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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 | 4,190 |
4. Along a circle, 123 points are placed at equal distances from each other. Anya and Borya take turns painting one point either blue or red (they can paint any previously unpainted point). The player loses if, after their move, two adjacent points of the same color appear. Who will win with correct play, if Anya goes ... | Answer: Borya.
Solution. Let's look at how the colored points are arranged when no new moves are possible. If there are two adjacent uncolored points, one of them can be colored. Therefore, in the final position, all points are divided into groups of consecutive colored points, separated by single uncolored points. Wi... | Borya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,191 |
5. (10 points) The distance from home to work is $s=3 \kappa$ km. At the moment Ivan left work, his favorite dog ran out of the house towards him. They met at a distance of a quarter of the entire path from work. The dog instantly turned around and ran back home. Upon reaching home, he instantly turned around again and... | Answer: 9 ki
Solution. By the time of the first meeting, Ivan had walked one fourth of the entire distance. Therefore, the dog ran three fourths of the entire distance. That is, the dog's speed is three times Ivan's speed: $v_{n}=3 v_{u}$.
On the way from work to home, Ivan walked 3 ki.
Therefore, the dog ran $l=3 \... | 9\kappa | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,192 |
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 | 4,193 |
7. (15 points) The structure consists of two weightless rods $AB$ and $CD$, a weightless pulley, and a load of mass $m=2$ kg, which is suspended at point $G$. All threads are weightless and inextensible. At point $A$, the rod is attached to a hinge that allows the rod to rotate in the plane of the diagram. Here, $CG=\f... | Answer: $T_{0}=10 \mathrm{H}$
Solution. The moment rule for rod $C D$ about point $C$: $m g \cdot \frac{1}{4} C D=T_{2} \cdot C D$.
Tension force in thread $B D: T_{2}=\frac{1}{4} m g=5 \mathrm{H}$.
Therefore, the tension force in thread $C F: T_{1}=\frac{3}{4} m g=15 \mathrm{H}$.
The moment rule for rod $A B$ abou... | 10\mathrm{H} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,194 |
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 | 4,195 |
1. Petya and Vasya competed in a 60 m race. When Petya finished, Vasya was 9 m behind him. During the second race, Petya started exactly 9 m behind Vasya. Who finished first in the second race and by how many meters did he outpace his opponent? (Assume that each boy ran at the same constant speed both times). | Answer: Petya outpaced Vasya by 1.35 m.
Solution. In the second race, Petya will eliminate the lag from Vasya by running 60 m (Vasya will run 51 m during this time). 9 m remain to the finish line - which is $20 / 3$ times less than at the start of the 1st race. Therefore, the lead will also be $20 / 3$ times less, i.e... | 1.35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,196 |
2. Twenty numbers are arranged in a circle. It is known that the sum of any six consecutive numbers is 24. What is the number in the 12th position if the number in the 1st position is 1? | Answer: 7.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 20)$. Fix 5 consecutive numbers. The numbers to the left and right of this quintet must match. Therefore, $a_{i}=a_{i+6}$. Let's go in a circle, marking the same numbers:
$$
a_{1}=a_{7}=a_{13}=a_{19}=a_{5}=a_{11}=a_{17}=a_{3}=a_{9}=a_{1... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,197 |
3. Let $d$ be the greatest common divisor of ten natural numbers whose sum is 1001. What is the largest value that $d$ can take? | Answer: 91.
Solution. If each of the numbers is divisible by $d$, then their sum is also divisible by $d$. Therefore, $d$ is a divisor of the number 1001. Let's factorize the latter into prime factors: $1001=7 \cdot 11 \cdot 13$ and list all its divisors:
$$
1,7,11,13,77,91,143,1001
$$
Each of the ten numbers (from ... | 91 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,198 |
4. Along the circumference, 33 points are placed at equal distances from each other. Anya and Borya take turns painting one point either blue or red (they can paint any previously unpainted point). The player loses if, after their move, two adjacent points of the same color appear. Who will win with correct play, if An... | Answer: Borya.
Solution. Let's look at how the colored points are arranged when no new moves are possible. If there are two adjacent uncolored points, one of them can be colored. Therefore, in the final position, all points are divided into groups of consecutive colored points, separated by single uncolored points. Wi... | Borya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,199 |
5. (10 points) The distance from home to work is $s=6 \kappa$ km. At the moment Ivan left work, his favorite dog ran out of the house and ran towards his owner. They met at a distance of one third of the entire path from work. The dog instantly turned around and ran back home. Upon reaching home, he instantly turned ar... | Answer: $12 \kappa n$
Solution. By the time of the first meeting, Ivan had walked one third of the entire distance. Therefore, the dog ran two thirds of the entire distance. That is, the dog's speed is twice Ivan's speed $v_{n}=2 v_{u}$.
(4 points)
On the way from work to home, Ivan walked $6 \kappa$.
Therefore, th... | 12\kappan | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,200 |
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 | 4,201 |
7. (15 points) The structure consists of two weightless rods $AB$ and $CD$, a weightless pulley, and a load of mass $m=3$ kg, which is suspended at point $G$. All threads are weightless and inextensible. At point $A$, the rod is attached to a hinge that allows the rod to rotate in the plane of the diagram. Here, $CG=\f... | Answer: $T_{0}=15 \mathrm{H}$
Solution. The moment rule for rod $C D$ about point $C$: $m g \cdot \frac{1}{4} C D=T_{2} \cdot C D$
Tension force in thread $B D: T_{2}=\frac{1}{4} m g=7.5 \mathrm{H}$.
Therefore, the tension force in thread $C F: T_{1}=\frac{3}{4} m g=22.5 \mathrm{H}$.
The moment rule for rod $A B$ a... | 15\mathrm{H} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,202 |
8. (15 points) The mass of a vessel that is completely filled with kerosene is 20 kg. If this vessel is completely filled with water, its mass will be 24 kg. Determine the mass of the empty vessel. The density of water $\rho_{W}=1000 \kappa g / \mu^{3}$, the density of kerosene $\rho_{K}=800 \kappa g / \mathrm{M}^{3}$. | Answer: 4 kg
Solution. The mass of the vessel filled with kerosene: $m_{1}=m_{c}+\rho_{K} V$, where $m_{c}$ - the mass of the empty vessel, $V$ - 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}}=\f... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,203 |
1. (12 points) Consider the quadratic functions $\mathrm{y}=x^{2}+p x+q$, for which $3 p+q=2023$. Show that the parabolas, which are the graphs of these functions, intersect at one point. | Answer: All graphs intersect at the point ( $(3 ; 2032)$.
