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4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?
|
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is
$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$
We need to find the value of $x$ for which the function
$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$
has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$
$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$
therefore, for all $x$
$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$
and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]
Second method. The point of maximum of the function $p(x)$ can be found using the derivative.
Answer: 60 seconds. The probability is $4 / 9$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
11. Toll Road (from 8th grade. 3 points). The cost of traveling along a section of a toll road depends on the class of the vehicle: passenger cars belong to the first class, for which the travel cost is 200 rubles, and light trucks and minivans belong to the second class, for which the cost is 300 rubles.
At the entrance to the toll booth, an automatic system is installed that measures the height of the vehicle. If the height of the vehicle is less than a certain threshold value \( h \), the system automatically classifies it as class 1; if it is higher than \( h \), it is classified as class 2. In this case, errors are possible. For example, a low minivan may be classified as class 1, and its driver will be pleased. A low SUV may be incorrectly classified as class 2, and its driver will not be happy. The driver can file a claim, and the operating company will have to refund 100 rubles.
The company's management tasked the engineers with the goal of configuring the system so that the number of errors is minimized.
For several weeks, the engineers collected data on the height of vehicles passing through the toll booth. Separately for class 1 vehicles and separately for class 2 vehicles (Fig. 5). On the x-axis, the height of the vehicle (in cm) is plotted, and on the y-axis, the average number of vehicles of such height passing through the toll booth per day. For clarity, the points are connected by a smooth line.
Fig. 5. Graphs - the number of class 1 and class 2 vehicles per day
Having gathered all this information and drawn the graphs, the engineers began to think about what to do next: how to determine the threshold value \( h \) so that the probability of error is the smallest possible? Solve this problem for them.
|
Solution. We will construct both graphs in the same coordinate system. Draw a vertical line $x=h$ through the point of intersection of the graphs. This value of $h-$ is the one we are looking for.
We will prove this geometrically. Let's call the point of intersection $C$. Draw any vertical line $x=h$ (for definiteness, to the right of the point of intersection of the graphs). Introduce the notation for a few more points, as shown
Fig. 3. as shown in Fig. 3.
There can be two types of errors. The first type is when the system incorrectly classifies a class 1 car as class 2. The number of such errors is represented by the area of the figure under the graph 1 to the right of the line $x=h$, that is, the area of the curvilinear triangle $G F D$. The second type of error: the system incorrectly classifies a class 2 car as class 1. The number of such errors is equal to the area of the curvilinear triangle $A E G$ under the graph 2 to the left of the line $x=h$. The figure composed of triangles $F G D$ and $A E G$ has the smallest area when the area of triangle $C E F$ is the smallest. This area is zero only if the line $x=h$ coincides with the line $B C$. Therefore, the value we are looking for is the abscissa of the common point of the graphs ${ }^{2}$: approximately 190 cm.
Answer: approximately 190 cm.
|
190
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Traffic Lights (from 9th grade. 2 points). Long Highway intersects with Narrow Street and Quiet Street (see fig.). There are traffic lights at both intersections. The first traffic light allows traffic on the highway for $x$ seconds, and for half a minute on Narrow St. The second traffic light allows traffic on the highway for two minutes, and for $x$ seconds on Quiet St. The traffic lights operate independently of each other. For what value of $x$ will the probability of driving through both intersections on Long Highway without stopping at the traffic lights be the greatest? What is this maximum probability?
|
Solution. First method. We will measure time in seconds. The probability of passing the intersection with Narrow St. without stopping is $\frac{x}{x+30}$. The probability of passing the intersection with Quiet St. without stopping is $\frac{120}{x+120}$. Since the traffic lights operate independently of each other, the probability of passing both intersections without stopping is
$$
p(x)=\frac{120 x}{(x+30)(x+120)}=120 \cdot \frac{x}{x^{2}+150 x+3600}
$$
We need to find the value of $x$ for which the function
$$
f(x)=\frac{120}{p(x)}=\frac{x^{2}+150 x+3600}{x}=x+\frac{3600}{x}+150
$$
has the smallest value on the ray $(0 ;+\infty)$. By the Cauchy inequality for means $^{4}$
$$
f(x) \geq 150+2 \sqrt{x \cdot \frac{3600}{x}}=150+2 \cdot 60=270
$$
therefore, for all $x$
$$
p(x)=\frac{120}{f(x)} \leq \frac{120}{270}=\frac{4}{9}
$$
and equality is achieved if $x=\frac{3600}{x}$, from which $x=60$.[^3]
Second method. The point of maximum of the function $p(x)$ can be found using the derivative.
Answer: 60 seconds. The probability is $4 / 9$.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
18. The figure shows a track scheme for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they want, returning to the starting point.
The young driver Yura spends one minute on the path from $A$ to $B$ or back. Yura also spends one minute on the loop. The loop can only be driven counterclockwise (arrows indicate possible directions of movement). Yura does not turn back halfway and does not stop. The race duration is 10 minutes. Find the number of possible different routes (sequences of passing sections). #
|
# Solution.
Let $M_{n}$ be the number of all possible routes of duration $n$ minutes. Each such route consists of exactly $n$ segments (a segment is the segment $A B, B A$ or the loop $B B$).
Let $M_{n, A}$ be the number of such routes ending at $A$, and $M_{n, B}$ be the number of routes ending at $B$.
A point $B$ can be reached in one minute either from point $A$ or from point $B$, so $M_{n, B}=M_{n-1}$.
A point $A$ can be reached in one minute only from point $B$, so $M_{n, A}=M_{n-1, B}=M_{n-2}$. Therefore,
$$
M_{n, A}=M_{n-2}=M_{n-2, A}+M_{n-2, B}=M_{n-4}+M_{n-3}=M_{n-2, A}+M_{n-1, A}
$$
Additionally, note that $M_{1, A}=0, M_{2, A}=1$. Thus, the numbers $M_{n, A}$ form the sequence $0,1,1,2,3,5,8,13,21,34, \ldots$
The number $M_{10, A}$ is 34 - the ninth Fibonacci number ${ }^{1}$.
Answer: 34.
|
34
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Ninth-grader Gavriil decided to weigh a basketball, but he only had 400 g weights, a light ruler with the markings at the ends worn off, a pencil, and many weightless threads at his disposal. Gavriil suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 9 cm. When a third weight was attached to the first two, and the pencil was moved another 5 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
|
2. Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\left(l_{1}+x\right)=2 m\left(l_{2}-x\right) \\
M\left(l_{1}+y\right)=3 m\left(l_{2}-y\right)
\end{gathered}
$$
Subtracting the first equation from the second and third, we get:
$$
\begin{gathered}
M x=m l_{2}-2 m x \\
M y=2 m l_{2}-3 m y
\end{gathered}
$$
From this,
$$
M=\frac{3 y-4 x}{2 x-y} m=600
$$
|
600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. The distances from three points lying in a horizontal plane to the base of a television tower are 800 m, 700 m, and 500 m, respectively. From each of these three points, the tower is visible (from base to top) at a certain angle, and the sum of these three angles is $90^{\circ}$. A) Find the height of the television tower (in meters). B) Round the answer to the nearest whole number of meters.
|
Solution. Let the given distances be denoted by $a, b$, and $c$, the corresponding angles by $\alpha, \beta$, and $\gamma$, and the height of the tower by $H$. Then $\operatorname{tg} \alpha=\frac{H}{a}, \operatorname{tg} \beta=\frac{H}{b}, \operatorname{tg} \gamma=\frac{H}{c}$. Since $\frac{H}{c}=\operatorname{tg} \gamma$ $=\operatorname{tg}\left(90^{\circ}-(\alpha+\beta)\right)=\frac{1}{\operatorname{tg}(\alpha+\beta)}=\frac{1-\operatorname{tg} \alpha \cdot \operatorname{tg} \beta}{\operatorname{tg} \alpha+\operatorname{tg} \beta}=\frac{1-\frac{H}{a} \cdot \frac{H}{b}}{\frac{H}{a}+\frac{H}{b}}$, we have $\frac{H}{c}\left(\frac{H}{a}+\frac{H}{b}\right)=1-\frac{H}{a} \cdot \frac{H}{b} \Leftrightarrow$ $\frac{H^{2}(a+b)}{a b c}=\frac{a b-H^{2}}{a b} \Leftrightarrow H^{2}(a+b)=c\left(a b-H^{2}\right) \Leftrightarrow H^{2}(a+b+c)=a b c \Leftrightarrow$ $H=\sqrt{\frac{a b c}{a+b+c}}$. For the given numerical data: $H=\sqrt{\frac{800 \cdot 500 \cdot 700}{800+500+700}}$ $=\sqrt{\frac{800 \cdot 350}{2}}=\sqrt{140000}=100 \sqrt{14}$.
Since $374=\sqrt{139876}<\sqrt{140000}<\sqrt{140250.25}=374.5$, we have $\sqrt{140000} \approx 374$.
A strict justification for the approximate answer (of the type indicated above or another) is required. Note that a conclusion of the form “since $140000-374^{2}<375^{2}-140000$, then $\sqrt{140000} \approx 374$” is unjustified.
Answer: A) $100 \sqrt{14}$ m; B) 374 m.
Answer for variant 212: A) $25 \sqrt{35}$; B) 148.
|
374
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. If the cold water tap is opened, the bathtub fills up in 5 minutes and 20 seconds. If both the cold water tap and the hot water tap are opened simultaneously, the bathtub fills up to the same level in 2 minutes. How long will it take to fill the bathtub if only the hot water tap is opened? Give your answer in seconds.
$\{192\}$
|
Solution. According to the condition: $\frac{16}{3} v_{1}=1,\left(v_{1}+v_{2}\right) 2=1$, where $v_{1}, v_{2}$ are the flow rates of water from the cold and hot taps, respectively. From this, we get: $v_{1}=3 / 16, v_{2}=5 / 16$.
Then the time to fill the bathtub from the hot tap is $\frac{16}{5}$. Answer: 3 minutes 12 seconds $=192$ seconds
|
192
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A weight with a mass of 200 grams stands on a table. It was flipped and placed on the table with a different side, the area of which is 15 sq. cm smaller. As a result, the pressure on the table increased by 1200 Pa. Find the area of the side on which the weight initially stood. Give your answer in sq. cm, rounding to two decimal places if necessary.
$\{25\}$
|
Solution. After converting to SI units, we get: $\frac{2}{S-1.5 \cdot 10^{-3}}-\frac{2}{S}=1200$.
Here $S-$ is the area of the original face.
From this, we get a quadratic equation: $4 \cdot 10^{5} S^{2}-600 S-1=0$.
After substituting the variable $y=200 S$, the equation becomes: $10 y^{2}-3 y-1=0$, the solution of which is easily found $y=1 / 2$. Thus, the area of the required face is $S=\frac{1}{400} \mathrm{~m}^{2}=25 \mathrm{~cm}^{2}$.
Answer: 25
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The villages of Arkadino, Borisovo, and Vadimovo are connected by straight roads. A square field adjoins the road between Arkadino and Borisovo, one side of which completely coincides with this road. A rectangular field adjoins the road between Borisovo and Vadimovo, one side of which completely coincides with this road, and the second side is 4 times longer. $\mathrm{K}$ road between Arkadino and Vadimovo adjoins a rectangular forest, one side of which completely coincides with this road, and the second side is 12 km. The area of the forest is 45 sq. km greater than the sum of the areas of the fields. Find the total area of the forest and fields in sq. km.
$\{135\}$
|
Solution. The condition of the problem can be expressed by the following relation:
$r^{2}+4 p^{2}+45=12 q$
where $p, q, r$ are the lengths of the roads opposite the settlements Arkadino, Borisovo, and Vadimovo, respectively.
This condition is in contradiction with the triangle inequality:
$r+p>q \Rightarrow 12 r+12 p>12 q \Rightarrow 12 r+12 p>r^{2}+4 p^{2}+45 \Rightarrow(r-6)^{2}+(2 p-3)^{2}<0$.
From this, it follows that all three settlements lie on the same straight line.
Moreover, $r=6, p=1.5, q=7.5$.
The total area is the sum of:
$r^{2}+4 p^{2}+12 q=36+9+90=135$
Answer: 135
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Alloy $A$ of two metals with a mass of 6 kg, in which the first metal is twice as much as the second, placed in a container with water, creates a pressure force on the bottom of $30 \mathrm{N}$. Alloy $B$ of the same metals with a mass of 3 kg, in which the first metal is five times less than the second, placed in a container with water, creates a pressure force on the bottom of $10 \mathrm{N}$. What pressure force (in newtons) will a third alloy, obtained by melting the initial alloys, create?
