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4. (15 points) During a walk, a pedestrian first walked $3 \mathrm{km}$ at a speed of $1.5 \mathrm{m} / \mathrm{s}$, and then another $3600 \mathrm{m}$ at a speed of $3.6 \kappa \mathrm{m} / \mathrm{h}$. Determine his average speed for the entire walk. | Answer: $\approx 1.18 \frac{M}{s}$
Solution. Time spent on the first part of the journey $t_{1}=\frac{s_{1}}{v_{1}}=\frac{3000}{1.5}=2000$ s. Time spent on the second part of the journey $t_{2}=\frac{s_{2}}{v_{2}}=\frac{3600}{1}=3600$ s. Average speed for the entire walk: $v_{av}=\frac{s_{1}+s_{2}}{t_{1}+t_{2}}=\frac{... | 1.18\frac{M}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,301 |
5. (20 points) The clock shows the time as four hours and fifteen minutes. Determine the angle between the minute and hour hands at this moment. | Answer: $37.5^{\circ}$
Solution. Five minutes is $\frac{5}{60}$ of the circumference, which is $30^{\circ}$. The minute hand shows fifteen minutes, which is $90^{\circ}$. The hour hand has moved a quarter of the distance between four $\left(120^{\circ}\right)$ and five $\left(150^{\circ}\right)$ hours in fifteen minut... | 37.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,302 |
6. (15 points) The radar received the signal reflected from the target after 15 microseconds. Determine the distance to the target, given that the speed of the radar signal is $300000 \mathrm{kM} / \mathrm{c}$. Note that one microsecond is one millionth of a second. | Answer: 2250 m
Solution. In 15 microseconds, the signal travels the distance from the radar to the target and back. Therefore, the distance to the target is:
$$
S=v \frac{t}{2}=3 \cdot 10^{8} \cdot \frac{15 \cdot 10^{-6}}{2}=2250 \mathrm{~m}
$$
## Preliminary Stage
7th grade
## Variant 2
## Problems, answers, and... | 2250\mathrm{~} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,303 |
1. (17 points) When walking uphill, the tourist walks 3 km/h slower, and downhill 3 km/h faster, than when walking on flat ground. Climbing the mountain takes the tourist 8 hours, while descending the mountain takes 4 hours. What is the tourist's speed on flat ground? | # Answer: 9
Solution. Let $x$ km/h be the tourist's speed on flat terrain. According to the problem, we get the equation $8(x-3)=4(x+3)$. From this, we find $x=9$. | 9 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,304 |
3. (16 points) Mitya, Anton, Gosha, and Boris bought a lottery ticket for 20 rubles. Mitya paid $24\%$ of the ticket's cost, Anton - 3 rubles 70 kopecks, Gosha - $0.21$ of the ticket's cost, and Boris contributed the remaining amount. The boys agreed to divide the winnings in proportion to their contributions. The tick... | Answer: 292
Solution. The ticket costs 2000 kop. Mitya paid 480 kop, Anton - 370 kop, Gosha - 420 kop, therefore, Boris had to pay an additional 730 kop. Since the prize is 40 times the cost of the ticket, Boris is entitled to 292 rubles. | 292 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,306 |
4. (15 points) During a walk, a pedestrian first walked $2.8 \kappa m$ at a speed of $1.4 m / s$, and then ran another 1800 m at a speed of 7.2 km/h. Determine his average speed for the entire walk. | Answer: $\approx 1.59 \frac{M}{c}$
Solution. Time spent on the first part of the journey $t_{1}=\frac{S_{1}}{v_{1}}=\frac{2800}{1.4}=2000 \mathrm{c}$. Time spent on the second part of the journey $t_{2}=\frac{s_{2}}{v_{2}}=\frac{1800}{2}=900 \mathrm{c}$. Average speed for the entire walk: $v_{c p}=\frac{s_{1}+s_{2}}{t... | 1.59\frac{\mathrm{M}}{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,307 |
5. (20 points) The clock shows the time as fifteen minutes past five. Determine the angle between the minute and hour hands at this moment.
---
Note: The translation maintains the original text's formatting and structure. | Answer: $67.5^{\circ}$
Solution. Five minutes is $\frac{5}{60}$ of the circumference, which is $30^{\circ}$. The minute hand shows fifteen minutes, which is $90^{\circ}$. The hour hand has moved a quarter of the distance between five ($150^{\circ}$) and six ($180^{\circ}$) hours in fifteen minutes, so the hour hand sh... | 67.5 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,308 |
6. (15 points) The radar received the signal reflected from the target after 6 microseconds. Determine the distance to the target, given that the speed of the radar signal is $300000 \mathrm{k} \nu / \mathrm{c}$. Note that one microsecond is one millionth of a second. | Answer: 900 m
Solution. In 6 microseconds, the signal travels the distance from the radar to the target and back. Therefore, the distance to the target is:
$$
S=v \frac{t}{2}=3 \cdot 10^{8} \cdot \frac{6 \cdot 10^{-6}}{2}=900 \mathrm{m}
$$ | 900\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,309 |
1. (16 points) The dividend is six times larger than the divisor, and the divisor is four times larger than the quotient. Find the dividend. | # Answer: 144
Solution. From the condition of the problem, it follows that the quotient is 6. Then the divisor is 24, and the dividend is 144. | 144 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,310 |
2. (17 points) Hooligan Vasily tore out a whole chapter from a book, the first page of which was numbered 231, and the number of the last page consisted of the same digits. How many sheets did Vasily tear out of the book?
# | # Answer: 41
Solution. The number of the last page starts with the digit 3 and must be even, so the last page has the number 312. Vasily tore out $312-231+1=82$ pages or 41 sheets. | 41 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,311 |
3. (17 points) Divide the number 90 into two parts such that $40\%$ of one part is 15 more than $30\%$ of the other part. Write the larger of the two parts in your answer. | Answer: 60
Solution. Let one part of the number be $x$, then the other part will be $90-x$. We get the equation $0.4 \cdot x = 0.3 \cdot (90 - x) + 15$, solving it we get $x = 60$, and the other part of the number is 30. | 60 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,312 |
4. (15 points) A tanker is being filled with oil at a rate of 3 barrels per minute. Given that 1 barrel equals 159 liters, determine the rate of filling the tanker in m ${ }^{3} /$ hour. | Answer: $28.62 \frac{M^{3}}{\varphi}$
Solution. $3 \frac{\text { barrels }}{\text{min}}=3 \frac{159 \text { liters }}{\frac{1}{60} \varphi}=3 \cdot 159 \cdot 10^{-3} \cdot 60 \frac{\mu^{3}}{\varphi}=28.62 \frac{\mu^{3}}{\varphi}$ | 28.62 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,313 |
6. (20 points) A one-kilogram model of a sports car body was made from aluminum at a scale of 1:10. What is the mass of the actual body if it is also entirely made of aluminum? | Answer: 1000 kg
Solution. All dimensions of the body are 10 times larger compared to the model. Therefore, the volume of the body is larger by $10 \cdot 10 \cdot 10=1000$ times. Mass is directly proportional to volume, therefore, the mass of the body:
$$
m_{\text {body }}=1000 m_{\text {model }}=1000 \text { kg. }
$$... | 1000 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,315 |
1. (16 points) The dividend is five times larger than the divisor, and the divisor is four times larger than the quotient. Find the dividend.
