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4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entra... | Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
. Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the ... | Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the... | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young dr... | # Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ ... | 34 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
2. Ninth-grader Gavriil decided to weigh a basketball, but he only had 400 g weights, a light ruler with the markings at the ends worn off, a pencil, and many weightless threads at his disposal. Gavriil suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. The... | 2. 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 |
Problem 2. The distances from three points lying in a horizontal plane to the base of a television tower are 800 m, 700 m, and 500 m, respectively. From each of these three points, the tower is visible (from base to top) at a certain angle, and the sum of these three angles is $90^{\circ}$. A) Find the height of the te... | Solution. Let the given distances be denoted by $a, b$, and $c$, the corresponding angles by $\alpha, \beta$, and $\gamma$, and the height of the tower by $H$. Then $\operatorname{tg} \alpha=\frac{H}{a}, \operatorname{tg} \beta=\frac{H}{b}, \operatorname{tg} \gamma=\frac{H}{c}$. Since $\frac{H}{c}=\operatorname{tg} \ga... | 374 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. If the cold water tap is opened, the bathtub fills up in 5 minutes and 20 seconds. If both the cold water tap and the hot water tap are opened simultaneously, the bathtub fills up to the same level in 2 minutes. How long will it take to fill the bathtub if only the hot water tap is opened? Give your answer in second... | Solution. According to the condition: $\frac{16}{3} v_{1}=1,\left(v_{1}+v_{2}\right) 2=1$, where $v_{1}, v_{2}$ are the flow rates of water from the cold and hot taps, respectively. From this, we get: $v_{1}=3 / 16, v_{2}=5 / 16$.
Then the time to fill the bathtub from the hot tap is $\frac{16}{5}$. Answer: 3 minutes ... | 192 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A weight with a mass of 200 grams stands on a table. It was flipped and placed on the table with a different side, the area of which is 15 sq. cm smaller. As a result, the pressure on the table increased by 1200 Pa. Find the area of the side on which the weight initially stood. Give your answer in sq. cm, rounding t... | Solution. After converting to SI units, we get: $\frac{2}{S-1.5 \cdot 10^{-3}}-\frac{2}{S}=1200$.
Here $S-$ is the area of the original face.
From this, we get a quadratic equation: $4 \cdot 10^{5} S^{2}-600 S-1=0$.
After substituting the variable $y=200 S$, the equation becomes: $10 y^{2}-3 y-1=0$, the solution of ... | 25 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 |
4. Alloy $A$ of two metals with a mass of 6 kg, in which the first metal is twice as much as the second, placed in a container with water, creates a pressure force on the bottom of $30 \mathrm{N}$. Alloy $B$ of the same metals with a mass of 3 kg, in which the first metal is five times less than the second, placed in a... | Solution. Due to the law of conservation of mass, in the resulting alloy, the mass of each metal is equal to the sum of the masses of these metals in the initial alloys. Thus, both the gravitational forces and the forces of Archimedes also add up. From this, it follows that the reaction force will be the sum of the rea... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. From a square steel sheet with a side of 1 meter, a triangle is cut off from each of the four corners so that a regular octagon remains. Determine the mass of this octagon if the sheet thickness is 3 mm and the density of steel is 7.8 g/cm ${ }^{3}$. Give your answer in kilograms, rounding to the nearest whole numbe... | Answer: $46.8(\sqrt{2}-1) \approx 19$ kg.
Solution. A regular octagon must have equal angles and sides. Therefore, four equal triangles with angles $45^{\circ}, 45^{\circ}$, and $90^{\circ}$ are cut off. If the legs of this triangle are equal to $x$, then the hypotenuse is $x \sqrt{2}$ - this will be the side of the o... | 19 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
3. 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... | Answer: In 75 days.
Solution. Let the pond have a volume of $a$ (conditional units). These units can be liters, buckets, cubic meters, etc. Let one cow drink $b$ (conditional units) of water per day, and the springs add $c$ (conditional units) of water per day. Then the first condition of the problem is equivalent to ... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. Gavriila 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 m/s against the wind at an angle to the horizon. After 1 second, the ball hit him in the stomach 1 m below the point of release. Determine the gre... | 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 \cdo... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
6. To lift a load, it is attached to the hook of a crane using slings made of steel cable. The calculated mass of the load is $M=20$ t, the number of slings $n=3$. Each sling forms an angle $\alpha=30^{\circ}$ with the vertical. All slings carry the same load during the lifting of the cargo. According to safety require... | Answer: 26 mm
Solution. For each of the $n$ lower tie-downs, the force of the cargo weight is $\frac{P}{n}$. Then the tension force in the tie will be $N=\frac{P}{n \cdot \cos \alpha}$. Therefore, the strength of the rope must be $Q \geq k T=\frac{k P}{n \cdot \cos \alpha}$.
