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3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 70, 1 \leqslant b \leqslant 50$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that... | Answer: 1260.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be d... | 1260 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Find the number of pairs of integers $(a ; b)$ such that $1 \leqslant a \leqslant 80,1 \leqslant b \leqslant 30$, and the area $S$ of the figure defined by the system of inequalities
$$
\left\{\begin{array}{l}
\frac{x}{a}+\frac{y}{b} \geqslant 1 \\
x \leqslant a \\
y \leqslant b
\end{array}\right.
$$
is such that ... | Answer: 864.
Solution. The given system of inequalities defines a triangle on the plane with vertices $(a ; 0),(0 ; b)$, and $(a ; b)$. This triangle is right-angled, and its doubled area is equal to the product of the legs, i.e., $a b$. According to the condition, $a b: 5$, so one of the numbers $a$ or $b$ must be di... | 864 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
10.2. There are 90 cards - 10 with the digit 1, 10 with the digit $2, \ldots, 10$ with the digit 9. From all these cards, two numbers were formed, one of which is three times the other. Prove that one of these numbers can be factored into four not necessarily distinct natural factors, greater than one. | Solution. Let these numbers be $A$ and $B=3A$. Then the sum of the digits of the number $B$ is divisible by 3. But the sum of the digits on all cards is divisible by 9 (and therefore by 3), so the sum of the digits of the number $A$ is divisible by 3. This means that the number $A$ is divisible by 3. But then the numbe... | 40 | Number Theory | proof | Yes | Yes | olympiads | false |
10.4. Quadrilateral $A B C D$ is inscribed in a circle. The perpendicular to side $B C$, drawn through its midpoint - point $M$, intersects side $A B$ at point $K$. The circle with diameter $K C$ intersects segment $C D$ at point $P(P \neq C)$. Find the angle between the lines $M P$ and $A D$. | Answer: $90^{\circ}$.
Solution. We will prove that the lines $M P$ and $A D$ are perpendicular. Let $\omega$ be the circle constructed with $K C$ as its diameter, then point $M$ lies on $\omega$, since angle $C M K$ is a right angle. Therefore, $\angle C P M = \angle C K M = \alpha$ (they subtend the arc $C M$ of circ... | 90 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
8. For each natural $n$, which is not a perfect square, all values of the variable $x$ are calculated for which both numbers $x+\sqrt{n}$ and $x^{3}+$ param1 $\sqrt{n}$ are integers. Find the total number of such values of $x$.
| param1 | answer |
| :---: | :---: |
| 1524 | |
| 1372 | |
| 1228 | |
| 1092 | |
| 964... | 8. For each natural $n$, which is not a perfect square, all values of the variable $x$ are calculated for which both numbers $x+\sqrt{n}$ and $x^{3}+$ param1 $\sqrt{n}$ are integers. Find the total number of such values of $x$.
| param1 | answer |
| :---: | :---: |
| 1524 | 39 |
| 1372 | 33 |
| 1228 | 35 |
| 1092 | 27... | 33 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
9. A regular param 1 -sided polygon is inscribed in a circle, with different natural numbers written at its vertices. A pair of non-adjacent vertices of the polygon $A$ and $B$ is called interesting if, on at least one of the two arcs $A B$, all the numbers written at the vertices of the arc are greater than the number... | 9. A regular param 1-gon is inscribed in a circle, with different natural numbers written at its vertices. A pair of non-adjacent vertices of the polygon $A$ and $B$ is called interesting if, on at least one of the two arcs $A B$, all the numbers written at the vertices of the arc are greater than the numbers written a... | 92 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) The vertices of the broken line $A B C D E F G$ have coordinates $A(-1 ; -7), B(2 ; 5), C(3 ; -8), D(-3 ; 4), E(5 ; -1), F(-4 ; -2), G(6 ; 4)$.
Find the sum of the angles with vertices at points $B, E, C, F, D$. | Answer: $135^{\circ}$.
Solution. The closed broken line $B C D E F B$ forms a five-pointed "star". The sum of the angles at the rays of this star is $180^{\circ}$.
We will prove that the sum of the angles at the rays of any five-pointed star $B C D E F B$ is $180^{\circ}$. Let $O$ be the point of intersection of the ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) The vertices of the broken line $A B C D E F G$ have coordinates $A(0 ; -5), B(3 ; 7), C(4 ; -6), D(-2 ; 6), E(6 ; 1), F(-3 ; 0), G(7 ; 6)$.
Find the sum of the angles with vertices at points $B, E, C, F, D$. | Answer: $135^{\circ}$.
Solution. The closed broken line $B C D E F B$ forms a five-pointed "star". The sum of the angles at the rays of this star is $180^{\circ}$.
We will prove that the sum of the angles at the rays of any five-pointed star $B C D E F B$ is $180^{\circ}$. Let $O$ be the point of intersection of the ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Nils has a goose farm. Nils calculated that if he sells 75 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 15 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-75)=(k-15) x(n+100) \\
k n=(k+20)(n-75)=(k-15)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Nils has a goose farm. Nils calculated that if he sells 50 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 10 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-50)=(k-10) x(n+100) \\
k n=(k+20)(n-50)=(k-10)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 7 (10 points)
A person and his faithful dog started moving along the perimeter of a block from point $A$ simultaneously at time $t_{0}=0$ min. The person moved with a constant speed clockwise, while the dog ran with a constant speed counterclockwise (see figure). It is known that they met for the first tim... | # Answer: in 14 min
## Solution and evaluation criteria:
In $t_{1}=2$ min, the person and the dog together covered a distance equal to the perimeter of the block, with the person moving 100 meters from the starting point of the journey.
That is, during each subsequent meeting, the person will be 100 meters away from... | 14 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem № 8 (15 points)
In some English-speaking countries, temperature is measured in degrees Fahrenheit. An English schoolboy, observing a thermometer in a glass of cooling water, noticed that it cooled by $30^{\circ} \mathrm{F}$. He became curious about how much heat was released. In books, he found the following... | # Answer: 140 kJ
## Solution and grading criteria:
The change in temperature in degrees Fahrenheit is related to the change in temperature in degrees Celsius:
That is, in degrees Celsius,... | 140 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. 20 balls of the same mass are moving along a chute towards a metal wall with the same speed. Coming towards them at the same speed are 16 balls of the same mass. When two balls collide, they fly apart with the same speed. After colliding with the wall, a ball bounces off it with the same speed. (The balls move only ... | # Answer: 510.