$\{40\}$
|
Solution. Due to the law of conservation of mass, in the resulting alloy, the mass of each metal is equal to the sum of the masses of these metals in the initial alloys. Thus, both the gravitational forces and the forces of Archimedes also add up. From this, it follows that the reaction force will be the sum of the reaction forces in the first two cases. Answer: $30+10=40$.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. From a square steel sheet with a side of 1 meter, a triangle is cut off from each of the four corners so that a regular octagon remains. Determine the mass of this octagon if the sheet thickness is 3 mm and the density of steel is 7.8 g/cm ${ }^{3}$. Give your answer in kilograms, rounding to the nearest whole number if necessary.
|
Answer: $46.8(\sqrt{2}-1) \approx 19$ kg.
Solution. A regular octagon must have equal angles and sides. Therefore, four equal triangles with angles $45^{\circ}, 45^{\circ}$, and $90^{\circ}$ are cut off. If the legs of this triangle are equal to $x$, then the hypotenuse is $x \sqrt{2}$ - this will be the side of the octagon. Therefore, the length of the side of the square is $2 x+x \sqrt{2}$, and we get the equation $2 x+x \sqrt{2}=100$. Hence, $x=\frac{100}{2+\sqrt{2}}=50(2-\sqrt{2})$ cm, and the area of the remaining octagon is $10000-4 \cdot \frac{x^{2}}{2}=10000-2 \cdot 2500 \cdot(2-\sqrt{2})^{2}=10000-5000 \cdot(6-4 \sqrt{2})=20000(\sqrt{2}-1)\left(\mathrm{cm}^{2}\right)$. Therefore, its volume is $20000(\sqrt{2}-1) \cdot 0.3=6000(\sqrt{2}-1)$ (cm ${ }^{3}$ ), and its mass is $6000(\sqrt{2}-1) \cdot 7.8$ $=46800(\sqrt{2}-1)$ g, which is $46.8(\sqrt{2}-1)$ kg.
Rounding to the nearest whole number can be done in different ways. One way: since $\sqrt{2}>1.4$, the mass is greater than 18.72. Since $\sqrt{2}<1.415$ (proven by squaring), the mass is less than 19.422. Therefore, the nearest whole number is 19 kg. The main requirement is that the proof must be based on strict estimates, not on approximate calculations without accuracy assessment. In the case of unjustified rounding, a score of 10 points was given. In the case of an incorrect answer (regardless of the reasons), the problem was scored 0 points.
Answer to variant 172: $43.2(\sqrt{2}-1) \approx 18$ kg.
Answer to variant 173: $31.2(\sqrt{2}-1) \approx 13$ kg.
Answer to variant 174: $64.8(\sqrt{2}-1) \approx 27$ kg.
|
19
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In the village where Glafira lives, there is a small pond that is filled by springs at the bottom. Curious Glafira found out that a herd of 17 cows completely drank the pond dry in 3 days. After some time, the springs refilled the pond, and then 2 cows drank it dry in 30 days. How many days would it take for one cow to drink the pond dry?
|
Answer: In 75 days.
Solution. Let the pond have a volume of $a$ (conditional units). These units can be liters, buckets, cubic meters, etc. Let one cow drink $b$ (conditional units) of water per day, and the springs add $c$ (conditional units) of water per day. Then the first condition of the problem is equivalent to the equation $a+3c=3 \cdot 17b$, and the second to the equation $a+30c=30 \cdot 2b$. We have obtained a system of two equations with three unknowns. The unknowns cannot be found from here, but the relationship between them can be established.
Subtracting the first equation from the second, we get $b=3c$. Substituting into one of the equations gives $a=150c$.
If one cow drinks the pond dry in $x$ days, then we get $a+xc=xb$, that is, $x=\frac{a}{b-c}=75$ days.
Answer to variant 172: In 140 days.
Answer to variant 173: In 80 days.
Answer to variant 174: In 150 days.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Gavriila was traveling in Africa. On a sunny and windy day, at noon, when the rays from the Sun fell vertically, the boy threw a ball from behind his head at a speed of 5 m/s against the wind at an angle to the horizon. After 1 second, the ball hit him in the stomach 1 m below the point of release. Determine the greatest distance the shadow of the ball moved away from Gavriila's feet. The force acting on the ball from the air is directed horizontally and does not depend on the position and speed. The acceleration due to gravity \( g \) is \( 10 \, \text{m} / \text{s}^2 \).
|
Answer: 75 cm.
Solution. In addition to the force of gravity, a constant horizontal force $F=m \cdot a$ acts on the body, directed opposite. In a coordinate system with the origin at the point of throw, the horizontal axis $x$ and the vertical axis $y$, the law of motion has the form:
$$
\begin{aligned}
& x(t)=V \cdot \cos \alpha \cdot t-\frac{a t^{2}}{2} \\
& y(t)=V \cdot \sin \alpha \cdot t-\frac{g t^{2}}{2}
\end{aligned}
$$
From the condition, the coordinates of the point at time $\tau$ are known:
$$
\begin{gathered}
0=V \cdot \cos \alpha \cdot \tau-\frac{a \tau^{2}}{2} \\
-H=V \cdot \sin \alpha \cdot \tau-\frac{g \tau^{2}}{2}
\end{gathered}
$$
From this system of two equations with two unknowns, we find the acceleration created by the force acting from the air and the angle at which the throw was made. From the second equation: $\sin \alpha=\frac{1}{V}\left(\frac{g \tau}{2}-\frac{H}{\tau}\right)=\frac{1}{5}\left(\frac{10 \cdot 1}{2}-\frac{1}{1}\right)=\frac{4}{5}$. Therefore, $\cos \alpha=\frac{3}{5}$, and from the first equation: $a=\frac{2 V \cdot \cos \alpha}{\tau}=\frac{2 \cdot 5 \cdot 3}{5 \cdot 1}=6 \mathrm{~m} / \mathrm{c}^{2}$.
The maximum distance of the shadow corresponds to the maximum value of $x$, which is achieved at $t=\frac{V \cos \alpha}{a}$ (the vertex of the parabola). This value is $L=\frac{V^{2} \cos ^{2} \alpha}{2 a}$ $=\frac{5^{2} \cdot 3^{2}}{5^{2} \cdot 2 \cdot 6}=\frac{3}{4} \mathrm{m}$.
For reference: the answer in general form $L=\frac{\tau}{4} \sqrt{V^{2}-\left(\frac{g \tau}{2}-\frac{H}{\tau}\right)^{2}}$ (in variants 171 and 173), $L=\frac{\tau}{4} \sqrt{V^{2}-\left(\frac{g \tau}{2}+\frac{H}{\tau}\right)^{2}}$ (in variants 172 and 173).
Answer to variant 172: 1.6 meters.
Answer to variant 173: 60 cm.
Answer to variant 174: 2 meters.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. To lift a load, it is attached to the hook of a crane using slings made of steel cable. The calculated mass of the load is $M=20$ t, the number of slings $n=3$. Each sling forms an angle $\alpha=30^{\circ}$ with the vertical. All slings carry the same load during the lifting of the cargo. According to safety requirements, the tensile strength of the cable should exceed the calculated load by a factor of $k=6$. The cable consists of a very large number of steel threads, each of which can withstand a maximum load of $q=10^{3} \mathrm{H} / \mathrm{mm}^{2}$. Indicate the smallest diameter of the cable that can be used to manufacture the slings, if the available cables have any integer diameter in millimeters. The acceleration due to gravity $g$ is considered to be $10 \mathrm{~m} / \mathrm{s}^{2}$.
|
Answer: 26 mm
Solution. For each of the $n$ lower tie-downs, the force of the cargo weight is $\frac{P}{n}$. Then the tension force in the tie will be $N=\frac{P}{n \cdot \cos \alpha}$. Therefore, the strength of the rope must be $Q \geq k T=\frac{k P}{n \cdot \cos \alpha}$.
Since the strength of the rope $Q$ is determined by the cross-sectional area of all the threads $S$ and the strength of each thread $q(Q=S \cdot q)$, then $S q \geq \frac{k P}{n q \cdot \cos \alpha} \Rightarrow S \geq \frac{k P}{n q \cdot \cos \alpha}$.
Let's establish the relationship between the cross-sectional area of all the threads $S$ and the cross-sectional area of the rope $A$, assuming the number of threads is very large.
Tile the cross-section of the rope with hexagons of area $\mathrm{S}_{6}$, in which we inscribe circles of area $\mathrm{S}_{0}$. Then $\frac{A}{S}=\frac{S_{6}}{S_{0}}=\frac{6 \cdot 0.5 R \cdot 2 R}{\pi R^{2} \sqrt{3}}=\frac{2 \sqrt{3}}{\pi}$.
For the diameter of the rope $D$, we get
$$
A=\frac{\pi D^{2}}{4}=\frac{2 \sqrt{3}}{\pi} \cdot \frac{k P}{n q \cdot \cos \alpha} \Rightarrow D=\frac{2}{\pi} \cdot \sqrt{\frac{2 \sqrt{3} \cdot k M g}{n q \cdot \cos \alpha}} .
$$
Substituting the numerical values gives $D=\frac{80}{\pi}$. Since this number is between 25 and 26, the smallest diameter of the rope is 26 mm.
Answer to variant 172: 13 mm
Answer to variant 173: 26 mm
Answer to variant 174: 13 mm
|
26
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. On the shores of a circular island (viewed from above), there are cities $A, B, C$, and $D$. The straight asphalt road $A C$ divides the island into two equal halves. The straight asphalt road $B D$ is shorter than road $A C$ and intersects it. The speed of a cyclist on any asphalt road is 15 km/h. The island also has straight dirt roads $A B, B C, C D$, and $A D$, on which the cyclist's speed is the same. The cyclist reaches each of the points $A, C$, and $D$ from point $B$ via a straight road in 2 hours. Find the area enclosed by the quadrilateral $A B C D$.
|
Answer: 450 sq. km. Solution. The condition of the problem means that a quadrilateral $ABCD$ is given, in which angles $B$ and $D$ are right (they rest on the diameter), $AB = BC$ (both roads are dirt roads, and the cyclist travels them in the same amount of time), $BD = 15 \frac{\text{km}}{\text{hour}} \cdot 2$ hours $= 30$ km. Drop two perpendiculars from point $B$: $BM$ to line $AD$, and $BN$ to line $CD$. Then $\triangle BMA = \triangle BNC$ (both are right triangles, the hypotenuses are equal, $\angle BCN = \angle BAM$ - each of these angles, together with $\angle BAD$, gives $180^{\circ}$). Therefore, quadrilateral $MBND$ is equal in area to quadrilateral $ABCD$. At the same time, $MBND$ is a square, the diagonal $BD$ of which is known. Therefore, its area is $\frac{30^2}{2} = 450$ sq. km.
Answer to variant 162: 800 sq. km.
Answer to variant 163: 200 sq. km.
Answer to variant 164: 50 sq. km.
|
450
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. One mole of a monatomic ideal gas undergoes a cyclic process $a b c a$. The diagram of this process in the $P-T$ axes represents a curvilinear triangle, the side $a b$ of which is parallel to the $T$ axis, the side $b c$ - a segment of a straight line passing through the origin, and the side $c a$ - an arc of a parabola passing through the origin, the axis of which is parallel to the $T$ axis. At points $a$ and $c$, the temperature of the gas is the same and equal to $T_{0}=320 \mathrm{~K}$, and the pressure at point $a$ is half the pressure at point $c$. Determine the work done by the gas over the cycle.
|
Answer: 664 J. Solution. Process $a b$ is an isobar, process $b c$ is an isochore, process $c a$ is described by the equation $T=P(d-k P)$, where $d, k$ are some constants. It is not difficult to see that in such a process, the volume turns out to be a linear function of pressure, that is, in the $P V$ axes, this cyclic process is represented by a right triangle with legs $a b$ and $b c$.
Let $P_{0}, V_{0}$ be the parameters of the gas at point $b$. Then the parameters of the gas at point $a: P_{0}, 2 V_{0}$, at point $c$: $2 P_{0}, V_{0}$. From this, we find: $A=\frac{1}{2} P_{0} V_{0}=\frac{1}{4} R T_{0}=664$ J.
Answer to variant 162: $\frac{2}{3} R T_{0}=1662$ J.
Answer to variant 163: $\frac{2}{3} R T_{0}=1994$ J.