Since the strength of the rope $Q$ is dete... | 26 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. On the shores of a circular island (viewed from above), there are cities $A, B, C$, and $D$. The straight asphalt road $A C$ divides the island into two equal halves. The straight asphalt road $B D$ is shorter than road $A C$ and intersects it. The speed of a cyclist on any asphalt road is 15 km/h. The island also h... | Answer: 450 sq. km. Solution. The condition of the problem means that a quadrilateral $ABCD$ is given, in which angles $B$ and $D$ are right (they rest on the diameter), $AB = BC$ (both roads are dirt roads, and the cyclist travels them in the same amount of time), $BD = 15 \frac{\text{km}}{\text{hour}} \cdot 2$ hours ... | 450 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
4. One mole of a monatomic ideal gas undergoes a cyclic process $a b c a$. The diagram of this process in the $P-T$ axes represents a curvilinear triangle, the side $a b$ of which is parallel to the $T$ axis, the side $b c$ - a segment of a straight line passing through the origin, and the side $c a$ - an arc of a para... | Answer: 664 J. Solution. Process $a b$ is an isobar, process $b c$ is an isochore, process $c a$ is described by the equation $T=P(d-k P)$, where $d, k$ are some constants. It is not difficult to see that in such a process, the volume turns out to be a linear function of pressure, that is, in the $P V$ axes, this cycli... | 664 | Other | math-word-problem | Yes | Yes | olympiads | false |
5. For moving between points located hundreds of kilometers apart on the Earth's surface, people in the future will likely dig straight tunnels through which capsules will move without friction, solely under the influence of Earth's gravity. Let points $A, B$, and $C$ lie on the same meridian, and the distance from $A$... | Answer: 42 min. Solution. Let point $O$ be the center of the Earth. To estimate the time of motion from $A$ to B, consider triangle $A O B$. We can assume that the angle $\alpha=90^{\circ}-\angle A B O$ is very small, so $\sin \alpha \approx \alpha$. Since the point in the tunnel $A B$ is attracted to the center by the... | 42 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. Gavriila 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). | 2.1. Gavriila 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).
Answer. $\{24000\}$. | 24000 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
2.2. 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). | 2.2. 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).
Answer. $\{28800\}$. | 28800 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.3. 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). | 2.3. Gavriila 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 |
2.4. Gavriila found out that the front tires of the car last for 21,000 km, while the rear tires last for 28,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km). | 2.4. Gavriila found out that the front tires of the car last for 21000 km, while the rear tires last for 28000 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 |
3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabili... | 3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabili... | 34 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
4. On the shores of a circular island (viewed from above), there are cities $A, B, C$, and $D$. A straight asphalt road $A C$ divides the island into two equal halves. A straight asphalt road $B D$ is shorter than road $A C$ and intersects it. The speed of a cyclist on any asphalt road is 15 km/h. The island also has s... | Answer: 450 sq. km. Solution. The condition of the problem means that a quadrilateral $ABCD$ is given, in which angles $B$ and $D$ are right (they rest on the diameter), $AB=BC$ (both roads are dirt roads, and the cyclist travels them in the same amount of time), $BD=15 \frac{\text { km }}{\text { h }} 2$ hours $=30$ k... | 450 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
1. Scientists have found a fragment of an ancient manuscript on mechanics. It was a piece of a book, the first page of which was numbered 435, and the last page was written with the same digits but in some other order. How many sheets were in this fragment? | Solution. Since the sheet has 2 pages and the first page is odd, the last page must be even. Therefore, the last digit is 4. The number of the last page is greater than the first. The only possibility left is 534. This means there are 100 pages in total, and 50 sheets.
Answer: 50.