Solution. Let $x=3$. Then $y=9+3 p+q=2032$.
Evaluation criteria. Correct solution 12 points. | (3;2032) | Algebra | proof | Yes | Yes | olympiads | false | 4,204 |
2. (13 points) In a hall, the floor has dimensions $4 \times 5 \mathrm{~m}^{2}$, and the ceiling height is 4 m. On the ceiling, in one corner, sits a fly named Masha, and in the opposite corner of the ceiling, a spider named Petya. Masha set off on foot to visit Petya by the shortest route, but with a stop on the floor... | Answer: $\sqrt{145}$ m.
Solution. Let the fly sit at vertex $M$, and the spider sit at vertex $\Pi$ (Fig. 1.).
There are 4 choices of paths for Masha.
Masha can start her journey from the... | \sqrt{145} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,205 |
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 | 4,206 |
4. (13 points) Sixteen people are standing in a circle: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle? | Answer: 10.
Solution. A truthful person can only be next to liars. Three liars in a row cannot stand, so between any two nearest truthful persons, there is one or two liars. Then, if there are 5 or fewer truthful persons, in the intervals between them, there can be no more than 10 liars in total, so there are no more ... | 10 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,207 |
5. (15 points) A glass is filled to the brim with saltwater. On the surface, there is a piece of fresh ice with a mass of \( m = 50 \) g. What volume \( \Delta V \) of water will spill out of the glass by the time the ice melts? Neglect surface tension. The density of fresh ice \( \rho_{\mathrm{n}} = 0.9 \mathrm{r} / \... | Answer: $\approx 2.63 \mathrm{~cm}^{3}$.
Solution. The condition for the floating of fresh ice in salt water:
$\rho_{\text {sw }} g V_{\text {subm }}=m g=\rho_{\text {f }} g V_{1}$,
where $V_{1}$ is the entire volume of fresh ice, $\rho_{\text {sw }}$ is the density of salt water.
The condition for the floating of ... | 2.63\mathrm{~}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,208 |
6. (10 points) In winter, at an ambient air temperature of $t_{\mathrm{o}}=-10{ }^{\circ} \mathrm{C}$, each square meter of the lake gives off 200 kJ of heat to the air per hour. Estimate how long after the formation of ice a fisherman will be able to walk on the surface of the lake, if the safe thickness of ice is 10 ... | Answer: $\approx 153.2$ hours.
Solution. Inside the formed ice, the temperature changes linearly from $0^{\circ} \mathrm{C}$ in the depth to $-10^{\circ} \mathrm{C}$ on the surface. That is, the heat released can be considered to result from the crystallization of ice and its subsequent cooling to an average of $-2^{\... | 153.2 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,209 |
7. (15 points) A rigid rod is moving on a horizontal table. At a certain moment, the speed of one end of the rod is \( v_{1}=5 \mathrm{~m} / \mathrm{c} \), and the speed of the other end is \( v_{2}=4 \) m / s, directed along the axis of the rod (see figure). Determine the speed of the midpoint of the rod at this momen... | Answer: $\approx 4.3 \mathrm{~m} / \mathrm{s}$.
Solution. All points on the rod have a velocity component directed to the right, equal to $v_{2}$.
(4 points)
Therefore, the velocity component directed upwards for point A: $v_{\text {vert A }}=\sqrt{v_{1}^{2}-v_{2}^{2}}=3 \mathrm{m} / \mathrm{s}$.
The velocity compo... | 4.3\mathrm{~}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,210 |
1. (12 points) Consider the quadratic functions $\mathrm{y}=x^{2}+p x+q$, for which $-2 p+q=2023$. Show that the parabolas, which are the graphs of these functions, intersect at one point. | Answer: All graphs intersect at the point ( $-2 ; 2027$ ).
Solution. Let $x=-2$. Then $y=4-2 p+q=2027$.
Evaluation criteria. Correct solution 12 points. | (-2;2027) | Algebra | proof | Yes | Yes | olympiads | false | 4,212 |
2. (13 points) The floor of the hall has dimensions $7 \times 8$ m$^2$, and the ceiling height is 4 m. On the ceiling, in one corner, sits a fly named Masha, and in the opposite corner of the ceiling, a spider named Petya. Masha set off on foot to visit Petya by the shortest route, but with a stop on the floor. Find th... | Answer: $\sqrt{265} \text{m}$.
Solution. Let the fly sit at vertex $M$, and the spider sit at vertex $\Pi$ (Fig. 1.).
Fig. 1.
There are 4 choices of paths for Masha.
Masha can start her ... | \sqrt{265} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,213 |
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 | 4,214 |
4. (13 points) In a circle, there are 17 people: each of them is either a truth-teller (he always tells the truth) or a liar (he always lies). Everyone said that both of their neighbors are liars. What is the maximum number of liars that can be in this circle? | Answer: 11.
Solution. A truth-teller can only be surrounded by liars. Three liars in a row cannot stand, so between any two nearest truth-tellers, there is one or two liars. Then, if there are 5 or fewer truth-tellers, there can be no more than 10 liars in the gaps between them, making a total of no more than 15 peopl... | 11 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,215 |
5. (15 points) A glass is filled to the brim with saltwater. On the surface, there is a piece of fresh ice with a mass of $m=100$ g. What volume $\Delta V$ of water will spill out of the glass by the time the ice melts? Neglect surface tension. The density of fresh ice $\rho_{\mathrm{n}}=0.9 \text{ g/cm}^{3}$, the dens... | Answer: $\approx 5.26 \mathrm{~cm}^{3}$.
Solution. The condition for the floating of fresh ice in salt water:
$\rho_{\text {sw }} g V_{\text {sub }}=m g=\rho_{\text {f }} g V_{1}$,
where $V_{1}$ is the entire volume of fresh ice, $\rho_{\text {sw }}$ is the density of salt water.
The condition for the floating of t... | 5.26\mathrm{~}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,216 |
6. (10 points) In winter, at an ambient air temperature of $t_{0}=-20^{\circ} \mathrm{C}$, each square meter of the lake gives off 300 kJ of heat to the air per hour. Estimate how long after the formation of ice a fisherman will be able to walk on the surface of the lake, if the safe thickness of ice is 10 cm? The temp... | Answer: 105.3 hours.
