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Option 1

Part B

Task B1.To paint the walls with a total area of ​​175 m 2, it is planned to purchase paint. The volume and cost of paint cans are shown in the table.

What minimum amount (in rubles) will be spent on the purchase of the required amount of paint, if its consumption is 0.2 l / m 2?

Solution.

Since on 1 m 2 takes 0.2 liters of paint, then 175 m 2 will require a volume of paint equal to 175 0.2 \u003d 35 liters.

Thus, the task is to find the minimum purchase price for 35 or more liters of paint.

Let's determine the cost of 1 liter of paint in each of the cans.

The price of a liter in a 2.5-liter can is: 75,000:2.5 = 30,000 rubles, and the price of a liter in a 10-liter can is 270,000:10 = 2,700 rubles.

Since paint is cheaper in large cans, it is advisable to collect 35 liters of paint using only large cans. However, you cannot get exactly 35 liters with the help of large cans, since each of the cans has a volume of 10 liters. There are two options here:

1. We buy 4 cans of paint, 10 liters each. As a result, we have 40 liters of paint, which exceeds the 35 liters we need. The price of paint in this case: 270,000 4 = 1,080,000 rubles.

2. We buy 3 cans of paint of 10 liters and 2 cans of paint of 2.5 liters. As a result, we have exactly 35 liters of paint. Paint price in this case: 3 270,000 + 2 75,000 = .960,000 rubles.

Since the second option is cheaper than the first, the minimum amount required to purchase the right amount of paint is 960,000 rubles.

Answer: 960 000.

Do you have any questions or comments about the solution to the problem? Ask them to the author, Anton Lebedev.

Task B2.Find the sum of the roots (root, e c is it the only one) equations

Solution.

First, note that squaring both sides of the equation is not a good idea in this task, as the result will be a 4th degree equation, which in general cannot be solved

In such situations, workarounds should be sought.

First, let's define the ODZ equation:

The resulting equation is equivalent to the system:

Comment.The first inequality of the system is necessary in order to avoid the appearance of extra roots: if we simply square both parts, then the roots of the equation will also be added to the roots of the equation.

So, we solve the equation from the written system:

Obviously, only the second of the found roots satisfies the inequality from the system.

Thus, the original equation has only one root, equal to 9.

Answer: 9.

Task B3.A circle is inscribed in an isosceles trapezoid whose area is . The sum of two angles of a trapezoid is 60°. Find the perimeter of the trapezoid.

Solution.

Let ABCD is a given trapezoid.

Since the trapezoid is isosceles, the angles at the base of the trapezoid are equal:

.

By convention, the sum of two angles of a trapezoid is 60°. Obviously, we are talking about two acute angles, since 60 °< 90° , which means that in our notation we are talking about the angles BAD and CDA . Since they are equal, and their sum is 60°, then each of them is equal to 30°.

As you know, not every trapezoid (and not every isosceles trapezoid) can be inscribed in a circle, which means that the fact that a circle is inscribed in our trapezoid gives us some additional information. A circle can only be inscribed in a trapezoid in which the sum of the bases is equal to the sum of the sides. In our case it should be:

Since the trapezoid is isosceles, then AB=CD. Let's denote the sides by x.

Then we get

where MN - median line of the trapezoid.

We also express the height of the trapezoid VK in terms of x. To do this, consider a right triangle ABK.

.

Then the sum of the sides is 2 x= 17, and the perimeter of a trapezoid is 34 (the sum of the bases is equal to the sum of the sides).

Answer: 34.

Task B4.Let (x, y)- solution of the system of equations

Find the value of an expression 5y-x.

Solution.

We transform the second equation of the system:

Taking into account the first equation, we get:

Calculate the value of the expression:

Answer: 23.

Task B5.Find the value of an expression

Solution.

Comment.The most common problems of applicants when solving such examples are the inability to get rid of irrationality in the denominator by multiplying by the conjugate and ignorance that the order of calculating successive roots does not matter (for example,).

Answer:-22.

Task B6.Find the sum of the roots of the equation.

Solution.

Before starting the solution, we say the magic phrase: "The product is equal to zero if at least one of the factors is equal to zero." After that, the equation miraculously breaks down into a set:

The first set equation has a single root x = 81.

Let's transform the second equation:

The further solution is carried out using a change of variable:

We get

(the roots are found using the inverse Vieta theorem).

The negative root does not suit us, so we get

This means that the original equation has two roots: 1 and 81.

