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4. Condition One mole of an ideal gas undergoes a cyclic process, which in terms of $T, V$ is described by the equation
$$
\left(\frac{V}{V_{0}}-a\right)^{2}+\left(\frac{T}{T_{0}}-b\right)^{2}=c^{2}
$$
where $V_{0}, T_{0}, a, b, c-$ are some constants, $c^{2}<a^{2}+b^{2}$. Determine the maximum pressure of the gas in... | Answer: $P_{\max }=\frac{R T_{0}}{V_{0}} \frac{a \sqrt{a^{2}+b^{2}-c^{2}}+b c}{b \sqrt{a^{2}+b^{2}-c^{2}}-a c}$
Solution In the axes $P / P_{0}, T / T_{0}$, the process is represented by a circle with center at point $(a, b)$ and radius $c$. For each point, the volume occupied by the gas is $R \frac{T}{T}$, that is, p... | P_{\max}=\frac{RT_{0}}{V_{0}}\frac{\sqrt{^{2}+b^{2}-^{2}}+}{b\sqrt{^{2}+b^{2}-^{2}}-} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,226 |
4. Condition: Two weightless cylindrical rollers lie on a horizontal surface parallel to each other and have radii $R=1$ m and $r=40$ cm. A heavy flat slab of mass $m=150$ kg is placed on them such that it is inclined to the horizontal by $\alpha=\arccos (0.68)$. Find the magnitude and direction of the slab's accelerat... | Answer: $4 \mathrm{m} / \mathrm{c} ; \arcsin 0.4$
Solution: Let's move to a coordinate system where the center (of the left wheel) is at rest. In this system, all points on the rim move at the same speed $v$. That is, the surface moves horizontally to the left at this speed. And the point of contact with the slab also... | 4\mathrm{}/\mathrm{}^2;\arcsin0.4 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,227 |
4. Condition: A certain farmer has created a system of channels $A B C D E F G H$, as shown in the diagram. All channels are the same, and their connections occur at nodes marked by the letters $B, D$, and $G$. Water enters at node $A$ and exits at node $E$. Let's call the flow rate the number of cubic meters of water ... | Answer: $\frac{4}{3} q_{0}, \frac{2}{3} q_{0}, \frac{7}{3} q_{0}$
Solution: First, due to symmetry, note that the flow in channel HG is also equal to $q_{0}$. Let the flow in channel $\mathrm{BC}$ and its symmetric channel $\mathrm{CD}$ be $x$, and the flow in channel $\mathrm{BG}$ and its symmetric channel GD be $y$.... | \frac{4}{3}q_{0},\frac{2}{3}q_{0},\frac{7}{3}q_{0} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,228 |
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $20 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 50 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude... | Answer: a) 75 km; b) yes. Solution. Let $a=20$ m/s$^2$, $\tau=50$ s. During the first part of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t$, $y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau$, $y_{0}=\frac{a \tau^{2}}{2}$ - this will b... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,229 |
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 8 meters long and 5 meters wide, loaded with grain. The surface of the grain has an angle of no more than 45 degrees with the base plane (otherwise the grains will spill), the density of the grain is 1200 kg/m³. Find the maximum mass of gra... | Answer: 47.5 t. Solution. The calculation shows that the maximum height of the grain pile will be half the width of the platform, that is, 2.5 m. The pile can be divided into a "horizontally lying along the platform" prism (its height is 3 m, and the base is a right-angled isosceles triangle with legs $\frac{5 \sqrt{2}... | 47500 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,230 |
3. Gavriil saw a squirrel sitting on a tree branch right in front of him at a distance of 3 m 75 cm through the window. He decided to feed the little animal and threw a nut horizontally at a speed of 5 m/s directly towards the squirrel. Will the squirrel be able to catch the nut if it can jump in any direction at a spe... | Answer: No. Solution: The coordinates of the squirrel at the initial moment: $x=a ; y=0$. The nut moves according to the law: $x(t)=V_{0} t ; y(t)=\frac{g t^{2}}{2}$. Therefore, the square of the distance from the sitting squirrel to the flying nut at time $t$ is: $r^{2}=\left(V_{0} t-a\right)^{2}+\left(\frac{g t^{2}}{... | \frac{5\sqrt{2}}{4}>\frac{17}{10} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,231 |
4. Condition One mole of an ideal gas undergoes a cyclic process, which in terms of $P, T$ is described by the equation
$$
\left(\frac{P}{P_{0}}-a\right)^{2}+\left(\frac{T}{T_{0}}-b\right)^{2}=c^{2}
$$
where $P_{0}, T_{0}, a, b, c-$ are some constants, $c^{2}<a^{2}+b^{2}$. Determine the minimum volume of the gas in t... | Answer: $V_{\text {min }}=\frac{R T_{0}}{P_{0}} \frac{a \sqrt{a^{2}+b^{2}-c^{2}}-b c}{b \sqrt{a^{2}+b^{2}-c^{2}}+a c}$
Solution In the axes $P / P_{0}, T / T_{0}$, the process is represented by a circle with center at point $(a, b)$ and radius $c$. For each point, the volume occupied by the gas is $R \frac{T}{P}$, whi... | V_{\text{}}=\frac{RT_{0}}{P_{0}}\frac{\sqrt{^{2}+b^{2}-^{2}}-}{b\sqrt{^{2}+b^{2}-^{2}}+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,232 |
4. Condition: Two weightless cylindrical rollers lie on a horizontal surface parallel to each other and have radii $R=1$ m and $r=75$ cm. A heavy flat slab of mass $m=75$ kg is placed on them such that it is inclined to the horizontal by $\alpha=\arccos (0.98)$. Find the magnitude and direction of the slab's accelerati... | Answer: 1 m/s; $\arcsin 0.1;$
Solution: Let's transition to a coordinate system where the center of the (left) wheel is at rest. In this system, all points on the rim move with the same speed $v$. That is, the surface moves horizontally to the left with this speed. And the point of contact with the plate also has this... | \sqrt{\frac{1-\cos\alpha}{2}} | Other | math-word-problem | Yes | Yes | olympiads | false | 11,233 |
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $30 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 20 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude... | Answer: a) 24 km; b) yes. Solution. Let $a=30 \mathrm{m} / \mathrm{s}^{2}, \tau=20$ s. During the first part of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t, y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,234 |
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 8 meters long and 4 meters wide, loaded with sand. The surface of the sand has an angle of no more than 45 degrees with the base plane (otherwise the sand grains will spill), the density of the sand is 1500 kg/m³. Find the maximum mass of s... | Answer: 40 t. Solution. The calculation shows that the maximum height of the sand pile will be equal to half the width of the platform, that is, 2 m. The pile can be divided into a "horizontally lying along the platform" prism (its height is 4 m, and the base is an isosceles right triangle with legs $2 \sqrt{2}$ and hy... | 40000 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,235 |
3. Gavriil saw a squirrel sitting on a tree branch right in front of him at a distance of 3 m 75 cm through the window. He decided to feed the little animal and threw a nut horizontally at a speed of 2.5 m/s directly towards the squirrel. Will the squirrel be able to catch the nut if it can jump in any direction at a h... | Answer: Yes. Solution: The coordinates of the squirrel at the initial moment: $x=a ; y=0$. The nut moves according to the law: $x(t)=V_{0} t ; y(t)=\frac{g t^{2}}{2}$.
