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Task 2. Development of a Method for Optimizing a Mathematical Problem
Problem Statement. At Unit 3 of an oil and gas company, the task is to optimize the operation of the navigation system elements. The task of this unit is to receive encrypted signals from Unit 1 and Unit 2, synthesize the incoming data packets, and ... | Solution to the problem. Solving the problem "head-on" by raising one number to the power of another is bound to exceed the computational power not only of a single computer but even a data center would require a certain amount of time to perform the calculations. Since we only need to send the last digit, let's focus ... | 6 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,417 |
11. It is necessary to clearly follow the requirements for the volume of work if it is specified in the assignment.
## Task 1 (Maximum 12 points)
Let there be 5 banks in the country Alpha. The probability that a bank will close is the same for all banks and is equal to 0.05. Banks close independently of each other. A... | # Solution:
Let's introduce the following events: event A - at least one bank in country Alpha has closed, event $\mathrm{A}_{\mathrm{i}}$ - the $\mathrm{i}$-th bank has closed, event $\mathrm{B}_{\mathrm{i}}$ - the $\mathrm{i}$-th bank has not closed. Then the opposite of event A (denoted as event B) states that no b... | 0.54 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,418 |
1. Let $a_{1}, \ldots, a_{2015}$ be integers, and $b_{1}, \ldots, b_{2015}$ be the same numbers in a different order. Can the equality $\left(a_{1}-b_{1}\right) \ldots\left(a_{2015}-b_{2015}\right)=2017^{2016}$ hold? | Justify your answer. (8 points)
## Solution:
The equality cannot hold, otherwise all $\left(a_{i}-b_{i}\right)$ would be odd integers and there would be an odd number of them. Consequently, their sum should be odd, but it equals $\mathbf{0}$.
Answer: No, it cannot.
| Points | Evaluation Criteria for Task № 1 |
| :-... | proof | Number Theory | proof | Yes | Yes | olympiads | false | 7,420 |
# 2. Solve the inequality:
$$
\{x\}([x]-1)<x-2,
$$
where $[x]$ and $\{x\}$ are the integer and fractional parts of the number $\boldsymbol{x}$, respectively (9 points).
# | # Solution:
Let $\boldsymbol{a}=[\boldsymbol{x}], \boldsymbol{b}=\{\boldsymbol{x}\}$, then $\boldsymbol{x}=\boldsymbol{a}+\boldsymbol{b}$. Considering this, we get:
$$
\begin{gathered}
b(a-1)0, \boldsymbol{a}>2,[x]>2$. The last inequality is equivalent to $x \geq 3$. This will be the desired solution.
Answer: $\math... | x\geq3 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,421 |
3. There are 48 matches. They are laid out in three unequal piles. Then three transfers of matches from pile to pile are performed. First, from the first pile to the second, as many matches are transferred as there were in the second pile. Then, from the second pile to the third, as many matches are transferred as ther... | # Solution:
We solve the problem from the end. As a result of all three transfers, the number of matches in all piles became the same, i.e., 48: 3 = 16 matches in each pile. During the third transfer, the number of matches in the first pile was doubled by adding as many matches as it already had. Therefore, before the... | 22;14;12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,422 |
4. In square $A B C D$ with side 2, point $A_{1}$ lies on $A B$, point $B_{1}$ lies on $B C$, point $C_{1}$ lies on $C D$, point $D_{1}$ lies on $D A$. Points $A_{1}, B_{1}, C_{1}, D_{1}$ are the vertices of the square of the smallest possible area. Find the area of triangle $A A_{1} D_{1} .(\mathbf{1 1}$ points) | # Solution:
We need to prove that the square of the smallest area is obtained under the condition that points $A_{1}, B_{1}, C_{1}, D_{1}$ are the midpoints of the sides of the original square. This is equivalent to proving that the minimum length of the segment $\boldsymbol{A}_{1} \boldsymbol{D}_{1}$ is obtained when... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,423 |
5. Does there exist a natural number that, when divided by the sum of its digits, gives both a quotient and a remainder of 2014? If there is more than one such number, write their sum as the answer. If no such numbers exist, write 0. (12 points)
# | # Solution:
Suppose there exists a natural number $\boldsymbol{n}$ with the sum of its digits $\boldsymbol{s}$, such that $n=2014 s+2014$, from which we get $n-s=2013 s+2014$. By the divisibility rule, $n-s$ is divisible by 3. However, the number $2013 s+2014$ is not divisible by 3, since the number $2013s$ is a multi... | 0 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,424 |
6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, under all other ... | # Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3... | 1.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,425 |
7. Let in the country of Wonder, only three goods are produced and consumed: good alpha, good beta, and good gamma. The volume of their consumption and prices in 2009 and 2015 are given in the table below. ( $\mathbf{\text { ba points) }}$
| | 2009 year | | 2015 year | |
| :--- | :--- | :--- | :--- | :--- |
| | Pr... | # Solution:
(a) Since 2009 is the base year, the real and nominal GDP in this year are the same and equal to:
$\mathrm{Y}_{\mathrm{n}}{ }^{09}=\Sigma \mathrm{p}_{\mathrm{i}}{ }^{0} \mathrm{q}_{\mathrm{i}}{ }^{0}=5 * 12+7 * 8+9 * 6=60+56+54=170$.
Nominal GDP for 2015: $\mathrm{Y}_{\mathrm{n}}{ }^{15}=\Sigma \mathrm{p... | -4.12,101.17 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,426 |
8. Calculating Samson loves to have lunch at the Italian restaurant "At Pablo's". During his next visit to the restaurant, Samson was offered to purchase a loyalty card for a period of 1 year at a price of 30000 rubles, which gives the client a $30 \%$ discount on the bill amount.
(a) Suppose that during the week Sams... | # Solution:
(a) After purchasing the card, one lunch will cost the client $900 * 0.3 = 270$ rubles less. Since there are 52 full weeks in a year $(365 / 7 = 52.14)$, Samson will save $270 * 3 * 52 = 42120$ rubles on discounts over the year, which is more than the cost of the card. Therefore, it is beneficial for him t... | 167 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,427 |
9. Consider two countries, A and B, which have the same arable land. On these lands, they can only grow eggplants and corn. The alternative costs of producing either crop in each country are constant. The yield of both crops, if the entire area is planted with only one of the crops, is given in the table below: (11 poi... | # Solution:
(a) Country B has an absolute advantage in growing each product, as it can grow more of each product on the same area of land than Country A.
The opportunity cost of growing one unit of eggplant in Country A is $8 / 10 = 0.8$ units of corn. That is, to increase the production of eggplant by 1 unit, one mu... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,428 |
10. Consider the labor market for doctors. The Ministry of Health proposes to reintroduce job placement for graduates of medical universities and make working in state medical institutions mandatory for all young doctors. (12 points)
(a) Suppose the government introduces a minimum number of years that young doctors mu... | # Solution:
Let's present a possible argumentation for the solution, which is not the only or absolute one.
(a)
When introducing an additional restriction, the demand for studying on state-funded places may initially decline, which will subsequently lead to a decrease in the supply of doctors in the public sector. M... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,429 |
6. In a perfectly competitive market, the demand function for a certain good is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=150-\mathrm{p}$, and the supply function for this good is: $\mathrm{Q}_{\mathrm{s}}(\mathrm{p})=3 \mathrm{p}-10$. As a result of a sharp increase in the number of consumers of this good, the demand for i... | # Solution:
The new demand function will be $Q_{d}^{\text {new }}(p)=a(150-p)$.
