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4. Let $f(x)=x^{3}+3 x^{2}+5 x+7$. Find the polynomial $g(x)$ of the smallest degree such that
$$
f(3)=g(3), \quad f(3-\sqrt{3})=g(3-\sqrt{3}), \quad f(3+\sqrt{3})=g(3+\sqrt{3}) .
$$ | Answer: $12 x^{2}-19 x+25$.
Solution. Consider the polynomial $h(x)=g(x)-f(x)$. It is equal to zero at points $x=3$ and $x=3 \pm \sqrt{3}$. Therefore, $h(x)$ is divisible by the polynomial
\[
\begin{gathered}
q(x)=(x-3)(x-(3+\sqrt{3}))(x-(3-\sqrt{3}))= \\
=(x-3)\left((x-3)^{2}-3\right)=(x-3)\left(x^{2}-6 x+6\right)=x... | 12x^{2}-19x+25 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,963 |
5. (10 points) Five identical non-ideal ammeters are connected as shown in the figure. Ideal power supply is connected to points $A$ and $B$. Determine the sum of the readings of all ammeters, given that the reading of the first ammeter $I_{1}=2 \mathrm{MA}$.
 A mathematical pendulum is suspended near a vertical wall and oscillates in a plane parallel to the wall. A thin nail is driven into the wall such that the midpoint of the pendulum string hits it every time the pendulum passes through the equilibrium position from right to left. Find the length of the st... | Answer: $2 m$
Solution. The period of oscillations of a mathematical pendulum of length $l$ is determined by the expression $T_{1}=2 \pi \sqrt{\frac{l}{g}}$.
And for a pendulum of length $\frac{l}{2}$, the period of oscillations is $T_{2}=2 \pi \sqrt{\frac{l}{2 g}}$.
From the condition of the problem, it follows tha... | 2\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,965 |
8. (10 points) There are two light springs of equal length but with different stiffnesses. The springs were placed vertically one on top of the other. A load of mass $m=100$ kg was placed on top. As a result, the construction was compressed by $x_{1}=12.5$ cm. Then the springs were placed side by side and the same load... | Answer: $k_{1}=400 \frac{\mathrm{H}}{\mathrm{M}}$ and $k_{2}=100 \frac{\mathrm{H}}{\mathrm{M}}$
Solution. In the first situation, the stiffness of the obtained structure is $k_{0}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}$. (2 points)
The equilibrium condition in this case is: $k_{0} x_{1}=m g$.
As a result: $\frac{k_{1} k_{2... | k_{1}=400\frac{\mathrm{H}}{\mathrm{M}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,967 |
2. Let
$$
\sqrt{25-a^{2}}-\sqrt{9-a^{2}}=3
$$
Calculate the value of the expression
$$
\sqrt{25-a^{2}}+\sqrt{9-a^{2}}
$$ | Answer: $16 / 3$.
Solution. Let
$$
\sqrt{25-a^{2}}+\sqrt{9-a^{2}}=x
$$
Multiplying this equality by the original one, we get $16=3 x$.
Evaluation. Full solution score: 11 points. | \frac{16}{3} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,968 |
3. Let $D$ be the midpoint of the hypotenuse $B C$ of the right triangle $A B C$. A point $M$ is chosen on the leg $A C$ such that $\angle A M B = \angle C M D$. Find the ratio $\frac{B M}{M D}$. | Answer: $2: 1$.
Solution. Let point $E$ be the intersection of rays $B A$ and $D M$.
Then angles $A M E$ and $C M D$ are equal as vertical angles. Therefore, angles $A M B$ and $A M E$ are ... | 2:1 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,969 |
4. Let $f(x)=x^{3}-3 x^{2}+5 x-7$. Find the polynomial $g(x)$ of the smallest degree such that
$$
f(2)=g(2), \quad f(2-\sqrt{2})=g(2-\sqrt{2}), \quad f(2+\sqrt{2})=g(2+\sqrt{2})
$$ | Answer: $3 x^{2}-5 x-3$.
Solution. Consider the polynomial $h(x)=g(x)-f(x)$. It is equal to zero at points $x=2$ and $x=2 \pm \sqrt{2}$. Therefore, $h(x)$ is divisible by the polynomial
\[
\begin{gathered}
q(x)=(x-2)(x-(2+\sqrt{2}))(x-(2-\sqrt{2}))= \\
=(x-2)\left((x-2)^{2}-2\right)=(x-2)\left(x^{2}-4 x+2\right)=x^{3... | 3x^{2}-5x-3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,970 |
6. (15 points) A mathematical pendulum is suspended near a vertical wall and oscillates in a plane parallel to the wall. A thin nail is driven into the wall such that the midpoint of the pendulum string hits it every time the pendulum passes through the equilibrium position from right to left. Find the period of oscill... | Answer: $\approx 2.4 s$
Solution. The period of oscillations of a mathematical pendulum of length $l$ is determined by the expression $T_{1}=2 \pi \sqrt{\frac{l}{g}}$.
And for a pendulum of length $\frac{l}{2}$, the period of oscillations is: $T_{2}=2 \pi \sqrt{\frac{l}{2 g}}$.
From the condition of the problem, it ... | 2.4\, | Other | math-word-problem | Yes | Yes | olympiads | false | 3,972 |
7. (15 points) In a cylindrical vessel, ice has frozen at the bottom. Its temperature is $0^{\circ} \mathrm{C}$. Above it, water is poured at the same temperature. The vessel is brought into a warm room. The dependence of the water level in the vessel on time is shown in the graph. Determine the initial masses of ice a... | Answer: mass of ice -675 g, mass of water - 1050 g
Solution. The change in the volume of water in the vessel $\Delta V=S \Delta h=15 \cdot 5=75 \mathrm{~cm}^{3}$. (2 points)
This change is the difference between the volumes of the initial ice and the water into which it has transformed. $\Delta V=V_{I}-V_{W}=\frac{m_... | mass\of\ice\-\675\,\mass\of\water\-\1050\ | Calculus | math-word-problem | Yes | Yes | olympiads | false | 3,973 |
8. (10 points) There are two light springs of equal length but with different stiffnesses. The springs were placed vertically one on top of the other. A load of mass \( m = 3 \) kg was placed on top. As a result, the construction was compressed by \( x_{1} = 40 \) cm. Then the springs were placed side by side and the s... | Answer: $k_{1}=300 \frac{H}{M}$ and $k_{2}=100 \frac{H}{M}$
Solution. In the first situation, the stiffness of the obtained structure is:
$k_{0}=\frac{k_{1} k_{2}}{k_{1}+k_{2}}$.
The equilibrium condition in this case is: $k_{0} x_{1}=m g$.