Their sum is 82.

Answer: 82.

Task B7.Find the area of ​​the lateral surface of a regular triangular pyramid if the length of the bisector of its base is equal and the flat angle at the top is equal to .

Solution.

Let SABC is a regular triangular pyramid.

Triangle ABC - the base of the pyramid, and this triangle is correct.

The bisector is also the height of the triangle ABC, so

The lateral surface area of ​​a regular pyramid is S=SK· p,

where

- half-perimeter of the base;

Apothem.

Then

S = 125 = 60 .

Answer: 60.

Task B8.Find the sum of the smallest and largest integer solutions of the inequality

Solution.

Considering that the logarithm is an increasing function if its base is greater than 1 and decreasing if its base is less than 1, and also that the sublogarithm expression must be positive, we get:

The smallest integer solution is -5, and the largest is 65. Their sum is 60.

Answer: 60.

Task B9.Find (in degrees) the sum of the roots of the equation 10sin5 x cos5 x+5sin10 x co18 x= 0 on the interval (110° ; 170° ).

Solution.

Using the double argument formula, we transform the first term of the left side:

Since from all the roots found, you need to choose those that lie on the interval (110 °; 170 °), then

We write out the corresponding roots:

126°; 144°; 162°

130°; 150°.

The sum of the found solutions is 712.

Answer: 712.

Task B10.Find the product of the smallest and largest integer solutions of the inequality

Solution.

Let's transform the original inequality:

The resulting inequality can be solved, for example, by the interval method. To do this, we first find the roots of the corresponding equation:

The found roots will be plotted on the numerical axis. These roots break the expression (| x + 5| - 4)(|x- 3| - 1) on intervals of sign constancy. Let's determine the sign of the written expression on each of the intervals by substituting any point from the given interval into the expression. For example, to determine the sign of the expression on the extreme right interval, take the point x= 5 and we get that the value of the expression at this point is positive, which means that the expression will be positive on the entire interval.

Now we can write down the solution of the inequality (the corresponding area is shaded in the figure):

.

The smallest integer from this area: x min = -8, and the largest integer x max = 3. The product of these numbers is -8 3 = -24. This number should be written in the answer.

Answer:-24.

Task B11.Point A moves along the perimeter of the triangle KMP. points K1 , M 1, P 1 lie on the medians of the triangle KMP and divide them in the ratio 11:3, counting from the tops. Along the perimeter of the triangle K 1 M 1 P 1 point B moves at a speed five times greater than the speed of point A. How many times does point B go around the perimeter of the triangle K 1 M 1 P 1 for the time it takes for point A to go around the perimeter of the triangle twice KMP.

Solution.

Let's make a drawing for the task. O is the point of intersection of the medians of the original triangle.

Intuitively, triangles KMP and K 1 M 1 P 1 should be similar. However, intuition only suggests a way to solve the problem, so the similarity of these triangles still needs to be proved.

To prove the similarity, consider the triangles COM and K 1 OM 1 .

MM' is the median of the triangle KMP , therefore , since the medians of a triangle are divided in a ratio of 2 to 1, counting from the top.

It follows from the condition of the problem that , since the point M 1 divides the median of MM' by a ratio of 11 to 3, counting from the top.

Then

Attitude

.

Similarly, one can show that

Moreover, like vertical.

So the triangles COM and K 1 OM 1 are similar in two sides and the angle between them with a similarity coefficient .

Then

Similarly

.

This means that triangles KMP and K 1 M 1 P 1 are similar with similarity coefficient and triangle perimeter KMP times the perimeter of the triangle K 1 M 1 P 1 .

Since point B moves at a speed 5 times greater than the speed of point A along a triangle, the perimeter of which is one time less than the perimeter of the triangle KMR, then during one revolution of point A, point B makes revolutions, and during two revolutions of point A, point B makes 56 revolutions.

Answer: 56.

Task B12.Volume of a cuboid ABCDA 1 B 1 C 1 D 1 is equal to 1728. The point P lies on the side edge CC 1 so that CP:PC 1 = 2:1. Through point P, vertex D and the middle of the lateral rib AA 1, a cutting plane is drawn, which divides the rectangular parallelepiped into two parts. Find the volume of the smaller part.

Solution.

Draw a parallelepiped in the drawing and construct the described section PDKEF. K- mid-rib AA 1 .