Criteria: 20 points - correct (not n... | 50 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. Usually, schoolboy Gavriil takes a minute to go up a moving escalator by standing on its step. But if Gavriil is late, he runs up the working escalator and thus saves 36 seconds. Today, there are many people at the escalator, and Gavriil decides to run up the adjacent non-working escalator. How much time will such a... | Solution. Let's take the length of the escalator as a unit. Let $V$ be the speed of the escalator, and $U$ be the speed of Gavrila relative to it. Then the condition of the problem can be written as:
$$
\left\{\begin{array}{c}
1=V \cdot 60 \\
1=(V+U) \cdot(60-36)
\end{array}\right.
$$
The required time is determined ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 40 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) 48 km; b) yes. Solution. Let $a=20$ m/s², $\tau=40$ s. On the first segment 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 working: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ - this will b... | 48 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
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 30 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) 54 km; b) yes. Solution. Let $a=30 \mathrm{m} / \mathrm{s}^{2}, \tau=30$ s. On the first segment 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: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ ... | 54 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 10 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 ... | Answer: 52 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 6 m, and the base is an isosceles right triangle with legs $2 \sqrt{2}$ and hy... | 52000 | Geometry | math-word-problem | Yes | Yes | olympiads | false |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
X=35 l, ... | 70 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
2. The mass of the first iron ball is $1462.5 \%$ greater than the mass of the second ball. By what percentage will less paint be needed to paint the second ball compared to the first? The volume of a sphere with radius $R$ is $\frac{4}{3} \pi R^{3}$, and the surface area of a sphere is $4 \pi R^{2}$. | 2. Answer: $84 \%$.
Let's denote the radii of the spheres as $R$ and $r$ respectively. Then the first condition means that
$$
\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 |
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 |
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 |
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:
$$
\begin{aligned}
& x(t)=V \cdot... | 75 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2.1. During the time it took for a slowly moving freight train to cover 1200 m, a schoolboy managed to ride his bicycle along the railway tracks from the end of the moving train to its beginning and back to the end. In doing so, the bicycle's distance meter showed that the cyclist had traveled 1800 m. Find the length o... | Answer: 500. Solution: Let $V$ and $U$ be the speeds of the cyclist and the train, respectively, and $h$ be the length of the train.
Then the conditions of the problem in mathematical terms can be written as follows:
$$
(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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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 |
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
$$
v\left(1-\frac{x}{100}\right)\left(1+\frac{0.5 x}{100}\right)=v\left(1-\frac{0.6 x}{100}\right) \Leftrightarrow\left(1-\frac{x}{100}\right)\left(1+\frac{x}{200}\right)=1-\frac{3 x}{500} \Leftrightarrow \frac{x^{2}}{20000}=\frac{... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Gavriil decided to weigh a football, but he only had weights of 150 g, a long light ruler with the markings at the ends worn off, a pencil, and many threads at his disposal. He suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second ... | 1. Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\le... | 600 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1.2.1 The time of the aircraft's run from the moment of start until 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 number if nece... | Solution. $v=a t, 100000 / 3600=a \cdot 15$, from which $a=1.85\left(\mathrm{~m} / \mathrm{s}^{2}\right)$. Then $S=a t^{2} / 2=208(\mathrm{m})$.)
Answer. 208 m | 208 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3.3.1 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 T... | Answer: 257 km.
Solution. Let's take the "capacity" of the smartphone battery as 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 ... | 257 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. A tractor is pulling a very long pipe on sled runners. Gavrila walked along the entire pipe at a constant speed in the direction of the tractor's movement and counted 210 steps. When he walked in the opposite direction at the same speed, the number of steps was 100. What is the length of the pipe if Gavrila's step i... | Answer: 108 m. 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 of $y$ (m). Then, if $m$ and $n$ are the number of steps Gavrila takes in one direction and the other, respectively, we get two equations: ${ }^{x=m(a-y)}, x=n(a+y)$. From th... | 108 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. Usually, schoolboy Gavriila takes a minute to go up a moving escalator by standing on its step. But if Gavriila is late, he runs up the working escalator and saves 36 seconds this way. Today, there are many people at the escalator, and Gavriila decides to run up the adjacent non-working escalator. How much time will... | Solution. Let's take the length of the escalator as a unit. Let $V$ be the speed of the escalator, and $U$ be Gavrila's speed relative to it. Then the condition of the problem can be written as:
$$
\left\{\begin{array}{c}
1=V \cdot 60 \\
1=(V+U) \cdot(60-36)
\end{array}\right.