Solution. We will assume that initially, each ball moving towards the wall has a red flag, and the rest of the balls have blue flags. Imagine that when the balls collide, they exchange flags. Then each blue flag moves at a constant speed in one direction (away from the wall), and each red flag reaches t... | 510 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Andrei was driving to the airport of a neighboring city. After an hour of driving at a speed of 60 km/h, he realized that if he did not change his speed, he would be 20 minutes late. Then he sharply increased his speed, as a result of which he covered the remaining part of the journey at an average speed of 90 km/h ... | Answer: 180 km.
Solution. Let the distance from Andrey's house to the airport be $s$ km, and the time he intended to spend on the road be $1+t$ hours. Then
$$
s=60+60\left(t+\frac{1}{3}\right)=60+90\left(t-\frac{1}{3}\right)
$$
From this,
$$
60 t+20=90 t-30, \quad t=\frac{5}{3}, \quad s=180
$$
Evaluation. Full sco... | 180 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $n$ is such that the number $36 n^{2}$ has exactly 51 different natural divisors. How many natural divisors does the number $5 n$ have? | Answer: 16.
Solution. Let $m=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{l}^{k_{l}}$, where $p_{1}, p_{2}, \ldots, p_{l}$ are pairwise distinct prime numbers. Then the number of natural divisors of the number $m$ is
$$
\tau(m)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{l}+1\right)
$$
Indeed,... | 16 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
1. Viktor was driving to the airport of a neighboring city. After half an hour of driving at a speed of 60 km/h, he realized that if he did not change his speed, he would be 15 minutes late. Then he increased his speed, as a result of which he covered the remaining part of the journey at an average speed of 80 km/h and... | Answer: 150 km.
Solution. Let the distance from Viktor's house to the airport be $s$ km, and the time he intended to spend on the road $\frac{1}{2}+t$ hours. Then
$$
s=30+60\left(t+\frac{1}{4}\right)=30+80\left(t-\frac{1}{4}\right)
$$
From this,
$$
60 t+15=80 t-20, \quad t=\frac{7}{4}, \quad s=150
$$
Evaluation. F... | 150 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. A natural number $n$ is such that the number $100 n^{2}$ has exactly 55 different natural divisors. How many natural divisors does the number 10n have? | Answer: 18.
Solution. Let $m=p_{1}^{k_{1}} \cdot p_{2}^{k_{2}} \cdot \ldots \cdot p_{l}^{k_{l}}$, where $p_{1}, p_{2}, \ldots, p_{l}$ are pairwise distinct prime numbers. Then the number of natural divisors of the number $m$ is
$$
\tau(m)=\left(k_{1}+1\right)\left(k_{2}+1\right) \ldots\left(k_{l}+1\right)
$$
Indeed,... | 18 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) The medians drawn from vertices $A$ and $B$ of triangle $ABC$ are perpendicular to each other. Find the area of the square with side $AB$, if $BC=28, AC=44$.
# | # Answer: 544
Solution. Let $D$ be the midpoint of $B C$, $E$ be the midpoint of $A C$, and $M$ be the point of intersection of the medians. Let $M D=a$, $M E=b$. Then $A M=2 a$, $B M=2 b$. From the right triangles $B M D$ and $A M E$, we have $a^{2}+4 b^{2}=B D^{2}=14^{2}$ and $4 a^{2}+b^{2}=A E^{2}=22^{2}$. Adding t... | 544 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $4 \times 5$ grid, 5 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 240
Solution. From the condition, it follows that in some row, two cells are marked (while in the others, only one each). This row can be chosen in 4 ways, and the two crosses in it can be chosen in $5 \cdot 4 / 2=10$ ways. The remaining three crosses can be chosen in $3 \cdot 2 \cdot 1=6$ ways. By the rule ... | 240 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) The medians drawn from vertices $A$ and $B$ of triangle $ABC$ are perpendicular to each other. Find the area of the square with side $AB$, if $BC=36, AC=48$.
# | # Answer: 720
Solution. Let $D$ be the midpoint of $B C$, $E$ be the midpoint of $A C$, and $M$ be the point of intersection of the medians. Let $M D=a$, $M E=b$. Then $A M=2 a$, $B M=2 b$. From the right triangles $B M D$ and $A M E$, we have $a^{2}+4 b^{2}=B D^{2}=18^{2}$ and $4 a^{2}+b^{2}=A E^{2}=24^{2}$. Adding t... | 720 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $3 \times 4$ grid, 4 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 36
Solution. From the condition, it follows that in some row, two cells are marked (while in the others, only one each). This row can be chosen in 3 ways, and the two crosses in it can be placed in $4 \cdot 3 / 2=6$ ways. The remaining two crosses can be chosen in 2 ways. By the multiplication rule, the tota... | 36 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. In a row, the numbers $1,2,3, \ldots, 2014,2015$ are written. We will call a number from this row good if, after its removal, the sum of all the remaining 2014 numbers is divisible by 2016. Find all the good numbers. | Answer: The only good number is 1008.
Solution. The remainder of the division of the sum of all 2015 numbers by 2016 is 1008:
$(1+2015)+(2+2014)+\ldots+(1007+1009)+1008=2016 \cdot 1007+1008$.
Therefore, only 1008 can be crossed out.
Scoring. 12 points for a correct solution. If it is shown that 1008 is a good numbe... | 1008 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Nils has a goose farm. Nils calculated that if he sells 75 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 15 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-75)=(k-15) x(n+100) \\
k n=(k+20)(n-75)=(k-15)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Nils has a goose farm. Nils calculated that if he sells 50 geese, the feed will last 20 days longer than if he doesn't sell any. If he buys an additional 100 geese, the feed will run out 10 days earlier than if he doesn't make such a purchase. How many geese does Nils have? | Answer: 300.