Answer to variant 164: $\frac{1}{4} R T_{0}=582$ J.
|
664
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. For moving between points located hundreds of kilometers apart on the Earth's surface, people in the future will likely dig straight tunnels through which capsules will move without friction, solely under the influence of Earth's gravity. Let points $A, B$, and $C$ lie on the same meridian, and the distance from $A$ to $B$ along the surface is to the distance from $B$ to $C$ along the surface as $m: n$. A capsule travels through the tunnel $A B$ in approximately 42 minutes. Estimate the travel time through the tunnel $A C$. Provide your answer in minutes.
|
Answer: 42 min. Solution. Let point $O$ be the center of the Earth. To estimate the time of motion from $A$ to B, consider triangle $A O B$. We can assume that the angle $\alpha=90^{\circ}-\angle A B O$ is very small, so $\sin \alpha \approx \alpha$. Since the point in the tunnel $A B$ is attracted to the center by the gravitational force $F$, the projection of this force onto the direction of motion is $F \sin \alpha \approx F \alpha$. Under the action of this force, the body will move along the channel with acceleration $a \approx R \ddot{\alpha}$. Therefore, the model of the point's motion from $A$ to $B$ can be considered as the model of a mathematical pendulum, for which the period does not depend on the distance from $B$ to $A$. This means that the time of motion will be the same.
Answer to variant 162: 42 min.
Answer to variant 163: 42 min.
Answer to variant 164: 42 min.
|
42
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. Gavriila found out that the front tires of the car last for 20000 km, while the rear tires last for 30000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
2.1. Gavriila found out that the front tires of the car last for 20000 km, while the rear tires last for 30000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{24000\}$.
|
24000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.2. Gavriila found out that the front tires of the car last for 24000 km, while the rear tires last for 36000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
2.2. Gavriila found out that the front tires of the car last for 24000 km, while the rear tires last for 36000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{28800\}$.
|
28800
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.3. Gavriila found out that the front tires of the car last for 42,000 km, while the rear tires last for 56,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
2.3. Gavriila found out that the front tires of the car last for 42000 km, while the rear tires last for 56000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{48000\}$.
|
48000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.4. Gavriila found out that the front tires of the car last for 21,000 km, while the rear tires last for 28,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
2.4. Gavriila found out that the front tires of the car last for 21000 km, while the rear tires last for 28000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{24000\}$.
|
24000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabilize in the first vessel if the valve is opened? The density of water is 1000 kg \(/ \text{m}^3\), and the density of oil is 700 kg \(/ \text{m}^3\). Neglect the volume of the connecting pipe. Provide the answer in centimeters.
|
3.2. Two identical cylindrical vessels are connected at the bottom by a small-section pipe with a valve. While the valve was closed, water was poured into the first vessel, and oil into the second, so that the level of the liquids was the same and equal to \( h = 40 \, \text{cm} \). At what level will the water stabilize in the first vessel if the valve is opened? The density of water is 1000 kg \(/ \text{m}^3\), and the density of oil is 700 kg \(/ \text{m}^3\). Neglect the volume of the connecting pipe. Give the answer in centimeters.
Answer. \{34\}.
|
34
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On the shores of a circular island (viewed from above), there are cities $A, B, C$, and $D$. A straight asphalt road $A C$ divides the island into two equal halves. A straight asphalt road $B D$ is shorter than road $A C$ and intersects it. The speed of a cyclist on any asphalt road is 15 km/h. The island also has straight dirt roads $A B, B C, C D$, and $A D$, on which the cyclist's speed is the same. The cyclist reaches each of the points $A, C$, and $D$ from point $B$ via a straight road in 2 hours. Find the area enclosed by the quadrilateral $A B C D$.
|
Answer: 450 sq. km. Solution. The condition of the problem means that a quadrilateral $ABCD$ is given, in which angles $B$ and $D$ are right (they rest on the diameter), $AB=BC$ (both roads are dirt roads, and the cyclist travels them in the same amount of time), $BD=15 \frac{\text { km }}{\text { h }} 2$ hours $=30$ km. Drop two perpendiculars from point $B$: $BM$ - to the line $AD$, and $BN$ - to the line $CD$. Then $\triangle \mathrm{BMA}=\triangle \mathrm{BNC}$ (both are right triangles, the hypotenuses are equal, $\angle BNC=\angle BAM$ - each of these angles, together with $\angle BAD$, gives $180^{\circ}$). Therefore, the quadrilateral $MBND$ is equal in area to the quadrilateral $ABCD$. At the same time, $MBND$ is a square, the diagonal $BD$ of which is known. Therefore, its area is $\frac{30^{2}}{2}=450$ sq. km.
|
450
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Scientists have found a fragment of an ancient manuscript on mechanics. It was a piece of a book, the first page of which was numbered 435, and the last page was written with the same digits but in some other order. How many sheets were in this fragment?
|
Solution. Since the sheet has 2 pages and the first page is odd, the last page must be even. Therefore, the last digit is 4. The number of the last page is greater than the first. The only possibility left is 534. This means there are 100 pages in total, and 50 sheets.
Answer: 50.
Criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; 15 points - correct solution, but the number of pages (100) is recorded in the answer instead of the number of sheets; 10 points - the number of the last page is correctly determined, but the solution either stops there or is incorrect; 5 points - there are some correct ideas in the solution; **0** points - everything else.
|
50
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Usually, schoolboy Gavriil takes a minute to go up a moving escalator by standing on its step. But if Gavriil is late, he runs up the working escalator and thus saves 36 seconds. Today, there are many people at the escalator, and Gavriil decides to run up the adjacent non-working escalator. How much time will such an ascent take him if he always exerts the same effort when running up the escalator?
|
Solution. Let's take the length of the escalator as a unit. Let $V$ be the speed of the escalator, and $U$ be the speed of Gavrila relative to it. Then the condition of the problem can be written as:
$$
\left\{\begin{array}{c}
1=V \cdot 60 \\
1=(V+U) \cdot(60-36)
\end{array}\right.
$$
The required time is determined from the relationship $1=U \cdot t$. From the system, we get $V=\frac{1}{60} ; U+V=\frac{1}{24}$; $U=\frac{1}{24}-\frac{1}{60}=\frac{1}{40}$. Therefore, $t=40$ seconds.
Answer: 40 seconds.
Criteria: 20 points - complete and correct solution, 15 points - correct approach to the solution, but an arithmetic error is made; $\mathbf{1 0}$ points - the system of equations is correctly set up, but the answer is not obtained; 0 points - everything else.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $20 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 40 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude of 45 km? Air resistance is not taken into account, and the acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
|
Answer: a) 48 km; b) yes. Solution. Let $a=20$ m/s², $\tau=40$ s. On the first segment of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t, y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop working: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ - this will be the "zero" moment for the second segment.
On the second segment: $V=V_{0}-g t=a \tau-g t, y=y_{0}+V_{0} t-\frac{g t^{2}}{2}=\frac{a \tau^{2}}{2}+a \tau t-\frac{g t^{2}}{2}$. At the maximum height, $V=a \tau-g t=0$, so the time when the rocket will be at the maximum height is: $t=\frac{a \tau}{g}$. Therefore, the maximum height the rocket can reach is: $y_{\max }=\frac{a \tau^{2}}{2}+a \tau \frac{a \tau}{g}-\frac{g}{2} \frac{a^{2} \tau^{2}}{g^{2}}=\frac{a \tau^{2}(g+a)}{2 g}$. Substituting the numbers gives $y_{\max }=48$ km.
Grading criteria: 20 points: correct solution and correct answers to both questions; 15 points: correct solution and correct answer to the first question, the answer to the second question is incorrect or missing; 10 points: correct equations and correct logic, but the answer is incorrect due to an arithmetic error; 5 points: correct equations and correct answer for the case where the acceleration on the first stage is reduced by $g$.
|
48
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $30 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 30 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude of 50 km? Air resistance is not taken into account, and the acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
|
Answer: a) 54 km; b) yes. Solution. Let $a=30 \mathrm{m} / \mathrm{s}^{2}, \tau=30$ s. On the first segment of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t, y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ - this will be the "zero" moment for the second segment.
On the second segment: $V=V_{0}-g t=a \tau-g t, y=y_{0}+V_{0} t-\frac{g t^{2}}{2}=\frac{a \tau^{2}}{2}+a \tau t-\frac{g t^{2}}{2}$. At the maximum height, $V=a \tau-g t=0$, so the time when the rocket will be at the maximum height is: $t=\frac{a \tau}{g}$. Therefore, the maximum height the rocket can reach is: $y_{\max }=\frac{a \tau^{2}}{2}+a \tau \frac{a \tau}{g}-\frac{g}{2} \frac{a^{2} \tau^{2}}{g^{2}}=\frac{a \tau^{2}(g+a)}{2 g}$. Substituting the numbers gives $y_{\max }=54$ km.
Grading criteria: 20 points: correct solution and correct answers to both questions; 15 points: correct solution and correct answer to the first question, incorrect or omitted answer to the second question; 10 points: correct equations and logic, but incorrect answer due to arithmetic error; 5 points: correct equations and correct answer for the case where the acceleration on the first stage is reduced by $g$.
|
54
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 10 meters long and 4 meters wide, loaded with sand. The surface of the sand has an angle of no more than 45 degrees with the base plane (otherwise the sand grains will spill), the density of the sand is 1500 kg/m³. Find the maximum mass of sand loaded onto the platform.
|
Answer: 52 t. Solution. The calculation shows that the maximum height of the sand pile will be equal to half the width of the platform, that is, 2 m. The pile can be divided into a "horizontally lying along the platform" prism (its height is 6 m, and the base is an isosceles right triangle with legs $2 \sqrt{2}$ and hypotenuse 4), and two identical pyramids at the "ends" of the platform (the base of each is a rectangle 4 x 2 and the height is 2).
The total volume is $\frac{1}{2} \cdot 4 \cdot 2 \cdot 6 + 2 \cdot \frac{1}{3} \cdot 4 \cdot 2 \cdot 2 = 24 + \frac{32}{3} = \frac{104}{3} \mathrm{m}^{3}$. Therefore, the mass is $\frac{104}{3} \cdot 1500 = 52000$ kg.
Grading criteria: 20 points: correct solution and correct answer; 15 points: with correct overall solution and correct answer, there are minor defects; 10 points: correct solution, but the answer is incorrect due to an arithmetic error.
|
52000
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $20 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 50 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude of 70 km? Air resistance is not taken into account, and the acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
|
Answer: a) 75 km; b) yes. Solution. Let $a=20$ m/s$^2$, $\tau=50$ s. During the first part of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t$, $y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau$, $y_{0}=\frac{a \tau^{2}}{2}$ - this will be the "zero" moment for the second part.
During the second part: $V=V_{0}-g t=a \tau-g t$, $y=y_{0}+V_{0} t-\frac{g t^{2}}{2}=\frac{a \tau^{2}}{2}+a \tau t-\frac{g t^{2}}{2}$. At the maximum height, $V=a \tau-g t=0$, so the time when the rocket will be at the maximum height is: $t=\frac{a \tau}{g}$. Therefore, the maximum height the rocket will reach is: $y_{\max }=\frac{a \tau^{2}}{2}+a \tau \frac{a \tau}{g}-\frac{g}{2} \frac{a^{2} \tau^{2}}{g^{2}}=\frac{a \tau^{2}(g+a)}{2 g}$. Substituting the numbers gives $y_{\max }=75$ km.
Grading criteria: 20 points: correct solution and correct answers to both questions; 15 points: correct solution and correct answer to the first question, incorrect or omitted answer to the second question; 10 points: correct equations and logic, but incorrect answer due to arithmetic error; 5 points: correct equations and correct answer for the case where $g$ is subtracted from the acceleration during the first stage.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 8 meters long and 5 meters wide, loaded with grain. The surface of the grain has an angle of no more than 45 degrees with the base plane (otherwise the grains will spill), the density of the grain is 1200 kg/m³. Find the maximum mass of grain loaded onto the platform.
|
Answer: 47.5 t. Solution. The calculation shows that the maximum height of the grain pile will be half the width of the platform, that is, 2.5 m. The pile can be divided into a "horizontally lying along the platform" prism (its height is 3 m, and the base is a right-angled isosceles triangle with legs $\frac{5 \sqrt{2}}{2}$ and hypotenuse 5), and two identical pyramids at the "ends" of the platform (the base of each is a rectangle 5 x 2.5 and the height is 2.5).
The total volume is $\frac{1}{2} \cdot 5 \cdot \frac{5}{2} \cdot 3 + 2 \cdot \frac{1}{3} \cdot 5 \cdot \frac{5}{2} \cdot \frac{5}{2} = \frac{75}{4} + \frac{125}{6} = \frac{25 \cdot 19}{12} \mathrm{m}^{3}$. Therefore, the mass is $\frac{25 \cdot 19}{12} \cdot 1200 = 47500$ kg.