# | # Answer: 100
Solution. From the condition of the problem, it follows that the quotient is 5. Then the divisor is 20, and the dividend is 100. | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,316 |
2. (17 points) Hooligan Vasily tore out a whole chapter from a book, the first page of which was numbered 241, and the number of the last page consisted of the same digits. How many sheets did Vasily tear out of the book?
# | # Answer: 86
Solution. The number of the last page starts with the digit 4 and must be even, so the last page has the number 412. Vasily tore out $412-241+1=172$ pages or 86 sheets. | 86 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,317 |
3. (17 points) Divide the number 80 into two parts such that $30\%$ of one part is 10 more than $20\%$ of the other part. Write the smaller of the two parts in your answer. | Answer: 28
Solution. Let one part of the number be $x$, then the other part will be $80-x$. We get the equation $0.3 \cdot x = 0.2 \cdot (80 - x) + 10$, solving it we get $x = 52$, and the other part of the number is 28. | 28 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,318 |
4. (15 points) A tanker is being filled with oil at a rate of 2 barrels per half minute. Given that 1 barrel equals 159 liters, determine the filling rate of the tanker in m ${ }^{3} /$ hour. | Answer: $38.16 \frac{M^{3}}{\psi}$
Solution. $\quad 2 \frac{\text { barrels }}{\frac{1}{2} \text { min }}=2 \frac{159 \text { liters }}{\frac{1}{120} \psi}=2 \cdot 159 \cdot 10^{-3} \cdot 120 \frac{\mathcal{M}^{3}}{\psi}=38.16 \frac{\mathcal{M}^{3}}{\psi}$. | 38.16\frac{\mathcal{M}^{3}}{\psi} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,319 |
5. (15 points) A train 360 m long, moving at a constant speed, passed a bridge 240 m long in 4 min. Determine the speed of the train. | Answer: $2.5 \frac{M}{c}$
Solution. The speed of the train: $v=\frac{360+240}{4 \cdot 60}=2.5 \frac{\mathrm{M}}{\mathrm{c}}$. | 2.5\frac{\mathrm{M}}{\mathrm{}} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,320 |
6. (20 points) A two-kilogram model of a sports car body was made from aluminum at a scale of $1: 8$. What is the mass of the actual body if it is also entirely made of aluminum? | Answer: 1024 kg
Solution. All dimensions of the body are 8 times larger compared to the model. Therefore, the volume of the body is larger by $8 \cdot 8 \cdot 8=512$ times. The mass is directly proportional to the volume, therefore, the mass of the body:
$m_{\text {body }}=512 m_{\text {model }}=1024$ kg | 1024 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,321 |
1. (17 points) Find the area of the triangle cut off by the line $y=3 x+1$ from the figure defined by the inequality $|x-1|+|y-2| \leq 2$. | Answer: 2.
Solution. The figure defined by the given inequality is a square.
The side of the square is $2 \sqrt{2}$ (this can be found using the Pythagorean theorem). The given line passes t... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,322 |
2. (16 points) Find the minimum value of the function
$$
f(x)=3 \sin ^{2} x+5 \cos ^{2} x+2 \cos x
$$ | Answer: 2.5.
Solution. Using the basic trigonometric identity, we get $f(x)=$ $2 \cos ^{2} x+2 \cos x+3$. Let's make the substitution $\cos x=t, t \in[-1 ; 1]$. After this, the problem reduces to finding the minimum value of the quadratic function $g(x)=2 t^{2}+$ $2 t+3$ on the interval $[-1 ; 1]$. The quadratic funct... | 2.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,323 |
3. (17 points) A section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the section if the side of the base is 6 and the height of the pyramid is 8. | Answer: 9.
Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.
 Identical gases are in two thermally insulated vessels with volumes $V_{1}=1$ l and $V_{2}=2$ l. The pressures of the gases are $p_{1}=2$ atm and $p_{2}=3$ atm, and their temperatures are $T_{1}=300$ K and $T_{2}=400$ K, respectively. The gases are mixed. Determine the temperature that will be establishe... | Answer $\approx \approx 369$ K.
Solution. The total internal energy of the gases before mixing is equal to the total internal energy after mixing $U_{1}+U_{2}=U$, then $\frac{i}{2} \vartheta_{1} R T_{1}+\frac{i}{2} \vartheta_{2} R T_{2}=\frac{i}{2}\left(\vartheta_{1}+\vartheta_{2}\right) R T$, where $\vartheta_{1}=\fr... | 369\mathrm{~K} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,325 |
5. (20 points) It is known that the resistances of the resistors in the presented circuit are \( R_{1}=4 \) Ohms, \( R_{2}=8 \) Ohms, and \( R_{3}=16 \) Ohms. Determine how and by how much the total resistance of the circuit will change when the switch \( K \) is closed.
 A capacitor with capacitance $C_{1}=10$ μF is charged to a voltage $U_{1}=15$ V. A second capacitor with capacitance $C_{2}=5$ μF is charged to a voltage $U_{2}=10$ V. The capacitors are connected with their oppositely charged plates. Determine the voltage that will establish across the plates. | Answer: 6.67 V.
Solution. The law of charge conservation $C_{1} U_{1}-C_{2} U_{2}=C_{1} U+C_{2} U$. We obtain:
$$
U=\frac{c_{1} U_{1}-C_{2} U_{2}}{C_{1}+C_{2}}=\frac{10 \cdot 15-5 \cdot 10}{10+5}=6.67 \mathrm{~V} .
$$
## Tasks, answers, and assessment criteria | 6.67\mathrm{~V} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,327 |
1. (17 points) Find the area of the triangle cut off by the line $y=2x+2$ from the figure defined by the inequality $|x-2|+|y-3| \leq 3$. | Answer: 3.
Solution. The figure defined by the given inequality is a square.
The side of the square is $3 \sqrt{2}$ (this value can be found using the Pythagorean theorem). The given line pa... | 3 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,328 |
2. (16 points) Find the minimum value of the function
$$
f(x)=7 \sin ^{2} x+5 \cos ^{2} x+2 \sin x
$$ | Answer: 4.5.
Solution. Using the basic trigonometric identity, we get $f(x)=$ $2 \sin ^{2} x+2 \cos x+5$. Let's make the substitution $\sin x=t, t \in[-1 ; 1]$. After this, the problem reduces to finding the minimum value of the quadratic function $g(x)=2 t^{2}+$ $2 t+5$ on the interval $[-1 ; 1]$. The quadratic funct... | 4.5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,329 |
3. (17 points) The cross-section of a regular triangular pyramid passes through the midline of the base and is perpendicular to the base. Find the area of the cross-section if the side of the base is 8 and the height of the pyramid is 12. | Answer: 18.
Solution. The section MNP passes through the midline of the base of the pyramid $MN$ and is perpendicular to the base. Therefore, the height $PH$ of the triangle $MNP$ is parallel to the height of the pyramid $DO$.
 Identical gases are in two thermally insulated vessels with volumes $V_{1}=2$ L and $V_{2}=3$ L. The pressures of the gases are $p_{1}=3$ atm and $p_{2}=4$ atm, and their temperatures are $T_{1}=400$ K and $T_{2}=500$ K, respectively. The gases are mixed. Determine the temperature that will be establishe... | Answer: $\approx 462$ K.