Find the new equilibrium price from the condition of equality of demand and supply: $3 \mathrm{p}-10=\mathrm{Q}_{s}(\mathrm{p})=\mathrm{Q}_{d}^{\text {new }}(\mathrm{p})=a(150-\mathrm{p}): \quad \mathrm{p}^{\text {new }}=\frac{150 a+10}{3... | 1.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,430 |
# Problem 1. Maximum 15 points
A company that produces educational materials for exam preparation incurs average costs per textbook of $100+\frac{100000}{Q}$, where $Q$ is the number of textbooks produced annually. What should be the annual production volume of the textbook to reach the break-even point if the planned... | # Solution
At the break-even point $\mathrm{P}=\mathrm{ATC}=\mathrm{MC}$
Form the equation $100+10000 / Q=300$
$100 \mathrm{Q}+100000=300 \mathrm{Q}$
$100000=200 \mathrm{Q}$
$\mathrm{Q}=100000 / 200=500$
## Evaluation Criteria
1. The correct answer is justified: 15 points | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,431 |
# Task 3. Maximum 20 points
The wizard has revealed the secret of wisdom. For those who wish to know exactly where the secret is hidden, he left a clue in his magic book: $5 \cdot$ BANK $=6 \cdot$ GARDEN
Each letter in this clue represents a certain digit. Find these digits and substitute them into the GPS coordinate... | # Solution
The hint is presented as an equation. Let's write it down and solve it in integers.
$$
5 * \mathrm{~S} * 1000+5 * \mathrm{~A} * 100+5 * \mathrm{H} * 10+5 * \mathrm{~K}=6 * \mathrm{C} * 100+6 * \mathrm{~A} * 10+6 * D
$$
Using the rules of multiplication and divisibility, we can say that D is either 5 or 0,... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,433 |
# Problem 4. Maximum 20 points
Svetlana is convinced that any feline can be interesting to her for two purposes - catching mice and conducting cat therapy. Based on these considerations, Svetlana has decided to acquire 2 kittens, each of which has an equal probability of being either a male or a female. The weekly pro... | # Solution:
(a) Males catch mice better than females. Two males can catch $40 \cdot 2=80$ mice in a week.
Answer: 160 (2 points).
(b) There are 3 options: one male and one female, two males, two females.
Answer: 3 (2 points).
(c) Two males with individual PPFs: $M=80-4K$. The opportunity costs for males are consta... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,434 |
# Task 5. Maximum 15 points
Today, many commercial banks implement and actively promote cashback services. The main idea of such services is that when making purchases, customers are partially refunded in rubles or bonus points to their account (cashback). Often, when paying for purchases with bonus points, one point ... | # Solution:
Attracting new customers (5 points); mutually beneficial cooperation with commercial organizations (5 points); if other banks implement cashback services, ignoring this segment can negatively affect customer interest in the bank's services (5 points).
For part (a), no more than 10 points.
(b) All other t... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,435 |
# Problem 6. Maximum 15 points
Find the values of the variable $x$ such that the four expressions: $2 x-6, x^{2}-4 x+5, 4 x-8, 3 x^{2}-12 x+11$ differ from each other by the same number. Find all possible integer values of the variable $x$ for any order of the expressions. | # Solution:
From the properties of numbers that differ from the following in a numerical sequence by the same number, we form two equations:
$$
\begin{gathered}
x^{2}-4 x+5-(2 x-6)=4 x-8-\left(x^{2}-4 x+5\right) \\
4 x-8-\left(x^{2}-4 x+5\right)=3 x^{2}-12 x+11-(4 x-8)
\end{gathered}
$$
Only \( x=4 \) satisfies both... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,436 |
# Task 1. Maximum 15 points
In a small town, three main straight highways converge pairwise at points A, B, and C. The town has a circular ring road that also passes through points A, B, and C and is represented by a circle. The lengths of the highways inside the circle are 7, 8, and 9 kilometers. At the center of thi... | # Solution.
Mathematical formulation of the problem:
The sides of the triangle are 7, 8, and 9. Find the distance between the centers of the inscribed and circumscribed circles.
The sides of triangle \( \mathrm{ABC} \) are: \( \mathrm{AB}=7, \mathrm{BC}=8, \mathrm{AC}=9 \).
\( \mathrm{O} 2 \) - center of the inscri... | 1024.7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,437 |
# Problem 4. Maximum 15 points
A trader sells stocks. It is known that stocks are sold once a month except during the summer, when the trader is on vacation. After economic analysis and entering the results of the past year into a table by months, it was found that the number of stocks sold corresponds to a $3 * 3$ ma... | # Solution:
Let's sum up the three rows in the square, then this sum is obviously equal to 306. On the other hand, since the numbers in the square form a sequence with a difference of one, we can use an arithmetic progression, where the first term is the minimum number in the square (let's call it a), and the maximum ... | \begin{pmatrix}37&32&33\\30&34&38\\35&36&31\\\end{pmatrix} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,439 |
# Problem 5. Maximum 15 points
The great-grandfather-banker left a legacy to his newborn great-grandson. According to the agreement with the bank, the amount in the great-grandson's account increases. Every year, on the day after the birthday, the current amount is increased by 1 million rubles more than in the previo... | # Solution:
Since the number consists of identical digits, it can be represented as 111 multiplied by a. According to the problem, the same number should be obtained as the sum of an arithmetic progression. The first element of the progression is 1, the last is \( \mathrm{n} \), and the number of elements in the progr... | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,440 |
3. Two females with PPF $M=40-2K$. PPFs are linear with identical opportunity costs. By adding individual PPFs, we get that the PPF of the two females: M=80-2K, $\mathrm{K} \leq 40$. | Answer: $\mathrm{M}=80-2 \mathrm{~K}, \mathrm{~K} \leq 40$. (1 point)
Two males with the production possibility frontier (PPF) M=64-K^2. The PPFs have monotonically increasing alternative costs, which coincide at each K_1=K_2. To obtain the maximum amount of M for each value of K, choose K_1=K_2=0.5K. The joint PPF of... | \mathrm{M}=104-\mathrm{K}^{\wedge}2,\mathrm{~K}\leq1\\\mathrm{M}=105-2\mathrm{~K},1<\mathrm{K}\leq21\\\mathrm{M}=40\mathrm{~K}-\mathrm{K}\wedge2-336,21<\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,441 |
# Task 7. Maximum 15 points
Some producers of luxury goods create an artificial shortage of their products. In particular, some brands of Swiss watches do this. Despite having sufficient labor and capital resources, they produce watches in smaller quantities than consumers are willing to buy at a certain price. For ex... | # Solution:
Companies can create artificial scarcity for luxury goods, which allows them to sell such goods at a higher added value in the future. If they sell all the watches to everyone who wants them today at the current price, there is a risk that in the next period, people may not be willing to pay a relatively h... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,443 |
# Task 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ mon... | # Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,444 |
# Task 2. Maximum 30 points
In a certain country, there are two regions - northern and southern. The northern region has 24 inhabitants, while the southern region has four times fewer. Both regions are capable of producing goods X in quantity $x$ and Y in quantity $y$. Due to climatic peculiarities, the production cap... | # Solution:
(a) In the northern region, there are 24 residents, and in the southern region, there are $\frac{24}{4}=\mathbf{6}$. Both regions operate independently of each other. Let's determine the production capabilities of each region. Given that the individual production possibility frontier (PPF) of each resident... | 0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,445 |
# Task 3. Maximum 20 points
At the conference "Economics of the Present," an intellectual tournament was held, in which more than 198 but fewer than 230 scientists, including doctors and candidates of sciences, participated. Within one match, participants had to ask each other questions and record correct answers with... | Solution: Let there be $\mathrm{n}$ scientists participating in the tournament, of which $\mathrm{m}$ are doctors and $\mathrm{n}-\mathrm{m}$ are candidates of science. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the doctors of science ... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,446 |
# Task 4. Maximum 15 points
A few years ago, the Russian national payment system "Mir" launched a special bonus program. To participate in this loyalty program, "Mir" cardholders of the bank must register on a special portal. After completing the registration, the card automatically becomes part of this bonus program,... | # Solution:
(a) (5 points) Possible answers: | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,447 |
# Task 5. Maximum 15 points
In the treasury of the Magic Kingdom, they would like to replace all old banknotes with new ones. There are a total of 3,628,800 old banknotes in the treasury. Unfortunately, the machine that prints new banknotes requires major repairs and each day it can produce fewer banknotes: on the fir... | # Solution:
(a) If a major repair is to be carried out, it is most effective to do so on the second day, as this will allow the production of new banknotes to increase from 604,800 to 1 million on that day, and the production will also be 1 million in subsequent days, which is more than the possibilities without major... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,448 |
# Task 1. Maximum 20 points
Let's say in country A there are only two firms engaged in research and development in a certain field. Currently, each of them independently decides whether to participate in the development of a new technology. It is known that if a firm develops a new technology, it will bring it $V$ mon... | # Solution and Grading Scheme:
(a) For each firm to decide to participate in the development, it is necessary that the expected profit from participation for each firm is greater than 0 (we use the formula for expected income in the case where both firms are involved in development):
$\alpha(1-\alpha) V+0.5 \alpha^{2... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,449 |
# Task 2. Maximum 30 points
In a certain country, there are two regions - northern and southern. The northern region has 24 inhabitants, while the southern region has four times fewer. Both regions are capable of producing goods X in quantity $x$ and Y in quantity $y$. Due to climatic peculiarities, the production cap... | # Solution:
(a) In the northern region, there are 24 residents, and in the southern region, there are $\frac{24}{4}=6$. Both regions operate independently of each other. Let's find the production capabilities of each region. Given that the individual production possibility frontier (PPF) of each resident in the northe... | 0.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,450 |
# Task 3. Maximum 20 points
At the "Economics and Law" congress, a "Tournament of the Best" was held, in which more than 220 but fewer than 254 delegates—economists and lawyers—participated. Within one match, participants had to ask each other questions within a limited time and record the correct answers. Each partic... | Solution: Let there be $\mathrm{n}$ delegates participating in the tournament, of which $\mathrm{m}$ are economists and $\mathrm{n}-\mathrm{m}$ are lawyers. All participants conducted $\mathrm{n}(\mathrm{n}-1) / 2$ matches and scored $\mathrm{n}(\mathrm{n}-1) / 2$ points. Among them, the economists competed in $\mathrm... | 105 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,451 |
# Task 4. Maximum 15 points
A few years ago, the Russian national payment system "Mir" launched a special bonus program. To participate in this loyalty program, Mir cards of bank clients must register on a special portal. After completing the registration, the card automatically becomes a participant in this bonus pro... | # Solution:
(a) (5 points) Possible answers: | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,452 |
# Task 5. Maximum 15 points
In the treasury of the Magic Kingdom, they would like to replace all old banknotes with new ones. There are a total of 3,628,800 old banknotes in the treasury. Unfortunately, the machine that prints new banknotes requires major repairs and each day it can produce fewer banknotes: on the fir... | # Solution:
(a) If a major repair is to be carried out, it is most effective to do so on the second day, as this will allow the production of new banknotes to increase from 604,800 to 1 million on that day, and the production will remain at 1 million in subsequent days, which is higher than the capacity without major ... | no | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,453 |
1. Find the minimum loss, which is EXPENSE - INCOME, where the letters $\boldsymbol{P}, \boldsymbol{A}, \boldsymbol{C}, \boldsymbol{X}, \boldsymbol{O}, \boldsymbol{D}$ represent digits forming an arithmetic progression in the given order. (2 points).
# | # Solution:
The difference in the progression is 1; otherwise, 6 digits will not fit (if the "step" is 2, then 6 digits will exceed the field of digits). The smaller the first digit, the smaller the loss. Therefore, $\boldsymbol{P}=1, \boldsymbol{A}=2, \boldsymbol{C}=3, \boldsymbol{X}=4, \boldsymbol{O}=5, D=6$. Thus, ... | 58000 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,454 |
3. In a certain state, only liars and economists live (liars always lie, while economists tell the truth). At a certain moment, the state decided to carry out monetary and credit, as well as budgetary and tax reforms. Since it was unknown what the residents expected, everyone was asked several questions (with only "yes... | # Solution:
Let $\boldsymbol{x}$ be the proportion of liars in the country, then (1-x) is the proportion of economists. Each economist answers affirmatively to one question, and each liar answers affirmatively to three. Therefore, we can set up the equation:
$3 x+1-x=0.4+0.3+0.5+0 ; 2 x=0.2 ; x=0.1$. Thus, in the cou... | 30 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,455 |
4. There are 2015 coins on the table. Two players play the following game: they take turns; on a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. How many coins should the first player ta... | # Solution:
The strategy of the first player: he takes 95 coins, and then on each move, he takes (101-x) coins, where $\boldsymbol{x}$ is the number of coins taken by the second player. Since $\boldsymbol{x}$ is even (by the condition), 101-x is odd. Then $2015-95=1920$, since 101-x+x=101 coins will be taken per move,... | 95 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,456 |
5. Let the number of people employed in city $N$ be 15 million. It is known that the unemployment rate in this city is $\mathbf{8 \%}$. Find the number of unemployed people (in million). Round your answer to the nearest thousandth. (12 points).
# | # Solution:
$\boldsymbol{L}=\boldsymbol{U}+\boldsymbol{E}, \boldsymbol{U}=\boldsymbol{L}-15$, since $\boldsymbol{E}=15$ (where $\boldsymbol{E}-$ employed (working), $\boldsymbol{L}$ - labour force, $\boldsymbol{U}$ unemployed (unemployed).