Solution. Let $A$ be the total amount of feed (in kg), $x$ be the amount of feed per goose per day (in kg), $n$ be the number of geese, and $k$ be the number of days the feed will last. Then
$$
\begin{gathered}
A=k x n=(k+20) x(n-50)=(k-10) x(n+100) \\
k n=(k+20)(n-50)=(k-10)(n+100)
\end{gathered}
$$
So... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (10 points) Five identical non-ideal ammeters are connected as shown in the figure. Ideal power supply is connected to points $A$ and $B$. Determine the sum of the readings of all ammeters, given that the reading of the first ammeter $I_{1}=2 \mathrm{MA}$.
$ is defined by the following relations: $a_{1}=1, a_{2}=2, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{2019}$. | Answer: 2020.
Solution. Let's list the first terms of the sequence:
$$
1,2,4,6,7,7,7,8,10,12,13,13,13,14,16,18,19,19,19,20,21,23, \ldots
$$
We can observe a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The sequence $\left(a_{n}\right)$ is defined by the following relations: $a_{1}=1, a_{2}=3, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{1000}$. | Answer: 1002.
Solution. Let's list the first terms of the sequence:
$$
1,3,5,6,6,6,7,9,11,12,12,12,13,15,17,18,18,18,19,21, \ldots
$$
We can see a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$ with $n+2... | 1002 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After reducing the entry fee, the number of spectators increased by $25 \%$, and the revenue increased by $12.5 \%$. What is the new price of the entry ticket after the price reduction? | Answer: 360
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 1.25 = 400 - 1.125$. From which $x = 360$. | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After increasing the entry fee, the number of spectators decreased by $20 \%$, but the revenue increased by $5 \%$. What is the new price of the entry ticket after the price increase? | Answer: 525
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 0.8 = 400 \cdot 1.05$. From which $x=525$. | 525 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) A chess player played 42 games and scored 30 points (1 point for each win, -0.5 points for each draw, 0 points for each loss). Find the difference between the number of his wins and the number of his losses. | Answer: 18
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(42-n-m)=30$. In the end, $n-m=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) An aluminum and a copper part have the same volume. The density of aluminum $\rho_{A}=2700 \mathrm{kg} / \mathrm{m}^{3}$, the density of copper $\rho_{M}=8900 \mathrm{kg} / \mathrm{m}^{3}$. Find the mass of copper, if it is known that the masses of the parts differ by $\Delta m=60 \mathrm{g}$. | Answer: 862
Solution. Volume of aluminum: $V=\frac{m_{M}-\Delta m}{\rho_{A}}$, volume of copper: $V=\frac{m_{M}}{\rho_{M}}$. We get: $\frac{m_{M}}{\rho_{M}}=\frac{m_{M}-\Delta m}{\rho_{A}} . \quad$ From this, the mass of aluminum: $m_{A}=\frac{\Delta m \cdot \rho_{M}}{\rho_{M}-\rho_{A}}=\frac{0.06 \cdot 8900}{8900-270... | 862 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A mowing crew mowed the entire meadow in two days. On the first day, half of the meadow and another 3 hectares were mowed, and on the second day, a third of the remaining area and another 6 hectares were mowed. What is the area of the meadow? | Answer: 24 hectares.
Solution. 6 hectares made up two-thirds of the remainder. Therefore, 9 hectares were mowed on the second day. Together with the 3 hectares from the first day, this totals 12 hectares, which constitute half of the meadow's area. Therefore, the total area of the meadow is 24 hectares.
Evaluation. 1... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. (16 points) After walking one-fifth of the way from home to school, Olya realized she had forgotten her notebook. If she does not go back for it, she will arrive at school 6 minutes before the bell, but if she returns, she will be 2 minutes late. How much time (in minutes) does the journey to school take? | # Answer: 20 min
Solution. The extra $2 / 5$ of the journey takes 8 min. Therefore, the entire journey to school will take 20 min. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $5 \times 5$ grid, 6 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 4200
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 5 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choo... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=500 \mathrm{M} / \mathrm{c}$, penetrates a massive board and exits with a speed of $v_{2}=200 \mathrm{M} / \mathrm{c}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 1050 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 500^{2}}{2}-\frac{0.01 \cdot 200^{2}}{2}=1050 \text { J. }
$$ | 1050 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) A light ray falls at an angle $\alpha=30^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.5$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $120^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (17 points) In a $4 \times 4$ grid, 5 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 432
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 4 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choos... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=400 \mathrm{m} / \mathrm{s}$, penetrates a massive board and exits with a speed of $v_{2}=100 \mathrm{m} / \mathrm{s}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 750 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 400^{2}}{2}-\frac{0.01 \cdot 100^{2}}{2}=750 \text { J. }
$$ | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) A light ray falls at an angle $\alpha=60^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.6$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $60^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \a... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
# Problem №4 (10 points)
A parallel beam of light falls on the base of a glass cone (refractive index $n=1.5$) along its axis (see fig.). The cross-section of the beam coincides with the base of the cone, the radius
$\operatorname{tg} \alpha=\frac{h}{R}=1.73$, i.e., $\alpha=60^{\circ}$
The law of refraction for rays passing through the cone:
$\frac{\sin \alpha}{\sin \beta}=\frac{1}{1.5}$
A... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 50 m ahead of Boris at the finish line. And Boris was 40 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished? | Answer: 88 m.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c$ m/s, respectively. From the condition, it follows that $b=0.95 a, c=0.96 b$. Therefore, $c=0.96 \cdot 0.95 a=0.912 a$. This means that when Andrey runs 1000 m, Valentin will cover 912 m. The lag will be 88 m.
Evaluation. 12 point... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 353 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 32 was added and a four-digit number was obtained, which is also a palindrome. Find $x$. | # Answer: 969.
Solution. Since $x+32$ is a four-digit number, the inequality $x+32 \geqslant 1000$ holds, from which $x \geqslant 968$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 6. By checking the four possible op... | 969 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $1.5 m$, width - $400 mm$, height - 80 cm. It is filled with water at a rate of 2 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points) | Answer: 14400 s
Solution. Volume of the aquarium $\quad V=1.5 \cdot 0.4 \cdot 0.8=0.48 \mathrm{~m}^{3}$.