Solution. The total internal energy of the gases before mixing is equal to the total internal energy after mixing $U_{1}+U_{2}=U$.
$$
\frac{i}{2} \vartheta_{1} R T_{1}+\frac{i}{2} \vartheta_{2} R T_{2}=\frac{i}{2}\left(\vartheta_{1}+\vartheta_{2}\right) R T, \text { where } \vartheta_{1}=\fra... | 462\mathrm{~K} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,331 |
5. (20 points) It is known that the resistances of the resistors in the presented circuit are \( R_{1}=1 \) Ohm, \( R_{2}=2 \) Ohms, and \( R_{3}=4 \) Ohms. Determine how and by how much the total resistance of the circuit will change when the switch \( K \) is closed.
 A capacitor with capacitance $C_{1}=20$ μF is charged to a voltage $U_{1}=20$ V. A second capacitor with capacitance $C_{2}=5$ μF is charged to a voltage $U_{2}=5$ V. The capacitors are connected with their oppositely charged plates. Determine the voltage that will establish across the plates. | Answer: 15 V.
Solution. Law of conservation of charge: $C_{1} U_{1}-C_{2} U_{2}=C_{1} U+C_{2} U$. We obtain:
$$
U=\frac{C_{1} U_{1}-C_{2} U_{2}}{C_{1}+C_{2}}=\frac{20 \cdot 20-5 \cdot 5}{20+5}=15 \mathrm{~V}
$$ | 15\mathrm{~V} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,333 |
1. (16 points) There are two circles: one with center at point $A$ and radius 6, and another with center at point $B$ and radius 3. Their common internal tangent touches the circles at points $C$ and $D$, respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $A E=10$. | Answer: 12
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of 2. From triangle $A C E$, using the Pythagorean theorem, we find $C E=8$. Therefore, $D E=4$, and $C D=12$. | 12 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,334 |
2. (17 points) Find the largest root of the equation
$$
\left|\cos (\pi x)+x^{3}-3 x^{2}+3 x\right|=3-x^{2}-2 x^{3}
$$ | # Answer: 1
Solution. It is obvious that 1 is a root of the equation (when $x=1$, both sides of the equation are equal to zero). If $x>1$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 1 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,335 |
3. (17 points) Find the smallest natural number that is simultaneously twice a perfect square and three times a perfect cube.
# | # Answer: 648
Solution. We have $k=3 n^{3}=2 m^{2}$. From this, the numbers $m$ and $n$ can be represented as $n=2 a, m=3 b$. After substitution, we get $4 a^{3}=3 b^{2}$. Further, we have $a=3 c, b=2 d, 9 c^{3}=d^{2}$. Here, the smallest solution is $c=1, d=3$. Then $a=3$, $b=6, n=6, m=18, k=648$. | 648 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,336 |
4. (15 points) The efficiency of an ideal heat engine is $40 \%$. What will it become if the temperature of the heater is increased by $40 \%$, and the temperature of the cooler is decreased by $40 \%$? | Answer: $\approx 74 \%$.
Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, initially the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.4=0.6$. After the changes:
$$
\eta_{2}=1-\frac{0.6 T_{X}}{1.4 T_{H}}=1-\frac{0.6 \cdot 0.6}{1.4} \appro... | 74 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,337 |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=5 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its im... | Answer: -10 Dptr $u \approx 7.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain that the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{5}{10}=5 \mathrm{~cm} . \quad$ The power of the lens: $D=\frac{1}{d}-\frac{1}{f}=\... | -10 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,338 |
1. (16 points) There are two circles: one with center at point $A$ and radius 5, and another with center at point $B$ and radius 15. Their common internal tangent touches the circles at points $C$ and $D$ respectively. Lines $A B$ and $C D$ intersect at point $E$. Find $C D$, if $B E=39$. | # Answer: 48
Solution. Triangles $A C E$ and $B D E$ are similar (they have vertical angles and a right angle each) with a similarity coefficient of $1 / 3$. Therefore, $A E=13$. From triangle $A C E$, using the Pythagorean theorem, we find $C E=12$. Hence, $D E=36$, and $C D=48$. | 48 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,340 |
2. (17 points) Find the smallest root of the equation
$$
\sin (\pi x)+\tan x=x+x^{3}
$$ | Answer: 0
Solution. Obviously, 0 is a root of the equation (when $x=0$, both sides of the equation are equal to zero). If $x<0$, the right side of the equation is negative, while the left side of the equation is always non-negative. | 0 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,341 |
3. (17 points) Find the smallest natural number that is simultaneously twice a perfect cube and three times a perfect square.
# | # Answer: 432
Solution. We have $k=2 n^{3}=3 m^{2}$. From this, the numbers $m$ and $n$ can be represented as $n=3 a, m=2 b$. After substitution, we get $9 a^{3}=2 b^{2}$. Further, we have $a=2 c, b=3 d, 4 c^{3}=d^{2}$. Here, the smallest solution is $c=1, d=2$. Then $a=2$, $b=6, n=6, m=12, k=432$. | 432 | Number Theory | math-word-problem | Yes | Yes | olympiads | false | 4,342 |
4. (15 points) The efficiency of an ideal heat engine is $50 \%$. What will it become if the temperature of the heater is increased by $50 \%$, and the temperature of the cooler is decreased by $50 \%$? | Answer: $\approx 83 \%$
Solution. The efficiency of an ideal heat engine: $\eta=1-\frac{T_{X}}{T_{H}}$. That is, at the beginning, the ratio of the temperatures of the refrigerator and the heater: $\frac{T_{X}}{T_{H}}=1-0.5=0.5$. After the changes:
$$
\eta_{2}=1-\frac{0.5 T_{X}}{1.5 T_{H}}=1-\frac{0.5 \cdot 0.5}{1.5}... | 83 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,343 |
5. (20 points) A point light source is located at an equal distance $x=10 \mathrm{~cm}$ from the lens and its principal optical axis. Its direct image is located at a distance $y=20 \mathrm{~cm}$ from the principal optical axis. Determine the optical power of the lens and the distance between the light source and its i... | Answer: 5 Dpt $i \approx 14.1$ cm
Solution. The image is upright, therefore, it is virtual. Magnification: $\Gamma=\frac{y}{x}=\frac{f}{d}$. We obtain the distance from the lens to the image: $f=d \cdot \frac{y}{x}=10 \cdot \frac{20}{10}=20 \mathrm{~cm} . \quad$ The optical power of the lens: $D=\frac{1}{d}-\frac{1}{f... | 5 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,344 |
6. (15 points) A tank with a mass of $m_{1}=3$ kg rests on a cart with a mass of $m_{2}=15$ kg, which is accelerated with an acceleration of $a=4 \mathrm{~m} / \mathrm{c}^{2}$. The coefficient of friction between the tank and the cart is $\mu=0.6$. Determine the frictional force acting on the tank from the cart. | Answer: $12 H$
Solution. Under the given conditions, we are talking about the force of static friction. According to Newton's second law: $F_{n p}=m_{1} a=12 H$. | 12H | Other | math-word-problem | Yes | Yes | olympiads | false | 4,345 |
1. How many times in a day does the angle between the hour and minute hands equal exactly $17^{\circ}$? | Answer: 44.
Solution. Over the time interval from 0:00 to 12:00, the hour hand will make one complete revolution, while the minute hand will make 12 such revolutions. Therefore, during this time, the minute hand will catch up with the hour hand 11 times. Between two consecutive meetings of the hands, the angle between... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,346 |
2. Solve the equation
$$
\sqrt{\frac{x-2}{11}}+\sqrt{\frac{x-3}{10}}=\sqrt{\frac{x-11}{2}}+\sqrt{\frac{x-10}{3}}
$$ | Answer: 13.