$u=U / L^{*} 100=8, \quad(L-15) / L * 100=8, \quad 8 L=100 L-1500, \quad 92 L=... | 1.304 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,457 |
6. The bank issued a loan to citizen $N$ on September 9 in the amount of 200 mln rubles. The repayment date is November 22 of the same year. The interest rate on the loan is $25 \%$ per annum. Determine the amount (in thousands of rubles) that citizen N will have to return to the bank. Assume that there are 365 days in... | # Solution:
Number of days of the loan: September - 21 days, October - 31 days, November - 21 days, i.e., $21+31+21=73$ days. The accrued debt amount is calculated using the formula:
$\boldsymbol{F V}=\boldsymbol{P V} \cdot(\boldsymbol{1}+\boldsymbol{t} \cdot \mathbf{Y}$ ), where $\boldsymbol{F} \boldsymbol{V}$ - the... | 210 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,458 |
7. You are the owner of a company selling Tesla Model S electric vehicles. The purchase price of one car is 2.5 million rubles. To transport one car across the border, you need to pay customs duties and fees totaling 2 million rubles per car. Monthly office rent is 50,000 rubles, and the total monthly salary of employe... | # Solution:
a) The sum of monthly costs to ensure the deal is: $(2.5+2) \cdot 30+(0.05+0.370+0.18)=135.6$ million rubles. The average cost per car will be $135.6 / 30=4.52$ million rubles. This is the minimum amount for which the company will be willing to sell one car. If the car's cost is lower, the company will inc... | 4.52 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,459 |
# Task 1. Maximum 20 points
In a certain city, all residents can be divided into three conditional, equal in number groups: poor, middle, and rich. The income of the poor group constitutes $x$ percent of the total income of all city residents, the income of the middle group constitutes $3 x$ percent, and that of the r... | # Solution:
First, we need to find 1 share $x$ by adding up all the shares of citizens.
We get: $6x + 3x + 1x = 10x$, from which we find that $1x$ equals $10\%$ ( $10x = 100\%, x = 10\%$)
Now we need to find the share of income of the citizens:
The share of income of the poor is: $1 * 10\% = 10\%$
The share of inc... | 22,36,42 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,460 |
# Task 2. Maximum 20 points
In recent times, banking products have been developing at a significant pace. Many banks offer their customers various loyalty programs, often including a cashback function, which is credited when paying for various categories of goods and services with bank cards. Typically, cashback for i... | # Solution:
(a) (10 points) By offering various loyalty programs, including cashback programs, banks have encountered the issue that these programs are often used by "too" financially savvy customers. When developing loyalty programs, banks typically calculate based on the average customer in a particular category of ... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,461 |
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and values the benefit of using the coffee machine at 10 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality, com... | # Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 10=20$ (1 point)... | 3 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,462 |
# Task 4. Maximum 20 points
A client of a brokerage firm deposited 10,000 rubles into a brokerage account at an exchange rate of 50 rubles per dollar, instructing the broker to invest the amount in foreign bank bonds with a guaranteed yield of $12\%$ per year in dollars.
(a) Determine the amount in rubles that the cl... | # Solution and Evaluation Criteria:
(a) The brokerage account received $10000 / 50 = 200$ dollars (1 point)
The stocks generated an income of $200 * 0.12 = 24$ dollars over the year (1 point)
At the end of the year, the account had $200 + 24 = 224$ dollars (1 point)
The broker's commission was $24 * 0.3 = 7.2$ doll... | 16476.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,463 |
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 150 matches can be played in the ... | # Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3 = 9$ games. The number of team pairs can be $150: 9 = 16.6 \ldots$ a maximum of 16. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs,... | 6 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,464 |
# Task 2. Maximum 20 points
In recent times, banking products have been developing at a significant pace. Many banks offer their customers various loyalty programs, often including a cashback function, which is credited when paying for various categories of goods and services with bank cards. As a rule, cashback for i... | # Solution:
(a) (10 points) By offering various loyalty programs, including cashback programs, banks have encountered the issue that these programs are often used by "too" financially savvy customers. When developing loyalty programs, banks typically calculate based on the average customer in a particular category of ... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,466 |
# Task 3. Maximum 20 points
Consider a consumer who plans to use a coffee machine. Let's assume the consumer lives for two periods and evaluates the benefit from using the coffee machine at 20 monetary units in each period. Coffee machines can be produced in two types: durable, working for two periods, and low-quality... | # Solution and Grading Scheme:
(a) When coffee machines are produced by a monopoly, it sets a selling price that will extract all consumer surplus:
1) when producing a durable machine, the price equals the consumer's benefit from using the coffee machine for 2 periods $\mathrm{p}_{\mathrm{L}}=2 \cdot 20=40$ (1 point)... | 6 | Other | math-word-problem | Yes | Yes | olympiads | false | 7,467 |
# Task 4. Maximum 20 points
A client of a brokerage firm deposited 12,000 rubles into a brokerage account at an exchange rate of 60 rubles per dollar, instructing the broker to invest the amount in bonds of foreign banks with a guaranteed yield of $12\%$ per year in dollars.
(a) Determine the amount in rubles that th... | # Solution and criteria for checking:
(a) The brokerage account received 12000 / $60=200$ dollars (1 point)
The stocks generated an income of $200 * 0.12=24$ dollars over the year (1 point)
At the end of the year, the account had $200+24=224$ dollars (1 point)
The broker's commission amounted to $24 * 0.25=6$ dolla... | 16742.4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,468 |
# Task 5. Maximum 20 points
In Moscow, a tennis tournament is being held. Each team consists of 3 players. Each team plays against every other team, with each participant of one team playing against each participant of the other exactly one match. Due to time constraints, a maximum of 200 matches can be played in the ... | # Solution:
Assume there are two teams - then each of the three members of one team plays against each of the other - i.e., $3 * 3=9$ games. The number of team pairs can be $200: 9=22.2 \ldots$ a maximum of 22. Two teams form only one pair; three teams form three pairs. If there is a fourth team, add 3 more pairs, so ... | 7 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,469 |
3. The remaining $60 \%$ of voters would not like to see any of the candidates as their deputy, but under equal conditions, they are willing to sell their vote to the candidate who offers them more money.
It is known that if a candidate offers 1 monetary unit for one voter's vote, they will receive only one additional... | # Solution and Grading | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,470 |
3. The remaining $40 \%$ of voters do not want any of the candidates to be their representative, but under equal conditions, they are willing to sell their vote to the candidate who offers them more money.
It is known that if a candidate offers 1 monetary unit for one voter's vote, they will receive only one additiona... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,471 | |
12. In the Tumba-Yumba tribe with a population of 30 people, a trader arrives. After studying the customs of the tribe, the trader proposes to play a game. For each natural exchange of goods conducted in the market by two tribespeople, the trader gives each participant one gold coin. If at the end of the day, two diffe... | Solution: In any company, there are at least two people who have the same number of acquaintances. Therefore, the natives had no chance. The chief proposed to distribute 270 coins, which means he knew that some natives would be removed. Let's say x people were removed. To distribute a different number of coins to every... | 24 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,472 |
2. Dima and Seryozha decided to have a competition around a circular lake. They start simultaneously from the same point. Seryozha drives a motorboat at a constant speed of 20 km/h and somehow crosses the lake (not necessarily along the diameter) in 30 minutes. During this time, Dima runs along the shore of the lake fo... | Solution: A trapezoid can be inscribed in a circle only if it is isosceles. The length of the segment resting on a chord of the same length in the same circle is the same. Dima runs 1.5 kilometers along the lake. This means they should run 1.5 kilometers together, at a closing speed of 12 km/h, which is 7.5 minutes. In... | 37.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,473 |
3. In recent years, the use of mortgage loans by young families has become quite popular. Consider the possibility of obtaining a mortgage with a constant, fixed interest rate. Assume that the repayment of such a loan is made through equal (annuity) payments at the end of each payment period stipulated by the contract.... | Solution: If the borrower immediately directs the funds to a partial prepayment, in this case, part of these funds will go towards paying the accrued interest for half the period, and the rest will go towards repaying the principal. As a result, due to the reduction in the principal amount, less interest will be accrue... | S-2T+rS-0.5rT+(0.5rS)^2<S-2T+rS | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,474 |
4. In the town of M, there are 3 types of people: fat, thin (and there are equal numbers of both), and well-fed, who make up a quarter of the town's population. One day, two old friends met at the train station - one thin and one fat. Starting to discuss old times, they, as often happens, moved on to discussing acute s... | # Solution:
The following are possible evaluation criteria.