As a result: $\frac{k_{1} k_{2}}{k_{1}+k_{2}}=\frac{m g}{x_{1}}=75 \frac{H}... | k_{1}=300\frac{H}{M} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,974 |
1. A column of infantry stretched out to 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked $4 / 3$ km during this time. How far did the sergeant travel during this time? | Answer: $8 / 3$ km
Solution. Let the speed of the column be $x$ km/h, and the corporal rides $k$ times faster, i.e., at a speed of $k x$ km/h. Kim rode to the end of the column for $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction for $t_{2}=\frac{1}{k x+x}$ hours (meeting head-on). During thi... | \frac{8}{3} | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 3,975 |
2. The sequence $\left(a_{n}\right)$ is defined by the following relations: $a_{1}=1, a_{2}=2, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{2019}$. | Answer: 2020.
Solution. Let's list the first terms of the sequence:
$$
1,2,4,6,7,7,7,8,10,12,13,13,13,14,16,18,19,19,19,20,21,23, \ldots
$$
We can observe a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$... | 2020 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,976 |
3. Given triangle $A B C$. The lengths of its sides are known: $A B=B C=80, A C=96$.
The circle $Q_{1}$ is inscribed in triangle $A B C$. The circle $Q_{2}$ is tangent to $Q_{1}$ and the sid... | Answer: 1.5.
Solution. Let $r_{i}$ be the radius of the circle $Q_{i}(i=1,2,3)$. It is easy to find $r_{1}$.
Let $B D$ be the height. From triangle $A B D$, we find its length:
$$
B D=\sq... | 1.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,977 |
4. Two vertices of a square lie on the parabola $y=x^{2}$, and one of its sides lies on the line $y=2 x-17$. What is the area of the square? | Answer: 80 or 1280.
Solution. It is easy to check that the parabola and the line from the condition of the problem do not intersect.
The line $l$, parallel to the line $2 x-y-17=0$, is given by the equation of the form $2 x-y+b=0$, where $b$ is some constant. The distance between these lines $d=\frac{|b+17|}{\sqrt{5}... | 80or1280 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,978 |
5. (15 points) The figure shows the dependence of gas pressure on its density in two processes. It is known that in the process $1-2$, work $A_{1-2}$ was done on the gas. Determine the work done on the gas in the process $3-4$.
 A stone with a mass of $m=6002$ was thrown at an angle $\alpha$ to the horizontal with an initial velocity of $v_{0}=20 \mathrm{M} / \mathrm{s}$. In the presence of air resistance proportional to the stone's velocity (proportionality coefficient $k=0.1(\mathrm{H} \cdot \mathrm{s}) / \mathrm{m})$, the max... | Answer: $\approx 10.1 \frac{M}{c^{2}}$
Solution. The law of conservation of energy in this situation is as follows: $\frac{m v_{0}^{2}}{2}=\frac{m v^{2}}{2}+m g h+A$.
As a result, we find the velocity at the highest point:
$$
v_{\kappa}=\sqrt{v_{0}^{2}-2 g h-\frac{2 A}{m}}=\sqrt{20^{2}-2 \cdot 10 \cdot 10-\frac{2 \c... | 10.1\frac{M}{^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 3,980 |
1. The infantry column stretched out over 1 km. Sergeant Kim, riding out on a gyro-scooter from the end of the column, reached its beginning and returned to the end. The infantrymen walked 2 km 400 m during this time. How far did the sergeant travel during this time? | Answer: 3 km $600 \mathrm{~m}$
Solution. Let the speed of the column be $x$ km/h, and the sergeant travels $k$ times faster, i.e., at a speed of $k x$ km/h. To reach the end of the column, Kim traveled $t_{1}=\frac{1}{k x-x}$ hours (catching up), and in the opposite direction, $t_{2}=\frac{1}{k x+x}$ hours (meeting he... | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 3,982 |
2. The sequence $\left(a_{n}\right)$ is defined by the following relations: $a_{1}=1, a_{2}=3, a_{n}=a_{n-1}-a_{n-2}+n$ (for $n \geqslant 3$). Find $a_{1000}$. | Answer: 1002.
Solution. Let's list the first terms of the sequence:
$$
1,3,5,6,6,6,7,9,11,12,12,12,13,15,17,18,18,18,19,21, \ldots
$$
We can see a pattern: $a_{n+6}=a_{n}+6$. Let's prove it. We have
$a_{n+1}=a_{n}-a_{n-1}+n+1=\left(a_{n-1}-a_{n-2}+n\right)-a_{n-1}+n+1=-a_{n-2}+2 n+1$.
By substituting $n$ with $n+2... | 1002 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,983 |
4. Two vertices of a square lie on the parabola $y=x^{2}$, and one of its sides lies on the line $y=2 x-22$. What is the area of the square? | Answer: 180 or 980.
Solution. It is easy to check that the parabola and the line from the condition of the problem do not intersect.
The line $l$, parallel to the line $2 x-y-22=0$, is given by the equation of the form $2 x-y+b=0$, where $b$ is some constant. The distance between these lines $d=\frac{|b+22|}{\sqrt{5}... | 180or980 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 3,985 |
7. (15 points) A stone with a mass of $m=8002$ was thrown at an angle $\alpha$ to the horizon with an initial velocity of $v_{0}=20 \mathrm{M} / \mathrm{s}$. In the presence of air resistance proportional to the stone's velocity (proportionality coefficient $k=0.2(H \cdot s) / M)$, the maximum height reached by the sto... | Answer: $\approx 10.3 \frac{M}{c^{2}}$
Solution. The law of conservation of energy in this situation is as follows: $\frac{m v_{0}^{2}}{2}=\frac{m v^{2}}{2}+m g h+A$.
As a result, we find the velocity at the highest point:
$$
v_{\kappa}=\sqrt{v_{0}^{2}-2 g h-\frac{2 A}{m}}=\sqrt{20^{2}-2 \cdot 10 \cdot 10-\frac{2 \c... | 10.3\frac{M}{^{2}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 3,987 |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After reducing the entry fee, the number of spectators increased by $25 \%$, and the revenue increased by $12.5 \%$. What is the new price of the entry ticket after the price reduction? | Answer: 360
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 1.25 = 400 - 1.125$. From which $x = 360$. | 360 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,989 |
2. (17 points) A tourist travels from point $A$ to point $B$ in 1 hour 56 minutes. The route from $A$ to $B$ goes uphill first, then on flat ground, and finally downhill. What is the length of the road on flat ground if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat ground is 5 km/h, and the tota... | # Answer: 3
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $9-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{9-x-y}{6}=\frac{29}{15}$. After transformations, $5 x+2 y=26$. It is obvious that $x$ must be even and $x+y \leq 9$. The only solution is $x=4, y=3$. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,990 |
3. (16 points) A chess player played 40 chess games and scored 25 points (1 point for each win, -0.5 points for a draw, 0 points for a loss). Find the difference between the number of his wins and the number of his losses.