Solution. If we perform the variable substitution $x=t+13$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}
$$
Now it is clear that for $t>0$ the right side of the equation is greater, and... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,347 |
3. Let in triangle $A B C$
$$
\cos (2 \angle A-\angle B)+\sin (\angle A+\angle B)=2 \text {. }
$$
Find the side $B C$, if $A B=4$. | Answer: 2.
Solution. Each term in the left part of the equation does not exceed 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$
$$
2 \alpha-\beta=2 \pi n, n \in \mathbb{Z} ; \quad \alpha+\beta=\frac{\pi}{2}+2 \pi k, ... | 2 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,348 |
4. Find the equation of a line $L$ such that the graph of the function
$$
y=x^{4}+4 x^{3}-26 x^{2}
$$
lies on one side of this line, having two common points with it. | Answer: $y=60 x-225$.
Solution. Let $y=a x+b$ be the equation of the line $L$. We need to find such $a$ and $b$ that the equation
$$
x^{4}+4 x^{3}-26 x^{2}-(a x+b)=0
$$
has two roots $x_{1}$ and $x_{2}$ of even multiplicity. For this, the polynomial of the fourth degree must have the following factorization:
$$
x^{... | 60x-225 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,349 |
5. A Stalker, to detect a gravitational anomaly (an area where the acceleration due to gravity changes sharply in magnitude), throws a small nut from the surface of the Earth at an angle $\alpha=30^{\circ}$ to the horizontal with a speed $v_{0}=10 \mathrm{M} / \mathrm{c}$. The normal acceleration due to gravity $g=10 \... | Answer: $250 \mu / c^{2}$
Solution. Equations of motion before the anomaly: $x=v_{0} \cos \alpha \cdot t=5 \sqrt{3} \cdot t$ (2 points), $y=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}=5 \cdot t-5 \cdot t^{2}$ (2 points), $v_{x}=v_{0} \cos \alpha=5 \sqrt{3}$ (2 points), $v_{y}=v_{0} \sin \alpha-g t=5-10 t$ (2 points). ... | 250/^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,350 |
6. The brakes of a car allow it to stand on an inclined asphalt surface with an angle at the base of no more than $15^{\circ}$. Determine the minimum braking distance of this car when moving at a speed of 20 m/s on a flat horizontal road with the same surface. The acceleration due to gravity $g=10 m / s^{2}, \cos 15^{\... | Answer: $74.6 m$
Solution. For the situation when the car is on an inclined plane:
$F_{n p}=m g \sin \alpha$ (2 points), $\mu m g \cos \alpha=m g \sin \alpha$ (2 points). Therefore, the coefficient of friction: $\mu=\frac{\sin 15^{\circ}}{\cos 15^{\circ}}=0.268 \quad(3 \quad$ points). For the situation when the car is... | 74.6 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,351 |
7. In the electrical circuit depicted in the figure, the resistances of the resistors are $R_{1}=1 \Omega$, $R_{2}=2 \Omega$, $R_{3}=3 \Omega$, and $R_{4}=4 \Omega$. Assume that the resistances of all diodes in the forward direction are negligibly small, and in the reverse direction are infinite. Determine the resistan... | Answer: 2.08 Ω
Solution. In the given situation, the following equivalent circuit is obtained (5 points):
Its total resistance: $R=\frac{R_{1} R_{3}}{R_{1}+R_{3}}+\frac{R_{2} R_{4}}{R_{2}+... | 2.08\Omega | Other | math-word-problem | Yes | Yes | olympiads | false | 4,352 |
1. How many times in a day does the angle between the hour and minute hands equal exactly $19^{\circ}$? | Answer: 44.
Solution. Over the time interval from 0:00 to 12:00, the hour hand will make one complete revolution, while the minute hand will make 12 such revolutions. Therefore, during this time, the minute hand will catch up with the hour hand 11 times. Between two consecutive meetings of the hands, the angle between... | 44 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,354 |
2. Solve the equation
$$
\sqrt{\frac{x-3}{11}}+\sqrt{\frac{x-4}{10}}=\sqrt{\frac{x-11}{3}}+\sqrt{\frac{x-10}{4}}
$$ | Answer: 14.
Solution. If we perform the variable substitution $x=t+14$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}
$$
Now it is clear that for $t>0$ the right side of the equation is greater, and... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,355 |
3. Let in triangle $A B C$
$$
\cos (\angle A-\angle B)+\sin (\angle A+\angle B)=2
$$
Find the side $B C$, if $A B=4$. | Answer: $2 \sqrt{2}$.
Solution. Each term in the left part of the equation is no more than 1. Therefore, the equality will hold only if each of them equals 1. We solve the corresponding equations, denoting $\alpha=\angle A, \beta=\angle B:$
$$
\alpha-\beta=2 \pi n, n \in \mathbb{Z} ; \quad \alpha+\beta=\frac{\pi}{2}+... | 2\sqrt{2} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,356 |
4. Find the equation of a line $L$ such that the graph of the function
$$
y=x^{4}-4 x^{3}-26 x^{2}
$$
lies on one side of this line, having two common points with it. | Answer: $y=-60 x-225$.
Solution. Let $y=a x+b$ be the equation of the line $L$. We need to find such $a$ and $b$ that the equation
$$
x^{4}-4 x^{3}-26 x^{2}-(a x+b)=0
$$
has two roots $x_{1}$ and $x_{2}$ of even multiplicity. For this, the polynomial of the fourth degree must have the following factorization:
$$
x^... | -60x-225 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,357 |
5. The Stalker, to detect a gravitational anomaly (an area where the acceleration due to gravity changes sharply in magnitude), throws a small nut from the surface of the Earth at an angle $\alpha=30^{\circ}$ to the horizontal with a speed of $v_{0}=20 \mathrm{M} / \mathrm{c}$. The normal acceleration due to gravity is... | Answer: $40 \mathrm{M} / \mathrm{c}^{2}$
Solution. Equations of motion before the anomaly: $x=v_{0} \cos \alpha \cdot t=10 \sqrt{3} \cdot t$ (2 points), $y=v_{0} \sin \alpha \cdot t-\frac{g t^{2}}{2}=10 \cdot t-5 \cdot t^{2} \quad\left(2\right.$ points), $v_{x}=v_{0} \cos \alpha=10 \sqrt{3}$ (2 points), $v_{y}=v_{0} \... | 40\mathrm{M}/\mathrm{}^{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,358 |
6. The brakes of a car allow it to stand on an inclined asphalt surface with an angle at the base of no more than $30^{\circ}$. Determine the minimum braking distance of this car when moving at a speed of $30 \, \text{m/s}$ on a flat horizontal road with the same surface. The acceleration due to gravity $g=10 \, \text{... | Answer: 78 m
Solution. For the situation when the car is on an inclined plane:
$F_{mp}=m g \sin \alpha$ (2 points), $\mu m g \cos \alpha=m g \sin \alpha$ (2 points). Therefore, the coefficient of friction: $\mu=\frac{\sin 30^{\circ}}{\cos 30^{\circ}} \approx 0.577$ (3 points). For the situation when the car is moving ... | 78\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,359 |
7. In the electrical circuit shown in the figure, the resistances of the resistors are $R_{1}=1$ Ohm, $R_{2}=2$ Ohms, $R_{3}=3$ Ohms, and $R_{4}=4$ Ohms. Assume that the resistances of all diodes in the forward direction are negligibly small, and in the reverse direction are infinite. Determine the resistance of the en... | Answer: 2.38 $O m$
Solution. In the given situation, the following equivalent circuit is obtained $(2$ points $):$
Its total resistance: $R=\frac{R_{1} R_{2}}{R_{1}+R_{2}}+\frac{R_{3} R_{4... | 2.38 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,360 |
1. The infantry column 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} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,362 |