Therefore, the square of the distance from the sitting squirrel to the flying nut at time $t$ is: $r^{2}=\left(V_{0} t-a\right)^{2}+\left(\frac{g t^{2}... | \frac{5\sqrt{5}}{4}<\frac{14}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,236 |
4. Condition One mole of an ideal gas undergoes a cyclic process, which in terms of $T, V$ is described by the equation
$$
\left(\frac{V}{V_{0}}-a\right)^{2}+\left(\frac{T}{T_{0}}-b\right)^{2}=c^{2}
$$
where $V_{0}, T_{0}, a, b, c-$ are some constants, $c^{2}<a^{2}+b^{2}$. Determine the minimum pressure of the gas in... | Answer: $P_{\min }=\frac{R T_{0}}{V_{0}} \frac{a \sqrt{a^{2}+b^{2}-c^{2}}-b c}{b \sqrt{a^{2}+b^{2}-c^{2}}+a c}$
Solution In the axes $P / P_{0}, T / T_{0}$, the process is represented by a circle with center at point $(a, b)$ and radius $c$. For each point, the volume occupied by the gas is $R \frac{T}{P}$, that is, p... | P_{\}=\frac{RT_{0}}{V_{0}}\frac{\sqrt{^{2}+b^{2}-^{2}}-}{b\sqrt{^{2}+b^{2}-^{2}}+} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,237 |
4. Condition: Two weightless cylindrical rollers lie on a horizontal surface parallel to each other and have radii $R=1$ m and $r=50$ cm. A heavy flat slab of mass $m=75$ kg is placed on them such that it is inclined to the horizontal by $\alpha=\arccos (0.82)$. Find the magnitude and direction of the slab's accelerati... | Answer: $3 \mathrm{~m} / \mathrm{c} ; \arcsin 0.2$.
Solution: Let's move to a coordinate system where the center (of the left wheel) is at rest. In this system, all points on the rim move with the same speed $v$. That is, the surface moves horizontally to the left at this speed. And the point of contact with the slab ... | 3\mathrm{~}/\mathrm{}^2;\arcsin0.2 | Other | math-word-problem | Yes | Yes | olympiads | false | 11,238 |
4. Condition: A certain farmer has created a system of channels $A B C D E F G H$, as shown in the diagram. All channels are the same, and their connections occur at nodes marked by the letters $B, D$, and $G$. Water enters at node $A$ and exits at node $E$. Let's call the flow rate the number of cubic meters of water ... | Answer: $\frac{4}{7} q_{0}, \frac{2}{7} q_{0}, \frac{3}{7} q_{0}$
Solution: First, due to symmetry, note that the flow in channel $\mathrm{BC}$ is equal to the flow in channel CD (denote it as $x$), and the flow in channel BG is equal to the flow in channel GD (denote it as $y$).
Then the total flow along the path $\... | \frac{4}{7}q_{0},\frac{2}{7}q_{0},\frac{3}{7}q_{0} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,239 |
1. A stone is thrown vertically upwards with an initial velocity $V$. Neglecting air resistance and assuming the acceleration due to gravity is $10 \mathrm{~m} / \mathbf{c}^{2}$, determine the values of $V$ for which all moments of reaching a height of 10 m will lie between: A) the first and second seconds after the st... | Solution. The dependence of height on time is $h(t)=V t-\frac{g t^{2}}{2}$. Therefore, the stone will be at a height of 10 m at the moments of time $10=V t-5 t^{2}$. This results in the equation $5 t^{2}-V t+10=0$, which for $V^{2} \geqslant 200$ (i.e., for $V \geqslant 10 \sqrt{2}$) has roots: $t_{1,2}=\frac{V \pm \sq... | A)V\in[10\sqrt{2};15)/;B)V\in\varnothing | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,241 |
3. In the cottage village where Gavrila spends his summers, there is a water supply with cold water. The boy's parents installed a water heater, which has a fixed power as long as the temperature of the water in it is below $100^{\circ} \mathrm{C}$. After the water pipe enters the house, a tee was installed so that par... | Solution. In both cases, the same amount of water passes through in the same time with the same flow rate, and the same amount of heat is transferred to it. Therefore, the outlet temperature is the same: $t_{3}=t_{2}=40^{\circ} \mathrm{C}$.
Answer: $40^{\circ} \mathrm{C}$.
Criteria: 20 points - correct solution, poss... | 40\mathrm{C} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,243 |
4. A ten-liter bucket was filled to the brim with currants. Gavrila immediately said that there were 10 kg of currants in the bucket. Glafira thought about it and estimated the weight of the berries in the bucket more accurately. How can this be done if the density of the currant can be approximately considered equal t... | Solution. In approximate calculations, the sizes of the berries can be considered the same and much smaller than the size of the bucket. If the berries are laid out in one layer, then in the densest packing, each berry will have 6 neighbors: the centers of the berries will be at the vertices of equilateral triangles wi... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,244 |
5. Two dumbbells, consisting of weightless rods of length $2 L$ and identical small balls, slide towards each other with the same speed $V$ as shown in the figure. Describe the motion of the dumbbells after the collision of the balls in two cases: A) the collision is perfectly elastic; B) the collision is perfectly ine... | Solution. When any two balls collide, the velocities of the others do not change, as the collision is instantaneous. In a perfectly elastic collision of two identical balls, they exchange velocities. Therefore, each dumbbell will begin to rotate around its center with an angular velocity of $\frac{V}{L}$. After half a ... | A)theywilldiverge,maintainingtheirinitialspeeds;B)theywillbegintorotatewithanangularvelocityof\frac{V}{2L} | Other | math-word-problem | Yes | Yes | olympiads | false | 11,245 |
6. Try to advance as far as possible in the analytical solution of the problem given below. In case $B$ is necessary, a computer may be used at the final stage.
Point $A$ is located on a meadow, point $B$ - on a sandy wasteland. The distance between the points is 24 km. The boundary between the wasteland and the meado... | Solution. Our task: find such a point $C$ on $A^{\prime} B^{\prime}$ (see the figure) so that the path along the trajectory $A C + C B$ takes the minimum possible time. The distance $A^{\prime} B^{\prime}$ is $12 \sqrt{3}$, let $A^{\prime} C = x$, where $x \in [0 ; 12 \sqrt{3}]$ (in fact, it is obvious that $x \in [8 \... | 4.89 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 11,246 |
1. From a square steel sheet with a side of 1 meter, identical rectangular isosceles triangles are cut from each of the four corners, leaving a smaller square. Determine its mass if the thickness of the sheet is 3 mm and the density of steel is 7.8 $g / \mathrm{cm}^{3}$. | Answer: 11.7 kg.
Solution. The area of the resulting square is half the area of the original square and is $\frac{10000}{2}=5000\left(\mathrm{~cm}^{2}\right)$, its volume is $5000 \cdot 0.3=1500\left(\mathrm{~cm}^{3}\right)$, and its mass is $1500 \cdot 7.8=11700(g)$. | 11.7 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,247 |
Variant 134. From 10 to 20 minutes (including 10 and 20).
Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 0}$ points - the solution is reduced to correct inequalities with respect to $t$, but further errors are made; 5 points - the equation and two ineq... | Solution. The range of a body thrown with an initial velocity $V_{0}$ at an angle $\alpha$ to the horizontal is $l=\frac{V_{0}^{2} \sin 2 \alpha}{g}$, and the time of flight is $\tau=\frac{2 V_{0} \sin \alpha}{g}$.
According to the condition $\frac{V_{0}^{2} \sin 2 \alpha}{g} \geqslant \frac{96}{100} \frac{V_{0}^{2} \... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,253 |
Variant 132. 9.6 meters. Variant 133. 2.0 seconds. Variant 134. 10.0 meters.
Evaluation criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - mostly correct solution, but with defects (for example, the choice of value $\alpha$ is not justified); 10 point... | Solution. 1) First, we solve the geometric problem of finding the side $O_{2}O_{3}$. According to the condition, $\sin \angle O_{1}=\frac{8}{10}=\frac{4}{5}$; therefore, $\cos \angle O_{1}= \pm \frac{3}{5}$. The third side is found using the cosine theorem and equals either 10 cm or $2 \sqrt{97}$ cm.
