 4. General term formula

a n = a 1 + d · ( n– 1) = 6 + 4 14 = 62.

Answer: 62.

Solution

Instead of immediately plugging numbers into this expression, let's first simplify it by writing it as a rational fraction:

Answer: 1,25.

Task 13

To convert the temperature on the Celsius scale to the Fahrenheit scale, use the formula tF = 1,8tC+ 32, where tC– temperature in degrees Celsius, tF– temperature in degrees Fahrenheit. How many degrees on the Fahrenheit scale does -25 degrees Celsius correspond to?

Solution

Let's substitute the value –25 into the formula

tF= 1.8 (–25) + 32 = –13

Answer: –13.

Specify the solution to the system of inequalities

Solution

Solving this system of inequalities, we obtain:

Consequently, the solution to the system of inequalities is the segment [–4; –2.6], which corresponds to Figure 2).

Answer: 2.

Solution

The figure shown in the figure is a rectangular trapezoid. The middle support is nothing more than the middle line of the trapezoid, the length of which is calculated by the formula

Where a, b– length of the bases. Let's make an equation:

b = 2,5.

Answer: 2,5.

In an isosceles triangle ABC with base AC the external angle at vertex C is 123°. Find the angle YOU. Give your answer in degrees.

Solution

Triangle ABC isosceles, so the angle YOU equal to angle BSA. But the angle BSA– adjacent with an angle of 123°. Hence

∠YOU = ∠BSA= 180° – 123° = 57°.

Answer: 57°.

Find the length of the chord of a circle of radius 13 if the distance from the center of the circle to the chord is 5.

Solution

Consider a triangle AOB(see picture).

It is isosceles ( JSC = OB) And HE it has a height (its length is equal to 5 according to condition). Means, HE– median according to the property of an isosceles triangle and AN = NV. We'll find AN from a right triangle ANO according to the Pythagorean theorem:

Means, AB = 2AN = 24.

Answer: 24.

Find the area of ​​the trapezoid shown in the figure.

Solution

The lower base of a trapezoid is 21. Let's use the formula for the area of ​​a trapezoid

Answer: 168.

Find the tangent of the acute angle shown in the figure.

Solution

Select a right triangle (see picture).

Tangent is the ratio of the opposite side to the adjacent side, from here we find

Answer: 2.

Which of the following statements are true?

1) Through a point not lying on a given line, you can draw a line parallel to this line.

2) A triangle with sides 1, 2, 4 exists.

3) Any parallelogram has two equal angles.

Solution

The first statement is the axiom of parallel lines. The second statement is incorrect, since for segments with lengths 1, 2, 4 the triangle inequality does not hold (the sum of the lengths of any two sides is less than the length of the third side)

1 + 2 = 3 > 4.

The third statement is true - in a parallelogram, opposite angles are equal.

Answer: 13 or 31.

Part 2

Solve the equation x 4 = (4x – 5) 2 .

Solution

Using the difference of squares formula, the original equation is reduced to the form:

(x 2 – 4x + 5)(x 2 + 4x – 5) = 0.

The equation x 2 – 4x+ 5 = 0 has no roots ( D < 0). Уравнение

x 2 + 4x – 5 = 0

has roots −5 and 1.

Answer: −5; 1.

At 5 o'clock in the morning, the fisherman set off from the pier against the flow of the river in a motor boat, after some time dropped anchor, fished for 2 hours and returned back at 10 o'clock in the morning of the same day. How far from the pier did he sail if the speed of the river is 2 km/h and the boat's own speed is 6 km/h?

Solution

Let the fisherman swim a distance equal to s. The time it took him to swim this distance is equal to hours (since the speed of the boat against the current is 4 km/h). The time he spent on the way back is equal to hours (since the speed of the boat along the current is 8 km/h). The total time, including parking, is 5 hours. Let’s create and solve the equation:

Answer: 8 kilometers.

Solution

The domain of definition of the function under consideration contains all real numbers, except for the numbers –2 and 3.

Let us simplify the form of the analytical relationship by factoring the numerator of the fraction:

Thus, the graph of this function is a parabola

y = x 2 + x – 6,

with two “punctured” points, the abscissas of which are equal to –2 and 3. Let’s construct this graph. Parabola vertex coordinates

(–0,5; –6,25).

Straight y = c has exactly one common point with the graph, either when it passes through the vertex of the parabola, or when it intersects the parabola at two points, one of which is punctured. Coordinates of the “punctured” points

(−2; −4) and (3; 6). That's why c = –6,25, c= –4 or c = 6.

Answer: c = –6,25; c = –4; c = 6.

In a right triangle ABC with right angle WITH known legs: AC = 6, Sun= 8. Find the median CK of this triangle.

Solution

In a right triangle, the median drawn to the hypotenuse is equal to half of it. That's why

Answer: 5.

In a parallelogram ABCD dot E– middle of the side AB. It is known that EU =ED. Prove that this parallelogram is a rectangle.

Solution

Consider triangles EBC and AED. They are equal on three sides. Indeed, A.E.= E.B., ED= E.C.(by condition), AD= B.C.(opposite sides of the parallelogram). Therefore, ∠ A = ∠B, but the sum of adjacent angles in a parallelogram is 180°, so ∠ A= 90° and ABCD- rectangle.

Base AC isosceles triangle ABC is equal to 12. A circle of radius 8 with a center outside this triangle touches the extensions of the sides of the triangle and touches the base AC. Find the radius of a circle inscribed in a triangle ABC.

Solution

Let O is the center of a given circle, and Q- center of a circle inscribed in a triangle ABC .

Since the point ABOUT equidistant from the sides of the angle ∠NVA, insofar as it lies on its bisector. At the same time, on the bisector of the angle ∠NVA lies the point Q and at the same time, due to the properties of an isosceles triangle, this bisector is both the median and altitude of the triangle ABC. From these considerations it is easy to deduce that the circles in question touch at one point M, point of contact M circles divides A.C. in half and OQ perpendicular A.C..

Let's draw the rays AQ And A.O.. It's easy to understand that AQ And A.O.- bisectors of adjacent angles, and therefore, the angle OAQ straight. From a right triangle OAQ we get:

AM 2 = MQ · M.O..

Hence,