2. Solve the inequality $\sqrt{9-x}-3 \geqslant x|x-3|+\ln (1+x)$. | Answer: $(-1 ; 0]$.
Solution. Domain: $(-1 ; 9]$. Let
$$
f(x)=\sqrt{9-x}-3, \quad g(x)=x|x-3|+\ln (1+x)
$$
At $x=0$, both functions are zero.
If $-1<x<0$, then $f(x)>0$, while $g(x)<0$. In this case, $f(x)>g(x)$.
If $0<x<9$, then $f(x)<0$, while $g(x)>0$ (the first term is non-negative, and the second term is posi... | (-1;0] | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,363 |
3. A regular quadrilateral pyramid is given. The side of the base is 6, the length of the lateral edge is 5. A sphere $Q_{1}$ is inscribed in the pyramid. A sphere $Q_{2}$ touches $Q_{1}$ and all the lateral faces of the pyramid. Find the radius of the sphere $Q_{2}$. | Answer: $\frac{3 \sqrt{7}}{49}$.
Solution. Let $r_{1}$ and $r_{2}$ be the radii of the spheres $Q_{1}$ and $Q_{2}$, respectively. Let $A B C D$ be the base of the pyramid, $M$ the apex of the pyramid, $E$ the center of the base, and $F$ the midpoint of $C D$. Using the Pythagorean theorem, we find the slant height of ... | \frac{3\sqrt{7}}{49} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,364 |
4. A circle is inscribed with 2019 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 2, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Answer: 6060.
Solution. Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 2$ and $y+z \geqslant 6$, we get $y \geqslant 4$. Then $x \geqslant y+2 \geqslant 6$. A number not less than 6 has been found. The remaining ... | 6060 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,365 |
5. (10 points) A small railway wagon with a jet engine is standing on the tracks. The tracks are laid out in the form of a circle with radius $R$. The wagon starts from rest, with the jet force having a constant value. What is the maximum speed the wagon will reach after one full circle, if its acceleration over this p... | Answer: $v_{\max }=\sqrt[4]{\frac{16 a^{2} R^{2} \pi^{2}}{\left(1+16 \pi^{2}\right)}}$
Solution. The acceleration of the trolley: $a_{1}=\frac{v^{2}}{2 s}=\frac{v^{2}}{4 \pi R}$.
In addition, the trolley has a centripetal acceleration:
$a_{2}=\frac{v^{2}}{R}$.
The total acceleration of the trolley: $a^{2}=a_{1}^{2}... | v_{\max}=\sqrt[4]{\frac{16^{2}R^{2}\pi^{2}}{(1+16\pi^{2})}} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,366 |
6. (10 points) A beam of light with a diameter of $d_{1}=10 \, \text{cm}$ falls on a thin diverging lens with an optical power of $D_{p}=-6$ Diopters. On a screen positioned parallel to the lens, a bright spot with a diameter of $d_{2}=20 \, \text{cm}$ is observed. After replacing the thin diverging lens with a thin co... | # Answer: 18 Dppr
Solution. The optical scheme corresponding to the condition:
(3 points)
The path of the rays after the diverging lens is shown in black, and after the converging lens in ... | 18\, | Other | math-word-problem | Yes | Yes | olympiads | false | 4,367 |
7. (15 points) Inside a cylinder, there are two pairs of identical supports. The distance between the lower supports and the bottom $l_{1}=10 \mathrm{~cm}$, between the lower and upper supports $l_{2}=15 \mathrm{~cm}$. On the lower supports lies a piston with the maximum possible mass $M=10$ kg that they can withstand.... | Answer: 127.5 D
Solution. During the heat transfer to the gas, three stages can be distinguished.
First, the isochoric increase in pressure to a value sufficient to move the piston from its position: $p_{1}=p_{0}+\frac{M g}{S}$.
The amount of heat received by the gas during this stage:
$$
Q_{1}=\Delta U_{1}=\frac{3... | 127.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,368 |
8. (15 points) A structure consisting of a rigidly connected light rod and a small load with mass \( m = 1 \) kg can oscillate under the action of two springs with stiffnesses \( k_{1} = 60 \frac{\mathrm{H}}{\mathrm{M}} \) and \( k_{2} = 10 \frac{\mathrm{H}}{\mathrm{M}} \), moving during rotation without friction aroun... | Answer: $\approx 1.9 s$
Solution. Let's deflect the rod from its equilibrium position by a small angle $\alpha$. In this position, the total mechanical energy of the system is:
$\frac{k_{1} x_{1}^{2}}{2}+\frac{k_{2} x_{2}^{2}}{2}+\frac{m v^{2}}{2}=$ const
where $x_{1}=\alpha \frac{l}{3}, x_{2}=\alpha \frac{2 l}{3}, ... | 1.9 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,369 |
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 $400 \mathrm{m}$ 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 | 4,370 |
3. A regular quadrilateral pyramid is given. The side of the base is 12, the length of the lateral edge is 10. A sphere \( Q_{1} \) is inscribed in the pyramid. A sphere \( Q_{2} \) touches \( Q_{1} \) and all the lateral faces of the pyramid. Find the radius of the sphere \( Q_{2} \). | Answer: $\frac{6 \sqrt{7}}{49}$.
Solution. Let $r_{1}$ and $r_{2}$ be the radii of the spheres $Q_{1}$ and $Q_{2}$, respectively. Let $A B C D$ be the base of the pyramid, $M$ be the apex of the pyramid, $E$ be the center of the base, and $F$ be the midpoint of $C D$. Using the Pythagorean theorem, we find the slant h... | \frac{6\sqrt{7}}{49} | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,372 |
4. A circle is inscribed with 1001 numbers. For any two adjacent numbers $x$ and $y$, the inequalities $|x-y| \geqslant 4, x+y \geqslant 6$ are satisfied. Find the smallest possible sum of the recorded numbers. | Answer: 3009.