Solution. Inside the formed ice, the temperature changes linearly from $0^{\circ} \mathrm{C}$ in depth to $-20^{\circ} \mathrm{C}$ on the surface. That is, the heat released can be considered to result from the crystallization of ice and its subsequent cooling to an average of $-10^{\circ} \mathrm... | 105.3 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,217 |
7. (15 points) A rigid rod is moving on a horizontal table. At a certain moment, the speed of one end of the rod is \( v_{1}=10 \mathrm{~m} / \mathrm{c} \), and the speed of the other end is \( v_{2}=6 \mathrm{~m} / \mathrm{c} \), directed along the axis of the rod (see figure). Determine the speed of the midpoint of t... | Answer: $\approx 7.2 \mathrm{~m} / \mathrm{s}$.
Solution. All points on the rod have a velocity component directed to the right, equal to $v_{2}$.
Therefore, the velocity component directed upwards for point $A: v_{\text {vert A }}=\sqrt{v_{1}^{2}-v_{2}^{2}}=8 \mathrm{~m} / \mathrm{s}$.
The velocity component of poi... | 7.2\mathrm{~}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,218 |
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 | 4,220 |
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 | 4,221 |
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 two acres. The cost of the land is 500,000 rubles per hectare. It is known that 1 acre = 4840 square yards, 1 yard = 0.9144 meters, 1 hectare = 10000 m². Calculate how much the En... | Answer: approximately 405000 rubles
Solution. 1 square yard $=0.9144 \cdot 0.9144=0.83612736 \mu^{2}$,
2 acres $=2 \cdot 4840=9680$ square yards $=9680 \cdot 0.83612739=8093.7128448 m^{2}$ (2 points), 8093.7128448 $\mu^{2}=\frac{8093.7128448 \mu^{2}}{10000} \approx 0.80942 a$,
$0.8094 \cdot 500000=404685.6$ rub. | 404685.6 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,223 |
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The smaller wheel makes 30 revolutions per minute. Determine how many seconds the larger wheel spends on one revolution?
The distances traveled by these points differ by three times.
The small wheel spends on one revolution: $t_{\text {small }}=\frac{1 \text { min }}{30 \text { revolutions }}=\frac{60 s}{30 \text { revolutions }}=2 s... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,224 |
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 | 4,225 |
1. Petya and Vasya competed in a 60 m race. When Petya finished, Vasya was 9 m behind him. During the second race, Petya started exactly 9 m behind Vasya. Who finished first in the second race and by how many meters did he outpace his opponent? (Assume that each boy ran at a constant speed both times). | Answer: Petya outpaced Vasya by 1.35 m.
Solution. In the second race, Petya will eliminate the lag from Vasya by running 60 m (Vasya will run 51 m during this time). 9 m remain to the finish line - which is $20 / 3$ times less than at the start of the 1st race. Therefore, the lead will also be $20 / 3$ times less, i.e... | 1.35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,226 |
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 | 4,227 |
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 | 4,228 |
4. Along a circle, 33 points are placed at equal distances from each other. Anya and Borya take turns painting one point either blue or red (they can paint any previously unpainted point). The player loses if, after their move, two adjacent points of the same color appear. Who will win with correct play if Anya goes fi... | Answer: Borya.
Solution. Let's look at how the colored points are arranged when no new moves are possible. If there are two adjacent uncolored points, one of them can be colored. Therefore, in the final position, all points are divided into groups of consecutive colored points, separated by single uncolored points. Wi... | Borya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,229 |
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 | 4,231 |
7. (10 points) Two wheels rotate, meshed with each other, around fixed axes passing through the centers of wheels $A$ and $B$. The radii of the wheels differ by a factor of three. The larger wheel makes 10 revolutions per minute. Determine how many seconds the smaller wheel spends on one revolution?
 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 | 4,233 |
1. $\left(12\right.$ points) Solve the equation $\left(x^{4}-2\right)\left(2^{\operatorname{tg} x}-1\right)+\left(3^{x^{4}}-9\right) \operatorname{tg} x=0$. | Answer: $\pm \sqrt[4]{2} ; \pi n, \quad n \in Z$.
Solution. The functions $y=2^{t}$ and $y=3^{t}$ are increasing, therefore, the expression $3^{x^{4}}-9=3^{x^{4}}-3^{2}$ has the same sign as $x^{4}-2$, and the expression $2^{\operatorname{tg} x}-1=2^{\operatorname{tg} x}-2^{0}$ has the same sign as $\operatorname{tg} ... | \\sqrt[4]{2};\pin,\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,234 |
2. (13 points) A triathlon competitor swam 1 km in the first stage. In the second stage, he cycled 25 km, and in the third stage, he ran 4 km. He completed the entire distance in 1 hour 15 minutes. Before the competition, he tried out the course: he swam for $1 / 16$ hours, cycled and ran for $1 / 49$ hours each, cover... | Answer: $5 / 7$ hour; 35 km $/$ hour.
Solution. Let $v_{1}, v_{2}, v_{3}$ be the speeds of the athlete on stages $1,2,3$ respectively. From the condition, we have: $\frac{1}{v_{1}}+\frac{25}{v_{2}}+\frac{4}{v_{3}}=\frac{5}{4}$ hour; $\frac{1}{16} v_{1}+\frac{1}{49} v_{2}+$ $\frac{1}{49} v_{3}=\frac{5}{4}$ km. Adding t... | \frac{5}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,235 |
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=3 \sin x, f_{n+1}(x)=\frac{9}{3-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{6}\right)$. | Answer: $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{9}{3-3 \sin x}$. Consequently, $f_{2023}\left(\frac{\pi}{6}\right)=6$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cy... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,236 |
4. (13 points) The lateral edges of a triangular pyramid are pairwise perpendicular, and the sides of the base are equal to $\sqrt{61}$, $\sqrt{52}$, and $\sqrt{41}$. The center of the sphere that touches all the lateral faces lies on the base of the pyramid. Find the radius of this sphere. | Answer: $\frac{60}{37}$.
Solution. Let the base of the pyramid be $A B C$, the apex of the pyramid be $D$, the center of the sphere be $O$, and the radius of the sphere be $r$. Let $A B=\sqrt{41}, B C=\sqrt{61}, A C=\sqrt{52}$. Denote $A D=x, B D=y, C D=z$.
