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6. Thirteen identical metal rods are connected as follows (see fig.). It is known that the resistance of the entire structure $R=8$ Ohms. Determine the resistance of one rod $R_{0}$, if this structure is connected to a current source at points $A$ and $B$.
$(10$ points)
,
where each resistor has a resistance of $R_{0}$. As a result, the total resistance is: $R=\frac{4}{10} R_{0}... | 20\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,076 |
7. The specific heat capacity of a body with a mass of $m=3$ kg depends on the temperature as follows: $c=c_{0}(1+\alpha t)$, where $c_{0}=200$ J/kg $ \cdot{ }^{\circ} \mathrm{C}$ is the specific heat capacity at $0^{\circ} \mathrm{C}$, $\alpha=0.05^{\circ} \mathrm{C}^{-1}$ is the temperature coefficient, and $t$ is th... | Answer: 112.5 kJ
Solution. Given that the specific heat capacity depends linearly on temperature, we can calculate its average value: $c_{c p}=\frac{c_{0}\left(1+\alpha t_{\mu}\right)+c_{0}\left(1+\alpha t_{\kappa}\right)}{2}=750 \frac{\text { J }}{\text { kg } \cdot{ }^{\circ} \mathrm{C}} \quad(5$ points $) . \quad$ ... | 112.5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,077 |
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 | 4,078 |
1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 100 m ahead of Boris at the finish line. And Boris was 50 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished? | Answer: $145 \mathrm{m}$.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c \mathrm{m} /$ s, respectively. From the condition, it follows that $b=0.9 a, c=0.95 b$. Therefore, $c=0.9 \cdot 0.95 a=0.855 a$. This means that when Andrey runs 1000 m, Valentin will cover 855 m. The lag will be 145 m.... | 145\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,079 |
3. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$ | Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the odd... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,081 |
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 | 4,082 |
5. A person is walking parallel to a railway track at a constant speed. A train also passes by him at a constant speed. The person noticed that depending on the direction of the train, it passes by him either in $t_{1}=1$ minute or in $t_{2}=2$ minutes. Determine how long it would take the person to walk from one end o... | Answer: 4 min
Solution. When the train and the person are moving towards each other:
$l=\left(v_{n}+v_{u}\right) \cdot t_{1}$ (3 points), where $l$ - the length of the train, $v_{n}$ - its speed, $v_{u}$ - the speed of the person. If the directions of the train and the person are the same, then:
$l=\left(v_{n}-v_{u}... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,083 |
6. A bent thin homogeneous rod $ABC$, with small loads $m_{1}=2$ kg and $m_{2}$ located at its ends, is in equilibrium relative to a support placed at point $B$. The mass per unit length of the rod $\lambda=2$ kg. It is known that $AB=7$ m, $BC=5$ m, $BO=4$ m, $OC=3$ m. Find $m_{2}$.
(10 points)
 and $m_{n}=\lambda \cdot B C=2 \cdot 5=10$ kg (2 points). The condition for equilibrium in this situation is: $m_{1} \cdot A B+m_{n} \cdot \frac{1}{2} A B=m_{2} \cdot B O+m_{n} \cdot \frac{1}{2} B O$ ... | 10.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,084 |
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 | 4,085 |
8. A cube consists of eight smaller identical cubes. Two of the smaller cubes were replaced with ones of the same size but with a density twice as high. Determine the ratio of the initial and final densities of the large cube. (10 points) | Answer: 0.8
Solution. The relationship between mass and volume: $m=\rho V$ (2 points), that is, the new cubes, with the same volume, are twice as heavy. Initial density: $\rho_{\text {initial }}=\frac{8 m_{0}}{8 V_{0}} \quad$ (3 points).
Final density: $\quad \rho_{\text {final }}=\frac{6 m_{0}+2 m_{1}}{8 V_{0}}=\fra... | 0.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,086 |
1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 100 m ahead of Boris at the finish line. And Boris was 60 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished? | Answer: $154 \mathrm{m}$.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c \mathrm{m} /$ s, respectively. From the condition, it follows that $b=0.9 a, c=0.94 b$. Therefore, $c=0.9 \cdot 0.94 a=0.846 a$. This means that when Andrey runs 1000 m, Valentin will cover 846 m. The lag will be 154 m.... | 154\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,087 |
3. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$ | Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: suppose the corner and central cells are black, and the rest are white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white cells, we get that the eve... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,089 |
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 | 4,090 |
6. A bent thin homogeneous rod $ABC$, with small loads $m_{1}$ and $m_{2}=20$ kg located at its ends, is in equilibrium relative to a support placed at point B. The mass per unit length of the rod is $\lambda=3$ kg. It is known that $AB=7$ m, $BC=5$ m, $BO=4$ m, $OC=3$ m. Find $m_{1}$.