# | # Answer: 10
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(40-n-m)=25$. In the end, $n-m=10$. | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,991 |
5. (15 points) Find the ratio $m_{1} / m_{2}$ of two hanging spheres, given that the tensions in the upper and lower threads differ by a factor of two. | Answer: $\frac{m_{1}}{m_{2}}=1$
Solution. For the lower ball: $m_{2} g=T_{H}$. For the upper ball: $m_{1} g+T_{H}=m_{1} g+m_{2} g=T_{B}$. At the same time, $T_{B}=2 T_{H}$. We obtain that
$... | \frac{m_{1}}{m_{2}}=1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,993 |
6. (15 points) From a homogeneous straight rod, a piece of length $s=60 \mathrm{~cm}$ was cut. By how much did the center of gravity of the rod move as a result? | Answer: $30 \, \text{cm}$
Solution. The center of gravity of the original rod was located at a distance of $\frac{l}{2}$ from its end, where $l$ is the length of the rod. After cutting off a piece of the rod, its center will be at a distance of $\frac{l-s}{2}$ from the other end. Therefore, the center of gravity of th... | 30\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,994 |
1. (17 points) The price of an entry ticket to the stadium is 400 p. After increasing the entry fee, the number of spectators decreased by $20 \%$, but the revenue increased by $5 \%$. What is the new price of the entry ticket after the price increase? | Answer: 525
Solution. Let the number of viewers before the ticket price reduction be 1 person. Then the revenue was 400 rubles. Let $x$ rubles be the new ticket price. We get the equation $x \cdot 0.8 = 400 \cdot 1.05$. From which $x=525$. | 525 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,995 |
2. (17 points) A tourist travels from point $A$ to point $B$ in 2 hours and 14 minutes. The route from $A$ to $B$ goes uphill first, then on flat terrain, and finally downhill. What is the length of the uphill road if the tourist's speed downhill is 6 km/h, uphill is 4 km/h, and on flat terrain is 5 km/h, and the total... | Answer: 6
Solution. Let $x$ km be the distance the tourist walks uphill, $y$ km - on flat ground, then $10-x-y$ km - downhill. We get $\frac{x}{4}+\frac{y}{5}+\frac{10-x-y}{6}=\frac{67}{30}$. After transformations, $5 x+2 y=34$. It is obvious that $x$ must be even and $x+y \leq 10$. The only solution is $x=6, y=2$. | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,996 |
3. (16 points) A chess player played 42 games and scored 30 points (1 point for each win, -0.5 points for each draw, 0 points for each loss). Find the difference between the number of his wins and the number of his losses. | Answer: 18
Solution. Let the chess player have $n$ wins and $m$ losses. Then we get $n+0.5 \cdot(42-n-m)=30$. In the end, $n-m=18$. | 18 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,997 |
4. (20 points) An aluminum and a copper part have the same volume. The density of aluminum $\rho_{A}=2700 \mathrm{kg} / \mathrm{m}^{3}$, the density of copper $\rho_{M}=8900 \mathrm{kg} / \mathrm{m}^{3}$. Find the mass of copper, if it is known that the masses of the parts differ by $\Delta m=60 \mathrm{g}$. | Answer: 862
Solution. Volume of aluminum: $V=\frac{m_{M}-\Delta m}{\rho_{A}}$, volume of copper: $V=\frac{m_{M}}{\rho_{M}}$. We get: $\frac{m_{M}}{\rho_{M}}=\frac{m_{M}-\Delta m}{\rho_{A}} . \quad$ From this, the mass of aluminum: $m_{A}=\frac{\Delta m \cdot \rho_{M}}{\rho_{M}-\rho_{A}}=\frac{0.06 \cdot 8900}{8900-270... | 862 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,998 |
5. (15 points) Find the ratio $m_{1} / m_{2}$ of two hanging spheres, given that the tensions in the upper and lower threads differ by a factor of three. | Answer: $\frac{m_{1}}{m_{2}}=2$
Solution. For the lower ball: $m_{2} g=T_{H}$. For the upper ball:
$m_{1} g+T_{H}=m_{1} g+m_{2} g=T_{B} . \quad$ Meanwhile, $T_{B}=3 T_{H}$. We get that $m_{... | \frac{m_{1}}{m_{2}}=2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 3,999 |
6. (15 points) From a homogeneous straight rod, a piece of length $s=80 \mathrm{~cm}$ was cut. By how much did the center of gravity of the rod move as a result? | Answer: $40 \mathrm{~cm}$
Solution. The center of gravity of the original rod was located at a distance of $\frac{l}{2}$ from its end, where $l$ is the length of the rod. After cutting off a piece of the rod, its center will be at a distance of $\frac{l-s}{2}$ from the other end. Therefore, the center of gravity of th... | 40\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,000 |
1. A mowing crew mowed the entire meadow in two days. On the first day, half of the meadow and another 3 hectares were mowed, and on the second day, a third of the remaining area and another 6 hectares were mowed. What is the area of the meadow? | Answer: 24 hectares.
Solution. 6 hectares made up two-thirds of the remainder. Therefore, 9 hectares were mowed on the second day. Together with the 3 hectares from the first day, this totals 12 hectares, which constitute half of the meadow's area. Therefore, the total area of the meadow is 24 hectares.
Evaluation. 1... | 24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,001 |
2. In the expression $1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9=100$, replace the asterisks with arithmetic operation signs to obtain a correct equation. | Solution. $1+2+3+4+5+6+7+8 \cdot 9=100$.
Evaluation. 12 points for the correct solution. | 1+2+3+4+5+6+7+8\cdot9=100 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,002 |
# Problem №1 (10 points)
The distance between Ivan's house and his grandmother's house is 12 km. At exactly $12 \stackrel{00}{}$ Ivan left his house and started walking straight to his grandmother's house at a speed of 1 m/s. At $12^{\frac{30}{}}$ Ivan's parents called his grandmother, told her that Ivan was on his wa... | # Solution:
In half an hour, Ivan will walk:
$S=vt=1 \cdot 30 \cdot 60=1800$ meters.
That is, between him and Tuzik remains:
$12000-1800=10200$ meters, which Ivan and Tuzik will overcome in:
$t=\frac{10200}{1+9}=1020 s=17$ min.