Since the radius drawn to the point of tangency of the spher... | \frac{60}{37} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,237 |
5. (10 points) Two stones are thrown simultaneously with the same initial speed $v_{0}$. The first stone is thrown horizontally from a height of $H=40$ m, and the second is thrown vertically upward from the surface of the Earth. It is known that the stones collided in the air. Determine the initial distance between the... | Answer: $\approx 56.6$ m.
Solution. Equations of motion for the first stone: $y_{1}=H-\frac{g t^{2}}{2}$.
$x_{1}=v_{0} t=L$
where $L$ is the initial horizontal distance between the bodies.
Equation of motion for the second body: $y_{2}=v_{0} t-\frac{g t^{2}}{2}$.
Since the stones met, $y_{1}=y_{2}$,
As a result: ... | 56.6\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,238 |
6. (15 points) The operation of a heat engine is based on a cycle consisting of an isochore, an isotherm, and a process with a directly proportional dependence of pressure on volume (see figure). An ideal monatomic gas is used as the working substance. It is known that the maximum and minimum temperatures differ by a f... | Answer: $8.8\%$.
Solution. Since the temperature changes by a factor of two during the process 3-1, it follows from the equation of state of an ideal gas $p V=\vartheta R T$ that the volume changes by a factor of $\sqrt{2}$.
The magnitude of the work done by the gas in the process 3-1 is equal to the area under the g... | 8.8 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,239 |
7. (10 points) An arc, with a central angle of $\alpha=30^{\circ}$, is cut from a circle with radius $R=50$ cm. A charge $q=2$ μC is uniformly distributed along the arc. Determine the electric field strength $E$ at the center of curvature of this arc. | Answer: $71 \mathrm{kB} / \mathrm{m}$.
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:
$d E=k \frac{d q}{R^{2}}$.
 In an ideal circuit consisting of a capacitor with capacitance \( C = 2 \) μF and an inductor with inductance \( L_{2} = 1 \) mH, undamped free harmonic oscillations of current with an amplitude \( I_{\max} = 5 \) mA occur. At the moment when the current through the inductor \( L_{2} \) is maximal, the s... | Answer: $90 \mathrm{mB}$.
Solution. After the key is closed, Kirchhoff's rule for the loop consisting of the coils: $L_{1} I_{1}^{I}+L_{2} I_{2}^{\mid}=0$.
We get that $L_{1} I_{1}+L_{2} I_{2}=$ const.
When the voltage across the capacitor is maximum, the current in the circuit segment with the capacitor is zero. Th... | 90\mathrm{mV} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,241 |
1. $\left(12\right.$ points) Solve the equation $\left(x^{3}-3\right)\left(2^{\operatorname{ctg} x}-1\right)+\left(5^{x^{3}}-125\right) \operatorname{ctg} x=0$. | Answer: $\sqrt[3]{3} ; \frac{\pi}{2}+\pi n, \quad n \in Z$.
Solution. The functions $y=2^{t}$ and $y=5^{t}$ are increasing, therefore, the expression $5^{x^{3}}-125=5^{x^{3}}-5^{3}$ has the same sign as $x^{3}-3$, and the expression $2^{\operatorname{ctg} x}-1=2^{\operatorname{ctg} x}-2^{0}$ has the same sign as $\ope... | \sqrt[3]{3};\frac{\pi}{2}+\pin,\quadn\inZ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,242 |
2. (13 points) A triathlon competitor swam 1 km in the first stage. In the second stage, he cycled 25 km, and in the third stage, he ran 4 km. He completed the entire distance in 1 hour 15 minutes. Before the competition, he tried out the course: he swam for $1 / 16$ hours, cycled and ran for $1 / 49$ hours each, cover... | Answer: $2 / 7$ hour; 14 km/h.
Solution. Let $v_{1}, v_{2}, v_{3}$ be the speeds of the athlete on stages $1,2,3$ respectively. From the condition, we have: $\frac{1}{v_{1}}+\frac{25}{v_{2}}+\frac{4}{v_{3}}=\frac{5}{4}$ hour; $\frac{1}{16} v_{1}+\frac{1}{49} v_{2}+$ $\frac{1}{49} v_{3}=\frac{5}{4}$ km. Adding these eq... | \frac{2}{7} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,243 |
3. (12 points) The sequence of functions is defined by the formulas
$$
f_{0}(x)=2 \cos x, f_{n+1}(x)=\frac{4}{2-f_{n}(x)}
$$
for any integer $n \geq 0$. Find $f_{2023}\left(\frac{\pi}{3}\right)$. | Answer: $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Solution. It is easy to compute: $f_{3}(x)=f_{0}(x)$, therefore $f_{2023}(x)=f_{1}(x)=$ $\frac{4}{2-2 \cos x}$. Consequently, $f_{2023}\left(\frac{\pi}{3}\right)=4$.
Remark. One can immediately compute the values of the functions at the given point. This results in a cy... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,244 |
4. (13 points) The lateral edges of a triangular pyramid are pairwise perpendicular, and the sides of the base are equal to $\sqrt{85}$, $\sqrt{58}$, and $\sqrt{45}$. The center of the sphere that touches all the lateral faces lies on the base of the pyramid. Find the radius of this sphere. | Answer: $\frac{14}{9}$.
Solution. Let the base of the pyramid be $A B C$, the apex of the pyramid be $D$, the center of the sphere be $O$, and the radius of the sphere be $r$. Let $A B=\sqrt{45}, B C=\sqrt{85}, A C=\sqrt{58}$. Denote $A D=x, B D=y, C D=z$.
Since the radius drawn to the point of tangency of the sphere... | \frac{14}{9} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,245 |
5. (10 points) Two stones are thrown simultaneously with the same initial speed $v_{0}$. The first stone is thrown horizontally from a height of $H=50$ m, and the second is thrown vertically upward from the surface of the Earth. It is known that the stones collided in the air. Determine the initial distance between the... | Answer: $\approx 70.7$ m.
Solution. Equations of motion for the first stone: $y_{1}=H-\frac{g t^{2}}{2}$.
$x_{1}=v_{0} t=L$,
where $L$ is the initial horizontal distance between the bodies.
Equation of motion for the second body: $y_{2}=v_{0} t-\frac{g t^{2}}{2}$.
Since the stones met, $y_{1}=y_{2}$,
As a result:... | 70.7\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,246 |
6. (15 points) The operation of a heat engine is based on a cycle consisting of an isobaric process, an isothermal process, and a process with a directly proportional dependence of pressure on volume (see figure). An ideal monatomic gas is used as the working substance. It is known that the maximum and minimum temperat... | Answer: $6.1\%$.