Filling rate $v=\frac{0.002 \mu^{3}}{60 s}$ (3 points). The aquarium will be completely filled in time $\quad t=\frac{V}{v}=\frac{0.48 \cdot 60}{0.002}=14400 ~ s \quad(4$ points $)$. | 14400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{s}$ and $v_{2}=4 \mathrm{~m} / \mathrm{s}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who turn... | Answer: 20 min
Solution. Let $S$ be the distance between points A and B. We get $S=v_{1} t_{1}+v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{2} t_{1}+v_{2} t_{1}+v_{2} t_{2}$ (5 points). As a result, we get $t_{1}=\frac{v_{2} t_{2}}{v_{1}-v_{2}}=\frac{4 \cdot 10}{6-4}=20 \t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=4$ m. It realized it was all too much and turned back. It crawled $l_{2}=3$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=40$ m. On which day of its journey will... | Answer: on the 37th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=4$ m (i.e., one transition) from its goal, it will be after 36 days of travel (4 points). Therefore, the goal will be reached on the 37th day of travel (3 points).... | 37 | Other | math-word-problem | Yes | Yes | olympiads | false |
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 333 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 22 was added and a four-digit number was obtained, which is also a palindrome. Find $x$. | # Answer: 979.
Solution. Since $x+22$ is a four-digit number, the inequality $x+22 \geqslant 1000$ holds, from which $x \geqslant 978$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 7. By checking the four possible op... | 979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $2 m$, width - 600 mm, height - 60 cm. It is filled with water at a rate of 3 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points) | Answer: 14400 s
Solution. Volume of the aquarium: $V=2 \cdot 0.6 \cdot 0.6=0.72 m^{3}$ (3 points). Filling rate: $v=\frac{0.003 m^{3}}{60 s}$ (3 points). The aquarium will be completely filled in: $t=\frac{V}{v}=\frac{0.72 \cdot 60}{0.003}=14400 s$ (4 points). | 14400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{c}$ and $v_{2}=4 \mathrm{~m} / \mathrm{c}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who did ... | Answer: 30 min
Solution. The distance from point $B$ to the meeting place $S=v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{1}\left(t_{1}-t_{2}\right)$ (5 points). As a result, we get: $t_{1}=\frac{v_{1} t_{2}}{v_{1}-v_{2}}=\frac{6 \cdot 10}{6-4}=30$ min (5 points). | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=5$ m. It realized it was all too much and turned back. It crawled $l_{2}=4$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=30$ m. On which day of its journey will... | Answer: on the 26th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=5$ m (i.e., one transition) from its goal, it will be after 25 days of travel (4 points). Therefore, the goal will be reached on the 26th day of travel. (3 points)... | 26 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2018. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$. | Answer: 673.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$$
2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4036+b+d \geqslant 4038 ; \qu... | 673 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. The specific heat capacity of a body with a mass of $m=2$ kg depends on the temperature as follows: $c=c_{0}(1+\alpha t)$, where $c_{0}=150$ J/kg ${ }^{\circ} \mathrm{C}-$ specific heat capacity at $0^{\circ} \mathrm{C}, \alpha=0.05^{\circ} \mathrm{C}^{-1}$ - temperature coefficient, $t$ - temperature in degrees Cel... | Answer: 96 kJ
Solution. Given that the specific heat capacity depends linearly on temperature, we can calculate its average value: $c_{c p}=\frac{c_{0}\left(1+\alpha t_{h}\right)+c_{0}\left(1+\alpha_{\kappa}\right)}{2}=600 \frac{\text { J }}{\text { kg } \cdot{ }^{\circ} \mathrm{C}} \quad(5$ points). The required amou... | 96 | Calculus | math-word-problem | Yes | Yes | olympiads | false |
8. A parallel beam of light falls normally on a thin lens. Behind the lens, at a distance of $80 \mathrm{~cm}$ from it, there is a screen on which a circular spot of a certain diameter is visible. If the screen is moved 40 cm, a spot of the same diameter will again be visible on the screen. Determine the focal length o... | Answer: 100 cm or 60 cm
Solution. A diagram explaining the situation described in the problem (5 points):
From this, it is clear that there are two possible scenarios: the screen can be mov... | 100 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2345. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$. | Answer: 782.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4690+b+d \geqslant 4692 ; \quad... | 782 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A parallel beam of light falls normally on a thin lens with a focal length of $F=150$ cm. Behind the lens, there is a screen on which a circular spot of a certain diameter is visible. If the screen is moved 40 cm, a spot of the same diameter will again be visible on the screen. Determine the initial distance from th... | Answer: 170 cm or 130 cm
Solution. A diagram explaining the situation described in the problem (5 points):
From this, we can see that there are two possible scenarios: the screen can be mov... | 170 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. On a circle, 40 red points and one blue point are marked. All possible polygons with vertices at the marked points are considered. Which type of polygons is more numerous, and by how many: those with a blue vertex, or those without it? | Answer: The number of polygons with a blue vertex is 780 more than the number of polygons without a blue vertex.
Solution. Let's call polygons with a blue vertex blue, and those without a blue vertex red. Take an arbitrary red polygon. Adding a blue vertex to it gives exactly one blue polygon. Thus, any blue polygon w... | 780 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
7. At the ends of a vertically positioned homogeneous spring, two small loads are fixed. Above is a load with mass $m_{1}$, and below is $-m_{2}$. A person grabbed the middle of the spring and held it vertically in the air. In this case, the upper half of the spring was deformed by $x_{1}=8 \mathrm{~cm}$, and the lower... | Answer: 16 cm
Solution. If the stiffness of the spring is $k$, then the stiffness of half the spring is $2 k$ (4 points). For the situation where the spring is held by a person: $m_{1} g=2 k x_{1}$ (4 points).
For the situation where the spring stands on a surface: $m_{1} g=k x$ (4 points). We obtain that $x=2 x_{1}=... | 16 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On a circle, 60 red points and one blue point are marked. All possible polygons with vertices at the marked points are considered. Which type of polygons is more numerous, and by how many: those with a blue vertex, or those without it? | Answer: The number of polygons with a blue vertex is 1770 more than the number of polygons without a blue vertex.