2) The perimeter ... | 32+4\pi | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,254 |
Variant 134. a) $36+8 \pi$ cm or $23+3 \sqrt{41}+8 \pi$ cm; b) not always.
Evaluation criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - mostly correct solution and correct answer, but there are defects (for example: comparison of numbers in the last ... | Solution. The efficiency of a heat engine is the ratio of useful work to the amount of heat received by the gas per cycle. To find the useful work, it is necessary to find the difference between the work done by the gas ($A_{+}$) and the work done on it ($A_{-}$). Each of these quantities can be found by integrating th... | \frac{1}{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 11,255 |
Variant 132. $\frac{1}{6}$. Variant 133. $\frac{5}{6}$. Variant 134. $\frac{2}{9}$.
Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; $\mathbf{1 5}$ points - an arithmetic error was made in the final calculation, provided the solution is otherwise correct; 5 points ... | Solution. 1) The equation $\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}-2013=x$ is equivalent to the equation $x^{5}-2013=x$.
Possible proof: a) If $x$ is a root of equation (2), then
$\left(\left(\left(x^{5}-2013\right)^{5}-2013\right)^{5}-2013\right)^{5}-2013=\left(\left(x^{5}-2013\right)... | [4;2017] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,256 |
Variant 132. $[-5 ;-2018]$. Variant 133. $[4 ; 2016]$. Variant 134. $[-5 ;-2017]$.
Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; 15 points - mostly correct solution and correct answer, but there are defects (for example: the transition from (1) to (2) is made wi... | Solution. The radii of the disks form a geometric progression with a common ratio $q<1$. Here $q=\frac{1}{2}, R=2$. Therefore, the masses of the disks also form a geometric progression with a common ratio $q^{2}<1$. The total mass of the system is finite, as is its size, so the position of the center of mass is defined... | \frac{6}{7} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,257 |
1. The time of the aircraft's run from the moment of start until the moment of takeoff is 15 seconds. Find the length of the run if the takeoff speed for this aircraft model is 100 km/h. Assume the aircraft's motion during the run is uniformly accelerated. Provide the answer in meters, rounding to the nearest whole num... | Answer: 208
$v=a t, 100000 / 3600=a \cdot 15$, from which $a=1.85\left(\mathrm{m} / \mathrm{s}^{2}\right)$. Then $S=a t^{2} / 2=208$ (m). | 208 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,258 |
2. Three liquids are given, the densities of which are in the ratio $\rho_{1}: \rho_{2}: \rho_{3}=6: 3: 2$. From these liquids, a mixture is prepared, in which the first liquid should be at least 3.5 times more than the second. In what ratio should the masses of these liquids be taken so that after mixing, the density ... | For example: 20:5:8 or 7:1:3 or any that satisfies the equation $4 x+15 y=7$ under the condition $\frac{1}{x} \geq 3.5$. | 20:5:8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,259 |
3. A covered football field of rectangular shape with a length of $90 \mathrm{m}$ and a width of 60 m is being designed, which should be illuminated by four spotlights, each hanging at some point on the ceiling. Each spotlight illuminates a circle, the radius of which is equal to the height at which the spotlight hangs... | Answer: 27.1
Let in rectangle $ABCD$ the diagonals intersect at point $O$. Place the projectors on the ceiling above the points that are the midpoints of segments $AO, BO, CO$, and $DO$, at a height equal to a quarter of the diagonal of the rectangle. Then the first projector will fully illuminate a circle, inside whi... | 27.1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,260 |
4. Hot oil at a temperature of $100^{\circ} \mathrm{C}$ in a volume of two liters is mixed with one liter of cold oil at a temperature of $20^{\circ} \mathrm{C}$. What volume will the mixture have when thermal equilibrium is established in the mixture? Heat losses to the external environment can be neglected. The coeff... | Answer: 3
Let $V_{1}=2$ L be the volume of hot oil, and $V_{2}=1$ L be the volume of cold oil. Then we can write $V_{1}=U_{1}\left(1+\beta t_{1}\right), V_{2}=U_{2}\left(1+\beta t_{2}\right)$, where $U_{1}, U_{2}$ are the volumes of the respective portions of oil at zero temperature; $t_{1}=100^{\circ} \mathrm{C}, t_{... | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,261 |
5. Gavriil got on the train with a fully charged smartphone, and by the end of the trip, his smartphone was completely drained. For half of the time, he played Tetris, and for the other half, he watched cartoons. It is known that the smartphone fully discharges in 3 hours of video watching or in 5 hours of playing Tetr... | Answer: 257
Let's assume the "capacity" of the smartphone battery is 1 unit (u.e.). Then the discharge rate of the smartphone when watching videos is $\frac{1}{3}$ u.e./hour, and the discharge rate when playing games is $\frac{1}{5}$ u.e./hour. If the total travel time is denoted as $t$ hours, we get the equation $\fr... | 257 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,262 |
Problem 1. Two runners started simultaneously in the same direction from one point of a circular distance, and the first runner, having moved ahead, caught up with the second again at the moment when the second had only run half a circle. From this moment, the second runner doubled his speed. Will the first runner catc... | Solution. From the first condition of the problem, it follows that the speed of the first runner is 3 times greater than the speed of the second (since he ran $3 / 2$ laps, while the second ran $1 / 2$ lap), that is, $V_{1}=3 V_{2}$. Therefore, even after doubling the speed of the second runner, the first will still mo... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,263 |
Problem 2. Grandma baked 19 pancakes. The grandchildren came from school and started eating them. While the younger grandson eats 1 pancake, the older grandson eats 3 pancakes, and during this time, grandma manages to cook 2 more pancakes. When they finished, there were 11 pancakes left on the plate. How many pancakes ... | Solution. In one "cycle", the grandsons eat $1+3=4$ pancakes, and the grandmother bakes 2 pancakes, which means the number of pancakes decreases by 2. There will be ( $19-11$ ) $/ 2=4$ such cycles. This means, in these 4 cycles, the younger grandson ate 4 pancakes, the older grandson ate 12 pancakes, and the grandmothe... | 12 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,264 |
Problem 3. Experimenters Glafira and Gavriil placed a triangle made of thin wire with sides of 30 mm, 40 mm, and 50 mm on a white flat surface. This wire is covered with millions of mysterious microorganisms. The scientists found that when an electric current is applied to the wire, these microorganisms begin to move c... | Solution. In one minute, the microorganism moves 10 mm. Since in a right triangle with sides $30, 40, 50$, the radius of the inscribed circle is 10, all points inside the triangle are no more than 10 mm away from the sides of the triangle. Therefore, the microorganisms will fill the entire interior of the triangle.
Wh... | 2114 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,265 |
Problem 4. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did... | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise... | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,266 |
Problem 5. An electric kettle heats water from room temperature $T_{0}=20^{\circ} \mathrm{C}$ to $T_{m}=100^{\circ} \mathrm{C}$ in $t=10$ minutes. How long will it take $t_{1}$ for all the water to boil away if the kettle is not turned off and the automatic shut-off system is faulty? The specific heat capacity of water... | Solution. The power $P$ of the kettle is fixed and equal to $P=Q / t$. From the heat transfer law $Q=c m\left(T_{m}-T_{0}\right)$ we get $P t=c m\left(T_{m}-T_{0}\right)$.
To evaporate the water, the amount of heat required is $Q_{1}=L m \Rightarrow P t_{1}=L m$.