That is, the moment when Tuzik reaches Ivan is $12^{47}$
# | 12^{47} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,004 |
# Problem №2 (15 points)
An empty glass jar is three times lighter than the same jar filled to the brim with milk. If it is known that the volume occupied by the glass is ten times less than the volume occupied by the milk, then compare the densities of glass and milk. The density of a material is defined as the ratio... | # Solution:
From the condition:
$m_{C T}=\frac{m_{C T}+m_{M}}{3}$, i.e. $m_{M}=2 m_{C T}$
(5 points)
$10 \cdot V_{C T}=V_{M}$
(5 points)
## Therefore, the density:
$\rho_{M}=\frac{m_{M}}{V_{M}}=\frac{2 m_{C T}}{10 V_{C T}}=0.2 \rho_{C T}$ | \rho_{M}=0.2\rho_{CT} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,005 |
1. (16 points) After walking one-fifth of the way from home to school, Olya realized she had forgotten her notebook. If she does not go back for it, she will arrive at school 6 minutes before the bell, but if she returns, she will be 2 minutes late. How much time (in minutes) does the journey to school take? | # Answer: 20 min
Solution. The extra $2 / 5$ of the journey takes 8 min. Therefore, the entire journey to school will take 20 min. | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,007 |
2. (17 points) Find the smallest root of the equation
$$
\sqrt{x+2}+2 \sqrt{x-1}+3 \sqrt{3 x-2}=10
$$ | Answer: 2
Solution. It is clear that 2 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,008 |
3. (17 points) In a $5 \times 5$ grid, 6 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 4200
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 5 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choo... | 4200 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,009 |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=500 \mathrm{M} / \mathrm{c}$, penetrates a massive board and exits with a speed of $v_{2}=200 \mathrm{M} / \mathrm{c}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 1050 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 500^{2}}{2}-\frac{0.01 \cdot 200^{2}}{2}=1050 \text { J. }
$$ | 1050 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,010 |
5. (15 points) A light ray falls at an angle $\alpha=30^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.5$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $120^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \... | 120 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,011 |
1. (16 points) Kolya, after walking a quarter of the way from home to school, realized he had forgotten his workbook. If he does not go back for it, he will arrive at school 5 minutes before the bell, but if he does go back, he will be one minute late. How much time (in minutes) does the journey to school take? | # Answer: 12 min
Solution. The extra $2 / 4$ of the journey takes 6 min. Therefore, the entire journey to school will take 12 min. | 12 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,013 |
2. (17 points) Find the largest root of the equation
$$
3 \sqrt{x-2}+2 \sqrt{2 x+3}+\sqrt{x+1}=11
$$ | Answer: 3
Solution. It is clear that 3 is a root of the equation. The function on the left side of the equation is increasing (as the sum of increasing functions). Therefore, there are no other roots. | 3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,014 |
3. (17 points) In a $4 \times 4$ grid, 5 crosses need to be placed such that each row and each column contains at least one cross. How many ways can this be done? | # Answer: 432
Solution. From the condition, it follows that in some row $a$ and some column $b$ there are two crosses (and in all other rows and columns - one each). Both row $a$ and column $b$ can be chosen in 4 ways. There are two possible cases.
First. At the intersection of $a$ and $b$, there is a cross. We choos... | 432 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,015 |
4. (20 points) A bullet with a mass of $m=10$ g, flying horizontally at a speed of $v_{1}=400 \mathrm{m} / \mathrm{s}$, penetrates a massive board and exits with a speed of $v_{2}=100 \mathrm{m} / \mathrm{s}$. Find the magnitude of the work done on the bullet by the resistance force of the board. | Answer: 750 J
Solution. From the law of conservation of energy, it follows that
$$
A=\frac{m v_{1}^{2}}{2}-\frac{m v_{2}^{2}}{2}=\frac{0.01 \cdot 400^{2}}{2}-\frac{0.01 \cdot 100^{2}}{2}=750 \text { J. }
$$ | 750 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,016 |
5. (15 points) A light ray falls at an angle $\alpha=60^{\circ}$ on the front surface of a parallel glass plate. The refractive index of the glass is $n=1.6$. By what angle is the ray, reflected from the back surface of the plate and exiting back through the front surface, deflected from the direction of the incident r... | Answer: $60^{\circ}$
Solution. The described situation is illustrated in the figure.
We need to find the angle between rays 1 and 2. It is clear that the required angle is $180^{\circ}-2 \a... | 60 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,017 |
4. Solve the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}=1 \\
x^{3}+y^{5}=1
\end{array}\right.
$$ | Answer: $(0 ; 1),(1 ; 0)$.
Solution. Subtract the first equation from the second:
$$
x^{2}(x-1)+y^{2}\left(y^{3}-1\right)=0 .
$$
From the first equation of the system, it follows that $x \leqslant 1$ and $y \leqslant 1$. Therefore, $x^{2}(x-1) \leqslant 0$ and $y^{2}\left(y^{3}-1\right) \leqslant 0$. The sum of two ... | (0;1),(1;0) | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,019 |
# Problem №1 (15 points)
Two identical cars are driving in the same direction. The speed of one is $36 \kappa \mu / h$, and the other is catching up at a speed of $54 \mathrm{km} / h$. It is known that the reaction time of the driver of the rear car to the activation of the brake lights of the front car is 2 seconds. ... | # Solution:
Convert all speeds to SI units:
$36 \kappa m / h = 10 m / s$
$54 \kappa m / h = 15 m / s$
$72 \kappa m / h = 20 m / s$
The acceleration of a car of this make:
$a=\frac{v^{2}}{2 S}=\frac{20^{2}}{2 \cdot 40}=5 \mathrm{M} / s^{2}$
Distance traveled by the first car before stopping:
$S_{1}=\frac{v_{2}^{... | 42.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,020 |
# Problem №2 (15 points)
A bar with a mass of $m=1$ kg is pressed against a vertical wall. The coefficient of friction between the bar and the wall is $\mu=0.1$. Determine the force that needs to be applied to the bar at an angle of $\alpha=60^{\circ}$ to the wall to keep the bar in equilibrium. | # Solution:
We need to consider two situations:
## First:
Newton's second law in projections on the axes:
$F \sin 60^{\circ}=N$
$F \cos 60^{\circ}=m g+F_{T P}$
Taking into account that $... | 17.05H\leqF\leq24.18H | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,021 |
# Problem №3 (10 points)
The circuit diagram shown is composed of identical resistors. If point $A$ is connected to point $C$, and point $B$ is connected to point $D$, the resistance of the circuit changes by 10 Ohms. Determine the resistance of one resistor. The connections between points were made using zero-resista... | # Solution:
If the resistance of one resistor is $R$, then the initial total resistance of the series connection is $3 R$
After the jumpers were activated, we get a parallel connection of three resistors.
In this case, their total resistance is $\frac{R}{3}$
According to the condition: $3 R-\frac{R}{3}=10$
Final r... | 3.75 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,022 |
# Problem №4 (10 points)
A parallel beam of light falls on the base of a glass cone (refractive index $n=1.5$) along its axis (see fig.). The cross-section of the beam coincides with the base of the cone, the radius
$\operatorname{tg} \alpha=\frac{h}{R}=1.73$, i.e., $\alpha=60^{\circ}$
The law of refraction for rays passing through the cone:
$\frac{\sin \alpha}{\sin \beta}=\frac{1}{1.5}$
A... | 34 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,023 |
1. (17 points) Solve the equation $12 x=\sqrt{36+x^{2}}\left(6+x-\sqrt{36+x^{2}}\right)$. | Answer: $-6 ; 0$.
Solution. Transform the equation to the form $x^{2}+12 x+36=\sqrt{36+x^{2}}(6+x)$. After completing the square and factoring, we get the equation
$$
(x+6)\left(x+6-\sqrt{36+x^{2}}\right)=0 \Leftrightarrow\left[\begin{array}{c}
x+6=0 \\
x+6=\sqrt{36+x^{2}}
\end{array}\right.
$$
from the obtained equ... | -6,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,024 |
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take... | # Answer: 6.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars m... | 6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,025 |
3. (17 points) In triangle $A B C \quad A B=4, B C=6$, angle $A B C$ is $30^{\circ}, B D-$ is the bisector of triangle $A B C$. Find the area of triangle $A B D$. | Answer: 2.4.