Solution. Let's call polygons with a blue vertex blue, and those without a blue vertex red. Take an arbitrary red polygon. Adding a blue vertex to it gives exactly one blue polygon. Thus, any blue polygon ... | 1770 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of all four-digit numbers in which the digits $0,3,6,9$ do not appear. | Answer: $9999 \cdot 6^{4} / 2=6479352$.
Solution. All four-digit numbers that meet the condition of the problem are divided into pairs of numbers of the form (1111,8888), (1112,8887), (1113,8886), ..., (4555,5444). In each pair, the sum of the numbers is 9999. Let's count the number of pairs. In total, there are $6^{4... | 6479352 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a square grid of size $8 \times 8$?
# | # Answer: 80.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken l... | 80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. Find the sum of all four-digit numbers in which the digits $0,4,5,9$ do not appear. | Answer: $9999 \cdot 6^{4} / 2=6479352$.
Solution. All four-digit numbers that meet the condition of the problem are divided into pairs of numbers of the form (1111,8888$),(1112,8887),(1113,8886)$, $\ldots$... (4555,5444$)$. In each pair, the sum of the numbers is 9999. Let's count the number of pairs. In total, there ... | 6479352 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
4. What is the greatest length that a closed, non-self-intersecting broken line can have, which runs along the grid lines of a $6 \times 10$ cell field? | Answer: 76.
Solution. We will color the nodes of the grid in a checkerboard pattern, with black and white colors. The length of a closed non-self-intersecting broken line is equal to the number of nodes it passes through. Each segment of the broken line connects a black and a white node. When traversing the broken lin... | 76 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. Given a parallelogram $A B C D$. It is known that the centers of the circles circumscribed around triangles $A B C$ and $C D A$ lie on the diagonal $B D$. Find the angle $D B C$, if $\angle A B D=40^{\circ}$. | Answer: $50^{\circ}$ or $40^{\circ}$.
Solution. The center of the circle circumscribed around triangle $A B C$ lies on the perpendicular bisector of segment $A C$. Therefore, either this center is the midpoint of $A C$ (and then $A B C D$ is a rectangle), or $D B \perp A C$ (and then $A B C D$ is a rhombus). In the fi... | 50 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Given a parallelogram $A B C D$. It is known that the centers of the circles circumscribed around triangles $A B C$ and $C D A$ lie on the diagonal $B D$. Find the angle $D B C$, if $\angle A B D=35^{\circ}$. | Answer: $55^{\circ}$ or $35^{\circ}$.
Solution. The center of the circle circumscribed around triangle $A B C$ lies on the perpendicular bisector of segment $A C$. Therefore, either this center is the midpoint of $A C$ (and then $A B C D$ is a rectangle), or $D B \perp A C$ (and then $A B C D$ is a rhombus). In the fi... | 55 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Three cyclists, Dima, Misha, and Petya, started a 1 km race simultaneously. At the moment Misha finished, Dima still had to cover one tenth of the distance, and at the moment Dima finished, Petya still had to cover one tenth of the distance. How far apart (in meters) were Petya and Dima when Misha finish... | Answer: 90.
Solution. Let $v_{1}$ be the speed of Misha, $v_{2}$ be the speed of Dima, and $v_{3}$ be the speed of Petya. Then $v_{2}=0.9 v_{1}$, since in the same amount of time Misha traveled 1000 m, while Dima traveled 900 m, and $v_{3}=0.9 v_{2}$, since in the same amount of time Dima traveled 1000 m, while Petya ... | 90 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) In a class, some students study only English, some study only German, and some study both languages. What percentage of the class studies both languages if 90% of all students study English and 80% study German? | Answer: 70.
Solution. From the condition, it follows that $10 \%$ do not study English and $20 \%$ do not study German. Therefore, $10 \%+20 \%=30 \%$ study only one language, and the remaining $70 \%$ study both languages. | 70 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
1. (17 points) Three cyclists, Dima, Misha, and Petya, started a 1 km race simultaneously. At the moment Misha finished, Dima still had to cover one tenth of the distance, and at the moment Dima finished, Petya still had to cover one tenth of the distance. How far apart (in meters) were Petya and Misha when Misha finis... | Answer: 190.
Solution. Let $v_{1}$ be the speed of Misha, $v_{2}$ be the speed of Dima, and $v_{3}$ be the speed of Petya. Then $v_{2}=0.9 v_{1}$, since in the same time Misha traveled $1000 \mathrm{m}$, while Dima traveled $900 \mathrm{m}$, and $v_{3}=0.9 v_{2}$, since in the same time Dima traveled 1000 m, while Pet... | 190 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (17 points) Point $M$ lies inside segment $A B$, the length of which is 60 cm. Points are chosen: $N$ at the midpoint of $A M, P$ at the midpoint of $M B, C$ at the midpoint of $N M, D$ at the midpoint of $M P$. Find the length of segment $C D$. | Answer: 15 cm.
## Solution.
$$
\begin{aligned}
& \quad \dot{N} \quad \dot{C} \quad \dot{M} \dot{D} \cdot B \\
& N P=N M+M P=\frac{1}{2} A M+\frac{1}{2} M B=\frac{1}{2} A B=30 \\
& C D=C M+M D=\frac{1}{2} N M+\frac{1}{2} M P=\frac{1}{2} N P=15
\end{aligned}
$$ | 15 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. (16 points) In a class, some students study only English, some study only German, and some study both languages. What percentage of the class studies both languages if 80% of all students study English and 70% study German? | Answer: 50.
Solution. From the condition, it follows that $20 \%$ do not study English and $30 \%$ do not study German. Therefore, $20 \%+30 \%=50 \%$ study only one language, and the remaining $50 \%$ study both languages. | 50 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (16 points) Aunt Masha decided to bake a cake. She kneaded the dough, which according to the recipe contains flour, butter, and sugar in the weight ratio of $3: 2: 1$, and mixed butter with sugar for the cream in the ratio of 2:3. Changing her mind about baking the cake, she combined both mixtures, added 200 g of fl... | Answer: 480.