$$
The required time is determined from ... | 40 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
1. The lieutenant is engaged in drill training with the new recruits. Upon arriving at the parade ground, he saw that all the recruits were lined up in several rows, with the number of soldiers in each row being the same and 5 more than the number of rows. After the training session, the lieutenant decided to line up t... | Solution. Let $n$ be the number of rows in the original formation. Then, there were originally $n+5$ soldiers in each row, and in the second formation, there were $n+9$ soldiers in each row. Let the age of the lieutenant be $x$. Then, according to the problem, we get the equation
$$
x=\frac{n(n+5)}{n+9} \Rightarrow x=... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. Among all six-digit natural numbers, the digits of which are arranged in ascending order (from left to right), numbers containing the digit 1 and not containing this digit are considered. Which numbers are more and by how many? | Solution. First, let's calculate how many six-digit natural numbers there are in total, with their digits arranged in ascending order. For this, we will write down all the digits from 1 to 9 in a row. To get six-digit numbers of the considered type, we need to strike out any three digits. Thus, the number of six-digit ... | 28 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
5. The Ivanovs' income as of the beginning of June:
$$
105000+52200+33345+9350+70000=269895 \text { rubles }
$$ | Answer: 269895 rubles
## Evaluation Criteria:
Maximum score - 20, if everything is solved absolutely correctly, the logic of calculations is observed, and the answer is recorded correctly.
## 20 points, including:
6 points - the final deposit amount is calculated correctly;
4 points - the size of the mother's sala... | 269895 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend
$$
4800 + 1500 - 4800 * 0.2 = 5340 \text{ rubles.}
$$
This is the most profitable... | # Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an allowable purchase is no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, you can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct ... | 84 | Other | math-word-problem | Yes | Yes | olympiads | false |
Task 14. (2 points)
Ivan opened a deposit in a bank for an amount of 100 thousand rubles. The bank is a participant in the state deposit insurance system. How much money will Ivan receive if the bank's license is revoked / the bank goes bankrupt?
a) Ivan will receive 100 thousand rubles and the interest that has been... | # Solution:
In accordance with the federal insurance law Federal Law No. 177-FZ of $23 \cdot 12.2003$ (as amended on 20.07.2020) "On Insurance of Deposits in Banks of the Russian Federation" (with amendments and additions, effective from 01.10.2020), compensation for deposits in a bank where an insurance case has occu... | 100 | Other | MCQ | Yes | Yes | olympiads | false |
4. Excursions (20,000 rubles for the whole family for the entire vacation).
The Seleznev family is planning their vacation in advance, so in January, the available funds for this purpose were calculated. It turned out that the family has 150,000 rubles at their disposal. Mr. Seleznev plans to set aside a certain amoun... | # Solution:
1) Calculate the vacation expenses
Flight expenses $=10200.00$ rubles * 2 flights * 3 people $=61200.00$
Hotel expenses $=6500$ rubles * 12 days $=78000.00$ rubles
Food expenses $=1000.00$ rubles * 14 days * 3 people $=42000.00$ rubles
Excursion expenses $=20000.00$ rubles
Total expenses $=201200.00$ ... | 47825 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
2. When insuring property, the insurance amount cannot exceed its actual value (insurance value) at the time of concluding the insurance contract.
Insurance tariff - the rate of the insurance premium or the insurance premium (insurance premium) expressed in rubles, payable per unit of the insurance amount, which is us... | # Solution:
In accordance with the instruction, the base rate is $0.2\%$ of the insurance amount, apply a reducing factor for the absence of a change in ownership over the past 3 years $(0.8)$ and an increasing factor for the absence of certificates from the PND and ND $(1.3)$.
In total: $0.2 * 0.8 * 1.3=0.208\%$
Th... | 31200 | Other | math-word-problem | Yes | Yes | olympiads | false |
3. Maria Ivanovna decided to use the services of an online clothing store and purchase summer clothing: trousers, a skirt, a jacket, and a blouse. Being a regular customer of this store, Maria Ivanovna received information about two ongoing promotions. The first promotion allows the customer to use an electronic coupon... | Solution:
(a) Maria Ivanovna can make one purchase, using only one of the promotions, or she can "split" the selected items into two purchases, using both promotions in this case.