Grading criteria: 20 points: correct solution and correct answer; 15 points: with minor defects in an otherwise correct solution and correct answer; 10 points: correct solution, but the answer is incorrect due to an arithmetic error.
|
47500
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The engines of a rocket launched vertically upward from the Earth's surface, providing the rocket with an acceleration of $30 \mathrm{~m} / \mathrm{c}^{2}$, suddenly stopped working 20 seconds after launch. To what maximum height will the rocket rise? Can this rocket pose a danger to an object located at an altitude of 20 km? Air resistance is not taken into account, and the acceleration due to gravity is considered to be $10 \mathrm{~m} / \mathrm{c}^{2}$.
|
Answer: a) 24 km; b) yes. Solution. Let $a=30 \mathrm{m} / \mathrm{s}^{2}, \tau=20$ s. During the first part of the motion, when the engines were working, the speed and the height gained are respectively: $V=a t, y=\frac{a t^{2}}{2}$. Therefore, at the moment the engines stop: $V_{0}=a \tau, y_{0}=\frac{a \tau^{2}}{2}$ - this will be the "zero" moment for the second part.
During the second part: $V=V_{0}-g t=a \tau-g t, y=y_{0}+V_{0} t-\frac{g t^{2}}{2}=\frac{a \tau^{2}}{2}+a \tau t-\frac{g t^{2}}{2}$. At the maximum height, $V=a \tau-g t=0$, so the time when the rocket will be at the maximum height is: $t=\frac{a \tau}{g}$. Therefore, the maximum height the rocket will reach is: $y_{\max }=\frac{a \tau^{2}}{2}+a \tau \frac{a \tau}{g}-\frac{g}{2} \frac{a^{2} \tau^{2}}{g^{2}}=\frac{a \tau^{2}(g+a)}{2 g}$. Substituting the numbers gives $y_{\max }=24$ km.
Grading criteria: 20 points: correct solution and correct answers to both questions; 15 points: correct solution and correct answer to the first question, incorrect or omitted answer to the second question; 10 points: correct equations and logic, but incorrect answer due to arithmetic error; 5 points: correct equations and correct answer for the case where the acceleration on the first stage is reduced by $g$.
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A mobile railway platform has a horizontal bottom in the form of a rectangle 8 meters long and 4 meters wide, loaded with sand. The surface of the sand has an angle of no more than 45 degrees with the base plane (otherwise the sand grains will spill), the density of the sand is 1500 kg/m³. Find the maximum mass of sand loaded onto the platform.
|
Answer: 40 t. Solution. The calculation shows that the maximum height of the sand pile will be equal to half the width of the platform, that is, 2 m. The pile can be divided into a "horizontally lying along the platform" prism (its height is 4 m, and the base is an isosceles right triangle with legs $2 \sqrt{2}$ and hypotenuse 4), and two identical pyramids at the "ends" of the platform (the base of each is a rectangle 4 x 2 and the height is 2).
The total volume is $\frac{1}{2} \cdot 4 \cdot 2 \cdot 4 + 2 \cdot \frac{1}{3} \cdot 4 \cdot 2 \cdot 2 = 16 + \frac{32}{3} = \frac{80}{3}$ m $^{3}$. Therefore, the mass is $\frac{80}{3} \cdot 1500 = 40000$ kg.
Grading criteria: 20 points: correct solution and correct answer; 15 points: with minor defects in an otherwise correct solution and correct answer; 10 points: correct solution, but the answer is incorrect due to an arithmetic error.
|
40000
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The time of the aircraft's run from the moment of start until the moment of takeoff is 15 seconds. Find the length of the run if the takeoff speed for this aircraft model is 100 km/h. Assume the aircraft's motion during the run is uniformly accelerated. Provide the answer in meters, rounding to the nearest whole number if necessary.
|
Answer: 208
$v=a t, 100000 / 3600=a \cdot 15$, from which $a=1.85\left(\mathrm{m} / \mathrm{s}^{2}\right)$. Then $S=a t^{2} / 2=208$ (m).
|
208
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Gavriil got on the train with a fully charged smartphone, and by the end of the trip, his smartphone was completely drained. For half of the time, he played Tetris, and for the other half, he watched cartoons. It is known that the smartphone fully discharges in 3 hours of video watching or in 5 hours of playing Tetris. What distance did Gavriil travel if the train moved half the distance at an average speed of 80 km/h and the other half at an average speed of 60 km/h? Give the answer in kilometers, rounding to the nearest whole number if necessary.
|
Answer: 257
Let's assume the "capacity" of the smartphone battery is 1 unit (u.e.). Then the discharge rate of the smartphone when watching videos is $\frac{1}{3}$ u.e./hour, and the discharge rate when playing games is $\frac{1}{5}$ u.e./hour. If the total travel time is denoted as $t$ hours, we get the equation $\frac{1}{3} \cdot \frac{t}{2} + \frac{1}{5} \cdot \frac{t}{2} = 1$, which simplifies to $\frac{(5+3) t}{2 \cdot 3 \cdot 5} = 1$, or $t = \frac{15}{4}$ hours. Then we have (where $S$ is the distance): $\frac{S}{2 \cdot 80} + \frac{S}{2 \cdot 60} = \frac{15}{4}$, which simplifies to $\frac{S}{40} + \frac{S}{30} = 15$,
$$
S = 15 \cdot \frac{40 \cdot 3}{4+3} = \frac{1800}{7} \approx 257 \text{ km. }
$$
## Lomonosov Olympiad in Mechanics and Mathematical Modeling - 2021/2022
9th grade
Each problem is worth 20 points. A score of 20 points is given for a correct and complete solution and the correct answer.
A score of 15 points is given for a solution with various deficiencies (inadequate justification, inaccuracies, etc.). In some problems, scores of 5 and 10 points were also given for partial progress in the solution.
## Grading Criteria for Problems:
|
257
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 3. Experimenters Glafira and Gavriil placed a triangle made of thin wire with sides of 30 mm, 40 mm, and 50 mm on a white flat surface. This wire is covered with millions of mysterious microorganisms. The scientists found that when an electric current is applied to the wire, these microorganisms begin to move chaotically on this surface in different directions at an approximate speed of $\frac{1}{6}$ mm/sec. As they move, the surface along their path is stained red. Find the area of the stained surface after 1 minute of current application. Round it to the nearest whole number of square millimeters.
|
Solution. In one minute, the microorganism moves 10 mm. Since in a right triangle with sides $30, 40, 50$, the radius of the inscribed circle is 10, all points inside the triangle are no more than 10 mm away from the sides of the triangle. Therefore, the microorganisms will fill the entire interior of the triangle.
When moving outward, points 10 mm away from the sides of the triangle and points 10 mm away from the vertices will be reached.
In the end, the total occupied area is: the area of the triangle + three strips 10 mm wide each, located outside the triangle, with a total length equal to the perimeter of the triangle + 3 circular sectors with a radius of 10, which together form a circle. We get:
$$
\frac{30 \cdot 40}{2} + 10 \cdot (30 + 40 + 50) + \pi \cdot 10^{2} = 600 + 1200 + 100 \pi = 1800 + 100 \pi \approx 2114 \mathrm{Mm}^{2}
$$
Answer: $1800 + 100 \pi \approx 2114$ mm $^{2}$.
|
2114
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 5. An electric kettle heats water from room temperature $T_{0}=20^{\circ} \mathrm{C}$ to $T_{m}=100^{\circ} \mathrm{C}$ in $t=10$ minutes. How long will it take $t_{1}$ for all the water to boil away if the kettle is not turned off and the automatic shut-off system is faulty? The specific heat capacity of water $c=4200$ J/kg $\cdot$ K. The specific latent heat of vaporization of water $L=2.3$ MJ/kg. Round the answer to the nearest whole number of minutes.
|
Solution. The power $P$ of the kettle is fixed and equal to $P=Q / t$. From the heat transfer law $Q=c m\left(T_{m}-T_{0}\right)$ we get $P t=c m\left(T_{m}-T_{0}\right)$.
To evaporate the water, the amount of heat required is $Q_{1}=L m \Rightarrow P t_{1}=L m$.
By comparing these relations, we obtain $\frac{t_{1}}{t}=\frac{L m}{c m\left(T_{m}-T_{0}\right)}=\frac{L}{c\left(T_{m}-T_{0}\right)} \Rightarrow$
$$
\frac{t_{1}}{t}=\frac{2.3 \cdot 10^{6}}{4200 \cdot 80}=\frac{2300}{336} \approx 6.845
$$
Thus, the water will boil away in $t_{1}=6.845 \cdot t=68.45$ minutes.
Answer: 68 minutes.
|
68
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Upon entering the Earth's atmosphere, the asteroid heated up significantly and exploded near the surface, breaking into a large number of fragments. Scientists collected all the fragments and divided them into groups based on size. It was found that one-fifth of all fragments had a diameter of 1 to 3 meters, another 26 were very large (more than 3 meters in diameter), and the rest were divided into several groups, each of which accounted for 1/7 of the total number of fragments. How many fragments did the scientists collect?
|
1. Answer: 70. Solution. Let $\mathrm{X}$ be the total number of fragments. The condition of the problem leads to the equation:
$\frac{x}{5}+26+n \cdot \frac{X}{7}=X$, where $n-$ is the unknown number of groups. From the condition of the problem, it follows that the number of fragments is a multiple of 35
$$
X=35 l, l \in \mathbb{N}
$$
Then the first equation can be rewritten in another form: $7 l+26+n \cdot 5 l=35 l$. From this, we express: $5 n=28-\frac{26}{l}$. It is clear that the natural variable $l$ can take the values $1,2,13$ and 26. By direct verification, we find that only one variant is possible $l=2, n=$ 3. Substituting into (1) we get the answer $X=70$.
|
70
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The mass of the first iron ball is $1462.5 \%$ greater than the mass of the second ball. By what percentage will less paint be needed to paint the second ball compared to the first? The volume of a sphere with radius $R$ is $\frac{4}{3} \pi R^{3}$, and the surface area of a sphere is $4 \pi R^{2}$.
|
2. Answer: $84 \%$.
Let's denote the radii of the spheres as $R$ and $r$ respectively. Then the first condition means that
$$
\frac{\frac{4}{3} \pi R^{3}-\frac{4}{3} \pi r^{3}}{\frac{4}{3} \pi r^{3}} \cdot 100=1462.5 \Leftrightarrow \frac{R^{3}-r^{3}}{r^{3}}=14.625 \Leftrightarrow \frac{R^{3}}{r^{3}}=\frac{125}{8} \Leftrightarrow R=\frac{5}{2} r . \text { The sought }
$$
percentage is: $\frac{4 \pi R^{2}-4 \pi r^{2}}{4 \pi R^{2}} \cdot 100=\frac{R^{2}-r^{2}}{R^{2}} \cdot 100=\left(1-\left(\frac{r}{R}\right)^{2}\right) \cdot 100$
$=\left(1-\left(\frac{2}{5}\right)^{2}\right) \cdot 100=\frac{2100}{25}=84$.
|
84
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Little Red Riding Hood is walking along a path at a speed of 6 km/h, while the Gray Wolf is running along a clearing perpendicular to the path at a speed of 8 km/h. When Little Red Riding Hood was crossing the clearing, the Wolf had 80 meters left to run to reach the path. But he was already old, his eyesight was failing, and his sense of smell was not as good. Will the Wolf notice Little Red Riding Hood if he can smell prey only within 45 meters?
|
3. Answer: No.
Solution. The problem can be solved in a moving coordinate system associated with Little Red Riding Hood. Then Little Red Riding Hood is stationary, and the trajectory of the Wolf's movement is a straight line. The shortest distance from a point to a line here is (by similarity considerations):
$80 \cdot \sin \alpha$, where $\sin \alpha=\frac{3}{5}$. This results in 48 m, which is greater than 45 m.