1) Coefficient $\mathrm{R} / \mathrm{P} 10 \%$ or decile coefficient: $K_{D}=\frac{85}{17}=5$ - the income of the top $10 \%$ richest is five times that of the bottom $10 \%$ poorest
2) Lorenz coefficient: $L=\frac{1}{2} \sum\left|w_{i}-d_{i}\right|$, where ... | 0.12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,475 |
5. In a certain kingdom, the workforce consists only of the clan of dwarves and the clan of elves. Historically, in this kingdom, dwarves and elves have always worked separately, and no enterprise has ever hired both at the same time. The aggregate supply of labor resources of the dwarves is represented by the function... | # Solution:
Before the law was introduced, the wages of elves and gnomes were determined by the condition of equality of supply and demand in each market:
$w_{\text {gnome }}^{S}=w_{\text {gnome }}^{D}$, from which $1+\frac{L}{3}=10-2 L / 3$ and $L_{\text {gnome }}=9$ and $w_{\text {gnome }}=4$
$w_{\text {elf }}^{S}... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,476 |
7. In trapezoid $ABCD$, points $G$ and $H$ are taken on the base $AD$, and points $E$ and $F$ are taken on the base $BC$. Segments $BG$ and $AE$ intersect at point $K$, segments $EH$ and $GF$ intersect at point $L$, and segments $FD$ and $HC$ intersect at point $M$. The area of quadrilateral $ELGK$ is 4, and the area o... | Solution. Let the areas of triangles $ABK, KEG$ be denoted by $x$, the areas of triangles $ELG, FHL$ by $y$, and the areas of triangles $FMH, CDM$ by $z$. Then $x+y=4, y+z=8$. From these equations and the fact that $z$ is an integer, it follows that $x, y$ are also integers. Since $y \geq 1$, the possible values for $x... | 57 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,477 |
3. In recent years, the use of mortgage loans by young families has become quite popular. Consider the possibility of obtaining a mortgage with a constant, fixed interest rate. Assume that the repayment of such a loan is made through equal (annuity) payments at the end of each payment period stipulated by the contract.... | Solution: If the borrower immediately directs the funds to a partial prepayment, in this case, part of these funds will go towards paying the accrued interest for half the period, and the rest will go towards repaying the principal. As a result, due to the reduction in the principal amount, less interest will be accrue... | S-2T+rS-0.5rT+(0.5rS)^2<S-2T+rS | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,478 |
1. Through the terminal for paying to a mobile phone, money can be transferred, with a commission - a positive integer percentage. Fedya put an integer amount of rubles on his mobile phone, and his account was credited with 847 rubles. How much money did Fedya put into the account, given that the commission is less tha... | # Solution:
The equation is written as: $\boldsymbol{X} \cdot(\mathbf{100 - n}) / 100 = 847$, where $\boldsymbol{X}$ is the positive amount of money that Fedya deposited, and $\mathbf{n}$ is a positive number of percentage points of the commission, less than 30. From the equation, we find: $X=847 \cdot 100 /(100-n)$. ... | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,479 | |
2. In the triangular pyramid $SABC$, points $\boldsymbol{D}$, $\boldsymbol{E}$, and $\boldsymbol{F}$ are chosen on the edges $SA$, $SB$, and $SC$ respectively, such that the area of triangle $\boldsymbol{ABE}$ is $1/3$ of the area of triangle $\boldsymbol{AES}$, the area of triangle $\boldsymbol{BCF}$ is $1/4$ of the a... | # Solution:
Let there be a pyramid $SABC$ and a point $D$ on side $SA$, such that $SD:SA = d$, on side
the volume ratio of $SDEF$ to $SABC$ is $\boldsymbol{def}$. Indeed, let's "lay down" the... | 0.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,480 |
3. Given a 2015-digit number divisible by 9. Let the sum of its digits be $\boldsymbol{a}$, the sum of the digits of $\boldsymbol{a}$ be $\boldsymbol{b}$, and the sum of the digits of $\boldsymbol{b}$ be $\boldsymbol{c}$. Find the number $\boldsymbol{c}$. (14 points). | Solution:
The sum of the digits of any number gives the same remainder when divided by 9 as the number itself.
The largest 2015-digit number consists of 2015 nines. The sum of its digits is $2015 * 9 = 18135$, i.e., it has 5 digits.
The sum of the digits of the largest 5-digit number is 45 (b).
Numbers less than 45... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 7,481 |
4. In how many ways can two knights, two bishops, two rooks, a queen, and a king be arranged on the first row of a chessboard so that the following conditions are met:
1) The bishops stand on squares of the same color;
2) The queen and the king stand on adjacent squares. (20 points). | # Solution:
Let's number the cells of the first row of the chessboard in order from left to right with numbers from **1** to **8** ( **1** - the first white cell, **8** - the last black cell). Since the queen and king are standing next to each other, they can occupy one of 7 positions: 1-2, 2-3, ..., 7-8. Additionally... | 504 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,482 |
5. Country Omega grows and consumes only vegetables and fruits. It is known that in 2014, Country Omega produced 1200 tons of vegetables and 750 tons of fruits. In 2015, 900 tons of vegetables and 900 tons of fruits were produced. Over the year, the price of one ton of vegetables increased from 90 to 100 thousand ruble... | # Solution:
$B B \Pi_{\text{no... } 2014}=1200 * 90+750 * 75=108000+56250=164250$
$B B \Pi_{\text{real. } 2015}=900 * 90+900 * 75=81000+67500=148500$
Then the real GDP of the country decreased by 100(1-148500/164250)=9.59\%.
Answer: $-9.59 \%$.
## Grading Criteria:
1) Correctly calculated nominal (real) GDP in 20... | -9.59 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,483 |
6. The supply of a certain good on a competitive market can be described by the function $\boldsymbol{Q}^{S}(\boldsymbol{p})=2+8 \boldsymbol{p}$ (where $\boldsymbol{Q}$ is the volume of sales, and $\boldsymbol{p}$ is the price per unit of the product). It is known that at a price of 2 monetary units, consumers were wil... | # Solution:
Let's find the demand function. Let $Q^{D}=\boldsymbol{k} \boldsymbol{p}+\boldsymbol{b}$. We have $8=2 \boldsymbol{k}+\boldsymbol{b}, 6=3 \boldsymbol{k}+\boldsymbol{b}$. We get $k=-2, b=12$, i.e., $Q^{D}(p)=-2 p+12$.
a) Market equilibrium: $Q^{D}(\boldsymbol{p})=Q^{S}(p),-2 p+12=2+8 p, p^{*=1,} Q^{*}=10$.... | )p^{*}=1,Q^{*}=10;b)1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,484 |
7. In his country of Milnlandia, Winnie-the-Pooh decided to open a company that produces honey. Winnie-the-Pooh sells honey only in pots, and it costs him 10 milnovs (the monetary units of Milnlandia) to produce any pot of honey. The inverse demand function for honey is given by $\boldsymbol{P}=310-3 \boldsymbol{Q}$ (w... | # Solution:
a) Profit $=P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q = 310Q - 3Q^2 - 10Q = 300Q - 3Q^2$.