Let's consider all possible options:
1) One purchase. In this case, Maria Ivanovna can save either 900 rubles by using the "third item fr... | 6265 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 11. (16 points)
The Dorokhov family plans to purchase a vacation package to Crimea. The family plans to travel with the mother, father, and their eldest daughter Polina, who is 5 years old. They carefully studied all the offers and chose the "Bristol" hotel. The head of the family approached two travel agencies, ... | # Solution:
Cost of the tour with the company "Globus"
$(3 * 25400) *(1-0.02)=74676$ rubles.
Cost of the tour with the company "Around the World"
$(11400+2 * 23500) * 1.01=58984$ rubles.
Answer: 58984 | 58984 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 12. (16 points)
The Vasilievs' family budget consists of the following income items:
- parents' salary after income tax deduction - 71000 rubles;
- income from renting out property - 11000 rubles;
- daughter's scholarship - 2600 rubles
The average monthly expenses of the family include:
- utility payments - 84... | # Solution:
family income
$71000+11000+2600=84600$ rubles
average monthly expenses
$8400+18000+3200+2200+18000=49800$ rubles
expenses for forming a financial safety cushion
$(84600-49800) * 0.1=3480$ rubles
the amount the Petrovs can save monthly for the upcoming vacation
$84600-49800-3480=31320$ rubles
## Ans... | 31320 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 13. (8 points)
Natalia Petrovna has returned from her vacation, which she spent traveling through countries in North America. She has a certain amount of money left in foreign currency.
Natalia Petrovna familiarized herself with the exchange rates at the nearest banks: "Rebirth" and "Garnet." She decided to take... | # Solution:
1) cost of currency at Bank "Vozrozhdenie":
$$
120 * 74.9 + 80 * 59.3 + 10 * 3.7 = 13769 \text{ RUB}
$$
2) cost of currency at Bank "Garant":
$$
120 * 74.5 + 80 * 60.1 + 10 * 3.6 = 13784 \text{ RUB}
$$
Answer: 13784 | 13784 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 14. (8 points)
To attend the section, Mikhail needs to purchase a tennis racket and a set of tennis balls. Official store websites have product catalogs. Mikhail studied the offers and compiled a list of stores where the items of interest are available:
| Item | Store | |
| :--- | :---: | :---: |
| | Higher Le... | # Solution:
1) cost of purchase in the store "Higher League":
$$
\text { 5600+254=5854 rub. }
$$
1) cost of purchase in the store "Sport-guru": $(2700+200)^{*} 0.95+400=6005$ rub.
## Answer: 5854 | 5854 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 15. (8 points)
The fast-food network "Wings and Legs" offers summer jobs to schoolchildren. The salary is 25000 rubles per month. Those who work well receive a monthly bonus of 5000 rubles.
How much will a schoolchild who works well at "Wings and Legs" earn per month (receive after tax) after the income tax is d... | Solution:
The total earnings will be 25000 rubles + 5000 rubles $=30000$ rubles
Income tax $13 \%-3900$ rubles
The net payment will be 30000 rubles - 3900 rubles $=26100$ rubles
## Correct answer: 26100
## 2nd Option | 26100 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 11. (16 points)
One way to save on utility bills is to use the night tariff (from 23:00 to 07:00). To apply this tariff, a multi-tariff meter needs to be installed.
The Romanov family is considering purchasing a multi-tariff meter to reduce their utility bills. The cost of the meter is 3500 rubles. The installat... | # Solution:
2) use of a multi-tariff meter:
$$
3500+1100+(230 * 3.4+(300-230) * 5.2) * 12 * 3=45856 \text { rub. }
$$
3) use of a typical meter
$$
300 * 4.6 * 12 * 3=49680 \text { rub. } \quad \text {. } \quad \text {. }
$$
the savings will be
$49680-45856=3824$ rub.
## Answer: 3824 | 3824 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 12. (16 points)
The budget of the Petrovs consists of the following income items:
- parents' salary after income tax deduction - 56000 rubles;
- grandmother's pension - 14300 rubles;
- son's scholarship - 2500 rubles
Average monthly expenses of the family include:
- utility payments - 9800 rubles;
- food - 210... | # Solution:
family income
$56000+14300+2500=72800$ rubles.
average monthly expenses
$9800+21000+3200+5200+15000=54200$ rubles.
expenses for forming a financial safety cushion
$(72800-54200) * 0.1=1860$ rubles.
the amount the Petrovs can save monthly for the upcoming vacation
$72800-54200-1860=16740$ rubles.