By comparing these relations, we obtain $\frac{t_{1}}{... | 68 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,267 |
2.4. A metal weight has a mass of 20 kg and is an alloy of four metals. The first metal in this alloy is one and a half times more than the second, the mass of the second metal is to the mass of the third as $3: 4$, and the mass of the third to the mass of the fourth - as $5: 6$. Determine the mass of the third metal. ... | 2.4. A metal weight has a mass of 20 kg and is an alloy of four metals. The first metal in this alloy is one and a half times more than the second, the mass of the second metal is to the mass of the third as $3: 4$, and the mass of the third to the mass of the fourth - as $5: 6$. Determine the mass of the third metal. ... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,275 |
4.4. From cities $A$ and $Б$, which are 240 km apart, two cars set off towards each other at the same time with speeds of 50 km/h and 90 km/h. At what distance from point $C$, located halfway between $A$ and $Б$, will the cars meet? Give the answer in kilometers, rounding to the hundredths if necessary. | 4.4. From cities $A$ and $Б$, which are 240 km apart, two cars set off towards each other at the same time with speeds of 50 km/h and 90 km/h. At what distance from point $C$, located halfway between $A$ and $Б$, will the cars meet? Give the answer in kilometers, rounding to the hundredths if necessary. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,282 |
4.6. From cities $A$ and $Б$, which are 245 km apart, two cars set off towards each other at the same time with speeds of 70 km/h and 90 km/h. At what distance from the midpoint $C$ between $A$ and $Б$ will the cars meet? Give the answer in kilometers, rounding to the nearest hundredth if necessary. | 4.6. From cities $A$ and $\delta$, which are 245 km apart, two cars set off towards each other at the same time with speeds of 70 km/h and 90 km/h. At what distance from point $C$, located halfway between $A$ and $B$, will the cars meet? Give the answer in kilometers, rounding to the hundredths if necessary. | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,284 |
1. Upon entering the Earth's atmosphere, the asteroid heated up significantly and exploded near the surface, breaking into a large number of fragments. Scientists collected all the fragments and divided them into groups based on size. It was found that one-fifth of all fragments had a diameter of 1 to 3 meters, another... | 1. Answer: 70. Solution. Let $\mathrm{X}$ be the total number of fragments. The condition of the problem leads to the equation:
$\frac{x}{5}+26+n \cdot \frac{X}{7}=X$, where $n-$ is the unknown number of groups. From the condition of the problem, it follows that the number of fragments is a multiple of 35
$$
X=35 l, ... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 11,295 |
2. The mass of the first iron ball is $1462.5 \%$ greater than the mass of the second ball. By what percentage will less paint be needed to paint the second ball compared to the first? The volume of a sphere with radius $R$ is $\frac{4}{3} \pi R^{3}$, and the surface area of a sphere is $4 \pi R^{2}$. | 2. Answer: $84 \%$.
Let's denote the radii of the spheres as $R$ and $r$ respectively. Then the first condition means that
$$
\frac{\frac{4}{3} \pi R^{3}-\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi r^{3}} \cdot 100=1462.5 \Leftrightarrow \frac{R^{3}-r^{3}}{r^{3}}=14.625 \Leftrightarrow \frac{R^{3}}{r^{3}}=\frac{125}{8} \L... | 84 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,296 |
3. Little Red Riding Hood is walking along a path at a speed of 6 km/h, while the Gray Wolf is running along a clearing perpendicular to the path at a speed of 8 km/h. When Little Red Riding Hood was crossing the clearing, the Wolf had 80 meters left to run to reach the path. But he was already old, his eyesight was fa... | 3. Answer: No.
Solution. The problem can be solved in a moving coordinate system associated with Little Red Riding Hood. Then Little Red Riding Hood is stationary, and the trajectory of the Wolf's movement is a straight line. The shortest distance from a point to a line here is (by similarity considerations):
$80 \cd... | 48 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,297 |
5. In the village where Glafira lives, there is a small pond that is filled by springs at the bottom. Curious Glafira found out that a herd of 17 cows completely drank the pond dry in 3 days. After some time, the springs refilled the pond, and then 2 cows drank it dry in 30 days. How many days would it take for one cow... | 5. Answer: In 75 days.
Solution. Let the pond have a volume of a (conditional units), one cow drinks b (conditional units) per day, and the springs add c (conditional units) of water per day. Then the first condition of the problem is equivalent to the equation $a+3c=3 \cdot 17 b$, and the second to the equation $a+30... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,299 |
6. Gavriil was traveling in Africa. On a sunny and windy day, at noon, when the rays from the Sun fell vertically, the boy threw a ball from behind his head at a speed of $5 \sim$ m/s against the wind at an angle to the horizon. After 1 s, the ball hit him in the stomach 1 m below the point of release. Determine the gr... | 6. Answer: 75 cm
Solution. In addition to the force of gravity, a constant horizontal force $F=m \cdot a$ acts on the body, directed opposite. In a coordinate system with the origin at the point of throw, the horizontal axis x, and the vertical axis y, the law of motion has the form:
$$
\begin{aligned}
& x(t)=V \cdot... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,300 |
1.1. The Eiffel Tower has a height of 324 m and weighs 10000 tons. How many kilograms will a copy with a height of 1.62 m weigh? | Answer: 1.25. Solution. The volumes of similar bodies are in the ratio of the cube of the similarity coefficient. Since the similarity coefficient is $\frac{324}{1.62}=200$, the weight of this copy of the Eiffel Tower is $\frac{10000}{200^{3}}$ tons, which equals $\frac{10^{7}}{8 \cdot 10^{6}}=\frac{10}{8}=\frac{5}{4}=... | 1.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,301 |
2.1. During the time it took for a slowly moving freight train to cover 1200 m, a schoolboy managed to ride his bicycle along the railway tracks from the end of the moving train to its beginning and back to the end. In doing so, the bicycle's distance meter showed that the cyclist had traveled 1800 m. Find the length o... | Answer: 500. Solution: Let $V$ and $U$ be the speeds of the cyclist and the train, respectively, and $h$ be the length of the train.
Then the conditions of the problem in mathematical terms can be written as follows:
$$
(V-U) t_{1}=h ; \quad(V+U) t_{2}=h ; \quad U\left(t_{1}+t_{2}\right)=l ; \quad V\left(t_{1}+t_{2}\... | 500 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,302 |
3.1. The friends who came to visit Gavrila occupied all the three-legged stools and four-legged chairs in the room, but there was no place left for Gavrila himself. Gavrila counted that there were 45 legs in the room, including the "legs" of the stools and chairs, the legs of the visiting guests (two for each!), and Ga... | Answer: 9. Solution. If there were $n$ stools and $m$ chairs, then the number of legs in the room is $3 n+4 m+2 \cdot(n+m)+2$, from which we get $5 n+6 m=43$. This equation in integers has the solution $n=5-6 p, m=3+5 p$. The values of $n$ and $m$ are positive only when $p=0$. Therefore, there were 5 stools and 3 chair... | 9 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,303 |
4.1. A train of length $L=600$ m, moving by inertia, enters a hill with an angle of inclination $\alpha=30^{\circ}$ and stops when exactly a quarter of the train is on the hill. What was the initial speed of the train $V$ (in km/h)? Provide the nearest whole number to the calculated speed. Neglect friction and assume t... | Answer: 49. Solution. The kinetic energy of the train $\frac{m v^{2}}{2}$ will be equal to the potential energy of the part of the train that has entered the hill $\frac{1}{2} \frac{L}{4} \sin \alpha \frac{m}{4} g$.
Then we get $V^{2}=\frac{L}{4} \frac{1}{8} *(3.6)^{2}=\frac{6000 *(3.6)^{2}}{32}=9 \sqrt{30}$.