Solution. Since the temperature changes by a factor of 2 during the process 2-3, it follows from the equation of state of an ideal gas $p V = \vartheta R T$ that the pressure changes by a factor of $\sqrt{2}$.
(1 point)
The magnitude of the work done by the gas in the process 2-3 is equal to the are... | 6.1 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,247 |
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:
$d E=k \frac{d q}{R^{2}}$.
 In an ideal circuit consisting of a capacitor with capacitance \( C=1 \) μF and an inductor with inductance \( L_{2}=2 \) mH, undamped free harmonic oscillations of current occur with an amplitude of \( I_{\max }=10 \) mA. At the moment when the current through the inductor \( L_{2} \) is maximal, the sw... | Answer: $516 \mathrm{mB}$.
Solution. After the key is closed, Kirchhoff's rule for the loop consisting of the coils: $L_{1} I_{1}^{\mid}+L_{2} I_{2}^{\mid}=0$.
We get that $L_{1} I_{1}+L_{2} I_{2}=$ const.
When the voltage across the capacitor is maximum, the current in the circuit segment with the capacitor is zero... | 516\mathrm{mV} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,249 |
1. (16 points) Solve the equation $x-5=\frac{3 \cdot|x-2|}{x-2}$. If the equation has multiple roots, write their sum in the answer.
# | # Answer: 8.
Solution. The equation has a restriction on the variable $x \neq 2$. We open the modulus: for $x>2, x-5=3, x=8$. For $x<2, \quad x-5=-3, x=2$ - an extraneous root. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,250 |
2. (17 points) 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 | 4,251 |
3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $50^{\circ}$. Find the angle $A B C$. | Answer: 115.
Solution. Complete triangle $A B C$ to parallelogram $A B C D$.
Then, $D B=2 N B=A B$. Therefore, triangle $A B D$ is isosceles and $\angle A D B=\frac{180^{\circ}-50^{\circ}}{... | 115 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,252 |
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 13 hours and 20 minutes. | Answer: $80^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{20}{60} \cdot 360=120^{0}$. The hour hand has moved away from twelve by $\frac{1}{12} \cdot 360+\frac{20}{60} \cdot \frac{1}{12} \cdot 360=40^{\circ}$. The angle between the hands $120^{\circ}-40^{\circ}=80^{\circ}$. | 80 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,254 |
6. (15 points) A metal bar with a temperature of $20{ }^{\circ} \mathrm{C}$ was submerged into water taken at a temperature of $80{ }^{\circ} \mathrm{C}$. After thermal equilibrium was established, the temperature was found to be $60{ }^{\circ} \mathrm{C}$. After this, without removing the first bar from the water, ano... | Answer: $50^{\circ} \mathrm{C}$.
Solution. The thermal balance equation in the first case is $c_{\mathrm{B}} m_{\mathrm{B}} 20=c_{\text {b }} m_{\text {b }} 40$. The thermal balance equation in the second case is
$$
c_{\mathrm{B}} m_{\mathrm{B}}(60-t)+c_{6} m_{6}(60-t)=c_{6} m_{6}(t-20) .
$$
We get $c_{6} m_{6} \cdo... | 50\mathrm{C} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,255 |
1. (16 points) Solve the equation $x-7=\frac{4 \cdot|x-3|}{x-3}$. If the equation has multiple roots, write their sum in the answer. | Answer: 11.
Solution. The equation has a restriction on the variable $x \neq 3$. We open the modulus: for $x>3, x-7=4, x=11$. For $x<3, \quad x-7=-4, x=3-$ extraneous root. | 11 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,256 |
2. (17 points) Given a rectangle $A B C D$. On two sides of the rectangle, different points are chosen, six points on $A B$ and seven - on $B C$. How many different triangles exist with vertices at the chosen points? | Answer: 231.
Solution. To form a triangle, one needs to choose two points on one side and one point on another. There are 6 ways to choose the first point on $AB$, 5 ways to choose the second, and since the triangle does not change with the permutation of its vertices, we divide $6 \cdot 5$ by 2. Thus, $\frac{6 \cdot ... | 231 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,257 |
3. (17 points) In triangle $A B C$, a point $N$ is taken on side $A C$ such that $A N=N C$. Side $A B$ is twice as long as $B N$, and the angle between $A B$ and $B N$ is $40^{\circ}$. Find the angle $A B C$. | Answer: 110.
## Solution.
Extend triangle $ABC$ to parallelogram $ABCD$. Then, $DB=2NB=AB$. Therefore, triangle $ABD$ is isosceles and $\angle ADB=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ... | 110 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,258 |
4. (15 points) The object consists of two parts. The density of one of them is 7800 kg $/ \mathrm{m}^{3}$. In addition, it is known that this part occupies $30 \%$ of the volume of the entire object and its mass is $60 \%$ of the total mass. Determine the density of the second part. | Answer: $2229 \kappa / / \mathrm{m}^{3}$.
Solution. From the condition: $m_{1}=0.6 m$, and $V_{1}=0.3 V$. We get $m_{2}=0.4 m$, and $V_{2}=0.7 V$. The density of the first part $\rho_{1}=\frac{m_{1}}{V_{1}}=\frac{0.6 m}{0.3 V}$. The density of the second part:
$$
\rho_{2}=\frac{m_{2}}{V_{2}}=\frac{0.4 m}{0.7 V}=\frac... | 2229\kappa/\mathrm{}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,259 |
5. (20 points) Determine the angle between the hour and minute hands at the moment when they show 15 hours and 40 minutes. | Answer: $130^{\circ}$.
Solution. The minute hand has moved away from twelve by $\frac{40}{60} \cdot 360=240^{\circ}$. The hour hand has moved away from twelve by $\frac{3}{12} \cdot 360+\frac{40}{60} \cdot \frac{1}{12} \cdot 360=110^{\circ}$. The angle between the hands: $240^{\circ}-110^{\circ}=130^{\circ}$. | 130 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,260 |
6. (15 points) A metal bar with a temperature of $20^{\circ} \mathrm{C}$ was submerged into water taken at a temperature of $100^{\circ} \mathrm{C}$. After thermal equilibrium was established, the temperature was found to be $80^{\circ} \mathrm{C}$. After this, without removing the first bar from the water, another ide... | Answer: $68^{\circ} \mathrm{C}$.