$(10$ points)
 and $m_{n}=\lambda \cdot B C=3 \cdot 5=15$ kg (2 points). The condition for equilibrium in this situation is: $m_{1} \cdot A B+m_{n} \cdot \frac{1}{2} A B=m_{2} \cdot B O+m_{n} \cdot \frac{1}{2} B O... | 5.2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,091 |
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: $30 \, \text{cm}$
Solution. If the stiffness of the spring is $k$, then the stiffness of half the spring is $2k$ (4 points). For the situation where the spring is held by a person: $m_{2} g=2 k x_{2}$ (4 points).
For the situation where the spring stands on a surface: $m_{2} g=k x$ (4 points). We obtain that ... | 30\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,092 |
8. A cube consists of eight smaller identical cubes. Three of the smaller cubes were replaced with ones of the same size but with a density three times greater. Determine the ratio of the final to the initial density of the large cube. (10 points) | Answer: 1.75
Solution. The relationship between mass and volume: $m=\rho V$ (2 points), that is, the new cubes, with the same volume, are three times heavier. Initial density: $\rho_{\text {initial }}=\frac{8 m_{0}}{8 V_{0}}$
(3 points). Final density: $\rho_{\text {final }}=\frac{5 m_{0}+3 m_{1}}{8 V_{0}}=\frac{5 m_... | 1.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,093 |
1. 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 | 4,094 |
2. 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 | 4,095 |
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 | 4,096 |
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 | 4,097 |
1. The cold water tap fills the bathtub in 19 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? | 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 | 4,098 |
2. 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 | 4,099 |
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 | 4,100 |
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 | 4,101 |
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 | 4,102 |
2. The teacher wrote a positive number $x$ on the board and asked Kolya, Petya, and Vasya to raise this number to the 3rd, 4th, and 12th power, respectively. It turned out that Kolya's number has at least 9 digits before the decimal point, and Petya's number has no more than 11 digits before the decimal point. How many... | Answer: 33.
Solution. From the condition, it follows that $x^{3} \geqslant 10^{8}$, and $x^{4}<10^{11}$. Therefore, $10^{32} \leqslant x^{12}<10^{33}$. This means that the integer part of Vasya's number is a 33-digit number.
Evaluation. 13 points for the correct solution. | 33 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,103 |
3. Solve the equation
$$
2 x+1+x \sqrt{x^{2}+1}+(x+1) \sqrt{x^{2}+2 x+2}=0 .
$$ | Answer: $-\frac{1}{2}$.
Solution. Let $f(x)=x\left(1+\sqrt{x^{2}+1}\right)$. The original equation can be rewritten as $f(x)+f(x+1)=0$. Note that the function $f(x)$ is odd. It increases on the positive half-axis (as the product of positive increasing functions). Due to its oddness, it increases on the entire real lin... | -\frac{1}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,104 |
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 | 4,106 |
2. The teacher wrote a positive number $x$ on the board and asked Kolya, Petya, and Vasya to raise this number to the 4th, 5th, and 20th power, respectively. It turned out that Kolya's number has at least 8 digits before the decimal point, and Petya's number has no more than 9 digits before the decimal point. How many ... | Answer: 36.
Solution. From the condition, it follows that $x^{4} \geqslant 10^{7}$, and $x^{5}<10^{9}$. Therefore, $10^{35} \leqslant x^{20}<10^{36}$. This means that the integer part of Vasya's number is a 36-digit number.
Evaluation. 13 points for the correct solution. | 36 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,107 |
3. Solve the equation
$$
2 x+2+x \sqrt{x^{2}+1}+(x+2) \sqrt{x^{2}+4 x+5}=0 .
$$ | Answer: -1.
Solution. Let $f(x)=x\left(1+\sqrt{x^{2}+1}\right)$. The original equation can be rewritten as $f(x)+f(x+2)=0$. Note that the function $f(x)$ is odd. It is increasing on the positive half-axis (as the product of positive increasing functions). Due to its oddness, it is increasing on the entire real line. F... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,108 |
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 | 4,109 |
2. (17 points) Point $M$ lies inside segment $A B$, the length of which is 40 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 $C D$ in cm. | Answer: 10.
## Solution.
$$
\begin{aligned}
& N P=N M+M P=\frac{1}{2} A M+\frac{1}{2} M B=\frac{1}{2} A B=20, \\
& C D=C M+M D=\frac{1}{2} N M+\frac{1}{2} M P=\frac{1}{2} N P=10 .