Solution. By the property of the angle bisector $A D: D C=4: 6=2: 3$. Therefore, $A D=\frac{2}{5}$. $A C$. The area $S_{A B D}=\frac{2}{5} S_{A B C}$, since triangles $A B C$ and $A B D$ have a common height.
 Two heaters are connected sequentially to the same DC power source. The water in the pot boiled after $t_{1}=120$ s from the first heater. The same water, taken at the same initial temperature, boiled after $t_{2}=180 \mathrm{s}$ from the second heater. How long would it take for the water to boil if the... | Answer: 72 s.
Solution. The amount of heat required for heating in the first case $Q=I^{2} R_{1} t_{1}$. The amount of heat required for heating in the second case $Q=I^{2} R_{2} t_{2}$. The amount of heat required for heating in the case of parallel connection $Q=I^{2} \frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} t$. As a r... | 72\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,027 |
5. (20 points) A car brakes to a complete stop. It is known that the speed at the midpoint of the distance was 40 km/h. Determine its initial speed. | Answer: 56.6 km/h.
Solution. The equations of motion for each half of the braking distance $\frac{s}{2}=\frac{v_{0}^{2}-v^{2}}{2 a}$, and $\frac{s}{2}=\frac{v^{2}-0^{2}}{2 a}$. We obtain $v_{0}=\sqrt{2} v=56.6$ km/h. | 56.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,028 |
6. (15 points) Three material points with masses $m_{1}=100$ g, $m_{2}=200$ g, and $m_{3}=400$ g are arranged sequentially on a straight line. The distance between points 1 and 2 is 50 cm. The distance between points 2 and 3 is 200 cm. Determine the distance from point 1 to the center of mass of this system of three po... | Answer: $1.57 \mathrm{~m}$.
Solution. By the definition of the center of mass $\boldsymbol{x}_{\text {center }}=\frac{m_{1} \cdot 0+m_{2} \cdot 0.5+m_{3} \cdot 2.5}{m_{1}+m_{2}+m_{3}}=1.57 \mathrm{M}$.
## Tasks, answers, and assessment criteria | 1.57\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,029 |
1. (17 points) Solve the equation $10 x=\sqrt{25+x^{2}}\left(5+x-\sqrt{25+x^{2}}\right)$. | Answer: $-5 ; 0$.
Solution. Transform the equation to the form $x^{2}+10 x+25=\sqrt{25+x^{2}}(5+x)$.
After completing the square and factoring, we get the equation $\quad(x+5)\left(x+5-\sqrt{25+x^{2}}\right)=0 \Leftrightarrow\left[\begin{array}{c}x+5=0, \\ x+5=\sqrt{25+x^{2}} .\end{array}\right.$ From the obtained eq... | -5,0 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,030 |
2. (16 points) A truck left the village of Mirny at a speed of 40 km/h. At the same time, a car left the city of Tikhaya in the same direction as the truck. In the first hour of the journey, the car traveled 50 km, and in each subsequent hour, it traveled 5 km more than in the previous hour. How many hours will it take... | # Answer: 7.
Solution. In the first hour of travel, the pursuit speed of the car relative to the truck was $50-40=10$ km/h. In each subsequent hour, the pursuit speed increases by 5 km/h. Thus, the pursuit speeds form an arithmetic progression $10,15,20 \ldots$ km/h. Let's find the number of hours $n$ until the cars m... | 7 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,031 |
3. (17 points) In triangle $ABC$, $AB=6$, $BC=9$, angle $ABC$ is $30^{\circ}$, and $BD$ is the bisector of triangle $ABC$. Find the area of triangle $ABD$.
# | # Answer: 5.4.
Solution. By the property of the angle bisector $A D: D C=6: 9=2: 3$. Therefore, $A D=\frac{2}{5}$. $A C$. The area $S_{A B D}=\frac{2}{5} S_{A B C}$, since triangles $A B C$ and $A B D$ have a common height.
 Two heaters are connected sequentially to the same DC power source. The water in the pot boiled after $t_{1}=3$ minutes from the first heater. The same water, taken at the same initial temperature, boiled after $t_{2}=6$ minutes from the second heater. How long would it take for the water to boil if the ... | Answer: 2 min.
Solution. The amount of heat required for heating in the first case $Q=I^{2} R_{1} t_{1}$. The amount of heat required for heating in the second case $Q=$ $I^{2} R_{2} t_{2}$. The amount of heat required for heating in the case of parallel connection $Q=I^{2} \frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}} t$. As... | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,033 |
5. (20 points) A car brakes to a complete stop. It is known that the speed at the midpoint of the distance was 100 km/h. Determine its initial speed. | Answer: 141.1 km/h.
Solution. Equations of motion for each half of the braking distance: $\frac{s}{2}=\frac{v_{0}^{2}-v^{2}}{2 a}$, and $\frac{s}{2}=\frac{v^{2}-0^{2}}{2 a}$. We obtain $v_{0}=\sqrt{2} v=141.4$ km/h. | 141.1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,034 |
6. $\left(15\right.$ points) Three material points with masses $m_{1}=2$ kg, $m_{2}=3$ kg, and $m_{3}=4$ kg are arranged sequentially on a straight line. The distance between points 1 and 2 is $25 \mathrm{~cm}$. The distance between points 2 and 3 is 75 cm. Determine the distance from point 1 to the center of mass of t... | Answer: 52.8 cm.
Solution. By the definition of the center of mass $x_{\text {center }}=\frac{m_{1} \cdot 0+m_{2} \cdot 25+m_{3} \cdot 100}{m_{1}+m_{2}+m_{3}}=52.8$ cm. | 52.8 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,035 |
1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 50 m ahead of Boris at the finish line. And Boris was 40 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished? | Answer: 88 m.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c$ m/s, respectively. From the condition, it follows that $b=0.95 a, c=0.96 b$. Therefore, $c=0.96 \cdot 0.95 a=0.912 a$. This means that when Andrey runs 1000 m, Valentin will cover 912 m. The lag will be 88 m.
Evaluation. 12 point... | 88 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,036 |
2. A new math teacher came to the class. He conducted a survey among the students of this class, asking if they love math. It turned out that $50 \%$ love math, and $50 \%$ do not. The teacher conducted the same survey at the end of the school year. This time, $70 \%$ of the students answered "yes," and $30 \%$ answere... | Answer: $20 ; 80$.
Solution. Without loss of generality, we can assume that there are 100 students in the class. Let
- $a$ students answered "yes" both times,
- $b$ students answered "no" both times,
- $c$ students changed their answer from "no" to "yes",
- $d$ students changed their answer from "yes" to "no".
We ne... | 20;80 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,037 |
4. In a chess tournament, two girls and several boys participated. Each participant played exactly one game with each other. The two girls together scored 8 points, and all the boys scored an equal number of points. How many boys could have participated in the tournament? (1 point is awarded for a win, $\frac{1}{2}$ po... | Answer: 7 or 14.