Solution. The heat balance equation in the first case is $c_{\mathrm{B}} m_{\mathrm{B}} 20=c_{\text {b }} m_{\text {b }} 60$. The heat balance equation in the second case:
$$
c_{\mathrm{B}} m_{\mathrm{B}}(80-t)+c_{6} m_{6}(80-t)=c_{6} m_{6}(t-20) .
$$
We get $c_{6} m_{6} \cdot 3 \cdo... | 68\mathrm{C} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,261 |
1. On the board, there are 99 ones written in a row. Is it possible to place + and - signs between some of them so that the value of the resulting expression is equal to 2017? | Answer: It is possible.
Solution. The signs can be arranged as follows:
$1111-111+1111-111+11+11-1-1-1-1-1+1-1+1-1+\ldots+1-1$.
Evaluation. 12 points for a correct example. | 1111-111+1111-111+11+11-1-1-1-1-1+1-1+1-1+\ldots+1-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,262 |
2. The cold water tap fills the bathtub in 17 minutes, and the hot water tap in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it? | Answer: in 3 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 8.5 minutes. Therefore, the hot water tap should be open for 3 minutes longer.
Evaluation. 12 points for the correct solution. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,263 |
3. A certain mechanism consists of 30 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 20 parts - at least one large part. How many of each type of part does the mechanism contain? | Answer: 11 large parts and 19 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 20 parts there is a large one, there are no more than 19 small parts. If there were fewer than 11 large parts or fewer than 19 small parts, there would be fewer tha... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,264 |
4. On a line, several points were marked, including points $A$ and $B$. All possible segments with endpoints at the marked points are considered. Vasya calculated that point $A$ is inside 40 of these segments, and point $B$ is inside 42 segments. How many points were marked? (The endpoints of a segment are not consider... | Answer: 14.
Solution. Let there be $a_{1}$ points on one side of point $A$ and $a_{2}$ points on the other side; $b_{1}$ points on one side of point $B$ and $b_{2}$ points on the other side. We can assume that $a_{1} \leqslant a_{2}, b_{1} \leqslant b_{2}$. Then $a_{1} a_{2}=40, b_{1} b_{2}=42$. At the same time, $a_{... | 14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,265 |
1. On the board, there are 79 ones written in a row. Is it possible to place + and - signs between some of them so that the value of the resulting expression is equal to 2017? | Answer: It is possible.
Solution. The signs can be arranged as follows:
$1111-111+1111-111+11+11-1-1-1-1-1+1-1+1-1+\ldots+1-1$.
Evaluation. 12 points for a correct example. | 1111-111+1111-111+11+11-1-1-1-1-1+1-1+1-1+\ldots+1-1 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,266 |
2. The cold water tap fills the bathtub in 19 minutes, while the hot water tap fills it in 23 minutes. The hot water tap was opened. After how many minutes should the cold water tap be opened so that by the time the bathtub is completely filled, there is an equal amount of cold and hot water in it? | Answer: in 2 minutes.
Solution. Half of the bathtub is filled with hot water in 11.5 minutes, and with cold water in 9.5 minutes. Therefore, the hot water tap should be open for 2 minutes longer.
Evaluation. 12 points for the correct solution. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,267 |
3. A certain mechanism consists of 25 parts, some of which are large, and some are small. It is known that among any 12 parts taken, there will be at least one small part, and among any 15 parts, there will be at least one large part. How many of each type of part does the mechanism contain? | Answer: 11 large parts and 14 small parts.
Solution. Since among any 12 parts there is a small one, there are no more than 11 large parts. Since among any 15 parts there is a large one, there are no more than 14 small parts. If there were fewer than 11 large parts or fewer than 14 small parts, the total number of part... | 11 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,268 |
4. On a line, several points were marked, including points $A$ and $B$. All possible segments with endpoints at the marked points are considered. Vasya calculated that point $A$ is inside 50 of these segments, and point $B$ is inside 56 segments. How many points were marked? (The endpoints of a segment are not consider... | Answer: 16.
Solution. Let there be $a_{1}$ points on one side of point $A$ and $a_{2}$ points on the other side; $b_{1}$ points on one side of point $B$ and $b_{2}$ points on the other side. We can assume that $a_{1} \leqslant a_{2}, b_{1} \leqslant b_{2}$. Then $a_{1} a_{2}=50, b_{1} b_{2}=56$. At the same time, $a_{... | 16 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,269 |
# Problem № 6 (15 points)
A body is thrown vertically upwards from a balcony. The dependence of the modulus of the body's acceleration on time is shown in the graph. Using this dependence, estimate the initial speed of the body. The acceleration due to gravity $g=10 \mathrm{~m} / \mathrm{s}^{2}$.
At time $t=2 s$, the acceleration equals the acceleration due to gravity. Therefore, at this moment, the ball was at the highest point of its trajectory and its velocity $v=0 m/s$. (4 poi... | 30\mathrm{~}/\mathrm{} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,271 |
# Problem No. 8 (10 points)
It is known that a heated body radiates energy from one square meter each second, which is determined by the expression $w=\sigma T^{4}$, where $\sigma=5.67 \cdot 10^{-8}$ J $/(c \cdot \mu^{2} \cdot K^{4})$. To what temperature will a piece of wire with a length of $L=50$ cm and a cross-sec... | # Solution and Evaluation Criteria:
Power of the current flowing through the wire: $P=U I$.
Power of thermal radiation: $P=w \cdot S=\sigma T^{4} L \cdot \pi D$
We obtain:
$\sigma T^{4} L \cdot \pi D=U I$. From which: $T=\sqrt[4]{\frac{U I}{L \pi D}}=\sqrt[4]{\frac{220 \cdot 5}{5.67 \cdot 10^{-8} \cdot 0.5 \cdot 3.... | 1576\mathrm{~K} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,273 |
# Problem № 6 (15 points)
A body is thrown vertically upwards from a high balcony. The dependence of the modulus of the body's acceleration on time is shown in the graph. Using this dependence, estimate the terminal velocity of the body. The acceleration due to gravity \( g = 10 \, \text{m} / \text{s}^2 \).
It is known that a heated body radiates energy from one square meter each second, which is determined by the expression $w=\sigma T^{4}$, where $\sigma=5.67 \cdot 10^{-8}$ J $/(c \cdot \mu^{2} \cdot K^{4})$. To what temperature will a piece of wire with a length $L=25$ cm and a cross-sectio... | # Solution and evaluation criteria:
Power of the current flowing through the wire: $P=U I$.