An... | 16740 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 13. (8 points)
Maxim Viktorovich returned from a trip to Asian countries. He has a certain amount of money in foreign currency left.
Maxim Viktorovich familiarized himself with the exchange rates at the nearest banks: "Voskhod" and "Elfa". He decided to take advantage of the most favorable offer. What amount wil... | # Solution:
1) cost of currency at "Voskhod" bank:
$110 * 11.7 + 80 * 72.1 + 50 * 9.7 = 7540$ rubles.
2) cost of currency at "Alpha" bank: $110 * 11.6 + 80 * 71.9 + 50 * 10.1 = 7533$ rubles.
Answer: 7540. | 7540 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 14. (8 points)
Elena decided to get a pet - a budgerigar. She faced the question of where to buy a cage and a bath more cost-effectively.
On the official websites of the stores, product catalogs are posted. Elena studied the offers and compiled a list of stores where the items she is interested in are available:... | # Solution:
2) cost of purchase in the "Zoimir" store: $4500+510=5010$ rubles
3) cost of purchase in the "Zooidea" store: $(3700+680) * 0.95+400=4561$ rubles
## Answer: 4561 | 4561 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
Task 15. (8 points)
Announcement: "Have free time and want to earn money? Write now and earn up to 2500 rubles a day working as a courier with the service 'Food.There-Here!'. Delivery of food from stores, cafes, and restaurants.
How much will a school student working as a courier with the service 'Food.There-Here!' e... | # Solution:
The total earnings will be (1250 rubles * 4 days) * 4 weeks = 20000 rubles
Income tax 13% - 2600 rubles
The amount of earnings (net pay) will be 20000 rubles - 2600 rubles = 17400 rubles
Correct answer: 17400 | 17400 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
5. There are several technologies for paying with bank cards: chip, magnetic stripe, paypass, cvc. Arrange the actions performed with a bank card in the order corresponding to the payment technologies.
1 - tap
2 - pay online
3 - swipe
4 - insert into terminal | Answer in the form of an integer, for example 1234.
Answer: 4312 | 4312 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (8 points)
Kolya's parents give him pocket money once a month, calculating it as follows: 100 rubles for each A in math, 50 rubles for a B, 50 rubles are deducted for a C, and 200 rubles are deducted for a D. If the amount turns out to be negative, Kolya simply gets nothing. The math teacher gives a grade... | # Solution:
If Kolya received a final grade of 2 for the quarter, then for each 5 he received more than 5 2s, for each 4 - more than 3 2s, and for each 3 - more than 1 2. This means that the number of 2s was greater than the total number of all other grades combined, so Kolya could receive money in at most one of the ... | 250 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 6. (10 points)
Vasily is planning to graduate from college in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known tha... | # Solution:
In 4 years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points).
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300 * $(1 / 5 * 60000 + 1 / 10 * 80$ $000 + 1 / 20 * 25000 + (1 -... | 45025 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Problem 1. (4 points)
In the run-up to the New Year, a fair is being held at the school where students exchange festive toys. As a result, the following exchange norms have been established:
1 Christmas tree ornament can be exchanged for 2 crackers, 5 sparklers can be exchanged for 2 garlands, and 4 Christmas tree ... | # Solution:
a) 10 sparklers $=4$ garlands = 16 ornaments = $\mathbf{32}$ crackers. (1 point)
b) Convert everything to crackers. In the first case, we have $\mathbf{11}$ crackers. In the second case, 2 sparklers $=4 / 5$ garlands $=16 / 5$ ornaments $=32 / 5$ crackers $=\mathbf{6.4}$ crackers. Answer: 5 ornaments and ... | 32 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
# Task 3. (8 points)
In the Sidorov family, there are 3 people: the father works as a programmer with an hourly rate of 1500 rubles. The mother works as a hairdresser at home and charges 1200 rubles per haircut, which takes her 1.5 hours. The son tutors in mathematics and earns 450 rubles per academic hour (45 minutes... | # Solution
In this problem, there are 2 possible interpretations, both of which were counted as correct.
In one case, it is assumed that 8 hours are spent on work on average over the month, in the other that no more than 8 hours are spent on work each day.
