Real tasks oge. Real variants of the OGE (GIA) in mathematics - File archive

Is it difficult to pass the OGE in mathematics? This question is asked by, perhaps, every graduate of the 9th grade. Let's figure it out together. The Main State Exam in Mathematics is one of the most difficult in the 9th grade - it's a fact. In addition, it is mandatory for every graduate of the basic school to pass in order to receive a certificate. Therefore, for all the difficulties of the OGE 2018 in mathematics, you should be prepared in advance.

We would like to draw your attention to the fact that at the Hodograph Training Center you will find qualified tutors for preparing for the OGE in mathematics for students, and. We practice individual and collective classes for 3-4 people, we provide discounts for training. Our students average 30 points more!

To begin with, it is worth noting the first OGE feature in mathematics, which distinguishes him from all exam tests not only in 9th, but also in 11th grade. This, of course, is a division into modules: "Algebra", "Geometry", " Real math". If you don't get through minimum threshold for each of them, it will negatively affect overall score for the exam.

That is, without scoring the required points in at least one of the modules (recall that in "Algebra" it is 3 points, in "Geometry" - 2, in "Real Mathematics" - 2), you can get an "unsatisfactory" mark for the entire examination work. Thus, the knowledge of students in all sections of the mathematics course of the basic school is checked. Therefore, sufficient time should be devoted to preparing for each block.

Tasks of the "Geometry" module in the OGE

So, traditionally in the OGE in mathematics, the largest percentage of unsolved tasks falls on the "Geometry" module. There are several reasons for this phenomenon.

First, on average, three times less time is spent on studying geometry at school than on algebra lessons. And the material, in fact, is perceived and assimilated more difficult and longer than algebraic.

Secondly, the skills of building and reading drawings for many children are poorly formed and require additional work at home, which most students, of course, do not do.

As a result, geometry assignments are often simply ignored by students. In other words, they don't even start doing them. The only advice here is to devote more time to geometry problems during the entire preparation time. Do not be lazy: look at the solution of similar problems on the Internet or ask the teacher, then over time the necessary solution skill will be formed and you will be fully equipped in the exam.

It is worth saying that there are simply no really difficult tasks in the OGE in mathematics, the exception is, perhaps, only tasks 25, 26, and even then not always. You can also learn how to solve these numbers: a few learned tricks for performing additional constructions and solution algorithms will allow you to cope with such tasks.

Tasks of the Algebra module in the OGE in mathematics

So, let's move on to the Algebra module. It probably makes no sense to stop at the first part, all tasks there are performed according to fairly simple algorithms, they do not require special ingenuity, every student is able to learn how to solve them secondary school. Much more interesting are the tasks of part 2. We will dwell on them in more detail.

Task 21 with the solution in the OGE in mathematics. Transform an expression, solve an equation, solve a system of equations

Fractional-rational or power expression. The solution requires attention at every step of the transformation. Consider an example:

Solve the inequality

1____ + __1____ + __1____ < 1 (х-3)(х-4) (х-3)(х-5) х²-9х+20 Решение: Для решения данного неравенства выполним следующее 1. Перенесем единицу в левую часть неравенства. 2. Знаменатель третьей дроби разложим на множители (х-4)(х-5) 3.

Since there is a variable in the denominator, it is necessary to indicate the ODZ - the area allowed values- those values of x for which the fraction does not make sense. x≠3; x≠4; x≠5 4. Add four fractions with different denominators(since an integer can be represented as a fraction with a denominator of 1) by multiplying the numerators. We get: (x-5) + (x-4) + (x-3) - (x-3) (x-4) (x-5)< 0 3х-12 - (х-3)(х-4)(х-5) < 0 3(х-4) - (х-3)(х-4)(х-5) < 0 Выносим common factor(x-4) bracket (x-4) 〈3 - (x-3)(x-5)〉< 0 (х-4) 〈3 - (х² - 8х + 15)〉 < 0 (х-4) (3 - х² + 8х - 15) < 0 Коэффициент при х² отрицательный. Меняем его на противоположный, умножая вторую скобку на (-1). При этом изменится знак неравенства на противоположный. (х - 4) (х² - 8х + 12) >0 (x - 4) (x - 6) (x - 2) > 0 Now we can solve the inequality using the interval method. Celebrate on numerical axis all the roots that we found in the numerator and all the roots of the ODZ from the denominator.