Let us depict in the drawing the lines along which the section plane intersects the planes of the three faces of the parallelepiped. Points where the section plane intersects lines BA, BC and BB 1 denoted by Z, Q, S.

Body SZBQ- a pyramid with a right triangle at its base ZBQ . This pyramid includes the volume of the lower part of the parallelepiped and the volumes of three pyramids SEB 1 F, QPCD, ZKAD.

To find the volume of the lower part of the parallelepiped, we find the volumes of the indicated pyramids.

For the convenience of calculations, we denote the sides of the parallelepiped through x, y and z, then the volume of the parallelepiped V = xyz = 1728.

Moreover,

.

The problem is to express the dimensions of these four pyramids in terms of x, y and z.

triangles FC 1 P and DAK are similar in two angles (all sides of these triangles are pairwise parallel).

Then

.

triangles PCD and KA 1 E are also similar, so

.

From the similarity of triangles SB 1 F and PC1 F follows:

.

Pyramid Volume SEB 1 F equals:

Pyramid QPCD like a pyramid SEB 1 F with similarity coefficient:

.

Then the volume of the pyramid QPCD equals:

similar pyramid ZKAD like a pyramid SEB 1 F with similarity coefficient

Then the volume of the pyramid ZKAD equals:

Finally, the pyramid SZBQ like a pyramid SEB 1 F with similarity coefficient

.

Then the volume of the pyramid SZBQ equals:

The volume of the lower part of the parallelepiped:

Then the volume of the upper part:

Since we need a smaller volume, the correct answer is 724.

Answer: 724.

Applicants of the Lyceum of BSU can get acquainted with the options for entrance examinations in 2019. The purpose of these options is to enable any participant in the entrance examinations to the Lyceum of the Belarusian State University to get an idea about the structure of the exam options, types of tasks and their levels of complexity. When reviewing the 2019 options, it should be borne in mind that the tasks included in them do not cover all the content elements that will be tested at the entrance examinations at the Lyceum of BSU in 2020. In addition to this, the tasks of the entrance examinations in 2020 will be drawn up in accordance with the new curricula. You can learn more about the structure of entrance examinations in 2020 by studying the specifications and solving posted on the LMS of the Lyceum of BSU.

Completing the variants of entrance examinations in 2019 will allow test takers to develop a strategy for preparing for admission to the Lyceum of BSU, systematize the studied material, prevent possible mistakes, as well as consolidate knowledge and effectively prepare for entrance examinations in 2020.

RIKZa has a specification for each subject of the DH for 2016. It explains what the structure of the test will be, how many tasks of each level of difficulty are in the test, and what program material will be used in them.

The photo is illustrative. Photo: Vadim Zamirovsky, TUT.BY

So, this year the Russian language test will consist of 40 tasks: 30 - in part A and 10 - in part B. The most tasks will be for spelling - 13, for punctuation - 9 tasks, the least for phonetics - one. There will be two tasks for the first level of difficulty, four tasks for the second, four tasks for the third and fourth, and six tasks for the fifth, the most difficult. You have 120 minutes to complete the test.

In the math test this year, there are 8 tasks in geometry (more than last year), 11 tasks in equations and inequalities, four tasks each in numbers and calculations and functions. The first level will have only two tasks, the second - eight, and the most tasks will be on the third level - 14. The fourth and fifth levels will have 4 and 2 tasks, respectively.

Alexander Nikolaevich, a math tutor since 2007 whose students are olympiad winners, BSU lyceum students and BSU students, believes that it is almost impossible to judge the complexity of a test from the specification.

Maybe there are more assignments on certain topics this year. But, in my opinion, this information does not greatly affect the preparation of the applicant. It's not about the number of jobs. One section of a math test might have some pretty strong items, but five or four isn't very important. Without seeing the tasks themselves, I would refrain from commenting that information about the specification will somehow affect the training of applicants.

Russian language tutor with 15 years of experience Lyudmila Grigorievna also does not believe that the specification somehow affects the process of preparing for the DT: “ The rules remain the same and you just need to know them. What proportion of tasks is not so important».

Recall that Belarus has already approved. Applicants take the first test on June 13 in the Belarusian language and on June 14 in the Russian language.

June 25— foreign language (English, German, French, Spanish, Chinese);

Each test starts at 11:00 am. Reserve day date - 5'th of July(Tuesday). The DT will be held at the Belarusian State University on this day, you can sign up for it from June 28 to July 1.