## First Case:
1) Determine the hourly wage for each family... | 19600 | Other | math-word-problem | Yes | Yes | olympiads | false |
# Problem 4. (10 points)
On December 31 at 16:35, Misha realized he had no New Year's gifts for his entire family. He wants to give different gifts to his mother, father, brother, and sister. Each of the gifts is available in 4 stores: Romashka, Odynachik, Nezabudka, and Lysichka, which close at 20:00. The journey fro... | # Solution:
Notice that in each of the stores, there is a "unique" gift with the lowest price.
If Misha managed to visit all 4 stores, he would spend the minimum amount of $980+750+900+800=3430$ rubles. However, visiting any three stores would take Misha at least $30 * 3+25+30+35=180$ minutes. Considering the additio... | 3435 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make t... | # Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an acceptable purchase would be no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, one can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and c... | 84 | Other | math-word-problem | Yes | Yes | olympiads | false |
2. A stationery store is running a promotion: there is a sticker on each notebook, and for every 5 stickers, a customer can get another notebook (also with a sticker). Fifth-grader Katya thinks she needs to buy as many notebooks as possible before the new semester. Each notebook costs 4 rubles, and Katya has 150 rubles... | Answer: 46.
1) Katya buys 37 notebooks for 148 rubles.
2) For 35 stickers, Katya receives 7 more notebooks, after which she has notebooks and 9 stickers.
3) For 5 stickers, Katya receives a notebook, after which she has 45 notebooks and 5 stickers.
4) For 5 stickers, Katya receives the last 46th notebook. | 46 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
3. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give the answer in rubles, rounding to the nearest whole number (the answer is a whole number). | Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997$. Since rounding was required, the answer is 24.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/1.html | 24 | Other | math-word-problem | Yes | Yes | olympiads | false |
5. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest numbe... | Answer: 15.
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let'... | 15 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 12. (6 points)
Victor received a large sum of money as a birthday gift in the amount of 45 thousand rubles. The young man decided to save this part of his savings in dollars on a currency deposit. The term of the deposit agreement was 2 years, with an interest rate of 4.7% per annum, compounded quarterly. On t... | Answer: 873 USD.
## Comment:
45000 RUB / 56.60 RUB $\times(1+4.7\% / 4 \text { quarters })^{2 \text { years } \times 4 \text { quarters }}=873$ USD. | 873 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
# Task 17-19. (2 points per Task)
Alena opened a multi-currency deposit at "Severny" Bank for 3 years. The deposit involves the deposit of funds in three currencies: euros, dollars, and rubles. At the beginning of the deposit agreement, Alena's account contained 3000 euros, 4000 dollars, and 240000 rubles. The interes... | Answer 17: $3280;
Answer 18: 4040 euros,
Answer 19: 301492 rubles.
## Comment:
1 year
Euros: 3000 euros $\times(1+2.1 \%)=3063$ euros.
Dollars: 4000 dollars $\times(1+2.1 \%)=4084$ dollars.
Rubles: 240000 rubles $\times(1+7.9 \%)=258960$ rubles.
2 year
Euros: (3063 euros - 1000 euros $) \times(1+2.1 \%)=2106$ ... | 3280 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 6. (4 points)
Ivan bought a used car from 2010 for 90,000 rubles with an engine power of 150 hp and registered it on January 29. On August 21 of the same year, the citizen sold his car and a month later bought a horse and a cart for 7,500 rubles. The transport tax rate is set at 20 rubles per 1 hp. What amount... | Solution: transport tax $=150 \times 20 \times 8 / 12=2000$ rubles. A horse and a cart are not subject to transport tax. | 2000 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 7. (6 points)
Sergei, being a student, worked part-time in a student cafe after classes for a year. Sergei's salary was 9000 rubles per month. In the same year, Sergei paid for his medical treatment at a healthcare facility in the amount of 100000 rubles and purchased medications on a doctor's prescription for... | Solution: the amount of the social tax deduction for medical treatment will be: $100000+20000=$ 120000 rubles. The possible tax amount eligible for refund under this deduction will be $120000 \times 13\% = 15600$ rubles. However, in the past year, Sergey paid income tax (NDFL) in the amount of $13\% \times (9000 \times... | 14040 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 8. (2 points)
After graduating from a technical university, Oleg started his own business producing water heaters. This year, Oleg plans to sell 5000 units of water heaters. Variable costs for production and sales of one water heater amount to 800 rubles, and total fixed costs are 1000 thousand rubles. Oleg wa... | Solution: the price of one kettle $=((1000000+0.8 \times 5000)+1500$ 000) / $(1000000$ + $0.8 \times 5000)=1300$ rub. | 1300 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 18. (4 points)
By producing and selling 4000 items at a price of 6250 rubles each, a budding businessman earned 2 million rubles in profit. Variable costs for one item amounted to 3750 rubles. By what percentage should the businessman reduce the production volume to make his revenue equal to the cost? (Provide... | Solution: fixed costs $=-2$ million, gap $+4000 \times 6250-3750 \times 4000$ million $=8$ million.