2_________3 _________ 4_________ 5_________ 6___________ - - In the record, where the coefficient at x is always positive, the interval method gives the right to apply next rule: to the right of the right root, the inequality sign is ALWAYS +! When passing through the root, the sign of inequality changes to the opposite.

In the event that the root has an even multiplicity (for example, x squared, to the fourth power, to the sixth power, etc.), as in our example with x = 4, the inequality sign does not change to the opposite. Hence the answer: (-∞, 2)∪(3,4)∪(4,5)∪(6,+∞).

At each step, a certain nuance of the solution is visible. But in general, the algorithm is clear and easy to learn.

Solving task 22 in the OGE in mathematics. Text task

There is not much to say here, the guys, as a rule, solve text problems. Errors can occur at the stage of drawing up an equation according to the condition of the problem. To avoid such problems, one should be able to correctly formalize text task, that is, translate from Russian into mathematical language. For this, it has been developed a large number of techniques: drawings, diagrams, tables, etc. The methods most often used in schools are the construction of tables in problems for movement and work, and schemes in problems for percentages. Mastering these methods is not difficult, you just need the desire to do it.

An example of task 23 in the OGE in mathematics. Construction of complex graphs of functions, expressions with a parameter

Many students say that the most difficult OGE task in mathematics is number 23. It’s hard to argue with them, such tasks usually look menacing, but in fact the whole solution comes down to transforming big expression into a compact fraction. Moreover, it is enough to know only the rules for factoring polynomials and be careful when reducing the resulting fractions. Building a graph should not be difficult, as a last resort, you can always “sketch” the graph point by point and understand what kind of function it turned out.

After completing the constructions, do not forget to complete the task itself: as a rule, you need to determine an unknown parameter (number) that ensures the fulfillment of such conditions as one, two, none, etc. common points with the graph of the constructed function. Constant training will help you gain confidence and solve this task without difficulty.

Thus, it cannot be categorically said that there are many difficult tasks in the OGE in mathematics. The only question is the correct and timely preparation. Make an effort, and even the most difficult tasks OGE in mathematics 2018 will seem frivolous to you! UC "Godograph" sincerely wishes you good luck in the exams!

When writing this work “OGE in Mathematics 2018. Option 1”, the manual “OGE 2018. Mathematics. 14 options. Typical test tasks from the developers of the OGE / I. R. Vysotsky, L. O. Roslova, L. V. Kuznetsova, V. A. Smirnov, A. V. Khachaturyan, S. A. Shestakov, R. K. Gordin, A. S. Trepalin, A. V. Semenov, P. I. Zakharov; edited by I. V. Yashchenko. - M .: Publishing house "Exam", MTSNMO, 2018 ″.

Part 1

Module "Algebra"

Show Solution

To add two fractions, they must be reduced to common denominator. IN this case is the number 100 :

Answer:

In several relay races that were held at the school, the teams showed the following results.

Team	I relay, points	II relay, points	III relay, points	IV relay, points
"Hit"	3	3	2	4
"Jerk"	1	4	4	2
"Takeoff"	4	2	1	3
"Spurt"	2	1	3	1

When summing up the scores of each team for all relay races are summed up. The team that gets the largest number points. Which team finished third?

"Hit"
"Jerk"
"Takeoff"
"Spurt"

Show Solution

First of all, we summarize the points scored by each team

"Strike" = 3 + 3 + 2 + 4 = 12
"Dash" = 1 + 4 + 4 + 2 = 11
« Takeoff» = 4 + 2 + 1 + 3 = 10
"Spurt" = 2 + 1 + 3 + 1 = 7

Judging by the result: the first place was taken by the "Strike" team, the second - by the "Spurt" team, and the third - by the "Rise" team.

Answer:

The third place was taken by the Vlet team, number 3.

On the coordinate line, points A, B, C and D correspond to the numbers: -0.74; -0.047; 0.07; -0.407.

Which point corresponds to the number -0.047?

Show Solution

On a coordinate line, positive numbers are to the right of the origin, and negative numbers are to the left. Means the only positive number 0.07 corresponds to point D. The largest a negative number- this is -0.74, which means it corresponds to point A. Considering that the remaining number is -0.047 more number-0.407, then they belong to points C and D, respectively. Let's show it on the drawing:

Answer:

The number -0.047 corresponds to point C, number 3.