\end{align... | 10 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,110 |
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 | 4,111 |
6. (15 points) On a ship, it was decided to determine the depth of the ocean at their location. The signal sent by the sonar was received on the ship after 8 s. The speed of sound in water is 1.5 km/s. Determine the depth of the ocean. | Answer: $6000 \mathrm{m}$.
Solution. The distance traveled by sound $s=v t=1500 \cdot 8=12000$ m. Considering that the sound travels from the ship to the bottom and back, we get the depth $h=\frac{s}{2}=6000 \mathrm{m}$.
## Tasks, answers and assessment criteria | 6000\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,114 |
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 | 4,115 |
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 | 4,116 |
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 | 4,117 |
4. (15 points) An oil tanker is being filled with oil at a rate of 5 barrels per minute. How much is this in cubic meters per hour? 1 barrel equals 159 liters.
# | # Answer: $47.7 \mathrm{~m}^{3} /$ h.
Solution. $5 \frac{\text { barrels }}{\text { min }}=5 \frac{159 \text { liters }}{\frac{1}{60} h}=5 \cdot \frac{159 \cdot 0.001 \mathrm{~m}^{3}}{\frac{1}{60} h}=47.7 \frac{\mathrm{m}^{3}}{h}$. | 47.7\mathrm{~}^{3}/ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,118 |
6. (15 points) On a ship, it was decided to determine the depth of the ocean at their location. The signal sent by the sonar was received on the ship after 5 s. The speed of sound in water is 1.5 km/s. Determine the depth of the ocean. | Answer: 3750 m.
Solution. The distance traveled by sound $s=v t=1500 \cdot 5=7500$ m. Considering that the sound travels from the ship to the bottom and back, we get the depth $h=\frac{s}{2}=3750 \mathrm{m}$. | 3750\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,120 |
# Problem № 5 (15 points)
A person was building a house. Only the roof was left to be done. The builder wanted to achieve the goal that raindrops falling on the roof would slide off it as quickly as possible. Determine the angle of inclination of the roof necessary to achieve this goal. Neglect the friction of the rai... | # Solution and Evaluation Criteria:
Newton's second law for a rolling drop: $\quad m g \sin \alpha=m a$.
Distance traveled by the drop: $s=\frac{1}{2} \frac{x}{\cos \alpha}$, where $x-$ is the width of the house.
(4 points)
Time of rolling: $t=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 \cdot \frac{1}{2} \frac{x}{\cos \alph... | \alpha=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,121 |
# Problem № 7 (15 points)
Consider the power supply and measuring instruments in the presented circuit to be ideal. Determine how the readings of the instruments will change if the voltmeter and ammeter are swapped. It is known that the voltage supplied by the power supply $U_{0}=45 V$, resistances $R=50$ Ohms and $r=... | Answer: The ammeter reading increased by $0.4 A$, and the voltmeter reading decreased by $7.14 V$.
## Solution and Evaluation Criteria:
In the original circuit, the total resistance: $R_{\text {total }}=\frac{R}{2}+r=45$ Ohms.
Therefore, the total current: $I_{\text {total }}=\frac{U_{0}}{R_{\text {total }}}=1 \math... | The\ammeter\reading\increased\\0.4\A,\\the\voltmeter\reading\decreased\\7.14\V | Other | math-word-problem | Yes | Yes | olympiads | false | 4,123 |
# Problem No. 8 (10 points)
A water heater with a power of \( P = 1000 \mathrm{W} \) is used to heat a certain amount of water. When the heater is turned on for \( t_{1} = 2 \) minutes, the temperature of the water increases by \( \Delta T = 2^{\circ} \mathrm{C} \), and after the heater is turned off, the temperature ... | # Solution and Evaluation Criteria:
The law of conservation of energy during the heating of water: $P \cdot t_{1}=c m \Delta T+P_{\text {loss }} \cdot t_{1} \cdot$
When the heating element is turned off: $P_{\text {loss }}=\frac{c m \Delta T}{t_{2}}$.
As a result, we get: $m=\frac{P \cdot t_{1} \cdot t_{2}}{c \Delta... | 4.76\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,124 |
# Problem № 5 (15 points)
A person was building a house. Only the roof was left to be done. The builder wanted to achieve the goal that raindrops falling on the roof would slide off it as quickly as possible. Determine the angle of inclination of the roof necessary to achieve this goal. Neglect the friction of the rai... | # Solution and Evaluation Criteria:
Newton's second law for a rolling drop: $m g \sin \alpha=m a$
Distance traveled by the drop: $S=\frac{1}{2} \frac{x}{\cos \alpha}$, where $x-$ is the width of the house.