Since the graph of the function $300Q - 3Q^2$ is a parabola opening downwards, its maximum is achieved at the vertex: $Q = -b / 2a = -300 / (-6) = 50$.
b) Winnie-the-Pooh maximizes the quantity $P(Q) \c... | )50;b)150 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,485 |
4. In all other cases - o points.
## Task 3
## Maximum 20 points
After long January holidays, on the first working day, 36 sad office employees decided to play with a ball. The rules of the game are as follows: a sad employee who hits another sad employee with the ball becomes cheerful and no longer feels sad. The e... | # Solution:
Assume that more energetic employees were eliminated. To become an energetic employee, each one must have hit a sad employee who would then be eliminated from the game. Therefore, at any point in time, the number of sad employees eliminated is not less than the number of energetic employees, including thos... | proof | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,487 |
4. In all other cases, $-\mathbf{0}$ points.
## Task 4
This year, Ivan Petrovich registered as a self-employed individual and now chooses how much time he will work daily. He teaches fine arts brilliantly to elementary school children and leads group activities for children at the cultural center. The cost of one les... | # Solution and Grading Scheme:
Ivan Petrovich works ( $24-8-2 L-k$ ) hours a day.
Then $24-8-2 L-k=L$, or $16-3 L=k$.
His daily earnings from lessons are $3 L=16-k$.
In thousands of rubles, Ivan Petrovich earns $3 L \times 21+14=21(16-k)+14$ per month.
At the same time, his monthly expenses amount to $70+\frac{21 ... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,488 |
6. The supply of a certain good on a competitive market can be described by the function $\boldsymbol{Q}^{S}(\boldsymbol{p})=2+8 \boldsymbol{p}$ (where $\boldsymbol{Q}$ is the volume of sales, and $\boldsymbol{p}$ is the price per unit of the product). It is known that at a price of 2 monetary units, consumers were wil... | # Solution:
Let's find the demand function. Let $Q^{D}=\boldsymbol{k} \boldsymbol{p}+\boldsymbol{b}$. We have $8=2 \boldsymbol{k}+\boldsymbol{b}, 6=3 \boldsymbol{k}+\boldsymbol{b}$. We get $k=-2, b=12$, i.e., $Q^{D}(p)=-2 p+12$.
a) Market equilibrium: $Q^{D}(\boldsymbol{p})=Q^{S}(p),-2 p+12=2+8 p, p^{*=1,} Q^{*}=10$.... | )p^{*}=1,Q^{*}=10;b)1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,489 |
7. In his country of Milnlandia, Winnie-the-Pooh decided to open a company that produces honey. Winnie-the-Pooh sells honey only in pots, and it costs him 10 milnovs (the monetary units of Milnlandia) to produce any pot of honey. The inverse demand function for honey is given by $\boldsymbol{P}=310-3 \boldsymbol{Q}$ (w... | # Solution:
a) Profit $=P(Q) \cdot Q - TC(Q) = (310 - 3Q) \cdot Q - 10 \cdot Q = 310Q - 3Q^2 - 10Q = 300Q - 3Q^2$.
Since the graph of the function $300Q - 3Q^2$ is a parabola opening downwards, its maximum is achieved at the vertex: $Q = -b / 2a = -300 / (-6) = 50$.
b) Winnie-the-Pooh maximizes the quantity $P(Q) \c... | 50 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,490 |
# Task 1.
Maximum 10 points
In the city of N, there is a correctional facility where there are 100 guards for every 1000 prisoners. Deputy Ivanov, in his pre-election campaign, promised to reduce the number of guards by exactly half, redirecting the freed-up resources to retraining personnel and developing agricultur... | # Solution:
a) Answer: Deputy Ivanov will be able to fulfill his promise. Indeed, mathematically, the problem boils down to whether we can find 50 two-digit numbers from which we can obtain all 1000 three-digit numbers by adding one digit at the beginning, in the middle, or at the end. (Obviously, within the problem, ... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,491 |
Task 2.
Maximum 15 points
Solve the inequality:
$$
\sqrt{2 x+8}+\sqrt{10-2 x} \geq \log _{2}\left(4 x^{2}-4 x+65\right)
$$ | # Solution:
It is not difficult to understand that
$$
\log _{2}\left(4 x^{2}-4 x+65\right)=\log _{2}\left((2 x-1)^{2}+64\right) \geq \log _{2} 64=6
$$
Let's estimate the expression $\sqrt{2 x+8}+\sqrt{10-2 x}$. Consider the vectors $\overrightarrow{\boldsymbol{a}}(1,1)$ and $\vec{b}(\sqrt{2 x+8}, \sqrt{10-2 x})$. We... | \frac{1}{2} | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,492 |
# Task 3.
Maximum 15 points
In an equilateral triangle with area $S_{1}$, a circle is inscribed, and in this circle, an equilateral triangle with area $S_{2}$ is inscribed. In the resulting new triangle, another circle is inscribed, and in this circle, another equilateral triangle with area $S_{3}$ is inscribed. The ... | Solution:
Step 1: $\Delta A_{1} B_{1} C_{1} \rightarrow \Delta A_{2} B_{2} C_{2}$.
$\Delta A_{1} B_{1} C_{1}$ is similar to $\Delta A_{2} B_{2} C_{2}$ with a similarity coefficient (ratio o... | S_{1}(n)=4^{n-1} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,493 |
# Task 4.
## Maximum 10 points.
Calculate using trigonometric transformations
$$
\sin \frac{\pi}{22} \cdot \sin \frac{3 \pi}{22} \cdot \sin \frac{5 \pi}{22} \cdot \sin \frac{7 \pi}{22} \cdot \sin \frac{9 \pi}{22}
$$
# | # Solution:
Transform each factor using the formula $\sin \alpha=\cos \left(\frac{\pi}{2}-\alpha\right)$.
We get:
$$
\begin{gathered}
\sin \frac{\pi}{22} \cdot \sin \frac{3 \pi}{22} \cdot \sin \frac{5 \pi}{22} \cdot \sin \frac{7 \pi}{22} \cdot \sin \frac{9 \pi}{22}= \\
=\cos \left(\frac{\pi}{2}-\frac{\pi}{22}\right)... | \frac{1}{32} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,494 |
# Task 5.
## Maximum 10 points
In the Kingdom of Consumerland, chips are sold in a perfectly competitive market. King Consumerland not only wants to reduce the consumption of chips in his country but also increase the treasury's revenue. To avoid making a mistake in choosing the tax for chip producers, the king order... | # Solution and Evaluation Criteria:
Let the inverse demand function for chips be: $P^{D}(Q)=a-b Q$, and the inverse supply function for chips: $P^{S}(Q)=c+d Q$.
Since with a per-unit tax of 4.5 monetary units, the tax revenue amounted to 22.5 monetary units, in equilibrium, 22.5 / 4.5 = 5 weight units of chips were c... | 40.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,495 |
# Task 6.
## Maximum 15 points
On January 23, 2019, Ivan Petrovich Sidorov went to work in South Africa. The day before, he activated a multi-currency bank card with a zero balance and converted 120,000 rubles into South African currency at the exchange rate of the Central Bank of the Russian Federation, with a 1% fe... | # Solution and evaluation criteria:
The text above has been translated into English, preserving the original text's line breaks and format. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,496 |
Task 7.