Find the value of an expression

Show Solution

IN this example you need to be smart. If the root of 64 is 8, since 8 2 = 64, then the root of 6.4 is quite difficult to find in a simple way. However, after finding the root of the number 6.4, it must be immediately squared. So two steps: finding square root and squaring cancel each other out. Therefore we get:

Answer:

The graph shows the relationship atmospheric pressure from height above sea level. On horizontal axis height above sea level in kilometers is marked, on the vertical - pressure in millimeters of mercury column. Determine from the graph at what height the atmospheric pressure is 140 millimeters of mercury. Give your answer in kilometers.

Show Solution

Let's find on the graph the line corresponding to 140 mmHg. Next, we determine the place of its intersection with the curve of dependence of atmospheric pressure on altitude above sea level. This intersection is clearly visible on the graph. Let's draw a straight line from the point of intersection down to the height scale. The desired value is 11 kilometers.

Answer:

Atmospheric pressure is 140 millimeters of mercury at an altitude of 11 kilometers.

Solve the Equation x 2 + 6 = 5X

If the equation has more than one root, write the smallest of the roots as your answer.

Show Solution

x 2 + 6 = 5X

Before us is the usual quadratic equation:

x 2 + 6 - 5X = 0

To solve it, you need to find the discriminant:

Answer:

smallest root given equation: 2

On sale in February mobile phone cost 2800 rubles. In September, it began to cost 2520 rubles. By what percent did the price of a mobile phone decrease between February and September?

Show Solution

So, 2800 rubles - 100%

2800 - 2520 \u003d 280 (p) - the amount by which the phone fell in price

280 / 2800 * 100 = 10 (%)

Answer:

The price of a mobile phone between February and September decreased by 10%

The diagram shows the seven largest countries in terms of area (in million km 2) of the world.

Which of the following statements wrong?

1) Canada is the largest country in the world by land area.
2) The territory of India is 3.3 million km 2.
3) The area of the territory of China more area territory of Australia.
4) The area of the territory of Canada is larger than the area of \u200b\u200bthe United States by 1.5 million km 2.

In response, write down the numbers of the selected statements without spaces, commas, or other additional characters.

Show Solution

Based on the graph, Canada is inferior in area to Russia, which means the first statement wrong .

An area of 3.3 million km 2 is indicated above the histogram of India, which corresponds to the second statement.

The area of China according to the graph is 9.6 million km 2, and the area of Australia is 7.7 million km 2, which corresponds to the statement in the third paragraph.

The territory of Canada is 10.0 million km 2, and the area of the USA is 9.5 million km 2, i.e. almost equal. And that means statement 4 wrong .

Answer:

In every twenty-fifth package of juice, according to the terms of the promotion, there is a prize under the lid. Prizes are distributed randomly. Vera buys a package of juice. Find the probability that Vera does not find the prize in her bag.

Show Solution

The solution of this problem is based on the classical formula for determining the probability:

where, m is the number of favorable outcomes of the event, and n is total outcomes

We get

So the chance of Vera not finding the prize is 24/25 or

Answer:

The probability that Vera will not find the prize is 0.96

Establish a correspondence between functions and their graphs.

In the table, under each letter, indicate the corresponding number.

Show Solution

The hyperbola shown in Figure 1 is located in the second and fourth quarters, therefore, this schedule function A can correspond. Let's check: a) at х = -6, y = -(12/-6) = 2; b) at x = -2, y = -(12/-2) = 6; c) at x = 2, y = -(12/2) = -6; d) at x = 6, y = -(12/6) = -2. Q.E.D.
The hyperbola shown in Figure 2 is located in the first and third quarters, therefore, function B can correspond to this graph. Perform the check yourself, by analogy with the first example.
The hyperbola shown in Figure 3 is located in the first and third quarters, therefore, function B can correspond to this graph. Let's check: a) at x = -6, y = (12/-6) = -2; b) at x = -2, y = (12/-2) = -6; c) at x = 2, y = (12/2) = 6; d) for x = 6, y = (12/6) = 2. As required.

Answer:

A - 1; B - 2; AT 3

Arithmetic progression(a n) is given by the conditions:

a 1 = -9, a n+1 = a n + 4.

Find the sum of the first six terms.