Time of rolling: $t=\sqrt{\frac{2 S}{a}}=\sqrt{\frac{2 \cdot \frac{1}{2} \frac{x}{\cos \alpha}}{g \sin \alpha}}... | \alpha=45 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,125 |
# Problem № 6 (10 points)
Five identical balls are rolling towards each other on a smooth horizontal surface. The speeds of the first and second are \( v_{1} = v_{2} = 0.5 \) m/s, while the others are \( v_{3} = v_{4} = v_{5} = 0.1 \) m/s. The initial distances between the balls are the same, \( l = 2 \) m. All collis... | # Solution and Evaluation Criteria:
In the case of a perfectly elastic collision, identical balls "exchange" velocities.
Therefore, the situation can be considered as if the balls pass through each other with unchanged speeds. The first collision occurs between the second and third balls. The last collision will occu... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,126 |
# Problem №7 (15 points)
Consider the power supply and measuring instruments in the presented circuit to be ideal. Determine how the readings of the instruments will change if the voltmeter and ammeter are swapped. It is known that the voltage supplied by the power supply $U_{0}=90 V$, resistances $R=50$ Ohms and $r=2... | Answer: The ammeter reading increased by $0.8 A$, and the voltmeter reading decreased by $14.3 V$
## Solution and grading criteria:
In the original circuit, the total resistance: $R_{\text {total }}=\frac{R}{2}+r=45$ Ohms.
Therefore, the total current: $I_{\text {total }}=\frac{U_{0}}{R_{\text {total }}}=2 \mathrm{~... | The\ammeter\reading\increased\\0.8\A,\\the\voltmeter\reading\decreased\\14.3\V | Other | math-word-problem | Yes | Yes | olympiads | false | 4,127 |
# Problem No. 8 (10 points)
A water heater with a power of \( P = 500 \mathrm{W} \) is used to heat a certain amount of water. When the heater is turned on for \( t_{1} = 1 \) minute, the temperature of the water increases by \( \Delta T = 2^{\circ} \mathrm{C} \), and after the heater is turned off, the temperature de... | # Solution and evaluation criteria:
The law of conservation of energy during the heating of water: $P \cdot t_{1}=c m \Delta T+P_{\text {loss }} \cdot t_{1}$.
When the heating plate is turned off: $P_{\text {loss }}=\frac{c m \Delta T}{t_{2}}$.
As a result, we get: $m=\frac{P \cdot t_{1} \cdot t_{2}}{c \Delta T\left... | 2.38 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,128 |
1. (17 points) Solve the equation ||$|x-1|+2|-3|=-2 x-4$.
| Answer: -4.
Solution. The set of possible solutions for the inequality $-2 x-4 \geq 0$, that is, $x \leq-2$. Under this condition, the inner absolute value is uniquely determined. Using the property of the absolute value $|-a|=|a|$, we get the equation $||x-3|-3|=-2 x-4$. Similarly, we expand the inner absolute value,... | -4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,129 |
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 | 4,130 |
3. (17 points) In a right triangle, the legs are equal to 3 and 4. Find the distance between the centers of the inscribed and circumscribed circles. In the answer, write the square of this distance. | Answer: 1.25.
Solution. By the Pythagorean theorem, the hypotenuse is 5, the area of the triangle is $S=\frac{1}{2} \cdot 3 \cdot 4=6$. The semiperimeter $p=\frac{3+4+5}{2}=6$, then the radius of the inscribed circle $r=\frac{s}{p}=1$. Let $O_{1}$ be the center of the inscribed circle, $O_{1} P$ and $O_{1} K$ be the r... | 1.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,131 |
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 | 4,132 |
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 | 4,133 |
6. (15 points) Two springs with stiffnesses of 3 kN/m and $6 \mathrm{kN} / \mathrm{m}$ are connected in series. How much work is required to stretch this system by 5 cm. | Answer: 2.5 J.
Solution. The total stiffness $k=\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{3 \cdot 6}{3+6}=2 \mathrm{kH} /$ m. Work:
$$
A=\Delta W=\frac{k x^{2}}{2}=\frac{2000 \cdot 0.05^{2}}{2}=2.5 \text { J. }
$$
## Tasks, answers, and evaluation criteria | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,134 |
1. (17 points) Solve the equation ||$|x-2|+3|-4|=-3 x-9$.
| Answer: -5.
Solution. The set of possible solutions for the inequality $-3 x-9 \geq 0$, that is, $x \leq-3$. Under this condition, the inner absolute value is uniquely determined. Using the property of absolute value $|-a|=|a|$, we get the equation $||x-5|-4|=-3 x-9$. Similarly, we expand the inner absolute value, obt... | -5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,135 |
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 | 4,136 |
4. (15 points) A ball was thrown vertically upwards from a balcony. It is known that it fell to the ground after 3 seconds. Knowing that the initial speed of the ball is 5 m/s, determine the height of the balcony. The acceleration due to gravity is $10 \mathrm{~m} / \mathrm{s}^{2}$. | Answer: 30 m.