Solution. Let $n$ be the number of boys participating in the tournament, and each of them scored $k$ points. Then all the players together scored $8 + kn$ points. On the other hand, there were $n + 2$ participants in the tournament, and they played $\frac{(n+2)(n+1)}{2}$ matches. In each match, exactl... | 7or14 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,039 |
6. The construction shown in the figure is in equilibrium. It is known that the mass of the load $m_{1}=1$ kg, the length of the uniform rod $l=50 \mathrm{~cm}$. The distance between the attachment points of the left thread to the rod $S=10$ cm. Determine the mass $m_{2}$ of the rod. All threads are weightless and inex... | Answer: 0.2 kg
Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread $T_{n}=\frac{m_{1} g}{2}$ (5 points). The equilibrium condition for the rod relative to the point of attachment of the right thread: $T_{n} \cdot l=T_{n} \cdot(l-S)+m_{2} g \cdot \frac{1}{2... | 0.2 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,040 |
7. Two identical resistors with resistance $R$ each are connected in series and connected to a source of constant voltage $U$. An ideal voltmeter is connected in parallel to one of the resistors. Its readings were $U_{v}=10 B$. After that, the voltmeter was replaced with an ideal ammeter. The ammeter readings were $-I_... | Answer: 2 Ohms
Solution. Voltage of the source: $U=U_{v}+U_{v}=20 V$ (4 points). The resistance of an ideal ammeter: $r_{A}=0$ Ohms (3 points). Therefore, the resistance of the resistor: $R=\frac{U}{I_{A}}=\frac{20}{10}=2$ Ohms (3 points). | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,041 |
8. An aluminum cube with edge length $l=10$ cm was heated to a temperature of $t_{1}=100^{\circ} \mathrm{C}$. After that, it was placed on ice, which has a temperature of $t_{2}=0^{\circ} \mathrm{C}$. Determine the maximum depth to which the cube can sink. The specific heat capacity of aluminum $c_{a}=900$ J $/$ kg ${ ... | Answer: 0.081 m
Solution. The mass of the cube: $m_{a}=\rho_{a} V=\rho_{a} l^{3}$ (2 points). The thermal balance equation for the cube and ice: $c_{a} m_{a} \Delta T=\lambda m_{n}$ (2 points). The mass of the melted ice: $m_{n}=\rho_{a} V_{n}=\rho_{s} S h=\rho_{a} l^{2} h$ (3 points), where $h$ is the maximum depth t... | 0.081 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,042 |
1. Andrei, Boris, and Valentin participated in a 1 km race. (We assume that each of them ran at a constant speed). Andrei was 60 m ahead of Boris at the finish line. And Boris was 50 m ahead of Valentin at the finish line. What was the distance between Andrei and Valentin at the moment Andrei finished? | Answer: $107 \mathrm{m}$.
Solution. Let the speeds of Andrey, Boris, and Valentin be $a, b$, and $c$ m/s, respectively. From the condition, it follows that $b=0.94 a, c=0.95 b$. Therefore, $c=0.94 \cdot 0.95 a=0.893 a$. This means that when Andrey runs 1000 m, Valentin will cover 893 m. The lag will be 107 m.
Evaluat... | 107\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,043 |
2. A new math teacher came to the class. He conducted a survey among the students of this class, asking if they love math. It turned out that $50 \%$ love math, and $50 \%$ do not. The teacher conducted the same survey at the end of the school year. This time, $60 \%$ of the students answered "yes," and $40 \%$ answere... | Answer: $10 ; 90$.
Solution. Without loss of generality, we can assume that there are 100 students in the class. Let
- $a$ students answered "yes" both times,
- $b$ students answered "no" both times,
- $c$ students changed their answer from "no" to "yes",
- $d$ students changed their answer from "yes" to "no".
We ne... | 10;90 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,044 |
4. In a chess tournament, two girls and several boys participated. Each participant played exactly one game with each other. The two girls together scored 6 points, and all the boys scored an equal number of points. How many boys could have participated in the tournament? (1 point is awarded for a win, $\frac{1}{2}$ po... | Answer: 5 or 10.
Solution. Let $n$ be the number of boys who participated in the tournament, and each of them scored $k$ points. Then all the players together scored $6 + k n$ points. On the other hand, there were $n + 2$ participants in the tournament, and they played $\frac{(n+2)(n+1)}{2}$ matches. In each match, ex... | 5or10 | Combinatorics | math-word-problem | Yes | Yes | olympiads | false | 4,046 |
5. A small load is suspended in the air by a spring. When this load is fully submerged in kerosene on the same spring, the magnitude of the spring's deformation remains the same. Determine the density of the material of the load. The density of kerosene $\rho=800 \mathrm{k} / \mathrm{m}^{3} .(15$ points $)$ | Answer: $400 \kappa g / \mathrm{M}^{3}$
Solution. In the air, the spring is stretched: $k x=m g$ (4 points). In kerosene, the spring is compressed: $m g+k x=F_{A}=\rho_{\text {kero }} g V$ (4 points). Since the mass of the load $m=\rho_{2} V$ (3 points), we get: $\rho_{2}=\frac{\rho_{\kappa \kappa}}{2}=\frac{800}{2}=4... | 400/^3 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,047 |
6. The structure shown in the figure is in equilibrium. It is known that the length of the homogeneous rod $l=50 \mathrm{~cm}$, and its mass $m_{2}=2$ kg. The distance between the attachment points of the left thread to the rod $S=10$ cm. Determine the mass $m_{1}$ of the load. All threads are weightless and inextensib... | Answer: 10 kg
Solution. From the equilibrium condition for the large block, it follows that the tension force in the left thread: $T_{n}=\frac{m g}{2}$ (5 points). The equilibrium condition for the rod relative to the attachment point of the right thread: $T_{n} \cdot l=T_{n} \cdot(l-S)+m_{2} g \cdot \frac{1}{2} l$ (5... | 10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,048 |
7. Two identical resistors with resistance $R$ each are connected in series and connected to a source of constant voltage $U$. An ideal voltmeter is connected in parallel to one of the resistors. Its readings were $U_{v}=15 B$. After that, the voltmeter was replaced with an ideal ammeter. The ammeter readings were $-I_... | Answer: 1.5 Ohms
Solution. Voltage of the source: $U=U_{v}+U_{v}=30 V$ (4 points). The resistance of an ideal ammeter: $r_{A}=0$ Ohms (3 points). Therefore, the resistance of the resistor: $R=\frac{U}{I_{A}}=\frac{30}{20}=1.5$ Ohms (3 points). | 1.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,049 |
8. A copper cube with edge length $l=5 \mathrm{~cm}$ was heated to a temperature of $t_{1}=100^{\circ} \mathrm{C}$. After that, it was placed on ice, which has a temperature of $t_{2}=0^{\circ} \mathrm{C}$. Determine the maximum depth to which the cube can sink. The specific heat capacity of copper $c_{\text {s }}=400$... | Answer: $0.06 . m$
Solution. The mass of the cube: $m_{n}=\rho_{w} V=\rho_{n} l^{3}$ (2 points). The thermal balance equation for the cube and ice: $c_{m} m_{»} \Delta T=\lambda m_{n}$ (2 points). The mass of the melted ice: $m_{n}=\rho_{a} V_{n}=\rho_{s} S h=\rho_{a} l^{2} h$ (3 points), where $h$ is the maximum dept... | 0.06\, | Other | math-word-problem | Yes | Yes | olympiads | false | 4,050 |
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 353 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 32 was added and a four-digit number was obtained, which is also a palindrome. Find $x$. | # Answer: 969.