$6.25 \times Q=3750 Q+8 \text{ million}$.
$Q=3200$ units.
Can be taken out of production $=4000-3200=800$ units, that is, $20\%$. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make t... | # Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an allowable purchase is no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, you can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct ... | 84 | Other | math-word-problem | Yes | Yes | olympiads | false |
2. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give your answer in rubles, rounding to the nearest whole number. | Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997 \approx 24$.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/23.html | 24 | Other | math-word-problem | Yes | Yes | olympiads | false |
7. Alexey plans to buy one of two car brands: "A" for 900 thousand rubles or "B" for 600 thousand rubles. On average, Alexey drives 15 thousand km per year. The cost of gasoline is 40 rubles per liter. The cars consume the same type of gasoline. The car is planned to be used for 5 years, after which Alexey will be able... | Answer: 160000.
Use of car brand "A":
$900000+(15000 / 100) * 9 * 5 * 40+35000 * 5+25000 * 5-500000=970000$
Use of car brand "B":
$600000+(15000 / 100) * 10 * 5 * 40+32000 * 5+20000 * 5-350000=810000$
Difference: $970000-810000=160000$ | 160000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. A family of 4, consisting of a mom, a dad, and two children, has arrived in city $\mathrm{N}$ for 5 days. They plan to make 10 trips on the subway each day. What is the minimum amount they will have to spend on tickets, given the following tariffs in city $\mathrm{N}$?
| Adult ticket for one trip | 40 rubles |
| :-... | Answer: 5200. The family will spend this amount if the parents buy a three-day pass for themselves, and for the remaining two days, they will buy a one-day pass. For this, they will spend ($900 + 350 * 2$) * $2 = 3200$ rubles.
For the children, it is most cost-effective to buy single-trip tickets for all 5 days, spend... | 5200 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 9. (12 points)
Andrey lives near the market, and during the summer holidays, he often helped one of the traders lay out fruits on the counter early in the morning. For this, the trader provided Andrey with a $10 \%$ discount on his favorite apples. But autumn came, and the price of apples increased by $10 \%$.... | # Answer: 99.
## Comment
Solution: the new price of apples at the market is 55 rubles per kg, with a discount of $10 \%$ applied to this price. Thus, the price for 1 kg for Andrei will be 49.5 rubles, and for 2 kg Andrei will pay 99 rubles monthly. | 99 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Task 10. (12 points)
The Ivanov family owns an apartment with necessary property worth 3 million rubles, a car that is currently valued at 900 thousand rubles on the market, and savings, part of which, amounting to 300 thousand rubles, is placed in a bank deposit, part is invested in securities worth 200 thousand rubl... | # Answer: 2300000
## Comment
Solution: equity (net worth) = value of assets - value of liabilities. Value of assets $=3000000+900000+300000+200000+100000=$ 4500000 rubles. Value of liabilities $=1500000+500000+200000=2200000$ rubles. Net worth $=4500000-2200000=2300000$ rubles
## MOSCOW FINANCIAL LITERACY OLYMPIAD ... | 2300000 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. (8 points)
Konstantin's mother opened a foreign currency deposit in the "Western" Bank for an amount of 10 thousand dollars for a term of 1 year. Literally 4 months later, the Bank of Russia revoked the license of the "Western" Bank. The exchange rate on the date of the license revocation was 58 rubles 15 k... | # Answer: b.
## Comment
$10000 \times 58.15$ RUB $=581500$ RUB. | 581500 | Other | MCQ | Yes | Yes | olympiads | false |
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