Solution. The equation of motion of the ball:
$y=0=h_{0}+v_{0} t-\frac{g t^{2}}{2}=h_{0}+5 \cdot 3-\frac{10 \cdot 3^{2}}{2}$. We get $h_{0}=30 \mathrm{m}$. | 30\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,137 |
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 | 4,138 |
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 | 4,139 |
1. (12 points) Solve the equation
$$
\sqrt[3]{(7-x)^{2}}-\sqrt[3]{(7-x)(9+x)}+\sqrt[3]{(9+x)^{2}}=4
$$ | Answer: -1
Solution. Let $a=7-x, b=9+x$. Notice that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $... | -1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,140 |
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 | 4,141 |
3. (13 points) On the hypotenuse $A B$ of the right triangle $A B C$, a square $A B D E$ is constructed outwardly, with $A C=1, B C=4$. In what ratio does the angle bisector of angle $C$ divide the side $D E$? | Answer: $1: 4$.
Solution. Let $O$ be the center of the square, $P$ be the intersection point of lines $A B$ and $C O$, and $Q$ be the intersection point of lines $D E$ and $C O$.
In quadril... | 1:4 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,142 |
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 | 4,143 |
5. (15 points) Two identical vessels of height $H$ are connected by a thin tube with a valve $K$ at a height of $0.2 H$. Water is poured into the left vessel, and gasoline into the right vessel. The levels of the liquids are the same and equal to $0.9 H$. Determine the levels of the liquids in the vessels that will be ... | Answer: $h_{\text {left }}=0.69 H ; h_{\text {right }}=H$.
Solution. When the tap is opened, water will begin to flow into the right vessel. Several situations are possible.
First: the water will not reach the tube, and the gasoline will not spill out of the vessel.
The equilibrium condition at the level of the tube... | h_{\text{left}}=0.69H;h_{\text{right}}=H | Other | math-word-problem | Yes | Yes | olympiads | false | 4,144 |
6. (10 points) Two loads with masses \( m_{1}=2 \) kg and \( m_{2}=3 \) kg are placed on two horizontal shelves, connected by a weightless inextensible string that passes over a weightless smooth pulley. The coefficients of friction between the loads and the shelves are \( \mu_{1}=0.1 \) and \( \mu_{2}=0.2 \). A horizo... | Answer: $a_{1}=1.5 \mathrm{M} / \mathrm{c}^{2}, a_{2}=0 \mathrm{M} / \mathrm{c}^{2}, a=0.6 \mathrm{~m} / \mathrm{c}^{2}$.
Solution. Since the pulley is weightless, the tension in the string $T=\frac{F}{2}=5 \mathrm{H}$.
The sliding friction force for the first body:
$F_{\text {fr } 1}=\mu_{1} N_{1}=\mu_{1} m_{1} g=0... | a_{1}=1.5\mathrm{~}/\mathrm{}^{2},a_{2}=0\mathrm{~}/\mathrm{}^{2},=0.6\mathrm{~}/\mathrm{}^{2} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,145 |
8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of $v=1 \mathrm{~m} / \mathrm{s}$ and at the end of its movement, it landed on the water surface. Five seconds before landing, it was at a height of $h=3$ m above the water surface. The cosine of the angle of incidence of the sun... | Answer: 0 m/s or $1.6 \mathrm{~m} / \mathrm{s}$.
Solution. The mosquito flew a distance of $s=v t=1 \cdot 5=5$ m before landing.
That is, it moved along the trajectory $A B$.
The cosine o... | 0\mathrm{~}/\mathrm{} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,147 |
1. (12 points) Solve the equation
$$
\sqrt[3]{(9-x)^{2}}-\sqrt[3]{(9-x)(7+x)}+\sqrt[3]{(7+x)^{2}}=4
$$ | Answer: 1.
Solution. Let $a=9-x, b=7+x$. Note that $a+b=16$. We obtain the system of equations
$$
\left\{\begin{array}{c}
\sqrt[3]{a^{2}}-\sqrt[3]{a b}+\sqrt[3]{b^{2}}=4 \\
a+b=16
\end{array}\right.
$$
Multiply the first equation by $\sqrt[3]{a}+\sqrt[3]{b}$, using the formula for the difference of cubes, we get $\s... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,148 |
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 | 4,149 |
3. (13 points) On the hypotenuse $A B$ of the right triangle $A B C$, a square $A B D E$ is constructed outwardly, with $A C=2, B C=5$. In what ratio does the angle bisector of angle $C$ divide the side $D E$? | Answer: $2: 5$.