Solution. Since $x+32$ is a four-digit number, the inequality $x+32 \geqslant 1000$ holds, from which $x \geqslant 968$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 6. By checking the four possible op... | 969 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,051 |
2. Anton, Boris, Vasya, and Grisha met. It is known that each of them is either from the tribe of knights (who always tell the truth) or from the tribe of liars (who always lie). Anton said that he and Grisha are from different tribes. Boris and Vasya called each other liars. And Grisha claimed that there are at least ... | Answer: One knight.
Solution. If Borya and Vasya were from the same tribe, they would have called each other knights. This did not happen. Therefore, one of them is a knight and the other is a liar. If Anton were right, there would be exactly two knights. In this case, Grisha would also be telling the truth, which con... | 1 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,052 |
3. Znayka knows that any triangle can be cut into 4 equal triangles. But does there exist a quadrilateral that can be cut into 5 equal triangles? | Answer: Yes.
Solution. One of the possible options is shown in the figure.
Grading. 13 points for a correct example. | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,053 |
4. In the cells of a $3 \times 3$ square, the numbers $1,2,3, \ldots, 9$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is $18?$
# | # Answer: 7.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that in cells of different colors, the numbers have different parities. Since there are five black cells and four white ones, we get that the odd numbers... | 7 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,054 |
5. Two cars are driving towards a single intersection on two mutually perpendicular roads at constant speeds. At the initial moment, the first car was at a distance of $s_{1}=500 \mathrm{M}$ from the intersection, and for the second car, the analogous distance was $S_{2}=700 \mathrm{M}$. The speed of the first car is $... | Answer: $12 \mathrm{~m} / \mathrm{c}$ or $16 \mathrm{~m} / \mathrm{c}$
## Solution.
The time of the first car's movement to the intersection $\quad t_{1}=\frac{s_{1}}{v_{1}}=\frac{500}{10}=50 \mathrm{c} . \quad$ (3 points) There are two possible scenarios. The second car has already passed the intersection or has not... | 12\mathrm{~}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,055 |
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $1.5 m$, width - $400 mm$, height - 80 cm. It is filled with water at a rate of 2 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points) | Answer: 14400 s
Solution. Volume of the aquarium $\quad V=1.5 \cdot 0.4 \cdot 0.8=0.48 \mathrm{~m}^{3}$.
Filling rate $v=\frac{0.002 \mu^{3}}{60 s}$ (3 points). The aquarium will be completely filled in time $\quad t=\frac{V}{v}=\frac{0.48 \cdot 60}{0.002}=14400 ~ s \quad(4$ points $)$. | 14400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,056 |
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{s}$ and $v_{2}=4 \mathrm{~m} / \mathrm{s}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who turn... | Answer: 20 min
Solution. Let $S$ be the distance between points A and B. We get $S=v_{1} t_{1}+v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{2} t_{1}+v_{2} t_{1}+v_{2} t_{2}$ (5 points). As a result, we get $t_{1}=\frac{v_{2} t_{2}}{v_{1}-v_{2}}=\frac{4 \cdot 10}{6-4}=20 \t... | 20 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,057 |
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=4$ m. It realized it was all too much and turned back. It crawled $l_{2}=3$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=40$ m. On which day of its journey will... | Answer: on the 37th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=4$ m (i.e., one transition) from its goal, it will be after 36 days of travel (4 points). Therefore, the goal will be reached on the 37th day of travel (3 points).... | 37 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,058 |
1. A palindrome is a number that reads the same from left to right and from right to left. For example, the numbers 333 and $4884-$ are palindromes. It is known that a three-digit number $x$ is a palindrome. To it, 22 was added and a four-digit number was obtained, which is also a palindrome. Find $x$. | # Answer: 979.
Solution. Since $x+22$ is a four-digit number, the inequality $x+22 \geqslant 1000$ holds, from which $x \geqslant 978$. From the fact that $x$ is a three-digit palindrome, we get that this number starts and ends with the digit 9, and its middle digit is not less than 7. By checking the four possible op... | 979 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,059 |
2. Anton, Boris, Vasya, and Grisha met. It is known that each of them is either from the tribe of knights (who always tell the truth) or from the tribe of liars (who always lie). Anton said that he and Grisha are from the same tribe. Boris and Vasya called each other knights. And Grisha claimed that there are no more t... | Answer: Borya is a liar.
Solution. Whoever Anton might be, from his statement it follows that Grisha is a knight. The statements of Borya and Vasya mean that they are from the same tribe. Since Grisha is truthful, Borya and Vasya cannot be knights. Therefore, they are liars.
Grading. Full solution: 12 points. If only... | Borya | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,060 |
3. Znayka knows that any triangle can be cut into 4 equal triangles. But does there exist a quadrilateral that can be cut into 7 equal triangles? | Answer: Yes.
Solution. One of the possible options is shown in the figure.
Grading. 13 points for a correct example. | Yes | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,061 |
4. In the cells of a $3 \times 3$ square, the numbers $0,1,2, \ldots, 8$ are arranged. It is known that any two consecutive numbers are located in adjacent (by side) cells. Which number can be in the central cell if the sum of the numbers in the corner cells is 18? | Answer: 2.
Solution. Let's color the cells in a checkerboard pattern: let the corner and central cells be black, and the rest white. From the condition, it follows that cells of different colors contain numbers of different parity. Since there are five black cells and four white cells, we get that the even numbers are... | 2 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false | 4,062 |
5. Two cars are driving towards a single intersection on two mutually perpendicular roads at constant speeds. At the initial moment, the first car was at a distance of $s_{1}=1600$ m from the intersection, and for the second car, the analogous distance was $s_{2}=800 \mathrm{M}$. The speed of the first car is $v_{1}=72... | Answer: 7.5 m/s or $12.5 \mathrm{~m} / \mathrm{s}$