Solution. Let the cake dough contain flour, butter, and sugar in the amounts of $3 x$, $2 x$, and $x$ respectively, and the cream contain butter and sugar in the amounts of $2 y$ and $3 y$ respectively. Using the ratios for the cookies, we get the system of equations $\left\{\begin{array}{l}\frac{3 x+200}... | 480 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (15 points) A ball was thrown vertically upwards from a balcony. It is known that it hit the ground after 6 seconds. Given that the initial speed of the ball is 20 m/s, determine the height of the balcony. The acceleration due to gravity is 10 m/s². | Answer: 60 m.
Solution. The equation of motion of the ball:
$$
y=0=h_{0}+v_{0} t-\frac{g t^{2}}{2}=h_{0}+20 \cdot 6-\frac{10 \cdot 6^{2}}{2}
$$
We get $h_{0}=60 m$ m. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface.
The refractive index of the glass $n=1.5$. The angle of refraction for the first ray $\beta=25^{\circ}$. Determine the angle between the refracted rays. | Answer: $56^{\circ}$.
Solution. The law of refraction for the first ray $\frac{\sin \alpha}{\sin 25^{\circ}}=1.5$. Therefore, the angle of incidence of the first ray $\alpha=39.34^{\circ}$. The angle of incidence of the second ray $\beta=90^{\circ}-39.34^{\circ}=$ $50.66^{\circ}$. The law of refraction for the second ... | 56 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
2. (16 points) Aunt Masha decided to bake a cake. She mixed the dough, which according to the recipe includes flour, butter, and sugar in the weight ratio of $3: 2: 1$, and mixed butter with sugar for the cream in the ratio of $2: 3$. Changing her mind about baking the cake, she combined both mixtures, added 300 g of f... | Answer: 1200.
Solution. Let the dough for the cake contain flour, butter, and sugar in the amounts of $3x$, $2x$, and $x$ respectively, and the cream contain butter and sugar in the amounts of $2y$ and $3y$ respectively. Using the ratios for the cookies, we get the system of equations $\left\{\begin{array}{l}\frac{3x+... | 1200 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (20 points) Two mutually perpendicular rays, propagating in the same vertical plane, fall from air onto a horizontal glass surface. The refractive index of the glass $n=1.6$. The angle of refraction for the first ray $\beta=30^{\circ}$. Determine the angle between the refracted rays. | Answer: $52^{\circ}$.
Solution. The law of refraction for the first ray $\frac{\sin \alpha}{\sin 30^{\circ}}=1.6$. Therefore, the angle of incidence of the first ray $\alpha=53.13^{0}$. The angle of incidence of the second ray $\beta=90^{0}-53.13^{0}=$ $36.87^{\circ}$. The law of refraction for the second ray $\frac{\... | 52 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) Two springs with stiffnesses of $6 \mathrm{kH} / \mathrm{m}$ and $12 \mathrm{kH} / \mathrm{m}$ are connected in series. How much work is required to stretch this system by 10 cm. | Answer: 20 J.
Solution. Total stiffness: $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{6 \cdot 12}{6+12}=4 \mathrm{kH} / \mathrm{m}$. Work:
$$
A=\Delta W=\frac{k x^{2}}{2}=\frac{4000 \cdot \cdot 0.1^{2}}{2}=20 \text { J. }
$$ | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to 1. Initially, the fly moved exactly along the parabola up to the point with an abscissa eq... | Answer: 6068.
Solution. Let the quadratic function be of the form $y=x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, on the other hand, $\left(3 ; 6069\right)$. Since $f(2)=4+2 b+c, f(4)=16+4 b+c$, then $20+6 b+2 c=12138$ or $3 b+c=6059$. Therefore, $f... | 6068 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (13 points) In a dance ensemble, there are 8 boys and 16 girls. Some of them form mixed (boy and girl) dance pairs. It is known that in each pair, at least one of the partners does not belong to any other pair. What is the maximum number of dance pairs that can be formed in this ensemble? | Answer: 22.
Solution. Example. Let Yana and Maxim be the best dancers in the ensemble. If Maxim dances with all the girls except Yana, and Yana dances with all the boys except Maxim, then the condition of the problem is satisfied and the number of pairs will be $15+7=22$.
Estimate. Let's call a pair in which the part... | 22 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) During the break, a fly flew into the math classroom and started crawling on a poster, on which the graph of a quadratic function $y=f(x)$ was depicted in the coordinate plane, with the leading coefficient equal to -1. Initially, the fly moved exactly along the parabola up to the point with an abscissa o... | Answer: 6070.
Solution. Let the quadratic function be of the form $y=-x^{2}+b x+c$. The midpoint of the line segment has coordinates $\left(\frac{2+4}{2} ; \frac{f(2)+f(4)}{2}\right)$, and on the other hand, $(3 ; 6069)$. Since $f(2)=-4+2 b+c, f(4)=-16+4 b+c$, then $-20+6 b+2 c=12138$ or $3 b+c=6079$. Therefore, $f(3)... | 6070 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. (13 points) In a dance ensemble, there are 8 boys and 20 girls. Some of them form mixed (boy and girl) dance pairs. It is known that in each pair, at least one of the partners does not belong to any other pair. What is the maximum number of dance pairs that can be formed in this ensemble? | Answer: 26.
Solution. Example. Let Yana and Maxim be the best dancers in the ensemble. If Maxim dances with all the girls except Yana, and Yana dances with all the boys except Maxim, then the condition of the problem is satisfied and the number of pairs will be $19+7=26$.
Estimate. Let's call a pair in which the part... | 26 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) The density of a body $\rho$ is defined as the ratio of the body's mass $m$ to its volume $V$. A unit of mass used in jewelry is the carat (1 carat equals 0.2 grams). A unit of length used in many countries is the inch (1 inch equals 2.54 centimeters). It is known that the density of diamond is $\rho=3.5... | Answer: $\approx 287$ carats/inch ${ }^{3}$.
Solution. 1 gram $=\frac{1}{0.2}$ carat $=5$ carats,
1 cm $=\frac{1}{2.54}$ inch.