## Maximum 15 points
In the modern world, every consumer often has to make decisions about replacing old equipment with more energy-efficient alternatives. Consider a city dweller who uses a 60 W incandescent lamp for 100 hours each month. The electricity tariff is 5 rubles/kWh.
The city dweller can buy a mo... | # Solution and Grading Scheme:
a) Expenses for 10 months when installing an energy-saving bulb independently: 120 rubles + 12 (W) * 100 (hours) / 1000 * 5 (rubles/kWh) * 10 (months) = 180 rubles.
Expenses for 10 months when turning to an energy service company:
$(12 + (60 - 12) * 0.75)($ W) * 100 (hours) $/ 1000 * 5... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,497 |
Task 8.
## Maximum 10 points
Imagine that you bought an apartment in a new building some time ago, and finally, the developer has called you for the apartment inspection. You know that you yourself will not be able to determine what defects the new apartment has and how serious they are.
(a) Explain how you will dec... | # Solution and Evaluation Criteria:
## Examples of Possible Arguments:
a) Since the services of a specialist for apartment acceptance are considered trust goods, there is a risk of not noticing defects, the correction of which in the future may cost much more than correction before finishing. Therefore, it is necessa... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,498 |
1. To climb from the valley to the mountain top, one must walk 4 hours on the road, and then -4 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 17 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour,... | # Solution:
The path along the road and the trail (there and back) takes 16 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 8 hours. Therefore, if you start moving along the trail immediately after the seco... | 38 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,499 |
2. Solve the inequality (9 points):
$$
\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x-1}}}} \geq \sqrt{1+\sqrt{2}}
$$ | # Solution:
The left side increases and at $x=1$ it is exactly equal to the right side.
Answer: $x \geq 1$.
| Points | Criteria for evaluating the completion of task № 2 |
| :---: | :--- |
| $\mathbf{9}$ | A correct and justified sequence of all steps of the solution is provided. The correct answer is obtained. |
| ... | x\geq1 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,500 |
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\... | # Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. The median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$,... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,501 |
4. Considering $\boldsymbol{x}$ and $\boldsymbol{y}$ as integers, solve the system of equations (11 points):
$$
\left\{\begin{array}{l}
4^{x^{2}+2 x y+1}=(z+2) 7^{|y|-1} \\
\sin \frac{3 \pi z}{2}=1
\end{array}\right.
$$ | # Solution:
From the second equation of the system, we obtain that $z=\frac{4 n+1}{3}$, where $n \in \mathbb{Z}$. The number $\mathbf{z}$ can have only the number 3 as its denominator. Therefore, the number 7, if it were in the numerator, would divide the right-hand side, but not the left-hand side. Therefore, $7^{1-|... | (1,-1,-1),(-1,1,-1) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,502 |
5. The power in the kingdom of gnomes was seized by giants. The giants decided to get rid of the gnomes and told them the following: "Tomorrow we will line you up so that each of you will see those who stand after and not see those who stand before (i.e., the 1st sees everyone, the last sees no one). We will put either... | # Solution:
The gnomes agree as follows: they risk the first gnome, telling him the following: "Denote a white hat as 1 and a black hat as 0, and count the sum of the remaining n-1 gnomes. If the sum is even, say 'white'; if it is odd, say 'black'. Under this condition, the first gnome dies with a probability of $1 / ... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,503 |
6. A group of friends decided to start a joint business. One of the tasks they faced was finding a warehouse to store their products. The friends liked two options for renting warehouses with identical characteristics. The monthly rent for the first warehouse is 800,000 rubles, and for the second - 200,000 rubles. The ... | # Solution:
(a) When renting the first warehouse for a year, the cost will be $80 * 12 = 960$ thousand rubles.
When renting the second warehouse in the worst-case scenario, where the bank takes over the warehouse, the cost for a year will be $20 * 12 + 80 * 7 + 150 = 950$ thousand rubles.
Thus, even in the worst-cas... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,504 |
7. In a certain country, the demand function for Moroccan tangerines is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=50-\mathrm{p}$, the marginal costs of production and supply of any tangerines are constant and equal to 5, and the tangerine market operates under conditions of perfect competition. An embargo, imposed on tanger... | # Solution:
Before the embargo, there was equilibrium in the competitive market for Moroccan mandarins. The price of the good should be equal to the marginal cost of its production, as the marginal revenue of any supplier is equal to the price of the good. Therefore, Moroccan mandarins were consumed in a volume of $\m... | 1.25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,505 |
9. Consider two countries, A and B, which have the same planting areas. On these areas, they can only grow zucchini and cauliflower. The alternative costs of producing either crop in each country are constant. The yield of both crops, if the entire area is planted with only one of the crops, is given in the table below... | # Solution:
(a) Country B has an absolute advantage in growing each product, as it can produce more of each product on the same area of land compared to Country A.
The opportunity cost of growing one unit of zucchini in Country A is $16 / 20 = 0.8$ units of broccoli. That is, to increase the production of zucchini by... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,507 |
10. Consider the labor market for teachers. The Ministry of Education and Science proposes to reintroduce job allocation for graduates of pedagogical universities based on geographical principles.
(a) Suppose the government introduces a restriction on the minimum number of years young teachers must work in a specific ... | # Solution:
We will present a possible argumentation for the solution, which is not the only or absolute one.
(a) When introducing an additional restriction, the demand for studying on state-funded places may initially decline, which will subsequently lead to a decrease in the supply of teachers in the public sector.... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,508 |
1. To climb from the valley to the mountain top, one must walk 6 hours on the road, and then - 6 hours on the path. On the mountain top, two fire-breathing dragons live. The first dragon spews fire for 1 hour, then sleeps for 25 hours, then spews fire for 1 hour again, and so on. The second dragon spews fire for 1 hour... | # Solution:
The path along the road and the trail (there and back) takes 24 hours. Therefore, if you start immediately after the first dragon's eruption, this dragon will not be dangerous. The path along the trail (there and back) takes 12 hours. Therefore, if you start moving along the trail immediately after the sec... | 80 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 7,509 |
2. Solve the inequality:
$$
\sqrt{x+\sqrt{x+\sqrt{x+\sqrt{x-4}}}} \geq \sqrt{4+\sqrt{6}}
$$ | # Solution:
The left side increases and at $\boldsymbol{x}=4$ it is exactly equal to the right side.
## Answer: $\mathbf{x} \geq 4$.
| Points | Criteria for evaluating the completion of task № 2 |
| :---: | :--- |
| $\mathbf{6}$ | A correct and justified sequence of all steps of the solution is provided. The correct... | x\geq4 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 7,510 |
3. On the extensions of sides $\boldsymbol{A B}, \boldsymbol{B C}, \boldsymbol{C D}$ and $\boldsymbol{A}$ of the convex quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$, points $\boldsymbol{B}_{1}, \boldsymbol{C}_{1}, \boldsymbol{D}_{1}$ and $\boldsymbol{A}_{1}$ are taken such that $\boldsymbol{B} \boldsymbol{B}_{1}=\... | # Solution:
Let the area of quadrilateral $\boldsymbol{A} \boldsymbol{B C D}$ be $\boldsymbol{S}$. A median divides the area of a triangle in half. Therefore, $S_{A B C}=S_{C B B 1}=S_{C B 1 C 1}$. Consequently, $S_{B B 1 C 1}=2 S_{A B C}$. Similarly, we have $S_{C C 1 D 1}=2 S_{B C D}$, $S_{D D 1 A 1}=2 S_{C D A}$, a... | 5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 7,511 |
4. Considering $\boldsymbol{x}$ and $\boldsymbol{y}$ as integers, solve the system of equations (11 points):
$$
\left\{\begin{array}{l}
8^{x^{2}-2 x y+1}=(z+4) 5^{|y|-1} \\
\sin \frac{3 \pi z}{2}=-1
\end{array}\right.