Power of thermal radiation: $P=w \cdot S=\sigma T^{4} L \cdot \pi D$
We obtain:
$$
\sigma T^{4} L \cdot \pi D=U I \text {. Therefore: } T=\sqrt[4]{\frac{U I}{L \pi D}}=\sqrt[4]{\frac{220 \cdot 5}{5.67 \cdot 10^{-8} \cdot 0.2... | 2229\mathrm{~K} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,277 |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 2 hours and 40 minutes, without it in 8 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to... | Answer: 288.
Solution. The tablet charges in 160 minutes on fast charging, and in 480 minutes on regular charging. Therefore, on fast charging, $\frac{1}{160}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{480}$ of the full charge is completed in 1 minute. Let $t-$ be the total chargi... | 288 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,278 |
2. (17 points) There are ten weights of different weights, each weighing an integer number of grams. It is known that the weight of the lightest and heaviest weight differs by 9 grams. One weight is lost. Find its weight if the total weight of the remaining weights is 2022 grams. | Answer: 223.
Solution. Let $x$ be the weight of the lightest weight. Denote the weight of the lost weight as $(x+y)$ $(0<y<9)$. Then $x+(x+1)+(x+2)+\cdots+(x+$ $9)-(x+y)=2022$. Combine like terms: $10 x+45-x-y=$ 2022 or $9 x=1977+y$. From this, $1977+y$ is divisible by 9. Considering the condition $0<y<9$, we get that... | 223 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,279 |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12 \%$. However, this was not enough, so it was decided to red... | Answer: 18.
Solution. Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $2(0.9 x + 0.8 y) = 0.88 \cdot 2(x + y)$ or $x = 4 y$. The original perimet... | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,280 |
4. (15 points) A one-and-a-half kilogram model of a sports car body was made from carbon fiber for aerodynamic studies at a scale of 1:10. What is the mass of the actual body if it is also entirely made of carbon fiber? | Answer: 1500 kg.
Solution. All dimensions of the body are 10 times larger compared to the model. Therefore, the volume of the body is larger by $10 \cdot 10 \cdot 10=1000$ times. Mass is directly proportional to volume, therefore, the mass of the body:
$$
m_{\text {body }}=1000 \cdot m_{\text {model }}=1500 \text { k... | 1500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,281 |
5. (20 points) A car traveled half of the distance at a speed 20 km/h faster than the average speed, and the second half of the distance at a speed 20% lower than the average. Determine the average speed of the car. | Answer: 60 km/h.
Solution. The average speed $v=\frac{s+s}{t_{1}+t_{2}}=\frac{s+s}{\frac{s}{v+20}+\frac{s}{0.8 v}}=\frac{2}{\frac{1}{v+20}+\frac{1}{0.8 v}}$. Solving this equation, we get $v=60$ km/h. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,282 |
6. (15 points) A pedestrian is moving towards a crosswalk along a straight path at a constant speed of 3.6 km/h. At the initial moment, the distance from the pedestrian to the crosswalk is 20 m. The length of the crosswalk is $5 \mathrm{~m}$. At what distance from the crosswalk will the pedestrian be after half a minut... | Answer: 5 m.
Solution. The pedestrian's speed is 3.6 km/h = 1 m/s. In half a minute, he walked $s = v t = 1 \cdot 30 s = 30$ m. From the crossing, he is at a distance of $l = s - 20 - 5 = 5$ m.
## Tasks, answers, and evaluation criteria | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,283 |
1. (17 points) Masha's tablet, which she needed for a presentation at school, was completely drained. Using additional equipment, the tablet can be fully charged in 3 hours, without it in 9 hours. Masha first put the discharged tablet on regular charging, and when she found the equipment, she switched to fast charging ... | Answer: 324.
Solution. The tablet charges in 180 minutes on fast charging, and in 540 minutes on regular charging. Therefore, on fast charging, $\frac{1}{180}$ of the full charge is completed in 1 minute, and on regular charging, $\frac{1}{540}$ of the full charge is completed in 1 minute. Let $t$ be the total chargin... | 324 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,284 |
2. (17 points) There are ten weights of different weights, each weighing an integer number of grams. It is known that the weight of the lightest weight and the heaviest differs by 9 grams. One weight is lost. Find the weight of the lightest weight if the total weight of the remaining weights is 2022 grams.
# | # Answer: 220.
Solution. Let $x$ be the weight of the lightest weight. Denote the weight of the lost weight as $(x+y)$ $(0<y<9)$. Then $x+(x+1)+(x+2)+\cdots+(x+$ 9) - $(x+y)=2022$. Combine like terms: $10 x+45-x-y=$ 2022 or $9 x=1977+y$. From this, $1977+y$ is divisible by 9. Considering the condition $0<y<9$, we get ... | 220 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,285 |
3. (16 points) In a garden plot, it was decided to create a rectangular flower bed. Due to a lack of space, the length of the flower bed was reduced by $10 \%$, and the width was reduced by $20 \%$. As a result, the perimeter of the flower bed decreased by $12.5 \%$. However, this was not enough, so it was decided to r... | Answer: 14.
Solution. Let $x$ be the length of the flower bed, $y$ be the width of the flower bed. After the reduction: $0.9 x$ - length of the flower bed, $0.8 y$ - width of the flower bed, $2(0.9 x + 0.8 y)$ - perimeter. We get the equation: $\quad 2(0.9 x + 0.8 y) = 0.875 \cdot 2(x + y) \quad$ or $\quad x = 3 y$. T... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,286 |
4. (15 points) A one-kilogram model of a sports car body was made from carbon fiber for aerodynamic studies at a scale of 1:11. What is the mass of the actual body if it is also entirely made of carbon fiber? | Answer: 1331 kg.
Solution. All dimensions of the body are 11 times larger compared to the model. Therefore, the volume of the body is larger by $11 \cdot 11 \cdot 11=1331$ times. The mass is directly proportional to the volume, therefore, the mass of the body:
$$
m_{\text {body }}=1331 \cdot m_{\text {model }}=1331 \... | 1331 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,287 |
5. (20 points) A car traveled half of the distance at a speed 30 km/h faster than the average speed, and the second half of the distance at a speed 30% lower than the average. Determine the average speed of the car. | Answer: 40 km/h.