We get: $\rho=3.5 \frac{\mathrm{r}}{\mathrm{cm}^{3}}=3.5 \frac{5 \text { carats }}{\left(\frac{1}{2.54} \text { inch }\right)^{3}}=3.5 \cdot 5 \cdot 2.54^{3} \frac{\text { carats }}{\text { ... | 287 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. (15 points) The density of a body $\rho$ is defined as the ratio of the body's mass $m$ to its volume $V$. A unit of mass used in jewelry is the carat (1 carat equals 0.2 grams). A unit of length used in many countries is the inch (1 inch equals 2.54 centimeters). It is known that the density of emerald is $\rho=2.7... | Answer: $\approx 221$ carats/inch ${ }^{3}$.
Solution. 1 gram $=\frac{1}{0.2}$ carat $=5$ carats,
1 cm $=\frac{1}{2.54}$ inch.
(2 points) | 221 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) Find a natural number $n$ such that the numbers $n+30$ and $n-17$ are squares of other numbers. | Answer: 546.
Solution. From the condition of the problem, it follows that $\left\{\begin{array}{l}n+30=k^{2} \\ n-17=m^{2}\end{array}\right.$. Subtracting the second equation from the first, we get $k^{2}-m^{2}=47(*)$ or $(k-m)(k+m)=47$. Since 47 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \... | 546 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) Every day, Ivan Ivanovich is taken to work by a company car. One day, Ivan Ivanovich decided to walk and left the house an hour earlier than usual. On the way, he met the company car and finished the journey in it. As a result, he arrived at work 10 minutes earlier than the usual time. How long did Ivan ... | Answer: 55 minutes.
Solution. Since Ivan Ivanovich saved the driver 10 minutes by walking, the car travels from Ivan Ivanovich's house to the meeting point in 5 minutes.
We get: $u T=5 v$,
(2 points)
where $u$ is Ivan Ivanovich's walking speed, $v$ is the car's speed, and $T$ is the time Ivan Ivanovich walked. The ... | 55 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. (12 points) Find a natural number $n$ such that the numbers $n+15$ and $n-14$ are squares of other numbers. | Answer: 210.
Solution. From the condition of the problem, it follows that $\left\{\begin{array}{l}n+15=k^{2} \\ n-14=m^{2}\end{array}\right.$ Subtracting the second equation from the first, we get $k^{2}-m^{2}=29(*)$ or $(k-m)(k+m)=29$. Since 29 is a prime number, the possible cases are $\left\{\begin{array}{l}k-m= \p... | 210 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (15 points) Every day, Ivan Ivanovich is taken to work by a company car. One day, Ivan Ivanovich decided to walk and left home one and a half hours earlier than usual. On the way, he met the company car and finished the journey in it. As a result, he arrived at work 20 minutes earlier than the usual time. How long d... | Answer: 80 minutes.
Solution. Since Ivan Ivanovich saved the driver 20 minutes by walking, the car takes 10 minutes to drive from Ivan Ivanovich's house to the meeting point.
We get: $u T=10 v$,
(2 points)
where $u$ is Ivan Ivanovich's walking speed, $v$ is the car's speed, and $T$ is the time Ivan Ivanovich walked... | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $d$ - the greatest common divisor of eight natural numbers, the sum of which is equal to 595. What is the largest value that $d$ can take? | Answer: 35.
Solution. If each of the numbers is divisible by $d$, then their sum is also a multiple of $d$. Therefore, $d$ is a divisor of the number 595. Let's factorize the latter into prime factors: $595=5 \cdot 7 \cdot 17$ and list all its divisors:
$$
1,5,7,17,35,85,119,595
$$
Each of the eight numbers (from th... | 35 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) A rigid board of mass $m$ and length $l=20 m$ is partially lying on the edge of a horizontal surface, hanging off it by three quarters of its length. To prevent the board from falling, a stone of mass $2 m$ is placed at the very edge of the board. How far from the stone can a person of mass $m / 2$ walk ... | Answer: $15 m$
Solution. The moment rule: $2 m g \cdot \frac{l}{4}=m g \cdot \frac{l}{4}+\frac{m}{2} g \cdot\left(x-\frac{l}{4}\right)$.
As a result, we get $x=\frac{3}{4} l=15$ m. | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (15 points) The mass of a vessel completely filled with kerosene is 31 kg. If this vessel is completely filled with water, its mass will be 33 kg. Determine the mass of the empty vessel. The density of water $\rho_{W}=1000 \kappa g / \mathrm{m}^{3}$, the density of kerosene $\rho_{K}=800$ kg $/ \mathrm{m}^{3}$. | Answer: 23 kg
Solution. The mass of the vessel filled with kerosene $m_{1}=m_{c}+\rho_{K} V$,
where $m_{c}$ is the mass of the empty vessel, $V$ is the volume occupied by the kerosene. The mass of the vessel filled with water: $m_{2}=m_{c}+\rho_{b} V$. (4 points)
We get that $V=\frac{m_{2}-m_{1}}{\rho_{B}-\rho_{K}}=... | 23 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Let $d$ be the greatest common divisor of ten natural numbers whose sum is 1001. What is the largest value that $d$ can take? | Answer: 91.
Solution. If each of the numbers is divisible by $d$, then their sum is also divisible by $d$. Therefore, $d$ is a divisor of the number 1001. Let's factorize the latter into prime factors: $1001=7 \cdot 11 \cdot 13$ and list all its divisors:
$$
1,7,11,13,77,91,143,1001
$$
Each of the ten numbers (from ... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) A rigid board of mass $m$ and length $l=24$ m is partially lying on the edge of a horizontal surface, hanging off it by two-thirds of its length. To prevent the board from falling, a stone of mass $2 m$ is placed at the very edge of the board. How far from the stone can a person of mass $m$ walk along th... | Answer: $20 m$
Solution. The moment rule: $2 m g \cdot \frac{l}{3}=m g \cdot \frac{l}{6}+m g \cdot\left(x-\frac{l}{3}\right)$.
As a result, we get $x=\frac{5}{6} l=20 m$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (12 points) There are four weights of different masses. Katya weighs the weights in pairs. As a result, she got 1800, 1970, 2110, 2330, and 2500 grams. How many grams does the sixth weighing variant weigh? | Answer: 2190.