Another way to solve it: the distance from the intersection point to Little Red Riding Hood, depending on time $t$, is $6 t$; the distance from the intersection point to the Wolf is $80-8 t$. Therefore, the distance from the Wolf to Little Red Riding Hood, by the Pythagorean theorem, is $\sqrt{36 t^{2}+(80-8 t)^{2}}=2 \sqrt{5\left(5 t^{2}-64 t+320\right)}$. The minimum of this value is achieved at $t=\frac{64}{2 \cdot 5}=\frac{32}{5}$, and is equal to 48.
|
48
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the village where Glafira lives, there is a small pond that is filled by springs at the bottom. Curious Glafira found out that a herd of 17 cows completely drank the pond dry in 3 days. After some time, the springs refilled the pond, and then 2 cows drank it dry in 30 days. How many days would it take for one cow to drink the pond dry?
|
5. Answer: In 75 days.
Solution. Let the pond have a volume of a (conditional units), one cow drinks b (conditional units) per day, and the springs add c (conditional units) of water per day. Then the first condition of the problem is equivalent to the equation $a+3c=3 \cdot 17 b$, and the second to the equation $a+30c=30 \cdot 2 b$. From this, we get that $b=3c, a=150c$. If one cow drinks the pond dry in x days, then we have $a+xc=xb$, that is, $x=\frac{a}{b-c}=75$.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Gavriil was traveling in Africa. On a sunny and windy day, at noon, when the rays from the Sun fell vertically, the boy threw a ball from behind his head at a speed of $5 \sim$ m/s against the wind at an angle to the horizon. After 1 s, the ball hit him in the stomach 1 m below the point of release. Determine the greatest distance the shadow of the ball moved away from Gavriil's feet. The force acting on the ball from the air is directed horizontally and does not depend on the position and speed. The acceleration due to gravity $\$ \mathrm{~g} \$$ is $\$ 10 \$ \sim \mathrm{m} / \mathrm{c} \$ \wedge 2 \$$.
|
6. Answer: 75 cm
Solution. In addition to the force of gravity, a constant horizontal force $F=m \cdot a$ acts on the body, directed opposite. In a coordinate system with the origin at the point of throw, the horizontal axis x, and the vertical axis y, the law of motion has the form:
$$
\begin{aligned}
& x(t)=V \cdot \cos \alpha \cdot t-\frac{a t^{2}}{2} \\
& y(t)=V \cdot \sin \alpha \cdot t-\frac{g t^{2}}{2}
\end{aligned}
$$
From the condition, the coordinates of the point at time $\tau$ are known:
$$
\begin{aligned}
0 & =V \cdot \cos \alpha \cdot \tau-\frac{a \tau^{2}}{2} \\
-H & =V \cdot \sin \alpha \cdot \tau-\frac{g \tau^{2}}{2}
\end{aligned}
$$
From this system of two equations with two unknowns, we find the acceleration created by the force acting from the air and the angle at which the throw was made. From the second equation: $\sin \alpha=\frac{1}{V}\left(\frac{g \tau}{2}-\frac{H}{\tau}\right)=\frac{1}{5}\left(\frac{10 \cdot 1}{2}-\frac{1}{1}\right)=\frac{4}{5}$. Therefore, $\cos \alpha=\frac{3}{5}$, and from the first equation: $a=\frac{2 V \cdot \cos \alpha}{\tau}=\frac{2 \cdot 5 \cdot 3}{5 \cdot 1}=6 \mathrm{M} / \mathrm{c}^{2}$.
The maximum distance of the shadow corresponds to the maximum value of $x$, which is achieved at $t=\frac{V \cos \alpha}{a}$ (the vertex of the parabola). This value is $L=\frac{V^{2} \cos ^{2} \alpha}{2 a}=\frac{5^{2} \cdot 3^{2}}{5^{2} \cdot 2 \cdot 6}=\frac{3}{4}$ m.
|
75
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2.1. During the time it took for a slowly moving freight train to cover 1200 m, a schoolboy managed to ride his bicycle along the railway tracks from the end of the moving train to its beginning and back to the end. In doing so, the bicycle's distance meter showed that the cyclist had traveled 1800 m. Find the length of the train (in meters).
|
Answer: 500. Solution: Let $V$ and $U$ be the speeds of the cyclist and the train, respectively, and $h$ be the length of the train.
Then the conditions of the problem in mathematical terms can be written as follows:
$$
(V-U) t_{1}=h ; \quad(V+U) t_{2}=h ; \quad U\left(t_{1}+t_{2}\right)=l ; \quad V\left(t_{1}+t_{2}\right)=L
$$
Here $t_{1}, t_{2}$ are the times the cyclist travels in the direction of the train and against it, respectively. Taking the ratio of the last two equations, it follows that the speed of the cyclist is $a=\frac{L}{l}=1.5$ times the speed of the train. From the first two equations, we express the time and substitute it into the third. Then for the length of the train, we get the formula
$$
h=\frac{l\left(\alpha^{2}-1\right)}{2 \alpha}=500
$$
|
500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4.1. A train of length $L=600$ m, moving by inertia, enters a hill with an angle of inclination $\alpha=30^{\circ}$ and stops when exactly a quarter of the train is on the hill. What was the initial speed of the train $V$ (in km/h)? Provide the nearest whole number to the calculated speed. Neglect friction and assume the acceleration due to gravity is $g=10 \mathrm{m} /$ sec $^{2}$.
|
Answer: 49. Solution. The kinetic energy of the train $\frac{m v^{2}}{2}$ will be equal to the potential energy of the part of the train that has entered the hill $\frac{1}{2} \frac{L}{4} \sin \alpha \frac{m}{4} g$.
Then we get $V^{2}=\frac{L}{4} \frac{1}{8} *(3.6)^{2}=\frac{6000 *(3.6)^{2}}{32}=9 \sqrt{30}$.
Since the following double inequality is true $49<9 \sqrt{30}<49.5$ (verified by squaring), the answer is 49.
|
49
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. 2. A car with a load traveled from one city to another at a speed of 60 km/h, and returned empty at a speed of 90 km/h. Find the average speed of the car for the entire route. Give your answer in kilometers per hour, rounding to the nearest whole number if necessary.
$\{72\}$
|
Solution. The average speed will be the total distance divided by the total time: $\frac{2 S}{\frac{S}{V_{1}}+\frac{S}{V_{2}}}=\frac{2 V_{1} \cdot V_{2}}{V_{1}+V_{2}}=\frac{2 \cdot 60 \cdot 90}{60+90}=72(\mathrm{km} / \mathrm{h})$.
|
72
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A tractor is pulling a very long pipe on sled runners. Gavrila walked along the entire pipe in the direction of the tractor's movement and counted 210 steps. When he walked in the opposite direction, the number of steps was 100. What is the length of the pipe if Gavrila's step is 80 cm? Round the answer to the nearest whole number of meters.
|
Solution. Let the length of the pipe be $x$ (meters), and for each step Gavrila takes of length $a$ (m), the pipe moves a distance $y$ (m). Then, if $m$ and $n$ are the number of steps Gavrila takes in each direction, we get two equations: $x=m(a-y), x=n(a+y)$. From this, $\frac{x}{m}+\frac{x}{n}=2 a$, and $x=\frac{2 a m n}{m+n}$. With the given numerical values, we get $x=\frac{2 \cdot 0.8 \cdot 210 \cdot 100}{210+100}=\frac{3360}{31} \approx 108.39$ (m).
Answer. 108 m.
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. At some point on the shore of a wide and turbulent river, 100 m from the bridge, Gavrila and Glafira set up a siren that emits sound signals at equal intervals. Glafira took another identical siren and positioned herself at the beginning of the bridge on the same shore. Gavrila got into a motorboat, which was located on the shore halfway between the first siren and the beginning of the bridge. The experimenters started simultaneously, with the speeds of the bicycle and the motorboat relative to the water being 20 km/h and directed perpendicular to the shore. It turned out that the sound signals from both sirens reached Gavrila simultaneously. Determine the distance from the starting point where Gavrila will be when he is 40 m away from the shore. Round your answer to the nearest whole number of meters. The riverbank is straight, and the current at each point is directed along the shore.
|
Solution. Let's introduce a coordinate system, with the $x$-axis directed along the shore, and the origin at Gavrila's starting point. The siren on the shore has coordinates $(L, 0), L=50$ m, and Glafira is traveling along the line $x=-L$. Since the experimenters are at the same distance from the shore, the equality of the times needed for the sound signal to travel gives the condition on the coordinates of Gavrila $(x, y)$:
$$
x+L=\sqrt{(x-L)^{2}+y^{2}}
$$
or
$$
y^{2}=4 x L,
$$
from which
$$
x=\frac{y^{2}}{4 L}
$$
which describes a parabola.
According to the problem, the value of $y=40$ m is known. The distance from Gavrila to the starting point can be found using the Pythagorean theorem:
$$
s=\sqrt{x^{2}+y^{2}}=\sqrt{\frac{y^{4}}{16 L^{2}}+y^{2}}=\frac{y \sqrt{y^{2}+16 L^{2}}}{4 L}=8 \sqrt{26}
$$
Since $40.5^{2}<(8 \sqrt{26})^{2}<41^{2}$, the nearest integer number of meters is 41.
Answer 41.
|
41
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 2. Experimenters Glafira and Gavriil placed a triangle made of thin wire with sides of 30 mm, 40 mm, and 50 mm on a white flat surface. This wire is covered with millions of mysterious microorganisms. The scientists found that when an electric current is applied to the wire, these microorganisms begin to move chaotically on this surface in different directions at an approximate speed of $\frac{1}{6}$ mm/sec. As they move, the surface along their path is stained red. Find the area of the stained surface after 1 minute of current application. Round it to the nearest whole number of square millimeters.
|
Solution. In one minute, the microorganism moves 10 mm. Since in a right triangle with sides $30, 40, 50$, the radius of the inscribed circle is 10, all points inside the triangle are at a distance from the sides of the triangle that does not exceed 10 mm. Therefore, the microorganisms will fill the entire interior of the triangle.
When moving outward, points will be reached that are 10 mm away from the sides of the triangle and points that are 10 mm away from the vertices.
In the end, the total occupied area is: the area of the triangle + three strips of 10 mm width each, located outside the triangle, with a total length equal to the perimeter of the triangle + 3 circular sectors of radius 10, which together form a circle. We get:
$$
\frac{30 \cdot 40}{2} + 10 \cdot (30 + 40 + 50) + \pi \cdot 10^{2} = 600 + 1200 + 100 \pi = 1800 + 100 \pi \approx 2114 \text{ mm}^2
$$
Answer: $1800 + 100 \pi \approx 2114$ mm².
|
2114
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. The villages of Arkadino, Borisovo, and Vadimovo are connected by straight roads. A square field adjoins the road between Arkadino and Borisovo, one side of which completely coincides with this road. A rectangular field adjoins the road between Borisovo and Vadimovo, one side of which completely coincides with this road, and the other side is 4 times longer. A rectangular forest adjoins the road between Arkadino and Vadimovo, one side of which completely coincides with this road, and the other side is 12 km. The area of the forest is 45 sq. km greater than the sum of the areas of the fields. Find the total area of the forest and the fields in sq. km.
$\{135\}$
|
Solution. The condition of the problem can be expressed by the following relation:
$r^{2}+4 p^{2}+45=12 q$
where $p, q, r$ are the lengths of the roads opposite the settlements Arkadino, Borisovo, and Vadimovo, respectively. This condition is in contradiction with the triangle inequality:
$r+p>q \Rightarrow 12 r+12 p>12 q \Rightarrow 12 r+12 p>r^{2}+4 p^{2}+45 \Rightarrow(r-6)^{2}+(2 p-3)^{2}<0$.
From this, it follows that all three settlements lie on the same straight line.
Moreover, $r=6, p=1.5, q=7.5$.
The total area is the sum: $r^{2}+4 p^{2}+12 q=36+9+90=135$.
Answer: 135
|
135
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Density is the ratio of the mass of a body to the volume it occupies. Since the mass did not change as a result of tamping, and the volume after tamping $V_{2}=$ $0.8 V_{1}$, the density after tamping became $\rho_{2}=\frac{1}{0.8} \rho_{1}=1.25 \rho_{1}$, that is, it increased by $25 \%$.
|
Answer: increased by $25 \%$.
|
25
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.1. Gavriil found out that the front tires of the car last for 20000 km, while the rear tires last for 30000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
1.1. Gavriil found out that the front tires of the car last for 20,000 km, while the rear tires last for 30,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{24000\}$.
|
24000
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.2. Gavriila found out that the front tires of the car last for 24000 km, while the rear tires last for 36000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
1.2. Gavriil found out that the front tires of the car last for 24,000 km, while the rear tires last for 36,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{28800\}$.
|
28800
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.3. Gavriila found out that the front tires of the car last for 42,000 km, while the rear tires last for 56,000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
|
1.3. Gavriil found out that the front tires of the car last for 42000 km, while the rear tires last for 56000 km. Therefore, he decided to swap them at some point to maximize the distance the car can travel. Find this distance (in km).
Answer. $\{48000\}$.
|
48000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Gavrila placed 7 smaller boxes into a large box. After that, Glafira placed 7 small boxes into some of these seven boxes, and left others empty. Then Gavrila placed 7 boxes into some of the empty boxes, and left others empty. Glafira repeated this operation and so on. At some point, there were 34 non-empty boxes. How many empty boxes were there at this moment?
|
Answer: 205.