$$ | # Solution:
From the second equation of the system, we obtain that $z=\frac{4 n+3}{3}$, where $n \in \mathbb{Z}$. The number $z$ can only have 3 as its denominator. Therefore, if the number 5 were in the numerator, it would divide the right-hand side but not the left-hand side. Therefore, $5^{1-|y|}$ is an integer, i.... | (-1;-1;-3),(1;1;-3) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,512 |
6. A group of friends decided to start a joint business. One of the tasks they faced was finding a warehouse to store their products. The friends liked two options for renting warehouses with identical characteristics. The monthly rent for the first warehouse is 50 thousand rubles, and for the second - 10 thousand rubl... | # Solution:
(a) When renting the first warehouse for a year, the cost will be $50 * 12 = 600$ thousand rubles.
When renting the second warehouse in the worst-case scenario, where the bank takes over the warehouse, the cost for a year will be $10 * 12 + 50 * 8 + 70 = 590$ thousand rubles.
Thus, even in the worst-case... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,513 |
7. In a certain country, the demand function for Moroccan tangerines is $\mathrm{Q}_{\mathrm{d}}(\mathrm{p})=100-\mathrm{p}$, the marginal costs of production and supply of any tangerines are constant and equal to 10, and the tangerine market operates under conditions of perfect competition. An embargo, imposed on tang... | # Solution:
Before the embargo, there was equilibrium in the competitive market for Moroccan mandarins. The price of the good should be equal to the marginal cost of its production, as the marginal revenue of any supplier is equal to the price of the good. Therefore, Moroccan mandarins were consumed in a quantity of $... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,514 |
9. Consider two countries, A and B, which have the same sown areas. On these areas, they can only grow eggplants and corn. The alternative costs of producing either crop in each country are constant. The yield of both crops, if the entire territory is planted with only one of the crops, is given in the table below:
| ... | Solution:
(a) Country B has an absolute advantage in growing each product, as it can produce more of each product on the same area of arable land than Country A.
The opportunity cost of growing one unit of eggplant in Country A is $8 / 10 = 0.8$ units of corn. That is, to increase the production of eggplant by 1 unit... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,516 |
10. Consider the labor market for doctors. The Ministry of Health proposes to reintroduce job placement for graduates of medical universities and make working in state medical institutions mandatory for all young doctors.
(a) Suppose the government introduces a restriction on the minimum number of years young doctors ... | # Solution:
We will present a possible argumentation for the solution, which is not the only or absolute one.
(a)
When introducing an additional restriction, the demand for studying on state-funded places may initially decline, which will subsequently lead to a decrease in the supply of doctors in the public sector.... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,517 |
4. In all other cases - o points.
## Task 2
Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$.
How many solutions of this equation satisfy the condition:
$0 \leq x \leq 2020 ?$ | Solution. Let $t=\frac{\pi x}{4}$. Then the equation takes the form
$2 \sqrt{2} \sin ^{3} t=\cos \left(\frac{\pi}{4}-t\right)$.
$2 \sqrt{2} \sin ^{3} t=\cos \frac{\pi}{4} \cos t+\sin \frac{\pi}{4} \sin t$
$2 \sqrt{2} \sin ^{3} t=\frac{\sqrt{2}}{2} \cos t+\frac{\sqrt{2}}{2} \sin t$
$4 \sin ^{3} t=\cos t+\sin t ; 4 \... | 505 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,518 |
4. In all other cases - $\mathbf{0}$ points.
## Task 2
## Maximum 15 points
Solve the equation $2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\sin \left(\frac{\pi}{4}(1+x)\right)$.
How many solutions of this equation satisfy the condition: $2000 \leq x \leq 3000$? | # Solution:
$\sin \left(\frac{\pi}{4}(1+x)\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$. The equation becomes
$2 \sqrt{2} \sin ^{3}\left(\frac{\pi x}{4}\right)=\cos \left(\frac{\pi}{4}(1-x)\right)$, i.e., we get problem 2 from option 1, the solution of which is:
$x=1+4 n, n \in Z . \quad 2000 \leq x \leq 3000,2000 \... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,522 |
4. In all other cases, $-\mathbf{0}$ points.
## Task 4
Maxim Andreevich, a former university lecturer, gives math lessons to groups of schoolchildren. The cost of one hour (60 minutes) of Maxim Andreevich's lesson with a group of schoolchildren is 3 thousand rubles (after all taxes). In addition to income from tutori... | # Solution and Grading Scheme:
(a) Maxim Andreevich works $\left(24-8-2 L_{i}-k_{i}\right)=16-2 L_{i}-k_{i}$ hours a day, where $L_{i}$ is the number of hours he works as a tutor on the $i$-th working day, and $k_{i}$ is the number of hours he spends on rest and household chores on the $i$-th working day. The number o... | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,524 |
12. Maximum 15 points. Let $[x]$ denote the integer part of the number $x$ (i.e., the greatest integer not exceeding $x$).
Solve the equation.
$$
[\sin x+\cos x]=1
$$
# | # Solution:
The given equation is equivalent to the double inequality
$$
1 \leq \sin x + \cos x \text{ correct answer } | 15 |
| The algorithm is provided, but one of the transitions is not justified, correct answer obtained | 10 |
| The algorithm is provided, correct answer obtained only for one trigonometric circle... | 1\leq\sinx+\cosx<2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 7,525 |
3. Maximum 15 points. In a chess tournament, 29 people are playing. One point is awarded for a win, half a point for a draw, and zero for a loss. If a player has no opponent, they receive 1 point, but according to the rules, a point without a game can be awarded to a player no more than once during the entire tournamen... | # Solution:
Since the total number of players is odd, someone must have been awarded 1 point due to the absence of an opponent. Since $29=2 * 14+1$, 15 people will receive 1 point after the first round, 15= $2 * 7+1$, eight people will have 2 points after two rounds, four will have 3 points after three rounds, two wil... | notfound | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 7,528 |
# 4. Maximum 15 points. Thousands of systems in the Old Republic have imposed sanctions on
the Trade Federation for attempting to blockade the peaceful planet of Naboo. The population of the Republic rapidly reduced the export of goods to the Trade Federation and imports from it. To prevent currency volatility, the Ce... | # Solution:
(a) The Trade Federation has experienced a decline in both exports and imports -> both demand for and supply of the national currency have decreased (ceteris paribus) -> it is impossible to determine the direction of the equilibrium price change.
Additional considerations may include:
- The decline in im... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,529 |
5. Maximum 15 points. Below is a fragment of a map of one of Moscow's districts (scale 1:50). Propose three different economically justified arguments explaining the reasons for the location of the "Perekrestok" stores. Explain what two disadvantages this strategy of store placement within one network might have. (Note... | # Solution:
The concentration of stores of one network in a relatively small area can be explained by several economically justified reasons.
1) The placement of "Perekrestoks" can prevent the emergence of other stores, which means that this can contribute to the monopolization of the supply of goods sold in such sup... | notfound | Other | math-word-problem | Yes | Yes | olympiads | false | 7,530 |
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