Solution. Let $O$ be the center of the square, $P$ be the intersection point of lines $A B$ and $C O$, and $Q$ be the intersection point of lines $D E$ and $C O$.
In quadril... | 2:5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,150 |
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 | 4,151 |
5. (15 points) Two identical vessels of height $H$ are connected by a thin tube with a valve $K$ at a height of $0.15 H$. Water is poured into the left vessel, and gasoline into the right vessel. The levels of the liquids are the same and equal to $0.9 H$. Determine the levels of the liquids in the vessels that will be... | Answer: $h_{\text {left }}=0.69 H ; h_{\text {right }}=H$.
Solution. When the tap is opened, water will begin to flow into the right vessel. Several situations are possible.
First: the water will not reach the tube, and the gasoline will not spill out of the vessel.
The equilibrium condition at the tube level:
$\rh... | h_{\text{left}}=0.69H;h_{\text{right}}=H | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,152 |
6. (10 points) Two weights with masses \( m_{1}=1 \) kg and \( m_{2}=10 \) kg are placed on two horizontal shelves, connected by a weightless, inextensible string that passes over a weightless, frictionless pulley. The coefficients of friction between the weights and the shelves are \( \mu_{1}=0.3 \) and \( \mu_{2}=0.1... | Answer: $a_{1}=7 \mathrm{M} / \mathbf{c}^{2}, a_{2}=0 \mathrm{M} / \mathrm{c}^{2}, a=0.64 \mathrm{M} / \mathrm{c}^{2}$.
Solution. Since the pulley is weightless, the tension in the string is: $T=\frac{F}{2}=10 \mathrm{H}$.
The sliding friction force for the first body:
$F_{\text {fr 1 }}=\mu_{1} N_{1}=\mu_{1} m_{1} ... | a_{1}=7\mathrm{M}/{}^{2},a_{2}=0\mathrm{M}/\mathrm{}^{2},=0.64\mathrm{M}/\mathrm{}^{2} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,153 |
8. (15 points) A mosquito was moving over the water in a straight line at a constant speed of \( v = 0.5 \) m/s and at the end of its movement, it landed on the water surface. 20 seconds before landing, it was at a height of \( h = 6 \) m above the water surface. The cosine of the angle of incidence of the sunlight on ... | Answer: 0 m/s or $0.8 \mathrm{~m} / \mathrm{s}$.
Solution. The mosquito flew a distance of $s=v t=0.5 \cdot 20=10$ m before landing.
That is, it moved along the trajectory $A B$.
The cosi... | 0 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,155 |
1. (13 points) What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $25 \%$ but less than $35 \%$, the number of sixth graders was more than $30 \%$ but less than $40 \%$, and the number of seventh graders was more than $35 \%$ but less th... | Answer: 11. Fifth-graders -3, sixth-graders -4, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.25 and l... | 11 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,156 |
2. (12 points) In a family, there are four children of different ages. Their total age is 31 years. Four years ago, the total age of all the children in the family was 16 years, 7 years ago it was 8 years, and 11 years ago it was 1 year. How old are the children at present? (Age is always expressed as a whole number of... | Answer: $3, 6, 10, 12$ years.
Solution. Let's start from the end. 11 years ago, there was one child who was 1 year old. We will call this child the oldest. 7 years ago, the oldest was 5 years old, and the second child was either 3 years old or the oldest was 5 years old, the second was 2 years old, and the third was 1... | 3,6,10,12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,157 |
3. (12 points) Four friends went to the forest to pick mushrooms. Upon returning, every pair of them counted the total number of mushrooms they had collected. The numbers obtained were $6,7,9,9,11,12$ How many mushrooms did each collect? | Answer: $2,4,5,7$.
Solution. Let $x_{1} \leq x_{2} \leq x_{3} \leq x_{4}$ be the number of mushrooms collected by the friends. Then $x_{1}+x_{2}=6, x_{1}+x_{3}=7, x_{2}+x_{3}=9$. From this, $x_{1}=2, x_{2}=4, x_{3}=5$. Therefore, $x_{4}=12-x_{3}=7$. Since there are unused conditions, a check must be performed.
Remark... | 2,4,5,7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,158 |
4. (13 points) In a large online chess tournament, each player had exactly three friends among the participants. Each player played one game with all the participants except for their three friends. Could exactly 804 games have been played? | Answer: No.
Solution. If there were $n$ players in the tournament, then the total number of games in the tournament is $\frac{n(n-4)}{2}$. Indeed, each player plays $n-4$ games. In the product $n(n-4)$, each game is counted twice, so the total number of games is half of that.