Since t... | 49 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,304 |
5.1. A grenade lying on the ground explodes into a multitude of small identical fragments, which scatter in a radius of $L=90$ m. Determine the time interval (in seconds) between the moments of impact on the ground of the first and the last fragment, if such a grenade explodes in the air at a height of $H=10 \mathrm{m}... | Answer: 6. Solution. From the motion law for a body thrown from ground level at an angle $\alpha$ to the horizontal, the range of flight is determined by the relation $L=\frac{V_{0}^{2}}{g} \sin 2 \alpha$. Therefore, the maximum range of flight is achieved at $\alpha=45^{\circ}$ and is equal to $L=\frac{V_{0}^{2}}{g}$.... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,305 |
6.1. A vertical cylindrical vessel with a cross-sectional area of $S=10 \mathrm{~cm}^{2}$, containing one mole of a monatomic ideal gas, receives an amount of heat $Q=500$ J per second. The vessel is closed at the top by a heavy piston weighing $P=100 \mathrm{N}$. At what speed (in meters per second) does this piston r... | Answer: 1. Solution. According to the first law of thermodynamics, we can write $Q \Delta t=\Delta A+\Delta U$ (2), where $\Delta t$ - time interval, $\Delta A$ - work done by the gas, $\Delta U-$ change in internal energy. Since the process is isobaric, the following relationships can be written for work and internal ... | \frac{Q}{2.5(p_{0}S+P)} | Other | math-word-problem | Yes | Yes | olympiads | false | 11,306 |
2. Two schoolchildren left two neighboring schools at the same time and headed towards each other. After ten minutes, the distance between the schoolchildren was $600 \mathrm{m}$, and after another five minutes, they met. Find the distance between the schools (in meters).
$\{=1800\}$
$:: 1.1::$ A weight with a mass o... | 1800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,307 | |
1. 2. A car with a load traveled from one city to another at a speed of 60 km/h, and returned empty at a speed of 90 km/h. Find the average speed of the car for the entire route. Give your answer in kilometers per hour, rounding to the nearest whole number if necessary.
$\{72\}$ | Solution. The average speed will be the total distance divided by the total time: $\frac{2 S}{\frac{S}{V_{1}}+\frac{S}{V_{2}}}=\frac{2 V_{1} \cdot V_{2}}{V_{1}+V_{2}}=\frac{2 \cdot 60 \cdot 90}{60+90}=72(\mathrm{km} / \mathrm{h})$. | 72 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,309 |
2.1. A metal weight has a mass of 20 kg and is an alloy of four metals. The first metal in this alloy is one and a half times more than the second, the mass of the second metal is to the mass of the third as $3: 4$, and the mass of the third to the mass of the fourth - as $5: 6$. Determine the mass of the fourth metal.... | Solution. The masses are in the ratio 45:30:40:48. The total sum $=163 x$. Therefore, $x=20 / 163$. The mass of the fourth metal is 48 times greater. | 5.89 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,310 |
4.1. From cities $A$ and $Б$, which are 240 km apart, two cars set off towards each other simultaneously at speeds of 60 km/h and 80 km/h. At what distance from point $C$, located halfway between $A$ and $Б$, will the cars meet? Give the answer in kilometers, rounding to the nearest hundredth if necessary.
## $\{17.14... | Solution. Time until meeting $=240 /(60+80)=12 / 7$ hours. In this time, the first car will travel $60 \cdot 12 / 7=720 / 7$ km. The required distance $=$ $120-60 \cdot 12 / 7=120 \cdot(1-6 / 7)=120 / 7 \approx 17.1428 \ldots$ (km). | 17.14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,311 |
7.1. A schoolboy with a mass of 70 kg, standing on smooth ice, throws a stone with a mass of 1 kg in a horizontal direction from a height of 2 m. The stone lands on the ice at a distance of 10 m from the point of throwing. How much work did the schoolboy do when throwing the stone? Assume the acceleration due to gravit... | Solution. Initial speed of the stone: $V_{\kappa}=L \cdot \sqrt{\frac{g}{2 h}}$, initial speed of the schoolboy (law of conservation of momentum): $V_{u}=\frac{m}{M} \cdot V_{k}$. Then the work done by the schoolboy in throwing the stone is equal to the total kinetic energy:
$$
A=\frac{m V_{\kappa}^{2}}{2}+\frac{M V_{... | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,312 | |
10.1. In the field, there are two trees: an aspen 20 meters tall, and a birch 30 meters tall. The distance between the trees is 50 meters. On the top of each tree, there is a crow. Find a place on the ground for the cheese so that the sum of the two distances the crows fly from their locations to the cheese is minimize... | Solution. Place the "stone" 30 meters underground symmetrically to the second crow. The required total distance for the two crows is the length of the two-segment broken line connecting the first crow and this "stone" with the cheese. The shortest distance is a straight line. This results in a point located 20 meters f... | 70.71 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,313 |
12.1. Two motorcyclists are moving along straight tracks (each on their own) at constant speeds. At 11:14, the distance between them was 40 km, at 11:46 - 30 km, at 12:10 - 30 km. Determine the magnitude of the relative speed of one motorcyclist relative to the other (in kilometers per hour). Provide the answer as a wh... | Solution. We solve the problem in a coordinate system associated with the first motorcyclist, denoted as $A$. The positions of the second motorcyclist: at the first moment of time $-B$; at the second moment $-C$; at the third $-D$. Then $A B=$ $40, A C=30, A D=30$.
Since $B C: C D=\frac{46-14}{70-46}=\frac{32}{24}=4: ... | 37.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,314 |
3.1. A transgalactic ship has encountered an amazing meteorite stream. Some of the meteorites are flying along a straight line, one after another, at equal speeds and at equal distances from each other. Another part is flying the same way but along another straight line parallel to the first, at the same speeds but in ... | 3.1. A transgalactic ship has encountered an amazing meteor shower. Some of the meteors are flying along a straight line, one after another, at the same speeds and at equal distances from each other. Another part is flying in the same way but along another straight line parallel to the first, at the same speeds but in ... | 9.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,317 |
1. Gavriil and Glafira took a glass filled to the brim with water and poured a little water into three ice cube trays, then placed them in the freezer. When the ice froze, they put the three resulting ice cubes back into the glass. Gavriil predicted that some water would spill out of the glass because ice expands in vo... | Solution. Let $V$ be the volume of water in the molds. Then the volume $W$ of ice in the molds can be determined from the law of conservation of mass $V \cdot \rho_{\text {water }}=W \cdot \rho_{\text {ice }}$. When ice of volume $W$ is floating, the submerged part of this volume $U$ can be determined from the conditio... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,318 |
2. First-grader Chukov runs one lap over rough terrain three minutes faster than his classmate Gekov (both run at a constant speed). If they start running simultaneously from one point of this lap but in opposite directions, they will not meet earlier than two minutes. If they start from the same point in the same dire... | Solution. If Chukov runs a lap in $t$ minutes, then it will take Gekov $t+3$ minutes. If the length of the lap is $L$ meters, then Chukov's speed $V_{1}=\frac{L}{t}$, and Gekov's speed $\begin{aligned} V_{2}= & \frac{L}{t+3} . \text { According to the condition } \frac{L}{V_{1}+V_{2}} \geqslant 2, \frac{L}{V_{1}-V_{2}}... | [3;6] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,319 |
3. From a point on the Earth's surface, a large number of small balls are launched in all directions at the same speed of 10 m/s. Among all the balls that land at a distance from the starting point not closer than $96 \%$ of the distance at which the farthest flying ball lands, find the one that will spend the most tim... | Solution. The range of a body thrown with an initial velocity $V_{0}$ at an angle $\alpha$ to the horizontal is $l=\frac{V_{0}^{2} \sin 2 \alpha}{g}$, and the flight time is $\tau=\frac{2 V_{0} \sin \alpha}{g}$.