Show Solution

a 1 = -9, a n+1 = a n + 4.

a n + 1 = a n + 4 ⇒ d = 4

a n = a 1 + d(n-1)

a 6 \u003d a 1 + d (n-1) \u003d -9 + 4 (6 - 1) \u003d -9 + 20 \u003d 11

S 6 \u003d (a 1 + a 6) ∙ 6 / 2

S 6 \u003d (a 1 + a 6) ∙ 3

S 6 \u003d (–9 + 11) ∙ 3 \u003d 6

Answer:

Find the value of an expression

Show Solution

We open the brackets. Don't forget that the first parenthesis is the square of the sum.

Answer:

The area of a quadrilateral can be calculated using the formula

where d 1 and d 2 are the lengths of the diagonals of the quadrilateral, a is the angle between the diagonals. Using this formula, find the length of the diagonal d 2 if

Show Solution

Remember the rule, if we have a three-story fraction, then the lower value is transferred to the top

Answer:

Specify the solution of the inequality

Show Solution

To solve this inequality, you need to do the following:

a) move the term 3x to the left side of the inequality, and 6 to right side, not forgetting to reverse the signs. We get:

b) Multiply both sides of the inequality by a negative number -1 and change the inequality sign to the opposite.

c) find the value of x

d) the set of solutions to this inequality will be a numerical interval from 1.3 to +∞, which corresponds to the answer 3)

Answer:
3

Geometry module

A 17 m long fire escape was attached to the window of the sixth floor of the house. The lower end of the ladder stands 8 m from the wall. How high is the window? Give your answer in meters.

Show Solution

In the figure, we see an ordinary right-angled triangle consisting of a hypotenuse (ladder) and two legs (the wall of the house and the ground. To find the length of the leg, we use the Pythagorean theorem:

IN right triangle square of the hypotenuse is equal to the sum squares of legs c 2 \u003d a 2 + b 2

So, the window is located at a height of 15 meters

Answer:

In the triangle ∆ ABC it is known that AB= 8, BC = 10, AC = 14. Find cos∠ABC

Show Solution

To solve this problem, you need to use the cosine theorem. The square of the side of a triangle is equal to the sum of the squares of the other 2 sides minus twice the product of these sides by the cosine of the angle between them:

a 2 = b 2 + c 2 – 2 bc cosα

AC² = AB² + BC² - 2 AB BC cos∠ABC
14² = 8² + 10² - 2 8 10 cos∠ABC
196 = 64 + 100 - 160 cos∠ABC

160 cos∠ABC = 164 - 196
160 cos∠ABC = - 32
cos∠ABC = - 32 / 160 = -0.2

Answer:

cos∠ABC = -0.2

On a circle centered at a point ABOUT points are marked A And B so that ∠AOB = 15 o. Lesser arc length AB is 48. Find the length of the larger arc AB.

Show Solution

We know that a circle is 360 o. Based on this, 15 about is:

360 o / 15 o \u003d 24 - the number of segments in a circle of 15 o

So, 15 o make up 1/24 of the entire circle, which means the rest of the circle:

those. remaining 345 o (360 o - 15 o \u003d 345 o) make up the 23rd part of the entire circle

If the length of the smaller arc AB is 48, then the length of the larger arc AB will be:

Answer:

in a trapeze ABCD it is known that AB = CD, ∠BDA= 35 o and ∠ bdc= 58 o. Find the angle ∠ ABD. Give your answer in degrees.

Show Solution

According to the condition of the problem, we have an isosceles trapezoid. Corners at the base isosceles trapezoid(upper and lower) are equal.

∠ADC = 35 + 58 = 93°
∠DAB = ∠ADC = 93°

Now consider the triangle ∆ABD as a whole. We know that the sum of the angles of a triangle is 180°. From here:

∠ABD = 180 - ∠ADB - ∠DAB = 180 - 35 - 93 = 52°.

Answer:

On checkered paper with a cell size of 1x1, a triangle is depicted. Find its area.

Show Solution

The area of a triangle is equal to the product of half the base of the triangle (a) and its height (h):

a - the length of the base of the triangle

h is the height of the triangle.

From the figure, we see that the base of the triangle is 6 (cells), and the height is 3 (cells). Based on what we get:

Answer:

Which of the following statements is correct?