Solution. Due to the odd number of total numbers, there will be three consecutive numbers $x, y$, and $z$ such that $x>y>z$. Adding the inequalities $y-z \geqslant 4$ and $y+z \geqslant 6$, we get $y \geqslant 5$. Then $x \geqslant y+4 \geqslant 9$. A number not less than 9 has been found. The remaining ... | 3009 | Inequalities | math-word-problem | Yes | Yes | olympiads | false | 4,373 |
6. (10 points) A beam of light with a diameter of $d_{1}=5 \mathrm{~cm}$ falls on a thin diverging lens with an optical power of $D_{p}=-6$ Diopters. On a screen positioned parallel to the lens, a bright spot with a diameter of $d_{2}=20 \mathrm{~cm}$ is observed. After replacing the thin diverging lens with a thin con... | # Answer: 10 Dptr
Solution. The optical scheme corresponding to the condition:
(3 points)
The path of the rays after the diverging lens is shown in black, and after the converging lens in ... | 10 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,374 |
7. (15 points) Inside a cylinder, there are two pairs of identical supports. The distance between the lower supports and the bottom $l_{1}=20 \mathrm{~cm}$, between the lower and upper supports $l_{2}=25 \mathrm{~cm}$. On the lower supports lies a piston with the maximum possible mass $M=10$ kg that they can withstand.... | Answer: 337.5 Joules
Solution. During the heat transfer to the gas, three stages can be distinguished.
Firstly, the isochoric increase in pressure to a value sufficient to move the piston: $p_{1}=p_{0}+\frac{M g}{S}$.
The amount of heat received by the gas during this stage:
$$
Q_{1}=\Delta U_{1}=\frac{5}{2} v R \D... | 337.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,375 |
8. (15 points) A structure consisting of a rigidly connected light rod and a small load with mass \( m = 1.6 \) kg can oscillate under the action of two springs with stiffnesses \( k_{1} = 10 \frac{\mathrm{H}}{\mathrm{M}} \) and \( k_{2} = 7.5 \frac{\mathrm{H}}{\mathrm{M}} \), moving during rotation without friction ar... | Answer: $\approx 3.8 s$
Solution. Let's deflect the rod from its equilibrium position by a small angle $\alpha$. In this position, the total mechanical energy of the system is:
$\frac{k_{1} x_{1}^{2}}{2}+\frac{k_{2} x_{2}^{2}}{2}+\frac{m v^{2}}{2}=$ const
where $x_{1}=\alpha \frac{l}{3}, x_{2}=\alpha \frac{2 l}{3}, ... | 3.8 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,376 |
Problem No. 5 (15 points)
With the same amount of monatomic ideal gas, two cyclic processes $1-2-3-1$ and $1-3-4-1$ are performed. Find the ratio of their efficiencies.
Answer: $\frac{13}{1... | # Solution and Evaluation Criteria:
For the cycle $1-2-3-1$: work done by the gas over the cycle: $A=\frac{1}{2} p_{0} V_{0}$
Heat received from the heater:
$$
Q=Q_{12}+Q_{23}=\Delta U_{123}+A_{123}=\frac{3}{2}\left(4 p_{0} V_{0}-p_{0} V_{0}\right)+2 p_{0} V_{0}=6.5 p_{0} V_{0}
$$
Efficiency of this cycle: $\eta_{1... | \frac{13}{12}\approx1.08 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,377 |
# Problem № 6 (10 points)
A cylinder with a mass of $M=1$ kg was placed on rails inclined at an angle $\alpha=30^{\circ}$ to the horizontal (the side view is shown in the figure). What is the minimum mass $m$ of the load that needs to be attached to the thread wound around the cylinder so that it starts rolling upward... | # Solution and evaluation criteria:
The moment of forces relative to the point of contact of the cylinder with the plane: $m g($ ( $-R \sin \alpha)=M g R \sin \alpha$ (5 points) $m\left(1-\f... | 1 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,378 |
# Problem № 7 (10 points)
In the electrical circuit shown in the diagram, the resistances of the resistors are $R_{1}=10$ Ohms and $R_{2}=20$ Ohms. A current source is connected to points A and B in the circuit. When the polarity of its connection is reversed, the ammeter readings change by one and a half times. Deter... | # Solution and Evaluation Criteria:
When the positive terminal of the power supply is connected to point $A$, the current flows only through resistor $R_{2}$, and in this case: $I_{1}=\frac{\varepsilon}{R_{2}+r}$.
## (3 points)
When the polarity is reversed, the current flows only through resistance $R_{1}$, and:
$... | 10\Omega | Other | math-word-problem | Yes | Yes | olympiads | false | 4,379 |
# Problem № 8 (15 points)
50 identical metal spheres with a radius of $R=1$ mm were connected by equal conducting segments into a chain, with the length of each connecting wire segment $l=30$ cm being much greater than the radius of the sphere. The resulting structure was then placed in a uniform electric field of kno... | Answer: $8.17 \cdot 10^{-11} C$
## Solution and Grading Criteria:
Since the spheres are far apart from each other, they can be considered isolated, i.e., the mutual influence on each other can be neglected. The potential of the field created by a sphere is:
$\varphi=k \frac{q}{R}$
The charges of the outermost spher... | 8.17\cdot10^{-11}C | Other | math-word-problem | Yes | Yes | olympiads | false | 4,380 |
# Problem № 6 (10 points)
A cylinder with a mass of $M=0.5$ kg was placed on rails inclined at an angle $\alpha=45^{\circ}$ to the horizontal (the side view is shown in the figure). What is the minimum mass $m$ of the load that needs to be attached to the thread wound around the cylinder so that it starts rolling upwa... | # Solution and evaluation criteria:
The moment of forces relative to the point of contact of the cylinder with the plane:
$m g(R - R \sin \alpha) = M g R \sin \alpha$
(5 points)
$m \left(... | 1.2 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,382 |
# Problem № 7 (10 points)
In the electrical circuit shown in the diagram, the resistances of the resistors are $R_{1}=10$ Ohms and $R_{2}=30$ Ohms. An ammeter is connected to points A and B in the circuit. When the polarity of the current source is reversed, the ammeter readings change by one and a half times. Determi... | # Solution and Evaluation Criteria:
When the positive terminal of the power supply is connected to point $A$, the current flows only through resistor $R_{2}$, and in this case: $I_{1}=\frac{\varepsilon}{R_{2}+r}$.
## (3 points)
When the polarity is reversed, the current flows only through resistance $R_{1}$, and:
$... | 30 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,383 |
# Problem № 8 (15 points)
100 identical metal balls with a radius of $R=1$ mm were connected by equal conducting segments into a chain, with the length of each connecting wire segment $l=50$ cm being much greater than the radius of the ball. The resulting structure was then placed in a uniform electric field of known ... | Answer: $2.75 \cdot 10^{-9}$ Cl
## Solution and Grading Criteria:
Since the balls are far apart from each other, they can be considered isolated, i.e., the mutual influence on each other can be neglected. The potential of the field created by a ball is:
$\varphi=k \frac{q}{R}$
The charges of the outermost balls are... | 2.75\cdot10^{-9}\mathrm{Cl} | Other | math-word-problem | Yes | Yes | olympiads | false | 4,384 |
1. Compute the area of the figure bounded on the plane $O x y$ by the lines $x=2, x=3, y=0, y=x^{1 / \ln x}$. | Answer: $e$.
Solution. Note that $\ln \left(x^{1 / \ln x}\right)=1$. Therefore, $x^{1 / \ln x}=e$ for permissible values of the variable $x$. Thus, the figure represents a rectangle of size $1 \times e$.