According to the condition $\frac{V_{0}^{2} \sin 2 \alpha}{g} \geqslant \frac{96}{100} \frac{V_{0}^{2} \sin... | 1.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,320 |
4. Three pulleys with parallel axes and identical radii $r=2$ cm must be connected by a flat belt drive. The distance between the axes of rotation of pulleys $O_{1}$ and $O_{2}$ is 12 cm, and the distance between the axes of rotation of pulleys $O_{1}$ and $O_{3}$ is 10 cm. The distance from the axis $O_{3}$ to the pla... | Solution. 1) First, we solve the geometric problem of finding the side $\mathrm{O}_{2} \mathrm{O}_{3}$. According to the condition, $\sin \angle O_{1}=\frac{8}{10}=\frac{4}{5}$; therefore, $\cos \angle O_{1}= \pm \frac{3}{5}$. The third side is found using the cosine theorem and equals either 10 cm or $2 \sqrt{97}$ cm.... | 32+4\pior22+2\sqrt{97}+4\pi;notalways | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,321 |
5. A vessel is filled with cold water. The following procedure is performed with this vessel: three quarters of the cold water are poured out and the vessel is refilled to its original volume with hot water. As a result, the temperature of the water in the vessel increases by $16^{\circ} \mathrm{C}$. After this, the pr... | Solution. Let $\alpha$ be the ratio of the volume of the added water to the volume of the vessel $\left(\alpha=\frac{3}{4}\right)$. Then, from the heat balance equation:
$$
t_{1}=(1-\alpha) t_{c}+\alpha t_{h}
$$
$$
t_{2}=(1-\alpha) t_{1}+\alpha t_{h} \Rightarrow t_{2}=t_{h}-(1-\alpha)^{2}\left(t_{h}-t_{c}\right)
$$
... | notfound | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,322 |
1. The law of motion of the first tourist: $S=\sqrt{1+6 t}-1$, and the second: $-S=$ $6\left(t-\frac{1}{6}\right)$ for $t \geq \frac{1}{6} ; S=0$ for $t<\frac{1}{6}$.
The required condition is obviously met when both tourists are "on the same side" of the sign. And it is not met when the sign is between them. Therefor... | Answer: $t \in\left[0, \frac{1}{2}\right] \cup\left[\frac{4}{3},+\infty\right)$. | \in[0,\frac{1}{2}]\cup[\frac{4}{3},+\infty) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,323 |
3. It is not difficult to show that for a fixed sum of two numbers, their product is maximal if and only if these numbers are equal. Thus, it is clear that it makes sense to consider suitcases whose base is a square.
In the case where the more significant condition is the sum of all three dimensions, the symmetry of t... | Answer: a) the shape of an elongated parallelepiped with sides of 44, 44, and 220 centimeters; b) the shape of a cube with a side length of 50 cm. | ) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,325 |
1. A tractor is pulling a very long pipe on sled runners. Gavrila walked along the entire pipe in the direction of the tractor's movement and counted 210 steps. When he walked in the opposite direction, the number of steps was 100. What is the length of the pipe if Gavrila's step is 80 cm? Round the answer to the neare... | Solution. Let the length of the pipe be $x$ (meters), and for each step Gavrila takes of length $a$ (m), the pipe moves a distance $y$ (m). Then, if $m$ and $n$ are the number of steps Gavrila takes in each direction, we get two equations: $x=m(a-y), x=n(a+y)$. From this, $\frac{x}{m}+\frac{x}{n}=2 a$, and $x=\frac{2 a... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,327 |
2. What is the greatest whole number of liters of water that can be heated to boiling temperature using the amount of heat obtained from the combustion of solid fuel, if in the first 5 minutes of combustion, 480 kJ is obtained from the fuel, and for each subsequent five-minute period, 25% less than the previous one. Th... | Answer: 5 liters
Solution: The amount of heat required to heat a mass $m$ of water under the conditions of the problem is determined by the relation $Q=4200(100-20) m=336 m$ kJ. On the other hand, if the amount of heat received in the first 5 minutes is $Q_{0}=480$ kJ. Then the total (indeed over an infinite time) amo... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,328 |
3. A three-stage launch vehicle consists of stages in the form of cylinders. All these cylinders are similar to each other. The length of the middle stage is two times less than the sum of the lengths of the first and third stages. In the fueled state, the mass of the middle stage is $13 / 6$ times less than the total ... | Answer: $\frac{7}{5}$
Solution. The mass of the fueled stage is proportional to the third power of the linear size $l_{2}=\frac{1}{2}\left(l_{1}+l_{2}\right), l_{2}^{3}=\frac{6}{13}\left(l_{1}^{3}+l_{2}^{3}\right)$. Solving this system, we get $\frac{l_{1}}{l_{3}}=\frac{7}{5}$ or $\frac{l_{1}}{l_{3}}=\frac{5}{7}$. Fro... | \frac{7}{5} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,329 |
4. At some point on the shore of a wide and turbulent river, 100 m from the bridge, Gavrila and Glafira set up a siren that emits sound signals at equal intervals. Glafira took another identical siren and positioned herself at the beginning of the bridge on the same shore. Gavrila got into a motorboat, which was locate... | Solution. Let's introduce a coordinate system, with the $x$-axis directed along the shore, and the origin at Gavrila's starting point. The siren on the shore has coordinates $(L, 0), L=50$ m, and Glafira is traveling along the line $x=-L$. Since the experimenters are at the same distance from the shore, the equality of... | 41 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,330 |
5. On a windless day, a polar bear found itself on a small piece of ice that had broken off from an iceberg in the middle of still water. Rescuers from a helicopter hovering above the ice floe used instruments to determine that the animal was walking in a circle with a diameter of 8.5 meters. How surprised they were wh... | Solution. Instruments and photographs were used to measure different things. The chain of tracks shows the radius of the bear's trajectory in a reference frame associated with the ice floe, while the rescuers' instruments show the radius of the trajectory in a reference frame associated with the Earth. The difference i... | notfound | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,331 |
Task 1. A vacuum robot is programmed to move on the floor according to the law:
$\left\{\begin{array}{l}x=(t-6)^{2} \\ y=0,0 \leq t \leq 7 ; y=(t-7)^{2}, t \geq 7\end{array}\right.$
where the axes are parallel to the walls. Time $t$ is measured in minutes, and coordinates in meters.
Find the path traveled by the rob... | Solution. For the first 7 minutes, the point moves with constant acceleration along the x-axis. The velocity of the point at $t \leq 7$ is $V_{x}=2(t-6)$ and becomes zero at $t_{1}=6$. The distance traveled will then be $L=x(0)+x(7)=1+36=37$.
The velocity vector $\bar{V}$ at any moment $t \geq 7$ is determined by its ... | 37;2\sqrt{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 11,333 |
Problem 2. Experimenters Glafira and Gavriil placed a triangle made of thin wire with sides of 30 mm, 40 mm, and 50 mm on a white flat surface. This wire is covered with millions of mysterious microorganisms. The scientists found that when an electric current is applied to the wire, these microorganisms begin to move c... | Solution. In one minute, the microorganism moves 10 mm. Since in a right triangle with sides $30, 40, 50$, the radius of the inscribed circle is 10, all points inside the triangle are at a distance from the sides of the triangle that does not exceed 10 mm. Therefore, the microorganisms will fill the entire interior of ... | 2114 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,334 |
Problem 4. Snowboarder Gavrila descended a hill 250 meters high and had a speed of 10 m/s at the foot of the hill. What fraction of the total loss of mechanical energy during the descent went to heating the snowboard weighing 6 kg, if it heated up by 1 degree? The specific heat capacity of the snowboard material is 300... | Solution. The law of change of mechanical energy during descent $\frac{m V^{2}}{2}=m g H-W$. From this, the energy loss $W=m g H-\frac{m V^{2}}{2}=72 \cdot 10 \cdot 250-36 \cdot 100=36 \cdot 4900$ J.