Let's determine if it is possible that $\... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,159 |
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 | 4,160 |
7. (10 points) An industrial robot travels from point $A$ to point $B$ according to a pre-determined algorithm. The diagram shows a part of its repeating trajectory. Determine how many times faster it would reach from point $A$ to point $B$ if it moved in a straight line at three times the speed?
 What is the minimum number of participants that could have been in the school drama club if the number of fifth graders was more than $22 \%$ but less than $27 \%$, the number of sixth graders was more than $25 \%$ but less than $35 \%$, and the number of seventh graders was more than $35 \%$ but less th... | Answer: 9. Fifth-graders - 2, sixth-graders - 3, seventh-graders - 4.
Solution. Let the required number of participants be
$$
n=a+b+c
$$
where $a$ is the number of fifth-graders, $b$ is the number of sixth-graders, and $c$ is the number of seventh-graders. Note: $a<c$ and all three numbers constitute more than 0.2 a... | 9 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,164 |
2. (12 points) In a family, there are four children of different ages. Their total age is 33 years. Three years ago, the total age of all the children in the family was 22 years, 7 years ago it was 11 years, and 13 years ago it was 1 year. How old are the children at present? (Age is always expressed as a whole number ... | Answer: $2, 6, 11, 14$ years.
Solution. Let's start from the end. 13 years ago, there was one child who was 1 year old. We will call this child the oldest. 7 years ago, the oldest was 7 years old, and the second child was 4 years old, or the oldest was 7 years old, the second was 3 years old, and the third was 1 year ... | 2,6,11,14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,165 |
3. (12 points) Four friends went to the forest to pick mushrooms. Upon returning, every pair of them counted the total number of mushrooms they had collected. The numbers obtained were $7,9,10,10,11,13$. How many mushrooms did each collect? | Answer: $3,4,6,7$.
Solution. Let $x_{1} \leq x_{2} \leq x_{3} \leq x_{4}$ be the number of mushrooms collected by the friends. Then $x_{1}+x_{2}=7, x_{1}+x_{3}=9, x_{2}+x_{3}=10$. From this, $x_{1}=3, x_{2}=4, x_{3}=6$. Therefore, $x_{4}=13-x_{3}=7$. Since there are unused conditions, a check is necessary.
Remark. Th... | 3,4,6,7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,166 |
4. (13 points) In a large online chess tournament, each player had exactly three friends among the participants. Each player played one game with all the participants except for their three friends. Could exactly 404 games have been played? | Answer: No.
Solution. If there were $n$ players in the tournament, then the total number of games in the tournament is $\frac{n(n-4)}{2}$. Indeed, each player plays $n-4$ games. In the product $n(n-4)$, each game is counted twice, so the total number of games is half of that.
Let's determine if it is possible that $\... | proof | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,167 |
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 | 4,168 |
7. (10 points) An industrial robot travels from point $A$ to point $B$ according to a pre-determined algorithm. The diagram shows a part of its repeating trajectory. Determine how many times faster it would reach from point $A$ to point $B$ if it moved in a straight line at twice the speed?
 A body moves along the Ox axis. The dependence of velocity on time is shown in the figure. Determine the distance traveled by the body in 6 seconds.
 | Answer: 5 meters.
Solution. In the first second, the body traveled 2 meters.
In the second - 1 meter.
In the fifth - 2 meters.
In total, the entire distance traveled is 5 meters. | 5 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,171 |
1. (12 points) Four friends went to the forest to pick mushrooms. Upon returning, every pair of them counted the total number of mushrooms they had collected. The numbers obtained were $6,7,9,9,11,12$. How many mushrooms did each collect? | Answer: $2,4,5,7$.
Solution. Let $x_{1} \leq x_{2} \leq x_{3} \leq x_{4}$ be the quantities of mushrooms collected by the friends. Then $x_{1}+x_{2}=6, x_{1}+x_{3}=7, x_{2}+x_{3}=9$. From this, $x_{1}=2, x_{2}=4, x_{3}=5$. Therefore, $x_{4}=12-x_{3}=7$. Since there are unused conditions, a check must be performed.
Re... | 2,4,5,7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,172 |
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 | 4,173 |
3. (13 points) Supercomputer Petya took a natural number $\boldsymbol{a}>2$, found the area of a rectangle with sides $\boldsymbol{a}-2$ and $\boldsymbol{a}+3$, and subtracted $\boldsymbol{a}$ from the result. He got an amazing number, in the decimal representation of which there were, in some order, only 2023 eights, ... | Answer: Petya made a mistake.