Solution. The average speed $v=\frac{s+s}{t_{1}+t_{2}}=\frac{s+s}{\frac{s}{v+30}+\frac{s}{0.7 v}}=\frac{2}{\frac{1}{v+30}+\frac{1}{0.7 v}}$. Solving this equation, we get $v=40$ km/h. | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,288 |
6. (15 points) A pedestrian is moving in a straight line towards a crosswalk at a constant speed of 3.6 km/h. At the initial moment, the distance from the pedestrian to the crosswalk is 40 m. The length of the crosswalk is 6 m. At what distance from the crosswalk will the pedestrian be after two minutes? | Answer: 74 m.
Solution. The pedestrian's speed is 3.6 km/h = 1 m/s. In two minutes, he walked $s=v t=1 \cdot 120 s=120$ m. From the crossing, he is at a distance of $l=s-40-$ $6=74 \mathrm{~m}$. | 74\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,289 |
# Problem № 5 (10 points)
When constructing this structure, a homogeneous wire of constant cross-section was used. It is known that points $B, D, F$ and $H$ are located equally at the midpoints of the corresponding sides of the square $A C E G$. The resistance of segment $A B$ is $R_{0}=1 \Omega$. Determine the resist... | # Answer: 0.94 Ohms
## Solution and grading criteria:
The resistance of a resistor is proportional to its length.
Its resistance: $\quad R=\frac{4+2 \sqrt{2}}{3+3 \sqrt{2}} R_{0} \approx 0.... | 0.94 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,290 |
# Problem No. 6 (10 points)
A pot was filled with $2 \pi$ liters of water, taken at a temperature of $t=0{ }^{\circ} C$, and brought to a boil in 10 minutes. After that, without removing the pot from the stove, ice at a temperature of $t=0{ }^{\circ} \mathrm{C}$ was added. The water began to boil again only after 15 m... | # Solution and Evaluation Criteria:
Mass of the initial water: $m_{B}=\rho V=2$ kg
Power of the stove in the first case: $P=\frac{c_{B} m_{B} \Delta T}{t_{1}}$.
And in the second case: $P=\frac{\lambda m_{J}+c_{B} m_{\pi} \Delta T}{t_{2}}$.
We obtain:
$$
\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B}... | 1.68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,291 |
# Problem No. 8 (15 points)
$100 \, \text{g} \, p$ of ice, taken at a temperature of $t_{n}=-5^{\circ} \text{C}$, was mixed with water taken at a temperature of $t_{B}=10^{\circ} \text{C}$. It is known that the final temperature in the container is $t=0^{\circ} \text{C}$. Determine the mass of the added water. The spe... | # Solution and Evaluation Criteria:
Two extreme situations possible in this problem:
First situation - only ice remains in the container at a temperature of $t=0{ }^{\circ} \mathrm{C}$
The heat balance equation in this case: $c_{J} m_{\pi} 5=c_{B} m_{B} 10+\lambda m_{B}$.
We get: $m_{B}=\frac{2100 \cdot 0.1 \cdot 5... | 0.0028\, | Other | math-word-problem | Yes | Yes | olympiads | false | 4,293 |
Problem No. 5 (10 points)
When constructing this structure, a homogeneous wire of constant cross-section was used. It is known that points $B, D, F$ and $H$ are located equally at the midpoints of the corresponding sides of the square ACEG. The resistance of segment $A B$ is $R_{0}=1 \Omega$. Determine the resistance ... | Answer: 2 Ohms
## Solution and grading criteria:
The resistance of a resistor is proportional to its length.
Taking this into account, the proposed circuit can be replaced by an equivalent one:
A pot was filled with $3 \pi$ liters of water at a temperature of $t=0{ }^{\circ} C$, and it was brought to a boil in 12 minutes. After that, without removing the pot from the stove, ice at a temperature of $t=0{ }^{\circ} \mathrm{C}$ was added. The water began to boil again only after 15 m... | # Solution and evaluation criteria:
Mass of the initial water: $m_{B}=\rho V=3$ kg
Power of the stove in the first case: $P=\frac{c_{B} m_{B} \Delta T}{t_{1}}$.
And in the second case: $P=\frac{\lambda m_{J}+c_{B} m_{J} \Delta T}{t_{2}}$.
We get:
$\frac{c_{B} m_{B} \Delta T}{t_{1}}=\frac{\lambda m_{J}+c_{B} m_{J} ... | 2.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,295 |
# Problem No. 7 (15 points)
For a body completely submerged in a liquid to be in equilibrium, a force $F=5 \mathrm{H}$ is applied to it. Determine the density of the body if its volume $V=1$ l, and the density of the liquid $\rho_{\text {ж }}=1000$ kg $/ \mathrm{m}^{3}$.
Answer: $1500 \mathrm{\kappa z} / \boldsymbol{... | # Solution and Evaluation Criteria:
We need to consider two situations:
The first is when the body tries to float. In this case, the equilibrium condition is:
$F_{a}=m g+F$.
As a result, we get: $\rho_{\text{ж}} g V=\rho_{T} V g+F$
$\rho_{T}=\frac{\rho_{\text{ж}} g V-F}{g V}=\frac{10-5}{10 \cdot 10^{-3}}=500 \mathr... | 1500\mathrm{k}/\mathrm{}^{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,296 |
# Problem No. 8 (15 points)
50 g of ice, taken at a temperature of $t_{\pi}=-10{ }^{\circ} \mathrm{C}$, was mixed with water taken at a temperature of $t_{B}=10{ }^{\circ} \mathrm{C}$. It is known that the final temperature in the container is $t=0{ }^{\circ} \mathrm{C}$. Determine the mass of the added water. The spe... | # Solution and Evaluation Criteria:
Two extreme situations possible in this problem:
First situation - only ice remains in the vessel at a temperature of $t=0{ }^{\circ} \mathrm{C}$
(2 points)
The heat balance equation in this case: $c_{L} m_{L} 10=c_{B} m_{B} 10+\lambda m_{B}$.
We get: $m_{B}=\frac{2100 \cdot 0.0... | 0.0028 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,297 |
1. (17 points) When walking uphill, the tourist walks 2 km/h slower, and downhill 2 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 10 hours, while descending the mountain takes 6 hours. What is the tourist's speed on flat ground? | # Answer: 8
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $10(x-2)=6(x+2)$. From this, we find $x=8$. | 8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,298 |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - 0.21 of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The ticket... | Answer: 365
Solution. The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 50 times the cost of the ticket, Boris is entitled to 365 rubles. | 365 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,300 |
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