Evaluation. 11 points for a correct solution. | e | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,385 |
2. The area of triangle $A B C$ is 1. On the rays $A B, B C$, $C A$, points $B^{\prime}, C^{\prime}, A^{\prime}$ are laid out respectively, such that
$$
B B^{\prime}=2 A B, \quad C C^{\{\prime}=3 B C, \quad A A^{\prime}=4 C A .
$$
Calculate the area of triangle $A^{\prime} B^{\prime} C^{\prime}$. | Answer: 36.
Solution. We will solve the problem in a general form, assuming that
$$
B B^{\prime}=q A B, \quad C C^{\{\prime}=r B C, \quad A A^{\prime}=p C A .
$$
We will calculate the ar... | 36 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,386 |
3. Solve the equation
$$
\sqrt{\frac{x-2}{11}}+\sqrt{\frac{x-3}{10}}+\sqrt{\frac{x-4}{9}}=\sqrt{\frac{x-11}{2}}+\sqrt{\frac{x-10}{3}}+\sqrt{\frac{x-9}{4}}
$$ | Solution. If we perform the variable substitution $x=t+13$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}+\sqrt{\frac{t}{9}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}+\sqrt{\frac{t}{4}+1}
$$
Now it is clear that for $t>0$ the right-hand si... | 13 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,387 |
4. For what values of the parameter $a$ does the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}+4 y=0 \\
x+a y+a z-a=0
\end{array}\right.
$$
have a unique solution | Answer: $a= \pm 2$.
Solution. Let's complete the square for $y$ in the first equation:
$$
x^{2}+(y+2)^{2}+z^{2}=2^{2}
$$
Now it is clear that this equation represents a sphere with center at the point $(0, -2, 0)$ and radius 2. The second equation represents a plane. The system of equations has a unique solution if ... | \2 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,388 |
5. The engine of a car traveling at a speed of $v_{0}=72 \mathrm{km} / \mathbf{h}$ operates with a power of $P=50$ kW. Determine the distance from the point of engine shutdown at which the car will stop, if the resistance force is proportional to the car's speed. The mass of the car is m=1500 kg. (15
## points) | Answer: $240 m$
Solution. The power of the engine: $P=F v=F_{\text {conp }} v_{0}=\alpha v_{0}^{2}$, that is, the coefficient of resistance to the car's movement: $\alpha=\frac{P}{v_{0}^{2}}$ (3 points). The projection of the second law of Newton, for a small time interval, when moving with the engine off: $m \frac{\D... | 240 | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,389 |
6. One mole of an ideal gas was expanded so that during the process, the pressure of the gas turned out to be directly proportional to its volume. In this process, the gas heated up by $\Delta T=100{ }^{\circ} \mathrm{C}$. Determine the work done by the gas in this process. The gas constant $R=8.31$ J/mol$\cdot$K. (15 ... | Answer: 415.5 J
Solution. From the condition: $p=\alpha V$ (3 points). The work of the gas is equal to the area under the graph of the given process, constructed in the coordinates $p-V$:
$A=\frac{p_{1}+p_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha V_{1}+\alpha V_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alp... | 415.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,390 |
7. Two small balls with charges $Q=-40 \cdot 10^{-6}$ C and $q=50 \cdot 10^{-6}$ C are located at the vertices $A$ and $B$ of a mountain slope (see figure). It is known that $AB=4$ m, $AC=5$ m. The masses of the balls are the same and equal to $m=100$ g each. At the initial moment of time, the ball with charge $q$ is r... | Answer: $7.8 \mathrm{M} / \mathrm{c}$
Solution. The law of conservation of energy for this situation:
$k \frac{Q q}{A B}+m g \cdot A B=\frac{m v^{2}}{2}+k \frac{Q q}{B C}(5$ points $)$.
As a result, we get:
$v=\sqrt{\frac{2 k Q q}{m \cdot A B}+2 \cdot g \cdot A B-\frac{2 k Q q}{m \cdot B C}}=\sqrt{\frac{2 \cdot 9 \... | 7.8\, | Other | math-word-problem | Yes | Yes | olympiads | false | 4,391 |
8. A thin beam of light falls normally on a plane-parallel glass plate. Behind the plate, at some distance from it, stands an ideal mirror (its reflection coefficient is equal to one). The plane of the mirror is parallel to the plate. It is known that the intensity of the beam that has passed through this system is 256... | Answer: 0.75
## Solution.
Let $k$ be the reflection coefficient, then we get $I_{1}=I_{0}(1-k)$ (2 points). Similarly, $I_{3}=I_{2}=I_{1}(1-k)=I_{0}(1-k)^{2}$ (2 points). As a result: $I_{k... | 0.75 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,392 |
1. Compute the area of the figure bounded on the plane Oxy by the lines $x=2, x=3, y=0, y=x^{2 / \ln x}$. | Answer: $e^{2}$.
Solution. Note that $\ln \left(x^{1 / \ln x}\right)=2$. Therefore, $x^{1 / \ln x}=e^{2}$ for permissible values of the variable $x$. Thus, the figure represents a rectangle of size $1 \times e^{2}$.
Evaluation. 11 points for the correct solution. | e^{2} | Calculus | math-word-problem | Yes | Yes | olympiads | false | 4,393 |
2. The area of triangle $A B C$ is 1. On the rays $A B, B C$, $C A$, points $B^{\prime}, C^{\prime}, A^{\prime}$ are laid out respectively, such that
$$
B B^{\prime}=A B, \quad C C^{\prime}=2 B C, \quad A A^{\prime}=3 C A
$$
Calculate the area of triangle $A^{\prime} B^{\prime} C^{\prime}$. | Answer: 18.
Solution. We will solve the problem in a general form, assuming that
$$
B B^{\prime}=q A B, \quad C C^{\prime}=r B C, \quad A A^{\prime}=p C A .
$$
We will calculate the area... | 18 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,394 |
3. Solve the equation
$$
\sqrt{\frac{x-3}{11}}+\sqrt{\frac{x-4}{10}}+\sqrt{\frac{x-5}{9}}=\sqrt{\frac{x-11}{3}}+\sqrt{\frac{x-10}{4}}+\sqrt{\frac{x-9}{5}}
$$ | Answer: 14.
Solution. If we perform the variable substitution $x=t+14$, then in all the expressions under the square roots, 1 will be factored out:
$$
\sqrt{\frac{t}{11}+1}+\sqrt{\frac{t}{10}+1}+\sqrt{\frac{t}{9}+1}=\sqrt{\frac{t}{2}+1}+\sqrt{\frac{t}{3}+1}+\sqrt{\frac{t}{4}+1} .(*)
$$
Now it is clear that for $t>0$... | 14 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,395 |
4. For what values of the parameter $a$ does the system of equations
$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z^{2}+2 y=0 \\
x+a y+a z-a=0
\end{array}\right.
$$
have a unique solution? | Answer: $a= \pm \frac{\sqrt{2}}{2}$.