Solution. The sum of the digits of the number $\boldsymbol{a}^{2}-\mathbf{6} (*)$ obtained by Petya is $2023 \cdot 8 +$ 2023 - 3. This number, when divided by 3, gives a remainder of 2, hence $\boldsymbol{a}^{2}-6$ also gives a remainder of 2 when divided by 3, which means $\boldsymbol{a}... | proof | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,174 |
4. (13 points) Is there a triangular pizza from which one can sequentially cut 11 identical triangular slices, with each slice being cut with a single straight cut? If yes, draw this triangular pizza and describe how to cut it. | Answer. Yes, see the figure.
Solution (example). Take an arbitrary square \(ABCD\) and cut it into 5 identical vertical rectangular strips, and each strip into 2 triangles along the diagona... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,175 |
5. (10 points) Determine the direction and magnitude of the velocity \( u \) of the left load, if the velocity of the right load \( v = 1 \) m/s. The threads are inextensible and weightless, and the lever is rigid.
 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 | 4,177 |
7. (10 points) Determine the force $F$ that needs to be applied to the plate to move a load of mass $m=2$ kg, lying on a horizontal smooth surface. All pulleys are smooth and weightless. The acceleration due to gravity $g=10 \mathrm{m} /$ kg.
We get: $F=4 T=m g=2 \cdot 10=20 \mathrm{H}$. | 20\mathrm{H} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,178 |
1. (12 points) Four friends went to the forest to pick mushrooms. Upon returning, every pair of them counted the total number of mushrooms they had collected. The numbers obtained were $7,9,10,10,11,13$. How many mushrooms did each collect? | Answer: $3,4,6,7$.
Solution. Let $x_{1} \leq x_{2} \leq x_{3} \leq x_{4}$ be the number of mushrooms collected by the friends. Then $x_{1}+x_{2}=7, x_{1}+x_{3}=9, x_{2}+x_{3}=10$. From this, $x_{1}=3, x_{2}=4, x_{3}=6$. Therefore, $x_{4}=13-x_{3}=7$. Since there are unused conditions, a check is necessary.
Remark. Th... | 3,4,6,7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,180 |
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 | 4,181 |
3. (13 points) Supercomputer Petya took a natural number $\boldsymbol{a}>\mathbf{3}$, found the area of a rectangle with sides $\boldsymbol{a}-\mathbf{3}$ and $\boldsymbol{a}+\mathbf{4}$, and subtracted $\boldsymbol{a}$ from the result. He got an amazing number, in the decimal representation of which there were, in som... | Answer: Petya made a mistake.
Solution. The number obtained by Petya, $\boldsymbol{a}^{2}-12$ (*), has a sum of digits equal to $2023 \cdot 8 + 2023 - 3$. This number, when divided by 3, gives a remainder of 2, hence $\boldsymbol{a}^{2}-12$ also gives a remainder of 2 when divided by 3, which means $\boldsymbol{a}^{2}... | proof | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,182 |
4. (13 points) Is there a triangular pizza from which one can sequentially cut 13 identical triangular slices, with each slice being cut with a single straight cut? If yes, draw this triangular pizza and describe how to cut it. | Answer. Yes, see the figure.
Solution (example). Take an arbitrary square \(ABCD\) and cut it into 6 identical vertical rectangular strips, and each strip into 2 triangles along the diagona... | proof | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,183 |
5. (10 points) Determine the direction and magnitude of the velocity \( u \) of the right load, if the velocity of the left load \( v = 0.5 \, \text{m/s} \). The threads are inextensible and weightless, and the lever is rigid.
 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 | 4,185 |
7. (10 points) Determine the force $F$ that needs to be applied to the plate to move a load of mass $m=3$ kg, lying on a horizontal smooth surface. All pulleys are smooth and weightless. The acceleration due to gravity $g=10 \mathrm{m} /$ kg.
We get: $F=4 T=2 \mathrm{mg}=2 \cdot 3 \cdot 10=60 \mathrm{H}$. | 60\mathrm{H} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,186 |
1. Pete and Vasya competed in a 100 m race. When Pete finished, Vasya was 10 m behind him. During the second race, Pete started exactly 10 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 m.
Solution. In the second race, Petya eliminates the lag from Vasya by running 100 m (during this time, Vasya will run 90 m). 10 m remain to the finish line - 10 times less than at the start of the 1st race. Therefore, the lead will be 10 times less, i.e., 1 m.
Scoring. 11 points fo... | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,188 |
2. Fifteen numbers are arranged in a circle. The sum of any six consecutive numbers is 50. Petya covered one of the numbers with a card. The two numbers adjacent to the card are 7 and 10. What number is under the card? | Answer: 8.
Solution. Let the number at the $i$-th position be $a_{i}(i=1, \ldots, 15$.) 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:
$$
a_{1}=a_{7}=a_{13}=a_{4}=a_{10}=a_{1} .
$$
Now it is clear th... | 8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,189 |
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