Solution. Let $x, y, z, t$ be the weight of each weight. Then the pairwise weighings will be $x+y, x+z, x+t, y+z, y+t, z+t$ grams. There are three pairs of given numbers: 1) $x+y$ and $z+t$, 2) $x+z$ and $y+t$, 3) $x+t$ and $y+z$, with the same total weight $x+y+z+t$. Let's find pairs with the same sum: ... | 2190 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
3. (12 points) There are four weights of different masses. Katya weighs the weights in pairs. As a result, she got 1700, 1870, 2110, 2330, and 2500 grams. How many grams does the sixth weighing variant weigh? | Answer: 2090.
Solution. Let $x, y, z, t$ be the weight of each weight. Then the pairwise weighings will be $x+y, x+z, x+t, y+z, y+t, z+t$ grams. There are three pairs: 1) $x+$ y and $z+t, 2) x+z$ and 3) $y+t, x+t$ and $y+z$ with the same weight $x+y+z+t$. Let's find pairs with the same sum: $1700+2500=$
$1870+2330$. N... | 2090 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2. The sofa cost 62500 rubles. Once a month, its price changed by $20 \%$ either increasing or decreasing. It is known that over the course of six months, the price increased three times and decreased three times (the order in which this happened is unknown). Can the final price of the sofa after six months be determin... | Answer: Yes; the sofa will cost 55296 rubles.
Solution. An increase in price by $20 \%$ means the current price is multiplied by $6 / 5$, and a decrease in price by $20 \%$ means the current price is multiplied by $4 / 5$. Therefore, regardless of the order in which the price increased or decreased, the price of the s... | 55296 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many five-digit numbers exist where the sum of the first two digits is half the sum of the last two digits? | Answer: 2250.
Solution. Let the first two digits of the number be $a$ and $b$, and the last two digits be $c$ and $d$. We need to find the number of combinations of these digits for which $2(a+b)=c+d$. Since $c+d \leqslant 18$, then $a+b \leqslant 9$.
Let $x_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the fi... | 2250 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
8. (15 points) Density is defined as the ratio of the mass of a body to its volume. There are two cubes. The second cube is made of a material with twice the density of the first, and the side length of the second cube is 100% greater than the side length of the first. By what percentage is the mass of the second cube ... | Answer: by $1500 \%$
Solution. From the condition, we get $\rho_{2}=2 \rho_{1}$, i.e., $\frac{m_{2}}{V_{2}}=2 \cdot \frac{m_{1}}{V_{1}}$.
Also, from the condition, it follows that the side length of the second cube $a_{2}=2 a_{1}$.
(2 points)
Therefore, their volumes are related by the ratio $V_{2}=a_{2}^{3}=\left(... | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. The item cost 64 rubles. Once a month, its price changed by $50\%$ either increasing or decreasing. It is known that over the course of six months, the price increased three times and decreased three times (the order in which this happened is unknown). Can the final price of the item after six months be determined u... | Answer: Yes; the item will cost 27 rubles.
Solution. An increase of $50\%$ means that the current price is multiplied by $3/2$, and a decrease of $50\%$ means that the current price is multiplied by $1/2$. Therefore, regardless of the order in which the price increased or decreased, the price of the sofa after six mon... | 27 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. How many five-digit numbers exist where the sum of the first two digits is twice the sum of the last two digits? | Answer: 2600.
Solution. Let the first two digits of the number be $a$ and $b$, and the last two digits be $c$ and $d$. We need to find the number of combinations of these digits for which $a+b=2(c+d)$. Since $a+b \leqslant 18$, then $c+d \leqslant 9$.
Let $x_{i}(i=1,2, \ldots, 9)$ denote the number of pairs of the la... | 2600 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
6. (10 points) An Englishman was the owner of a plot of land in Russia. He knows that, in the units familiar to him, the size of his plot is three acres. The cost of the land is 250000 rubles per hectare. It is known that 1 acre $=4840$ square yards, 1 yard $=0.9144$ meters, 1 hectare $=10000 m^{2}$. Calculate how much... | Answer: approximately 303514 rubles
Solution. 1 square yard $=0.9144 \cdot 0.9144=0.83612736 \mu^{2}$,
3 acres $=3 \cdot 4840=14520$ square yards $a=14520 \cdot 0.83612739=12140.57 \mu^{2}$ (2 points), 12140.57 $\mu^{2}=\frac{12140.57 \mu^{2}}{10000} \approx 1.214057 ~ 2 a$,
$$
1.214057 \cdot 250000=303514.25 \text ... | 303514 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. (15 points) Density is defined as the ratio of the mass of a body to its volume. There are two cubes. The second cube is made of a material with half the density of the first, and the side length of the second cube is 100% greater than the side length of the first. By what percentage is the mass of the second cube g... | Answer: by $300 \%$
Solution. From the condition, it follows that $\rho_{2}=\frac{1}{2} \rho_{1}$, i.e., $\frac{m_{2}}{V_{2}}=\frac{1}{2} \cdot \frac{m_{1}}{V_{1}}$.
It also follows from the condition that the side length of the second cube $a_{2}=2 a_{1}$.
(2 points)
Therefore, their volumes are related by the rat... | 300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
7. $\left(10\right.$ points) An arc, with a central angle of $\alpha=60^{\circ}$, is cut from a circle with radius $R=40 \mathrm{~cm}$. A charge $q=5$ μC is uniformly distributed along the arc. Determine the electric field strength $E$ at the center of curvature of this arc. | Answer: 269 kV/m.
Solution. Consider a small element of the arc of length $d l$, on which charge $d q$ is located. It creates a field strength at the point of interest:
$d E=k \frac{d q}{R^{2}}$.
 Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, five points on $A B$ and six on $B C$. How many different triangles exist with vertices at the chosen points? | Answer: 135.
Solution. To form a triangle, one needs to choose two points on one side and one point on another. We have: 5 ways to choose the first point on $AB$, 4 ways - the second, and since the triangle does not change with the permutation of its vertices, we divide $5 \cdot 4$ by 2. Thus, $\frac{5 \cdot 4}{2}=10$... | 135 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
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