The area of a rhombus is equal to the product of its two adjacent parties by the sine of the angle between them.
Each of the bisectors isosceles triangle is its median.
The sum of the angles of any triangle is 360 degrees.

In response, write down the number of the selected statement.

Show Solution

This task is not a task. The questions listed here must be known by heart and be able to answer them.

This statement is absolutely right.
Wrong, because according to the properties of an isosceles triangle, it can have only one median - this is the bisector drawn to the base. It is also the height of the triangle.
Wrong because the sum of the angles of any triangle is 180°.

Answer:

Part 2

Module "Algebra"

Solve the Equation

Show Solution

We transfer the expression √6-x from right side to the left

We reduce both expressions √6-x

Move 28 to the left side of the equation

Before us is the usual quadratic equation.

The range of acceptable values in this case is: 6 - x ≥ 0 ⇒ x ≤ 6

To solve the equation, you need to find the discriminant:

D \u003d 9 + 112 \u003d 121 \u003d 11 2

x 1 = (3 + 11)/2 = 14/2 = 7 - not a solution

x 2 \u003d (3 - 11) / 2 \u003d -8 / 2 \u003d -4

Answer:

The ship passes along the river to the destination for 210 km and after parking returns to the point of departure. Find the speed of the ship in still water, if the speed of the current is 4 km/h, the stay lasts 9 hours, and the ship returns to the point of departure 27 hours after leaving it.

Show Solution

x is own speed ship, then

x + 4 - speed of the ship downstream

x - 4 - the speed of the ship against the current

27 - 9 = 18 (h) - the time of the ship's movement from the point of departure to the point of destination and back, excluding parking

210 * 2 \u003d 420 (km) - the total distance traveled by the ship

Based on the above, we get the equation:

reduce to a common denominator and solve:

For further decision equation, you need to find the discriminant:

y = x 2 + 4x +4 (red line plot)

y = -45/x (graph depicted by the blue line)

Consider both functions:

y=x 2 +4x+4 on the interval [–5;+∞) is quadratic function, the graph is a parabola, and = 1 > 0 - the branches are directed upwards. If we reduce it according to the formula of the square of the sum of two numbers, we will get: y \u003d (x + 2) 2 - the graph shifts to the left by 2 units, which can be seen from the graph.
y=-45/x is inverse proportionality, graph hyperbola, branches are located in the 2nd and 4th quadrants.

The graph clearly shows that the line y \u003d m has one common point with the graph at m \u003d 0 and m > 9 and two common points at m=9, i.e. answer: m=0 and m≥9, check:
One common point at the top of the parabola y = x 2 + 4x +4

x 0 \u003d -b / 2a \u003d -4 / 2 \u003d -2

y 0 \u003d -2 2 + 4 (-2) + 4 \u003d 4 - 8 +4 \u003d 0 ⇒ c \u003d 0

Two common points at x \u003d - 5; y = 9 ⇒ c = 9

Answer:

Segments AB And CD are chords of the circle. Find the length of the chord CD, If AB = 24, and the distance from the center of the circle to the chords AB And CD are 16 and 12, respectively.

Show Solution

Triangles ∆AOB and ∆COD are isosceles.

AK=BK=AB/2=24/2=12

The segments OK and OM are heights and medians.

By the Pythagorean theorem: the square of the hypotenuse is equal to the sum of the squares of the legs, we have

OB 2 = OK 2 + BK 2

OB 2 = 16 2 + 12 2 = 256 + 144 = 400

Given that OB is the radius, we have:

OB=OA=OC=OD=20

From the triangle ∆COM, according to the Pythagorean theorem, we obtain:

CM 2 = OC 2 - OM 2

CM 2 = 20 2 - 12 2 = 400 - 144 = 256

CD=CM*2=16*2=32

The chord length CD is 32.

Answer:

in a trapeze ABCD with grounds AD And BC diagonals intersect at point O. Prove that the areas of triangles ∆ AOB and ∆ COD equal

Show Solution

Let AD be the bottom base of the trapezoid and BC the top, then AD>BC.