The amount of heat $Q=c \cdot m \cdot\left(T-T_{0}\right)=300 \cdot 6 \cdot 1=1800$.
The required value $k=\frac{Q}{W}... | \frac{1}{98} | Other | math-word-problem | Yes | Yes | olympiads | false | 11,335 |
Problem 5. Three beads with masses $m_{1}=150$ g, $m_{3}=30$ g, $m_{2}=1$ g (see figure) can slide along a horizontal rod without friction.
Determine the maximum speeds of the large beads if at the initial moment of time they were at rest, while the small bead was moving with a speed of $V=10$ m/s.
Consider the colli... | Solution. After each collision, the magnitude of the velocity of the bead with mass $m_{2}$ decreases. After a certain number of collisions, its velocity will be insufficient to catch up with the next bead $m_{1}$ or $m_{3}$. After such a final collision, the velocities of the beads will no longer change. Let $V_{1}$ a... | V_{1}\approx0.28\mathrm{~}/\mathrm{};V_{3}\approx1.72\mathrm{~}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,336 |
1. A certain man hired a worker for a year, promising to give him 12 rubles and a coat. But by chance, after working for 7 months, the worker wanted to leave and asked for a fair payment including the coat. He was given 5 rubles and the coat as fair compensation. What was the value of the coat? (An old problem) Give th... | Solution. This is problem E. D. Voityakhovsky from "A Course in Pure Mathematics" (1811). If the annual fee is $12+K$ (where $K$ is the cost of the caftan), then for 7 months, the payment should be $\frac{7(12+K)}{12}$. We get the equation $\frac{7(12+K)}{12}=5+K$, from which $\frac{5 K}{12}=2, K=\frac{24}{5}=4.8$ rubl... | 4.8 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,337 |
3. The villages of Arkadino, Borisovo, and Vadimovo are connected by straight roads. A square field adjoins the road between Arkadino and Borisovo, one side of which completely coincides with this road. A rectangular field adjoins the road between Borisovo and Vadimovo, one side of which completely coincides with this ... | Solution. The condition of the problem can be expressed by the following relation:
$r^{2}+4 p^{2}+45=12 q$
where $p, q, r$ are the lengths of the roads opposite the settlements Arkadino, Borisovo, and Vadimovo, respectively. This condition is in contradiction with the triangle inequality:
$r+p>q \Rightarrow 12 r+12 ... | 135 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,338 |
1. Density is the ratio of the mass of a body to the volume it occupies. Since the mass did not change as a result of tamping, and the volume after tamping $V_{2}=$ $0.8 V_{1}$, the density after tamping became $\rho_{2}=\frac{1}{0.8} \rho_{1}=1.25 \rho_{1}$, that is, it increased by $25 \%$. | Answer: increased by $25 \%$. | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,339 |
3. The law of motion of the first tourist: $S=\sqrt{1+6 t}-1$, and the second tourist: $-S=$ $6\left(t-\frac{1}{6}\right)$ for $t \geq \frac{1}{6} ; S=0$ for $t<\frac{1}{6}$. The required condition is met if and only if one of the tourists is on one side of the road sign, and the other is on the other side (one or the ... | Answer: from half an hour to 1 hour and 20 minutes after the first tourist exits. | [\frac{1}{2},\frac{4}{3}] | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,341 |
4. Let's switch to a reference frame falling with acceleration $g$ and zero initial velocity. The time count starts at the moment the apple is thrown. In the specified reference frame, both the apple and the arrow move uniformly and rectilinearly due to the law of acceleration addition. Therefore, the point where the a... | Answer: $T=\frac{L}{V_{0}} \frac{\sin \beta}{\sin (\alpha+\beta)}=\frac{3}{4} \mathrm{c}$. | \frac{3}{4} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,342 |
1.1. Gavriil found out that the front tires of the car last for 20000 km, while the rear tires last for 30000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km). | 1.1. Gavriil found out that the front tires of the car last for 20,000 km, while the rear tires last for 30,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{24000\}$. | 24000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,343 |
1.2. Gavriila found out that the front tires of the car last for 24000 km, while the rear tires last for 36000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km). | 1.2. Gavriil found out that the front tires of the car last for 24,000 km, while the rear tires last for 36,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{28800\}$. | 28800 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,344 |
1.3. Gavriila found out that the front tires of the car last for 42,000 km, while the rear tires last for 56,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km). | 1.3. Gavriil found out that the front tires of the car last for 42000 km, while the rear tires last for 56000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{48000\}$. | 48000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,345 |
4.1. A smooth sphere with a radius of 1 cm was dipped in red paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a red trail. During its mov... | 4.1. A smooth sphere with a radius of 1 cm was dipped in red paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a red trail. During its mov... | 83.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,355 |
4.2. A smooth sphere with a radius of 1 cm was dipped in blue paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a blue trail. During its m... | 4.2. A smooth sphere with a radius of 1 cm was dipped in blue paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a blue trail. During its m... | 60.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,356 |
4.3. A smooth sphere with a radius of 1 cm was dipped in red paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a red trail. During its mov... | 4.3. A smooth sphere with a radius of 1 cm was dipped in red paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a red trail. During its mov... | 105.75 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,357 |
4.4. A smooth sphere with a radius of 1 cm was dipped in blue paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a blue trail. During its m... | 4.4. A smooth sphere with a radius of 1 cm was dipped in blue paint and launched between two perfectly smooth concentric spheres with radii of 4 cm and 6 cm, respectively (this sphere ended up outside the smaller sphere but inside the larger one). Upon touching both spheres, the sphere leaves a blue trail. During its m... | 38.25 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,358 |
5.2. A boy presses a vertical rod against a rough horizontal surface with his thumb. Then he gradually tilts the rod, keeping the component of the force directed along the rod applied to its end unchanged. At an angle of inclination of the rod to the horizontal $\alpha=70^{\circ}$, the rod starts to slide along the sur... | 5.2. A boy presses a vertical rod against a rough horizontal surface with his thumb. Then he gradually tilts the rod, keeping the component of the force directed along the rod applied to its end unchanged. At an angle of inclination of the rod to the horizontal $\alpha=70^{\circ}$, the rod starts to slide along the sur... | 0.35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,360 |
1. A new model car travels $4 \frac{1}{6}$ kilometers more on one liter of gasoline compared to an old model car. At the same time, its fuel consumption per 100 km is 2 liters less. How many liters of gasoline does the new car consume per 100 km? | Answer: 6 liters.
Instructions. The fuel consumption of the new car is $x$ liters, and the consumption of the old car is $x+2$
liters. Equation: $\frac{100}{x}-\frac{100}{x+2}=\frac{25}{6} \Leftrightarrow \frac{4(x+2-x)}{x(x+2)}=\frac{1}{6} \Leftrightarrow x^{2}+2 x-48=0 \Leftrightarrow x=-8 ; x=6$. Therefore, $x=6$... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,363 |
2. Gavriil and Glafira took a glass filled to the brim with water and poured a little water into three ice cube trays, then placed them in the freezer. When the ice froze, the three resulting ice cubes were put back into the glass. Gavriil predicted that some water would spill out of the glass because ice expands in vo... | Answer: No one is right. The water will fill the glass exactly to its edges.
Instructions. Let $V$ be the volume of water in the molds. Then the volume $W$ of ice in the molds can be determined from the law of conservation of mass $V \cdot \rho_{\text {water }}=W \cdot \rho_{\text {ice }}$. When the ice of volume $W$ ... | U | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,364 |
3. Schoolchildren Chukov and Gekov are skating at constant speeds around a closed circular running track of an ice stadium. If Chukov skates clockwise and Gekov skates counterclockwise, their meetings occur four times more frequently than when they overtake each other while skating in the same direction. The speed of o... | Answer: Either $10 \mathrm{m} /$ sec, or $3.6 \mathrm{~m} /$ sec.