Instructions. Filling one box increases the number of empty boxes by 7-1=6, and the number of non-empty boxes by 1. Therefore, after filling $n$ boxes (regardless of the stage), the number of boxes will be: empty $-1+6 n$; non-empty $-n$. Thus, $n=34$, and the number of non-empty boxes will be $1+6 \cdot 34=205$.
|
205
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A car was moving at a speed of $V$. Upon entering the city, the driver reduced the speed by $x \%$, and upon leaving the city, increased it by $0.5 x \%$. It turned out that this new speed was $0.6 x \%$ less than the speed $V$. Find the value of $x$.
|
Answer: 20. Solution. The condition of the problem means that the equation is satisfied
$$
v\left(1-\frac{x}{100}\right)\left(1+\frac{0.5 x}{100}\right)=v\left(1-\frac{0.6 x}{100}\right) \Leftrightarrow\left(1-\frac{x}{100}\right)\left(1+\frac{x}{200}\right)=1-\frac{3 x}{500} \Leftrightarrow \frac{x^{2}}{20000}=\frac{3 x}{500}-\frac{x}{200} .
$$
$$
\Leftrightarrow x=0 ; x=20 \text {. The value } x=0 \text { contradicts the condition. Therefore, } x=20 \text {. }
$$
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Gavriil decided to weigh a football, but he only had weights of 150 g, a long light ruler with the markings at the ends worn off, a pencil, and many threads at his disposal. He suspended the ball from one end of the ruler and the weight from the other, and balanced the ruler on the pencil. Then he attached a second weight to the first, and to restore balance, he had to move the pencil 6 cm. When a third weight was attached to the first two, and the pencil was moved another 4 cm, balance was restored again. Calculate the mass of the ball, as Gavriil did.
|
1. Let the distances from the pencil to the ball and to the weight be $l_{1}$ and $l_{2}$ respectively at the first equilibrium. Denote the magnitude of the first shift by $x$, and the total shift over two times by $y$. Then the three conditions of lever equilibrium will be:
$$
\begin{gathered}
M l_{1}=m l_{2} \\
M\left(l_{1}+x\right)=2 m\left(l_{2}-x\right) \\
M\left(l_{1}+y\right)=3 m\left(l_{2}-y\right)
\end{gathered}
$$
Subtracting the first equation from the second and third, we get:
$$
\begin{gathered}
M x=m l_{2}-2 m x \\
M y=2 m l_{2}-3 m y
\end{gathered}
$$
From this,
$$
M=\frac{3 y-4 x}{2 x-y} m=600 \text { g. }
$$
In the variant with the football, the formula is the same, the answer is 450 g.
|
600
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1.2.1 The time of the aircraft's run from the moment of start until takeoff is 15 seconds. Find the length of the run if the takeoff speed for this aircraft model is 100 km/h. Assume the aircraft's motion during the run is uniformly accelerated. Provide the answer in meters, rounding to the nearest whole number if necessary.
|
Solution. $v=a t, 100000 / 3600=a \cdot 15$, from which $a=1.85\left(\mathrm{~m} / \mathrm{s}^{2}\right)$. Then $S=a t^{2} / 2=208(\mathrm{m})$.)
Answer. 208 m
|
208
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3.3.1 Gavriil got on the train with a fully charged smartphone, and by the end of the trip, his smartphone was completely drained. For half of the time, he played Tetris, and for the other half, he watched cartoons. It is known that the smartphone fully discharges in 3 hours of video watching or in 5 hours of playing Tetris. What distance did Gavriil travel if the train moved half of the distance at an average speed of 80 km/h and the other half at an average speed of 60 km/h? Provide the answer in kilometers, rounding to the nearest whole number if necessary.
|
Answer: 257 km.
Solution. Let's take the "capacity" of the smartphone battery as 1 unit (u.e.). Then the discharge rate of the smartphone when watching videos is $\frac{1}{3}$ u.e./hour, and the discharge rate when playing games is $\frac{1}{5}$ u.e./hour.
If the total travel time is denoted as $t$ hours, we get the equation $\frac{1}{3} \cdot \frac{t}{2}+\frac{1}{5} \cdot \frac{t}{2}$, from which $\frac{(5+3) t}{2 \cdot 3 \cdot 5}=1$, that is, $t=\frac{15}{4}$ hours.
Then we get (where $\mathrm{S}$ is the distance): $\frac{S}{2 \cdot 80}+\frac{S}{2 \cdot 60}$, that is, $\frac{S}{40}+\frac{S}{30}=15$,
$S=15 \cdot \frac{40 \cdot 3}{4+3}=\frac{1800}{7} \approx 257 \mathrm{km}$.
|
257
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. A tractor is pulling a very long pipe on sled runners. Gavrila walked along the entire pipe at a constant speed in the direction of the tractor's movement and counted 210 steps. When he walked in the opposite direction at the same speed, the number of steps was 100. What is the length of the pipe if Gavrila's step is 80 cm? Round the answer to the nearest whole number of meters. The speed of the tractor is constant.
|
Answer: 108 m. Solution. Let the length of the pipe be $x$ (meters), and for each step Gavrila takes of length $a$ (m), the pipe moves a distance of $y$ (m). Then, if $m$ and $n$ are the number of steps Gavrila takes in one direction and the other, respectively, we get two equations: ${ }^{x=m(a-y)}, x=n(a+y)$. From this, $\frac{x}{m}+\frac{x}{n}=2 a, \quad x=\frac{2 a m n}{m+n}$. With the given numerical values, we get $x=\frac{2 \times 0.8810160}{210+100}$ $=\frac{3360}{31} \approx 108.387$ (m).
Other methods of solving are possible. For example: With the same unknowns as above, let's also introduce Gavrila's speed $V$ (m/min) and the tractor's speed $U$ (m/min). Recording the time in both cases, we get the system:
$$
\left\{\begin{array}{l}
\frac{x}{V-U}=\frac{a m}{V} \\
\frac{x}{V+U}=\frac{a n}{V}
\end{array} \Leftrightarrow \left\{\begin{array}{l}
\frac{x}{a m}=1-\frac{U}{V} \\
\frac{x}{a n}=1+\frac{U}{V}
\end{array}\right.\right.
$$
Adding these, we get the same answer $x=\frac{2 a m n}{m+n}$.
Answer to variant 192: 98 m. Comment. $\frac{1463}{15} \approx 97.533$ (m).
Answer to variant 193: 89 m. Comment. $\frac{1152}{13} \approx 88.615$ (m).
Answer to variant 194: 82 m. Comment. $\frac{1071}{13} \approx 82.3846$ (m).
|
108
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. Usually, schoolboy Gavriila takes a minute to go up a moving escalator by standing on its step. But if Gavriila is late, he runs up the working escalator and saves 36 seconds this way. Today, there are many people at the escalator, and Gavriila decides to run up the adjacent non-working escalator. How much time will such an ascent take him if he always exerts the same effort when running up the escalator?
|
Solution. Let's take the length of the escalator as a unit. Let $V$ be the speed of the escalator, and $U$ be Gavrila's speed relative to it. Then the condition of the problem can be written as:
$$
\left\{\begin{array}{c}
1=V \cdot 60 \\
1=(V+U) \cdot(60-36)
\end{array}\right.
$$
The required time is determined from the relationship $1=U \cdot t$. From the system, we get $V=\frac{1}{60} ; U+V=\frac{1}{24}$; $U=\frac{1}{24}-\frac{1}{60}=\frac{1}{40}$. Therefore, $t=40$ seconds.
Answer: 40 seconds.
Criteria: 20 points - complete and correct solution, 15 points - correct approach to the solution, but an arithmetic error is made; $\mathbf{1 0}$ points - the system of equations is correctly set up, but the answer is not obtained; 0 points - everything else.
|
40
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The lieutenant is engaged in drill training with the new recruits. Upon arriving at the parade ground, he saw that all the recruits were lined up in several rows, with the number of soldiers in each row being the same and 5 more than the number of rows. After the training session, the lieutenant decided to line up the recruits again but couldn't remember how many rows there were. So he ordered them to line up with as many rows as his age. It turned out that the number of soldiers in each row was again equal, but there were 4 more soldiers in each row than in the initial formation. How old is the lieutenant?
|
Solution. Let $n$ be the number of rows in the original formation. Then, there were originally $n+5$ soldiers in each row, and in the second formation, there were $n+9$ soldiers in each row. Let the age of the lieutenant be $x$. Then, according to the problem, we get the equation
$$
x=\frac{n(n+5)}{n+9} \Rightarrow x=n-4+\frac{36}{n+9}
$$
The number $n+9$ must be a divisor of the number 36. Considering that $n$ and $x$ are natural numbers, we get the following solutions
$$
\left\{\begin{array}{l}
n+9=12 \\
x=2
\end{array},\left\{\begin{array}{l}
n+9=18 \\
x=7
\end{array},\left\{\begin{array}{l}
n+9=36 \\
x=24
\end{array}\right.\right.\right.
$$
From a common sense perspective, the last answer fits, although the other two answers were also considered correct.
Answer: 24 years
|
24
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Among all six-digit natural numbers, the digits of which are arranged in ascending order (from left to right), numbers containing the digit 1 and not containing this digit are considered. Which numbers are more and by how many?
|
Solution. First, let's calculate how many six-digit natural numbers there are in total, with their digits arranged in ascending order. For this, we will write down all the digits from 1 to 9 in a row. To get six-digit numbers of the considered type, we need to strike out any three digits. Thus, the number of six-digit natural numbers, with their digits arranged in ascending order, is $C_{9}^{3}=84$.
Now let's calculate how many of these numbers contain 1. For this, we fix the digit 1 in our row of digits and strike out any three digits from the remaining 8. We get that the number of numbers containing 1 is $C_{8}^{3}=56$. As a result, we find that the numbers containing 1 are 28 more than the numbers that do not contain 1.
Answer: the numbers containing 1 are 28 more.
|
28
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. The Ivanovs' income as of the beginning of June:
$$
105000+52200+33345+9350+70000=269895 \text { rubles }
$$
|
Answer: 269895 rubles
## Evaluation Criteria:
Maximum score - 20, if everything is solved absolutely correctly, the logic of calculations is observed, and the answer is recorded correctly.
## 20 points, including:
6 points - the final deposit amount is calculated correctly;
4 points - the size of the mother's salary and bonus is calculated correctly;
4 points - the cost of the monetary gift, taking into account the exchange rate of the dollar, is calculated correctly;
2 points - the size of the pension after indexing is calculated correctly;
4 points - the family's income on the relevant date is calculated correctly.
If the logic of calculations is observed, the student thinks correctly, but arithmetic errors are made, then -1 point for each error.
## Problem 3
To attract buyers, sales advertisements often state: "Discount up to 50%!" When a customer comes to the store, they often see that the advertisement was honest, but at the same time, it did not meet their expectations.
Explain the reason for this discrepancy?
## Suggested Answer:
A customer might choose to shop at this store, expecting a significant discount and the ability to save a lot of money.
However, by stating a discount "up to 50%," the seller can actually reduce the price by any amount (even just 1%), which might clearly not satisfy the customer.
## Maximum 15 points
15 points - fully correct reasoning, the key phrase "up to" is mentioned
10 points - the participant leads the reasoning, several options are mentioned, but without the key phrase "up to"
5 points - one option about the intentional price increase is mentioned
## Problem 4
Maria Ivanovna, a single pensioner, has two small kitchen appliances that have broken down: a blender and a meat grinder. She plans to purchase two new appliances in the coming month and has already decided on the models. In the store "Technic-City," a blender costs 2000 rubles, and a meat grinder costs 4000 rubles. However, the store offers a 10% discount on the entire purchase when buying two or more small household appliances. The store "Technomarket" offers to credit 20% of the purchase amount to the customer's store card, which can be used in full within a month for the next purchase of household appliances. In this store, a blender is sold for 1500 rubles, and a meat grinder for 4800 rubles. Suggest the most cost-effective option for Maria Ivanovna and calculate how much money she will spend.
## Solution:
Maria Ivanovna should consider the following purchase options:
|
269895
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend
$$
4800 + 1500 - 4800 * 0.2 = 5340 \text{ rubles.}
$$
This is the most profitable way to make the purchase.
## Maximum 30 points
## Problem 5
Schoolboy Vanechka informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use an additional card issued to her bank account. Vanechka needed to buy 100 large chocolate bars for organizing and conducting a quiz at the New Year's party. However, the card has a protection system that analyzes the last 5 purchases, determines their average value, and compares it with the planned purchase amount. If the planned purchase amount exceeds the average by 3x, the bank blocks the transaction and requires additional verification (for example, a call from his mother to the call center). Over the past month, his mother made purchases in the following amounts: 785 rubles, 2033 rubles, 88 rubles, 3742 rubles, 1058 rubles.
Calculate whether Vanechka will be able to buy the chocolate bars for the New Year's party in one transaction without calling his mother to the call center if the cost of one chocolate bar is 55 rubles?
|
# Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an allowable purchase is no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, you can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct answer.
7 points - correct calculation of the average value of purchases
3 points - partial solution
## Appendix to Problem No. 1
Insurance is a system of relationships between insurers (insurance company) and insured individuals (for example, any citizen), which allows reducing property risks by insuring, for example, property against possible adverse events...