Find the areas of triangles ∆ABD and ∆DCA:

S ∆ABD = 1/2 AD ∙ h1

S ∆DCA = 1/2 AD ∙ h2

Given that the size of the AD base and the height of both triangles are the same, we conclude that the areas of these triangles are equal:

S ∆ABD = S ∆DCA

Each of the triangles ∆ABD and ∆DCA consists of two other triangles:

S ∆ABO + S ∆AOD = S ∆ABD

S ∆DCO + S ∆AOD = S ∆DCA

If the areas of triangles S ∆ABD and S ∆DCA are equal, then the sum of the areas of their interior triangles is also equal. From here we get:

S ∆ABO + S ∆AOD = S ∆DCO + S ∆AOD

in this equality, the same triangle appears on both sides - S ∆AOD, which allows us to reduce it. We get the following equality:

S ∆ABO = S ∆DCO

Q.E.D.

Answer:

S ∆ABO = S ∆DCO

on the side BC acute triangle ABC how a semicircle is constructed on the diameter that intersects the height AD at the point M, AD = 9, MD=6, H- the point of intersection of the heights of the triangle ABC. Find AH.

Show Solution

To begin with, let's draw a triangle and a semicircle, as stated in the condition of the problem (Fig. 1).

We mark the point of intersection of the circle with the AC side with the letter F (Fig. 2)

BF - is the height of the triangle ∆ABC, since for a circle ∠BFC is the inscribed angle that is supported by the 180° arc (BC is the diameter), therefore:

∠BFC=180°/2=90°

According to the "two secant" theorem, we have: AF * AC = AM * AK

Now consider the chord MK.

Segment BC is the perpendicular to segment MK passing through the center of the circle, so BC is the perpendicular bisector.

This means that BC bisects the chord MK, i.e. MD = KD = 6 (see problem statement)

Consider the triangles ∆AHF and ∆ACD.

The angle ∠DAC is common for both triangles.

And the angles ∠AFH and ∠ADC are equal, moreover, they are right angles.

Therefore, according to the first criterion for the similarity of triangles, these triangles are similar.

From here, by definition of similarity, we can write: AC / AH = AD / AF => AC * AF = AD * AH

Earlier, we considered the equality (by the two-secant theorem) AF * AC = AM * AK, from which we obtain

AM * AK = AD * AH

AH = (AM * AK) / AD

From the figure we find:

AM=AD-MD=9-6=3

AK \u003d AD + KD \u003d 9 + 6 \u003d 15

AH = 3 * 15 / 9 = 45 / 9 = 5

Answer: AH = 5

State final examination 2019 in Algebra (Mathematics) for Grade 9 Graduates educational institutions is carried out to assess the level general education graduates in this discipline. Basic verifiable requirements for mathematical training students:

Be able to perform calculations and transformations.
enjoy basic units length, mass, time, speed, area, volume; express more large units through smaller ones and vice versa.
Describe various functions using functions. real dependencies between quantities; interpret graphs of real dependencies.
Be able to solve equations, inequalities and their systems.
Solve simple practical calculation problems.
Analyze real numerical data presented in tables, charts, graphs.
Decide practical tasks, requiring a systematic enumeration of options using the apparatus of probability and statistics.
Be able to build and read graphs of functions.
Carry out practical calculations using formulas, draw up simple formulas expressing relationships between quantities.
Describe real situations in the language of geometry, explore the constructed models using geometric concepts and theorems, to solve practical problems related to finding geometric quantities.
Be able to perform actions with geometric shapes, coordinates and vectors.
Conduct evidence-based reasoning when solving problems, evaluate the logical correctness of reasoning, recognize erroneous conclusions.
Be able to build and explore the simplest mathematical models.

IN this section you will find online tests to help you prepare for passing the OGE(GIA) in algebra (mathematics). We wish you success!

Standard OGE test(GIA-9) of the 2019 format consists of two modules: "Algebra" and "Geometry". Each module consists of two parts corresponding to testing at the basic and advanced levels. Part 2 of the modules "Algebra" and "Geometry" are aimed at testing the knowledge of the material on elevated level, they contain challenging tasks, which are not amenable to test assessment, since the inspector grades based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, in this test only the first part (the first 20 tasks) is presented. Among them, according to the current structure of the exam, answer options are offered only in a few problems. However, for the convenience of passing the tests, the site administration site offers several answers for each of the tasks. Naturally, for tasks in which the answer options by the compilers of real control measuring materials(KIM) are not provided, we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter in the exam.