Instructions. When moving towards each other, the time between meetings is $t_{1}=\frac{L}{V_{1}+V_{2}}$, when moving in the same direction, the time between overtakes: $t_{2}=\frac{L}{V_{1}-V_{2}}$ (here $L-$ is the length of one lap, $V_{1}, V_{2}$ - ... | 10\mathrm{}/ | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,365 |
4. Gavrila placed 7 smaller boxes into a large box. After that, Glafira placed 7 small boxes into some of these seven boxes, and left others empty. Then Gavrila placed 7 boxes into some of the empty boxes, and left others empty. Glafira repeated this operation and so on. At some point, there were 34 non-empty boxes. Ho... | Answer: 205.
Instructions. Filling one box increases the number of empty boxes by 7-1=6, and the number of non-empty boxes by 1. Therefore, after filling $n$ boxes (regardless of the stage), the number of boxes will be: empty $-1+6 n$; non-empty $-n$. Thus, $n=34$, and the number of non-empty boxes will be $1+6 \cdot ... | 205 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 11,366 |
5. A vessel is filled with cold water. The following procedure is performed with this vessel: three quarters of the cold water are poured out and the vessel is refilled to its original volume with hot water. As a result, the temperature of the water in the vessel increases by $24^{\circ} \mathrm{C}$. After this, the pr... | Answer: A) $n=4$; B) impossible.
Instructions. From the heat balance equation:
$$
\begin{aligned}
& t_{1}-t_{c}=t_{h}-t_{1} \Rightarrow 2 t_{1}=t_{c}+t_{h} \\
& t_{2}-t_{1}=t_{h}-t_{2} \Rightarrow 2 t_{2}=t_{h}+t_{1} \Rightarrow t_{2}=t_{h}+\frac{t_{h}+t_{c}}{2} \Rightarrow t_{2}=\frac{t_{c}+3 t_{h}}{4}
\end{aligned}... | A)n=4;B)impossible | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,367 |
1. From the condition at $12:00$, it follows that the tracks intersect at an angle of $60^{\circ}$. Since they crossed the intersection after this, they are moving towards it. Since the cyclist spent twice as much time on this road, his speed is half the speed of the motorcyclist, i.e., 36 km/h.
If we start the time c... | Answer: 09:00 and $17:00$. | 09:0017:00 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,368 |
3. The heat engine "Lomonosov" 12341 receives heat on the segment $1-2$ and gives it off on the segment $3-4$. Therefore,
$$
\eta_{0}=1-\frac{Q_{34}}{Q_{12}}
$$
(here and below, all heats are assumed to be positive). The engine "Avogadro" 1231 also receives heat on 12, but gives it off on 31, so
$$
\eta_{1}=1-\frac{... | Answer: $\eta_{2}=\frac{\alpha}{100-(100-\alpha) \eta_{0}} ;$ for all. | \eta_{2}=\frac{\alpha}{100-(100-\alpha)\eta_{0}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,370 |
1. A car was moving at a speed of $V$. Upon entering the city, the driver reduced the speed by $x \%$, and upon leaving the city, increased it by $0.5 x \%$. It turned out that this new speed was $0.6 x \%$ less than the speed $V$. Find the value of $x$. | Answer: 20. Solution. The condition of the problem means that the equation is satisfied
$$
v\left(1-\frac{x}{100}\right)\left(1+\frac{0.5 x}{100}\right)=v\left(1-\frac{0.6 x}{100}\right) \Leftrightarrow\left(1-\frac{x}{100}\right)\left(1+\frac{x}{200}\right)=1-\frac{3 x}{500} \Leftrightarrow \frac{x^{2}}{20000}=\frac{... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,374 |
2. A barrel 1.5 meters high is completely filled with water and covered with a lid. The mass of the water in the barrel is 1000 kg. A long thin tube with a cross-sectional area of $1 \mathrm{~cm}^{2}$ is inserted vertically into the lid of the barrel, which is completely filled with water. Find the length of the tube i... | Answer: 1.5 m. Solution. To solve the problem, you only need to know how pressure changes with depth: $p=\rho g$. From this relationship, it follows that to double the pressure, you need to double the height of the liquid column. This means that the tube should be the same height as the barrel - 1.5 m. | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,375 |
3. The Field of Wonders in the Land of Fools has the shape of a rectangle with sides of 6 km and 2.5 km. Malvina and Buratino started moving towards each other from two of its non-adjacent vertices along the diagonal at speeds of 4 km/h and 6 km/h, respectively. At the same moment, the poodle Artemon started running at... | Answer: 7.8 km. Solution. The diagonal of the rectangle is $\sqrt{6^{2}+2.5^{2}}=\sqrt{36+\frac{25}{4}}$ $=\sqrt{\frac{169}{4}}=\frac{13}{2}$ (km). Therefore, Malvina and Pinocchio will meet after $\frac{13}{2}:(4+6)$ $=\frac{13}{2 \cdot 10}=\frac{13}{20}$ (hour). During this time, the poodle Artemon ran at a speed of ... | 7.8 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 11,376 |
4. Tourists from the USA, when traveling to Europe, often use an approximate formula to convert temperatures in degrees Celsius $C$ to the familiar degrees Fahrenheit $F$: $F=2 C+30$. Indicate the range of temperatures (in degrees Celsius) for which the deviation of the temperature in degrees Fahrenheit, obtained using... | Answer: $1 \frac{11}{29} \leq C \leq 32 \frac{8}{11}$. Solution. Both temperature scales are uniform, so they are related by a linear law: $F=k C+b$. The constants $a$ and $b$ are determined from the condition. The exact formula is obtained: $F=\frac{9}{5} C+32$.
The deviation of temperatures calculated by the two for... | 1\frac{11}{29}\leqC\leq32\frac{8}{11} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,377 |
1. A stone is thrown vertically upwards with an initial velocity $V$. Neglecting air resistance and assuming the acceleration due to gravity is $10 \mathrm{~m} / \mathrm{c}^{2}$, determine the values of $V$ for which all moments of reaching a height of 10 m will lie between: A) the first and second seconds after the st... | Solution. The dependence of height on time is $h(t)=V t-\frac{g t^{2}}{2}$. Therefore, the stone will be at a height of 10 m at the moments of time $10=V t-5 t^{2}$. This results in the equation $5 t^{2}-V t+10=0$, which for $V^{2} \geqslant 200$ (i.e., for $V \geqslant 10 \sqrt{2}$) has roots: $t_{1,2}=\frac{V \pm \sq... | A)V\in[10\sqrt{2};15)/;B)V\in\varnothing | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,378 |
1. Gavriil decided to weigh a football, but he only had weights of 150 g, a long light ruler with the markings at the ends worn off, a pencil, and many threads at his disposal. He suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second ... | 1. Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\le... | 600 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,379 |
2. The cover of a vertical shaft 160 m deep periodically opens and closes instantly, so that the shaft is open for 4 seconds and closed for 4 seconds. From the bottom of the shaft, a pneumatic gun fires a bullet vertically upwards with an initial velocity $V$, exactly 2 seconds before the next opening of the cover. For... | 2. Let $h$ be the depth of the well, $g$ be the acceleration due to gravity, and $\tau$ be the interval during which the lid is open. The data is chosen such that $h /\left(g \tau^{2}\right)=1$.
Note that the ball cannot take too long to rise from the well. The maximum time for the ball to rise is $\sqrt{2 h / g}=\tau... | (\frac{17}{12},\frac{33}{20})\cup(\frac{57}{28},\frac{9}{4}) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 11,380 |
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