Property insurance is one of the types of insurance according to the Civil Code of the Russian Federation. Real estate is one type of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
84
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 14. (2 points)
Ivan opened a deposit in a bank for an amount of 100 thousand rubles. The bank is a participant in the state deposit insurance system. How much money will Ivan receive if the bank's license is revoked / the bank goes bankrupt?
a) Ivan will receive 100 thousand rubles and the interest that has been accrued on the deposit in full
b) Ivan will receive 100 thousand rubles, as according to the legislation, only the initial investments are subject to return. Interest on deposits is not returned
c) Ivan will not receive any funds, as in the deposit insurance system, only deposits of 500 thousand rubles or more are insured
|
# Solution:
In accordance with the federal insurance law Federal Law No. 177-FZ of $23 \cdot 12.2003$ (as amended on 20.07.2020) "On Insurance of Deposits in Banks of the Russian Federation" (with amendments and additions, effective from 01.10.2020), compensation for deposits in a bank where an insurance case has occurred is paid to the depositor in the amount of 100 percent of the deposit amount in the bank, but not more than 1,400,000 rubles. The deposit amount includes not only the funds deposited by the depositor but also accrued interest. Therefore, Ivan will receive back 100 thousand rubles and the interest that has been accrued on the deposit, in full.
|
100
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
4. Excursions (20,000 rubles for the whole family for the entire vacation).
The Seleznev family is planning their vacation in advance, so in January, the available funds for this purpose were calculated. It turned out that the family has 150,000 rubles at their disposal. Mr. Seleznev plans to set aside a certain amount from his salary over the six months before the vacation and suggests placing the available funds in a bank by opening a deposit. Mr. Seleznev tasked his daughter Alice with gathering information about the deposit conditions offered by banks and choosing the best ones. Alice found the following:
1) Rebs-Bank accepts funds from the population for a period of six months with monthly interest accrual at an annual rate of 3.6% (interest is capitalized). The deposit is non-replenishable.
2) Gamma-Bank accepts funds from the population for a period of six months with a single interest accrual at the end of the term at an annual rate of 4.5% (simple interest). The deposit is non-replenishable.
3) Tisi-Bank accepts funds from the population for a period of six months with quarterly interest accrual at an annual rate of 3.12% (interest is capitalized). The deposit is non-replenishable.
4) Btv-Bank accepts funds from the population for a period of six months with monthly interest accrual at a rate of 0.25% per month (interest is capitalized). The deposit is non-replenishable.
There is an option to open a deposit on January 1, 2021, and close it on June 30, 2021.
Alice needs to calculate the interest for each deposit and the amount that Mr. Seleznev needs to set aside from his salary, given the choice of each deposit.
Task: Match the conditions of the deposits offered by the banks with the amount that Mr. Seleznev needs to set aside from his salary, in the case of choosing the corresponding deposit as a savings instrument.
#
|
# Solution:
1) Calculate the vacation expenses
Flight expenses $=10200.00$ rubles * 2 flights * 3 people $=61200.00$
Hotel expenses $=6500$ rubles * 12 days $=78000.00$ rubles
Food expenses $=1000.00$ rubles * 14 days * 3 people $=42000.00$ rubles
Excursion expenses $=20000.00$ rubles
Total expenses $=201200.00$ rubles.
2) Calculate the interest amount for each deposit (the amount of savings at the end of the deposit minus the principal amount of the deposit).
Rebs-bank: $152720.33-150000=2720.33$ rubles
Gamma-bank: $153375-150000=3375$ rubles
Tisi-bank: $152349.13-150000=2349.13$ rubles
Btv-bank: $152264.11-150000=2264.11$ rubles
3) Calculate the amount that needs to be set aside from the salary when choosing:
Rebs-bank deposit: 201200-150000-2720.33=48479.67 rubles
Gamma-bank deposit: 201200-150000-3375=47825.00 rubles
Tisi-bank deposit: 201200-150000-2349.13=48850.87 rubles
Btv-bank deposit: 201200-150000-2264.11=48935.89 rubles
Answer:
| Deposit conditions (column set by the administrator) | The amount that Mr. Selyanin needs to set aside from his salary in case of choosing the corresponding deposit as a savings instrument (the correct ratio is indicated) |
| :--- | :--- |
| Rebs-bank deposit conditions | 48479.67 rubles |
| Gamma-bank deposit conditions | 47825.00 rubles |
| Tisi-bank deposit conditions | 48850.87 rubles |
| Btv-bank deposit conditions | 48935.89 rubles |
|
47825
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. When insuring property, the insurance amount cannot exceed its actual value (insurance value) at the time of concluding the insurance contract.
Insurance tariff - the rate of the insurance premium or the insurance premium (insurance premium) expressed in rubles, payable per unit of the insurance amount, which is usually 100 rubles. In simpler terms, it is the percentage of the insurance value of the property established by the insurance company within the framework of laws concerning insurance activities.
An underwriter of an insurance company is a person authorized by the insurance company to analyze, accept for insurance (reinsurance), and reject various types of risks, as well as classify the selected risks to obtain the optimal insurance premium for them.
Title insurance (title insurance) is protection against the risk of losing the right of ownership of real estate as a result of the transaction for the acquisition of real estate being declared invalid or illegal. A transaction can be declared invalid by a court in the following situations:
- the presence of unaccounted property interests of minors, heirs, persons serving time in places of deprivation of liberty;
- the discovery of errors or fraud in the current or previous transactions with the real estate;
- in the case of the incapacity of one of the parties to the transaction, etc.
The appraised value of the property is the final result of the appraisal study, calculated by a licensed appraiser.
The cadastral value of housing is the value of the real estate object established through state valuation, primarily for the calculation of taxes.
## Instruction:
In the event of a discrepancy between the insurance amount and the actual value of the property, the following consequences are established:
a) if the insurance amount is set below the actual value of the property in the Insurance Contract, the Insurer, in the event of an insurance case, compensates the Insured (Beneficiary) for the damage incurred within the limits of the insurance amount established by the Insurance Contract without applying the ratio of the insurance amount and the actual value of the property (first risk payment condition).
b) if the insurance amount specified in the Insurance Contract exceeds the actual value of the property, the Contract is void to the extent that the insurance amount exceeds the actual value of the property.
Gross insurance tariffs (title insurance)
| Group of insured objects | Number of previous transactions with the real estate object | Annual tariff |
| :--- | :--- | :---: |
| Apartments and parts of apartments, consisting of one or several isolated rooms, non-residential buildings | Primary market (one transaction) | 0.18 |
| Secondary market (no more than 3 transactions) | One transaction preceding the current transaction | 0.20 |
| No more than 3 transactions*** | | |
**** in the case of more than 3 transactions over the past 3 years preceding the current transaction, the tariff is agreed individually.
## List of corrective coefficients (title insurance)
| Factors | Coefficient |
| :--- | :---: |
| The presence in the history of the real estate object of one or more of the following facts: - a time gap of less than one year between two preceding (or between the current and the preceding) transfers of ownership of the real estate object; - the presence of lease transactions; - transfer of rights as a result of inheritance or gift; - transactions carried out by persons on the basis of a power of attorney; - more than 2 transfers of ownership of the real estate object. | |
| - the last transaction preceding the current one was more than 3 years ago | 1.2 |
| Absence of one of the necessary documents | 0.8 |
## MOSCOW OLYMPIAD OF SCHOOL STUDENTS IN FINANCIAL LITERACY FINAL STAGE $10-11$ GRADES 2nd variant Answers and solutions
## Problem 1
You are an underwriter of an insurance company. According to the instructions for title insurance (Appendix to this problem), based on the data provided below, you need to set the tariff and determine the amount of the insurance premium. Be sure to keep the glossary in front of you.
Data:
Insured Ostrozhnov Konstantin Petrovich is purchasing an apartment on the secondary market with the participation of credit funds. The loan amount secured by real estate is 20 million rubles. Konstantin Petrovich has a good credit history and easily obtained approval for such an amount. The appraised value of the apartment is 14,500,000 rubles, the cadastral value is 15,000,000 rubles. The bank-lender requires Ostrozhnov to purchase a title insurance policy to protect its financial interests.
Seller - Ivanov G.L., born in 1952, the sole owner of the apartment, purchased it more than 5 years ago under a construction investment agreement. Married, has adult children, no one is registered in the apartment. At the time of purchasing the apartment, he was married. Provided all necessary documents except for certificates from the psychiatric and narcological dispensaries. Does not drive, does not have a driver's license, and is not subject to military service.
|
# Solution:
In accordance with the instruction, the base rate is $0.2\%$ of the insurance amount, apply a reducing factor for the absence of a change in ownership over the past 3 years $(0.8)$ and an increasing factor for the absence of certificates from the PND and ND $(1.3)$.
In total: $0.2 * 0.8 * 1.3=0.208\%$
The loan amount exceeds both the appraised and the cadastral value. Therefore, we choose the maximum value to minimize the difference between the loan amount (bank requirement) and the allowable amount by law - 15000000 and take it as the insurance amount. $15000000 * 0.00208=31200$
Answer: rate $0.208\%$, SP=31,200 rubles.
Maximum 20 points
Evaluation criteria:
2 points for each correctly found coefficient and 4 points in total for the correct explanation of the use of these coefficients (if there is a correct explanation of only one coefficient, only 1 point out of 4 is given, and if there is a correct explanation of only two coefficients, only 3 points out of 4 are given). In total, 10 points for this part of the correct solution.
2 points for the correct rate.
A total of 8 points for the correct insurance premium amount.
- Participants receive only 2 points if the insurance premium is calculated without arithmetic errors based on an incorrect appraised value of the apartment.
- Participants receive only 4 points if the insurance premium is correctly determined but lacks a complete correct justification for its application.
## Task 2
The following information is available for JSC «Total Trade».
The annual revenue of JSC «Total Trade» is 2,500,000 rubles (after VAT payment). It is known that the ratio of operating expenses to revenue is 3/5 annually. The authorized capital of JSC «Total Trade» consists of the nominal value of the company's shares acquired by shareholders and is eight times the minimum authorized capital of a public company established under Russian law. The authorized capital is made up of 1600 ordinary shares, $35\%$ of which belong to the general director.
|
31200
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Maria Ivanovna decided to use the services of an online clothing store and purchase summer clothing: trousers, a skirt, a jacket, and a blouse. Being a regular customer of this store, Maria Ivanovna received information about two ongoing promotions. The first promotion allows the customer to use an electronic coupon worth 1000 rubles, which can cover no more than $15 \%$ of the cost of the ordered items. The second promotion allows the customer to get every third item, which has the lowest price, for free when making a purchase. The online store offers the option to pick up the items from any pickup point located in the city where Maria Ivanovna lives, or to have them delivered to any address within the city. Delivery from a pickup point costs the customer 100 rubles, while delivery to a specified address costs 350 rubles if the order amount does not exceed 5000 rubles. If the order amount exceeds 5000 rubles, delivery to the specified address is free. Maria Ivanovna has selected the following set of clothing:
| Trousers | 2800 rubles |
| :--- | :--- |
| Skirt | 1300 rubles |
| Jacket | 2600 rubles |
| Blouse | 900 rubles |
a) What is the smallest amount of money Maria Ivanovna will have to pay for her purchases, including delivery, if the promotions of the online store cannot be combined (used simultaneously)?
b) Explain why Maria Ivanovna's actual expenses for this purchase in this online store may be higher than the monetary amount found in the previous question. (20 points)
|
Solution:
(a) Maria Ivanovna can make one purchase, using only one of the promotions, or she can "split" the selected items into two purchases, using both promotions in this case.
Let's consider all possible options:
1) One purchase. In this case, Maria Ivanovna can save either 900 rubles by using the "third item free" promotion, or $((2800+1300+2600+900) * 0.85=1140>1000) \quad 1000$ rubles by using the coupon. Delivery will be free in both of these cases. Thus, if Maria Ivanovna buys all the items in one purchase, the maximum savings will be 1000 rubles, and the cost of the purchase will be $2800+1300+2600+900-1000=6600$ rubles.
2) Two purchases. The most expensive of the cheapest third items in the purchase can be either the skirt or the blouse. If the free item is the skirt, then the coupon should be used for the blouse and it should be ordered as a separate purchase. If the free item is the blouse, then the coupon is most profitable to use for the most expensive item - the trousers.
Let's consider the cost of each such purchase.
1st option:
Cost of purchase 1 (trousers, jacket, skirt) = 2800 + 2600 = 5400 rubles. Delivery of this purchase will be free.