Solution. Let's complete the square for $y$ in the first equation:
$$
x^{2}+(y+1)^{2}+z^{2}=1
$$
Now it is clear that this equation represents a sphere with center at the point $(0, -1, 0)$ and radius 1. The second equation represents a plane. The system of equations has a unique... | \\frac{\sqrt{2}}{2} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,396 |
5. The car engine operates with a power of $P=60 \kappa B m$. Determine the speed $v_{0}$ at which the car was traveling, if after the engine is turned off, it stops after $s=450$ m. The resistance force to the car's motion is proportional to its speed. The mass of the car is $m=1000$ kg. (15 points) | Answer: $30 \mathrm{m} / \mathrm{s}$
Solution. The power of the engine: $P=F v=F_{\text {comp }} v_{0}=\alpha v_{0}^{2}$, that is, the resistance coefficient of the car's movement: $\alpha=\frac{P}{v_{0}^{2}}$ (3 points). The projection of the second law of Newton, for a small time interval, when moving with the engin... | 30\mathrm{}/\mathrm{} | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,397 |
6. An ideal gas was expanded in such a way that during the process, the pressure of the gas turned out to be directly proportional to its volume. As a result, the gas heated up by $\Delta T=100^{\circ} \mathrm{C}$, and the work done by the gas was $A=831$ J. Determine the amount of substance that participated in this p... | Answer: 2 moles
Solution. From the condition $p=\alpha V$ (3 points). The work of the gas is equal to the area under the graph of the given process, constructed in coordinates $p-V$:
$A=\frac{p_{1}+p_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha V_{1}+\alpha V_{2}}{2} \cdot\left(V_{2}-V_{1}\right)=\frac{\alpha}{2... | 2 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,398 |
7. Two small balls with charges $Q=-20 \cdot 10^{-6}$ C and $q=50 \cdot 10^{-6}$ C are located at the vertices $A$ and $B$ of a mountain slope (see figure). It is known that $AB=2$ m, $AC=3$ m. The masses of the balls are the same and equal to $m=200$ g each. At the initial moment of time, the ball with charge $q$ is r... | Answer: $5 \mathrm{M} / \mathrm{c}$
Solution. The law of conservation of energy for this situation:
$k \frac{Q q}{A B}+m g \cdot A B=\frac{m v^{2}}{2}+k \frac{Q q}{B C}(5$ points $)$.
As a result, we get: $v=\sqrt{\frac{2 k Q q}{m \cdot A B}+2 \cdot g \cdot A B-\frac{2 k Q q}{m \cdot B C}}=\sqrt{\frac{2 \cdot 9 \cdo... | 5 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,399 |
8. A thin beam of light falls normally on a plane-parallel glass plate. Behind the plate, at some distance from it, stands an ideal mirror (its reflection coefficient is equal to one). The plane of the mirror is parallel to the plate. It is known that the intensity of the beam that has passed through this system is 16 ... | Answer: 0.5
## Solution.
Let $k$ be the reflection coefficient, then we get $I_{1}=I_{0}(1-k)$ (2 points). Similarly, $I_{3}=I_{2}=I_{1}(1-k)=I_{0}(1-k)^{2}$ (2 points). As a result: $I_{k}=I_{0}(1-k)^{4}$ (2 points). According to the condition $I_{0}=16 \cdot I_{k}=16 \cdot I_{0}(1-k)^{4}$ (2 points). In the end, we... | 0.5 | Other | math-word-problem | Yes | Yes | olympiads | false | 4,400 |
Problem No. 5 (15 points)
The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg, and the load is located at a distance of one quarter of the length of the rod from its left end. Determine the mass $m$ of the second load suspended f... | Answer: 100 kg
# Solution and grading criteria:
Diagram with forces correctly placed.
Tension force in the thread: $T=\frac{M g}{2}$
The lever rule, written relative to point v:
$m g \cdot \frac{1}{2} A B+m g \cdot \frac{3}{4} A B+T \cdot \frac{3}{4} A B=T \cdot A B$
As a result, we get: $M=10 m=100$ kg.
A pers... | 100 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,401 |
# Problem №8 (15 points)
A load was decided to be weighed on unequal-arm scales. When the load was placed on one of the pans of these scales, a weight of mass \( m_{1}=0.5 \) kg had to be placed on the other side to balance it. In the situation where the load was placed on the other pan of the scales, it had to be bal... | Solution and evaluation criteria:
From the condition, it can be concluded that the scales have their own mass, which must be taken into account.
The lever rule for the first weighing:
$m_... | 0.875 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,402 |
Problem No. 5 (15 points)
The system shown in the figure is in equilibrium. It is known that the uniform rod $AB$ and the load lying on it have the same mass $m=10$ kg. The load is located exactly in the middle of the rod. The thread, passing over the pulleys, is attached to one end of the rod and at a distance of one... | # Solution and Evaluation Criteria:
Diagram with forces correctly placed.
Tension force in the string: $T=\frac{M g}{2}$
The lever rule, written relative to point v:
$m g \cdot \frac{1}{2} A B+m g \cdot \frac{1}{2} A B+T \cdot \frac{3}{4} A B=T \cdot A B$
As a result, we get: $M=8 m=80$ kg. | 80 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,403 |
# Problem No. 6 (10 points)
A person started walking slowly along a straight line with a speed of $v_{1}=0.5 \mathrm{~m} / \mathrm{c}$. At time $t_{1}$, he turned strictly to the right and walked with a speed twice as high. After another time interval $t_{1}$, he turned strictly to the right again, and his speed becam... | # Solution and Evaluation Criteria:
From the problem statement, we can conclude that the person's trajectory is a rectangle.
(2 points)
Moreover, \( S_{2} = 2 S_{1} \)
(2 points)
Average... | 1.1\, | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,404 |
# Problem №8 (15 points)
A load was decided to be weighed on unequal-arm scales. When the load was placed on one of the pans of these scales, a weight of mass $m_{1}=1$ kg had to be placed on the other side for balance. In the situation where the load was placed on the other pan of the scales, it had to be balanced by... | # Solution and evaluation criteria:
From the condition, we can conclude that the scales have their own mass, which must be taken into account.
The lever rule for the first weighing:
$m_{z}... | 1.333 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,405 |
2. In triangle $A B C$, the median $B K$ is twice as small as side $A B$ and forms an angle of $32^{\circ}$ with it. Find the angle $A B C$. | Answer: $106^{\circ}$.
Solution. Let $K$ be the midpoint of segment $B D$. Then $A B C D$ is a parallelogram. In triangle $A B D$, we have the equality of sides $A B$ and $B D$. Therefore,
$$
\angle B D A=\frac{1}{2}\left(180^{\circ}-32^{\circ}\right)=74^{\circ}
$$
Angles $A D B$ and $C B D$ are equal as alternate i... | 106 | Geometry | math-word-problem | Yes | Yes | olympiads | false | 4,406 |
1. There are 4 kg of a copper-tin alloy, in which $40\%$ is copper, and 6 kg of another copper-tin alloy, in which $30\%$ is copper. What masses of these alloys need to be taken to obtain 8 kg of an alloy containing $p\%$ copper after melting? Find all $p$ for which the problem has a solution. | Answer: $0.8 p-24$ kg; $32-0.8 p$ kg; $32.5 \leqslant p \leqslant 35$.
Solution. If the first alloy is taken $x$ kg, then the second one is $(8-x)$ kg. The conditions of the problem impose restrictions on the possible values of $x$:
$$
\left\{\begin{array}{l}
0 \leqslant x \leqslant 4 ; \\
0 \leqslant 8-x \leqslant 6... | 32.5\leqslantp\leqslant35 | Algebra | math-word-problem | Yes | Yes | olympiads | false | 4,408 |
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