The standard OGE test (GIA-9) of the 2019 format consists of two modules: "Algebra" and "Geometry". Each module consists of two parts corresponding to testing at the basic and advanced levels. Part 2 of the "Algebra" and "Geometry" modules are aimed at checking the knowledge of the material at an advanced level, they contain complex tasks that cannot be assessed by the test, as the inspector grades based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among them, according to the current structure of the exam, answer options are offered only in a few tasks. However, for the convenience of passing the tests, the site administration site offers several answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (CMM), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter in the exam.

The standard OGE test (GIA-9) of the 2018 format consists of two modules: "Algebra" and "Geometry". Each module consists of two parts corresponding to testing at the basic and advanced levels. Part 2 of the "Algebra" and "Geometry" modules are aimed at checking the knowledge of the material at an advanced level, they contain complex tasks that cannot be assessed by the test, as the inspector grades based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among them, according to the current structure of the exam, answer options are offered only in a few problems. However, for the convenience of passing the tests, the site administration site offers several answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (CMM), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter in the exam.

The standard OGE test (GIA-9) of the 2017 format contains two parts. In the first part there are 3 modules: Algebra (8 problems), Geometry (5 problems), Real Mathematics (7 problems). In the second part there are 2 modules: Algebra (3 problems) and Geometry (3 problems). The second part contains complex tasks and is not amenable to test evaluation. The inspector gives an assessment based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among the 20 tasks according to the current structure of the exam, answer options are offered only in a few tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter at the end of the school year.

The standard OGE test (GIA-9) of the 2016 format contains two parts. In the first part there are 3 modules: Algebra (8 problems), Geometry (5 problems), Real Mathematics (7 problems). In the second part there are 2 modules: Algebra (3 problems) and Geometry (3 problems). The second part contains complex tasks and is not amenable to test evaluation. The inspector gives an assessment based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among the 20 tasks according to the current structure of the exam, answer options are offered only in a few tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter at the end of the school year.

The standard OGE test (GIA-9) of the 2015 format contains two parts. In the first part there are 3 modules: Algebra (8 problems), Geometry (5 problems), Real Mathematics (7 problems). In the second part there are 2 modules: Algebra (3 problems) and Geometry (3 problems). The second part contains complex tasks and is not amenable to test evaluation. The inspector gives an assessment based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among the 20 tasks according to the current structure of the exam, answer options are offered only in a few tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter at the end of the school year.

The standard GIA test of the 2014 format contains two parts. In the first part there are 3 modules: Algebra (8 problems), Geometry (5 problems), Real Mathematics (7 problems). In the second part there are 2 modules: Algebra (3 problems) and Geometry (3 problems). The second part contains complex tasks and is not amenable to test evaluation. The inspector gives an assessment based on complex criteria and analysis of the sufficiency of the justifications given by the student. In this regard, only the first part (the first 20 tasks) is presented in this test. Among the 20 tasks according to the current structure of the exam, answers are offered only in four tasks. However, for the convenience of passing the tests, the site administration decided to offer answers for each of the tasks. Naturally, for tasks in which answer options are not provided by the compilers of real control and measurement materials (KIMs), we decided to significantly increase the number of these answer options in order to bring our test as close as possible to what you will encounter at the end of the school year.

When completing tasks A1-A14, select only one correct option.

When completing tasks A1-A16, select only one correct option.

There is not much time left for ninth grade graduates before they have to take the main state exam. This is very milestone in life, since many students will go to study at technical schools and colleges, and to enter the desired budget place you need to do well on your tests. I will solve the OGE grade 9 - just an indispensable site. It will help you prepare for testing much faster than with self-study to pass it on the most appreciated"5".

How to prepare for exams?

To prepare for exams, students use different methods. It concerns the study additional literature, classes with a professional tutor, as well as additional lessons with a school teacher.

Still the most effective method is considered, of course, the use of specialized sites, such as "I will solve the OGE." It helps to prepare both children from the fifth and 9th grade.

Reshu OGE website

Why is this service so popular? It makes it possible to feel the same as in the case of the exam itself. For preparation, tests are given from previous years, because according to statistics, most of the “new” tasks will be very similar to those in previous years.

An important advantage is that you do not need to solve tickets in a complex every time, if this is not necessary. You can perform separate tasks on a specific topic, which will be very convenient if you need to prepare for specific knowledge.

How to find the necessary information on the site?