## Solution.
The time of the first car's movement to the intersection $t_{1}=\frac{S_{1}}{v_{1}}=\frac{1600}{20}=80 \mathrm{s}$
There are two possible scenarios. The second car has already passed the intersection or has not yet reached it. Therefore, the distance it... | 7.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,063 |
6. An aquarium in the shape of a rectangular parallelepiped has dimensions: length $2 m$, width - 600 mm, height - 60 cm. It is filled with water at a rate of 3 liters/minute. How many seconds after the start of filling will the aquarium be completely filled? (10 points) | Answer: 14400 s
Solution. Volume of the aquarium: $V=2 \cdot 0.6 \cdot 0.6=0.72 m^{3}$ (3 points). Filling rate: $v=\frac{0.003 m^{3}}{60 s}$ (3 points). The aquarium will be completely filled in: $t=\frac{V}{v}=\frac{0.72 \cdot 60}{0.003}=14400 s$ (4 points). | 14400 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,064 |
7. Two people are walking towards each other from points $A$ and $B$ with speeds $v_{1}=6 \mathrm{~m} / \mathrm{c}$ and $v_{2}=4 \mathrm{~m} / \mathrm{c}$. At the moment of their meeting, one of them turned around and walked in the opposite direction, while the other continued in the same direction. The person who did ... | Answer: 30 min
Solution. The distance from point $B$ to the meeting place $S=v_{2} t_{1}$ (5 points). In addition, the same distance can be expressed as follows: $S=v_{1}\left(t_{1}-t_{2}\right)$ (5 points). As a result, we get: $t_{1}=\frac{v_{1} t_{2}}{v_{1}-v_{2}}=\frac{6 \cdot 10}{6-4}=30$ min (5 points). | 30 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,065 |
8. A snail is crawling from one tree to another. In half a day, it crawled $l_{1}=5$ m. It realized it was all too much and turned back. It crawled $l_{2}=4$ m. It got tired. It fell asleep. The next day, everything repeated. And so every day. The distance between the trees is $s=30$ m. On which day of its journey will... | Answer: on the 26th day
Solution. In one day, the snail advances towards the other tree by:
$\Delta l=l_{1}-l_{2}=1$ m (3 points). At a distance of $l_{1}=5$ m (i.e., one transition) from its goal, it will be after 25 days of travel (4 points). Therefore, the goal will be reached on the 26th day of travel. (3 points)... | 26 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,066 |
3. A triangle $ABC$ with side lengths $AB=6 \text{~cm}, BC=5 \text{~cm}, CA=4 \text{~cm}$ has been cut out of paper. It was folded along a line so that vertex $C$ ended up at point $C_{1}$ on side $AB$. Moreover, in the resulting quadrilateral $AKMB$, two angles adjacent to the fold line $KM$ turned out to be equal. Fi... | Answer: $\frac{10}{3}$ cm and $\frac{8}{3}$.
Solution. Points $C$ and $C_{1}$ are symmetric with respect to the fold line. Therefore, $C C_{1} \perp K M$. From the equality of angles $A K M$ and $K M B$, it follows that the adjacent angles are also equal. Therefore, triangle $C K M$ is isosceles. Its height, lying on ... | \frac{10}{3} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,067 |
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2018. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$. | Answer: 673.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$$
2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4036+b+d \geqslant 4038 ; \qu... | 673 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,068 |
5. A small ball was released without initial speed from a height of $h=20 \text{ m}$. The impact with the horizontal surface of the Earth is perfectly elastic. Determine at what moment of time after the start of the fall the average path speed of the ball will be equal to its instantaneous speed. The acceleration due t... | Answer: $2.83 s$
Solution. Obviously, while the ball is flying down, the instantaneous speed will be greater than the average path speed. The moment of time when the ball hits the ground: $t_{1}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \cdot 20}{10}}=2 s$ (2 points). The corresponding speed: $v_{1}=g t_{1}=10 \cdot 2=20 m /... | 2.83 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,069 |
6. Thirteen identical metal rods are connected as follows (see fig.). It is known that the resistance of one rod \( R_{0}=10 \) Ohms. Determine the resistance of the entire structure if it is connected to a current source at points \( A \) and \( B \). (10 points)
,
where each of the resistors has a resistance of $R_{0}=10$ Ohms. As a result, the total resistance is: $R=\frac... | 4 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,070 |
7. The specific heat capacity of a body with a mass of $m=2$ kg depends on the temperature as follows: $c=c_{0}(1+\alpha t)$, where $c_{0}=150$ J/kg ${ }^{\circ} \mathrm{C}-$ specific heat capacity at $0^{\circ} \mathrm{C}, \alpha=0.05^{\circ} \mathrm{C}^{-1}$ - temperature coefficient, $t$ - temperature in degrees Cel... | Answer: 96 kJ
Solution. Given that the specific heat capacity depends linearly on temperature, we can calculate its average value: $c_{c p}=\frac{c_{0}\left(1+\alpha t_{h}\right)+c_{0}\left(1+\alpha_{\kappa}\right)}{2}=600 \frac{\text { J }}{\text { kg } \cdot{ }^{\circ} \mathrm{C}} \quad(5$ points). The required amou... | 96 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,071 |
8. A parallel beam of light falls normally on a thin lens. Behind the lens, at a distance of $80 \mathrm{~cm}$ from it, there is a screen on which a circular spot of a certain diameter is visible. If the screen is moved 40 cm, a spot of the same diameter will again be visible on the screen. Determine the focal length o... | Answer: 100 cm or 60 cm
Solution. A diagram explaining the situation described in the problem (5 points):
From this, it is clear that there are two possible scenarios: the screen can be mov... | 100 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,072 |
3. A triangle $ABC$ with side lengths $AB=7 \text{~cm}, BC=6 \text{~cm}, CA=5 \text{~cm}$ has been cut out of paper. It was folded along a line so that vertex $C$ ended up at point $C_{1}$ on side $AB$. Moreover, in the resulting quadrilateral $AKMB$, two angles adjacent to the fold line $KM$ turned out to be equal. Fi... | Answer: $\frac{42}{11}$ cm and $\frac{35}{11}$.
Solution. Points $C$ and $C_{1}$ are symmetric with respect to the fold line. Therefore, $C C_{1} \perp K M$. From the equality of angles $A K M$ and $K M B$, it follows that the adjacent angles are also equal. Therefore, triangle $C K M$ is isosceles. Its height, lying ... | \frac{42}{11} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,073 |
4. Let $a, b, c, d, e$ be positive integers. Their sum is 2345. Let $M=\max (a+b, b+c, c+d, d+e)$. Find the smallest possible value of $M$. | Answer: 782.
Solution. Estimation. We have the inequalities
$$
a+b \leqslant M ; \quad b+c \leqslant M ; \quad c+d \leqslant M ; \quad d+e \leqslant M
$$
Multiply the first and last inequalities by 2 and add the other two. We get:
$2(a+b+c+d+e)+b+d \leqslant 6 M ; \quad 6 M \geqslant 4690+b+d \geqslant 4692 ; \quad... | 782 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,074 |
5. A small ball was released without initial speed from a height of $h=45$ m. The impact with the horizontal surface of the Earth is perfectly elastic. Determine at what moment of time after the start of the fall the average path speed of the ball will be equal to its instantaneous speed. The acceleration due to gravit... | Answer: $4.24 s$
Solution. Obviously, while the ball is flying down, the instantaneous speed will be greater than the average path speed. The moment of time when the ball hits the ground: $t_{1}=\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2 \cdot 45}{10}}=3 s$ (2 points). The corresponding speed: $v_{1}=g t_{1}=10 \cdot 3=30 m /... | 4.24 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,075 |
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