What does any visitor see as soon as he enters the portal? At the very top of the page is the site header, and below it, in convenient icons, are the names of those subjects that you can choose for the exam. First of all, there are the following:

mathematics;
physics;
chemistry;
Russian language;
Informatics.

Disciplines

This list is not complete, because in order to find required item, for which you need to prepare, you just need to go to the site. You can choose right away necessary discipline and then the portal will show all the information on this subject.

Under the list of items are fifteen popular tickets selected by the moderators as indicative.

Test Options

If the student passes only them, and then analyzes his mistakes together with the teacher, then this will increase his chances of success by several times. successful solution OGE for grade 9.

Option No. 6561231

New User Registration

Such a desire to solve the OGE for grade 9 is natural for any student. For this you need good preparation. To use the entire service with already solved tasks in in full You must go through the registration process. This will make it possible not only to pass as many tests as you want, but also to keep your statistics.

Statistics in personal account

It will allow you to understand what tasks you need to work on in order to significantly raise the level of knowledge to required level. You can also share this data with a teacher or tutor so that they can determine what topics are best to draw the student's attention to and what to work on further.

Registration data

To register for the site Reshu OGE Grade 9, it is important to indicate certain user data, including the following:

address Email;
password;
teacher or student.

The most important thing in this case will be to specify the email. Since the registered address will begin to receive helpful information for the user. Additionally, it is worth noting the possibility that if the student forgets his password, then using e-mail it will be possible to restore this information. This means that a new temporary code will be sent to the address, which can then be replaced.

Catalog of popular tasks

Job catalog

After the user has successfully registered on the website Reshu OGE Grade 9, namely, the students of this class will be fully prepared for the exams. In the list on the left, you can find a button that says "Task Catalog" and then click on it.

There, all the tasks are already divided by topic, and you can safely go to the place with which information you need to work out further. For example, select "Actions with ordinary fractions." By clicking on this link, the student will get acquainted with the list of tasks that he may have on the exam.

Useful information for experts

School of Experts

This site is visited not only by students, but also by teachers, who will subsequently be engaged in checking assignments. Since each form must be checked in the same way as hundreds of thousands of others without biased attitude to the student.

To learn more about the information, it is important to go to the "Expert" tab. There are specific guidelines to check each job. Also, for training, you can start checking specifically selected tasks, and then get comments on grading: how to do it right, and how to avoid mistakes next time.

The unique site "I will solve the OGE" will help you prepare more effectively for the main state exam. Each student will know exactly what to expect on the test, and all examiners will be familiar with the requirements for checking papers.

Evaluation

The work consists of two modules: "Algebra and geometry". There are 26 tasks in total. Module "Algebra" "Geometry"

3 hours 55 minutes(235 minutes).

as a single digit

, squarecompass Calculators on the exam not used.

passport), pass and capillary or! Allowed to take with myself water(in a transparent bottle) and food

The work consists of two modules: "Algebra and geometry". There are 26 tasks in total. Module "Algebra" contains seventeen tasks: in part 1 - fourteen tasks; in part 2 - three tasks. Module "Geometry" contains nine tasks: in part 1 - six tasks; in part 2 - three tasks.

For execution examination work in mathematics 3 hours 55 minutes(235 minutes).

Answers to tasks 2, 3, 14 write down in the answer form No. 1 as a single digit, which corresponds to the number of the correct answer.

For the remaining tasks of part 1 the answer is a number or sequence of digits. Write your answer in the answer field in the text of the work, and then transfer it to the answer sheet No. 1. If the response received common fraction, convert it to decimal.

When performing work, you can use the containing basic formulas course of mathematics issued together with the work. You are allowed to use a ruler, square, other templates for building geometric shapes (compass). It is forbidden to use tools with reference materials. Calculators on the exam not used.

You must have an identity document with you for the exam. passport), pass and capillary or gel pen with black ink! Allowed to take with myself water(in a transparent bottle) and food(fruit, chocolate, buns, sandwiches), but may be asked to leave in the hallway.

Portal for the student. Self-training

Tasks of the "Geometry" module in the OGE

Tasks of the Algebra module in the OGE in mathematics

Part 1

Module "Algebra"

Geometry module

Part 2

Module "Algebra"

How to prepare for exams?

How to find the necessary information on the site?

New User Registration

Registration data

Catalog of popular tasks

Useful information for experts

Evaluation

RELATED ARTICLES