Cost of purchase 2 (blouse) $= 900 * 0.85 = 765$. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Total cost of the option: $5400 + 765 + 100 = 6265$
2nd option:
Cost of purchase 1 (jacket, skirt, blouse) = 2600 + 1300 = 3900 rubles. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Cost of purchase 2 (trousers) $= 2800 * 0.85 = 2380$. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Total cost of the option: $3900 + 2380 + 200 = 6480$
Thus, Maria Ivanovna will spend the least amount if she orders two purchases from the online store, one of which will include trousers, jacket, and skirt, and the other - blouse. The cost of such a purchase, including delivery, will be 6265 rubles.
(b) Attention should be paid to at least two main points that can increase the actual cost of the purchase for Maria Ivanovna.
1) The conditions of self-pickup imply that Maria Ivanovna (or the person she entrusts to pick up her order) will spend time and, possibly, money to get to the pickup point. Despite the low cost of this delivery method, the pickup point may be far from the customer. Considering this fact, many people choose delivery to the address specified by the customer, and some may even prefer (in this case) not to split the purchase into two, but to buy all the selected items in one purchase.
2) When placing two purchases instead of one, the customer will also spend more time, the cost of which for him may exceed the benefit of "double" order placement.
Grading criteria (total 20 points):
(a) (total 15 points)
The idea of dividing orders into 1 or 2 purchases is present - 2 points.
Explicit monetary expenses for 1 purchase are correctly calculated - 3 points.
Explicit expenses for 2 purchases are correctly calculated - 4 points for each option.
Comparison is made and the correct answer about the lowest cost of orders is given - 2 points.
A penalty for each arithmetic error is 1 point. However, in the case of a conceptually correct approach to the solution, the penalty for arithmetic errors, if their number exceeds 1 error in each critical section, cannot exceed 50% of the maximum score assigned to it.
(b) (total 5 points)
The idea that implicit (opportunity) costs of purchasing items in an online store are not taken into account is present - 2 points.
At least 2 examples of implicit costs are provided - 3 points.
|
6265
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11. (16 points)
The Dorokhov family plans to purchase a vacation package to Crimea. The family plans to travel with the mother, father, and their eldest daughter Polina, who is 5 years old. They carefully studied all the offers and chose the "Bristol" hotel. The head of the family approached two travel agencies, "Globus" and "Around the World," to calculate the cost of the tour from July 10 to July 18, 2021.
The first agency offered the tour on the following terms: 11,200 rubles per person under 5 years old and 25,400 rubles per person over 5 years old. In addition, the "Globus" agency provides the Dorokhov family with a discount of 2% of the total cost of the tour as a loyal customer.
The terms of purchasing the tour from the "Around the World" agency are as follows: 11,400 rubles per person under 6 years old and 23,500 rubles per person over 6 years old. When purchasing a tour from the "Around the World" agency, a commission of 1% of the total cost of the tour is charged.
Determine which agency is more cost-effective to purchase the tour from. In your answer, indicate the minimum expenses of the Dorokhov family for the vacation in Crimea.
In your answer, provide only the number without units of measurement!
|
# Solution:
Cost of the tour with the company "Globus"
$(3 * 25400) *(1-0.02)=74676$ rubles.
Cost of the tour with the company "Around the World"
$(11400+2 * 23500) * 1.01=58984$ rubles.
Answer: 58984
|
58984
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 12. (16 points)
The Vasilievs' family budget consists of the following income items:
- parents' salary after income tax deduction - 71000 rubles;
- income from renting out property - 11000 rubles;
- daughter's scholarship - 2600 rubles
The average monthly expenses of the family include:
- utility payments - 8400 rubles;
- food - 18000 rubles;
- transportation expenses - 3200 rubles;
- tutor services - 2200 rubles;
- other expenses - 18000 rubles.
10 percent of the remaining amount is transferred to a deposit for the formation of a financial safety cushion. The father wants to buy a car on credit. Determine the maximum amount the Vasilievs family can pay monthly for the car loan.
In your answer, provide only the number without units of measurement!
|
# Solution:
family income
$71000+11000+2600=84600$ rubles
average monthly expenses
$8400+18000+3200+2200+18000=49800$ rubles
expenses for forming a financial safety cushion
$(84600-49800) * 0.1=3480$ rubles
the amount the Petrovs can save monthly for the upcoming vacation
$84600-49800-3480=31320$ rubles
## Answer: 31320
|
31320
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 13. (8 points)
Natalia Petrovna has returned from her vacation, which she spent traveling through countries in North America. She has a certain amount of money left in foreign currency.
Natalia Petrovna familiarized herself with the exchange rates at the nearest banks: "Rebirth" and "Garnet." She decided to take advantage of the most favorable offer. What amount will she receive in rubles for exchanging 120 US dollars, 80 Canadian dollars, and 10 Mexican pesos at one of the two banks?
| Type of Currency | Exchange Rate | |
| :--- | :---: | :---: |
| | Rebirth | Garnet |
| US Dollar | 74.9 rub. | 74.5 rub. |
| Canadian Dollar | 59.3 rub. | 60.1 rub. |
| Mexican Peso | 3.7 rub. | 3.6 rub. |
In your answer, provide only the number without units of measurement!
|
# Solution:
1) cost of currency at Bank "Vozrozhdenie":
$$
120 * 74.9 + 80 * 59.3 + 10 * 3.7 = 13769 \text{ RUB}
$$
2) cost of currency at Bank "Garant":
$$
120 * 74.5 + 80 * 60.1 + 10 * 3.6 = 13784 \text{ RUB}
$$
Answer: 13784
|
13784
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 14. (8 points)
To attend the section, Mikhail needs to purchase a tennis racket and a set of tennis balls. Official store websites have product catalogs. Mikhail studied the offers and compiled a list of stores where the items of interest are available:
| Item | Store | |
| :--- | :---: | :---: |
| | Higher League | Sport-Guru |
| Tennis racket | 5600 rub. | 5700 rub. |
| Set of tennis balls (3 pcs.) | 254 rub. | 200 rub. |
Mikhail plans to use delivery for the items. The delivery cost from the store "Higher League" is 550 rub., and from the store "Sport-Guru" it is 400 rub. If the purchase amount exceeds 5000 rub., delivery from the store "Higher League" is free.
Mikhail has a discount card from the store "Sport-Guru," which provides a 5% discount on the purchase amount.
Determine the total expenses for purchasing the sports equipment, including delivery costs, assuming that Mikhail chooses the cheapest option.
In your answer, provide only the number without units of measurement!
|
# Solution:
1) cost of purchase in the store "Higher League":
$$
\text { 5600+254=5854 rub. }
$$
1) cost of purchase in the store "Sport-guru": $(2700+200)^{*} 0.95+400=6005$ rub.
## Answer: 5854
|
5854
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 15. (8 points)
The fast-food network "Wings and Legs" offers summer jobs to schoolchildren. The salary is 25000 rubles per month. Those who work well receive a monthly bonus of 5000 rubles.
How much will a schoolchild who works well at "Wings and Legs" earn per month (receive after tax) after the income tax is deducted?
In your answer, provide only the number without units of measurement!
|
Solution:
The total earnings will be 25000 rubles + 5000 rubles $=30000$ rubles
Income tax $13 \%-3900$ rubles
The net payment will be 30000 rubles - 3900 rubles $=26100$ rubles
## Correct answer: 26100
## 2nd Option
|
26100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11. (16 points)
One way to save on utility bills is to use the night tariff (from 23:00 to 07:00). To apply this tariff, a multi-tariff meter needs to be installed.
The Romanov family is considering purchasing a multi-tariff meter to reduce their utility bills. The cost of the meter is 3500 rubles. The installation cost of the meter is 1100 rubles. On average, electricity consumption is 300 kWh per month, with 230 kWh used from 23:00 to 07:00.
The cost of electricity when using a multi-tariff meter: from 07:00 to 23:00 - 5.2 rubles per kWh, from 23:00 to 07:00 - 3.4 rubles per kWh.
The cost of electricity when using a standard meter is 4.6 rubles per kWh.
Determine how much the Romanov family will save by using a multi-tariff meter over three years.
In your answer, provide only the number without units of measurement!
|
# Solution:
2) use of a multi-tariff meter:
$$
3500+1100+(230 * 3.4+(300-230) * 5.2) * 12 * 3=45856 \text { rub. }
$$
3) use of a typical meter
$$
300 * 4.6 * 12 * 3=49680 \text { rub. } \quad \text {. } \quad \text {. }
$$
the savings will be
$49680-45856=3824$ rub.
## Answer: 3824
|
3824
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 12. (16 points)
The budget of the Petrovs consists of the following income items:
- parents' salary after income tax deduction - 56000 rubles;
- grandmother's pension - 14300 rubles;
- son's scholarship - 2500 rubles
Average monthly expenses of the family include:
- utility payments - 9800 rubles;
- food - 21000 rubles;
- transportation expenses - 3200 rubles;
- leisure - 5200 rubles;
- other expenses - 15000 rubles
10 percent of the remaining amount is transferred to a deposit for the formation of a financial safety cushion. Determine the amount that the Petrovs can save monthly for an upcoming vacation.
In the answer, indicate only the number without units of measurement!
|
# Solution:
family income
$56000+14300+2500=72800$ rubles.
average monthly expenses
$9800+21000+3200+5200+15000=54200$ rubles.
expenses for forming a financial safety cushion
$(72800-54200) * 0.1=1860$ rubles.
the amount the Petrovs can save monthly for the upcoming vacation
$72800-54200-1860=16740$ rubles.
Answer: 16740.
|
16740
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 13. (8 points)
Maxim Viktorovich returned from a trip to Asian countries. He has a certain amount of money in foreign currency left.
Maxim Viktorovich familiarized himself with the exchange rates at the nearest banks: "Voskhod" and "Elfa". He decided to take advantage of the most favorable offer. What amount will he receive as a result of exchanging 110 Chinese yuan, 80 Japanese yen, and 50 Hong Kong dollars into rubles at one of the two banks?
| Type of Currency | Exchange Rate | |
| :--- | :---: | :---: |
| | Voskhod | Elfa |
| Chinese yuan | 11.7 rub. | 11.6 rub. |
| Japanese yen | 72.1 rub. | 71.9 rub. |
| Hong Kong dollar | 9.7 rub. | 10.1 rub. |
In your answer, indicate only the number without units of measurement!
|
# Solution:
1) cost of currency at "Voskhod" bank:
$110 * 11.7 + 80 * 72.1 + 50 * 9.7 = 7540$ rubles.
2) cost of currency at "Alpha" bank: $110 * 11.6 + 80 * 71.9 + 50 * 10.1 = 7533$ rubles.
Answer: 7540.
|
7540
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 14. (8 points)
Elena decided to get a pet - a budgerigar. She faced the question of where to buy a cage and a bath more cost-effectively.
On the official websites of the stores, product catalogs are posted. Elena studied the offers and compiled a list of stores where the items she is interested in are available:
| Item | Store | |
| :--- | :---: | :---: |
| | ZooWorld | ZooIdea |
| Cage | 4500 rub. | 3700 rub. |
| Bath | 510 rub. | 680 rub. |
Elena plans to use delivery. The delivery cost from the store "ZooWorld" is 500 rub., and from the store "ZooIdea" it is 400 rub. If the purchase amount exceeds 5000 rub., delivery from the store "ZooWorld" is free.
Elena has a discount card from the store "ZooIdea," which provides a 5% discount on the purchase amount.
Determine the minimum total cost for purchasing a cage and a bath for the budgerigar, including delivery costs.
In your answer, provide only the number without units of measurement!
|
# Solution:
2) cost of purchase in the "Zoimir" store: $4500+510=5010$ rubles
3) cost of purchase in the "Zooidea" store: $(3700+680) * 0.95+400=4561$ rubles
## Answer: 4561
|
4561
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 15. (8 points)
Announcement: "Have free time and want to earn money? Write now and earn up to 2500 rubles a day working as a courier with the service 'Food.There-Here!'. Delivery of food from stores, cafes, and restaurants.
How much will a school student working as a courier with the service 'Food.There-Here!' earn in a month (4 weeks), working 4 days a week with a minimum load for 1250 rubles a day after the income tax deduction?
In your answer, provide only the number without units of measurement! #
|
# Solution:
The total earnings will be (1250 rubles * 4 days) * 4 weeks = 20000 rubles
Income tax 13% - 2600 rubles
The amount of earnings (net pay) will be 20000 rubles - 2600 rubles = 17400 rubles
Correct answer: 17400
|
17400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. There are several technologies for paying with bank cards: chip, magnetic stripe, paypass, cvc. Arrange the actions performed with a bank card in the order corresponding to the payment technologies.
1 - tap
2 - pay online
3 - swipe
4 - insert into terminal
|
Answer in the form of an integer, for example 1234.
Answer: 4312
|
4312
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. (8 points)
Kolya's parents give him pocket money once a month, calculating it as follows: 100 rubles for each A in math, 50 rubles for a B, 50 rubles are deducted for a C, and 200 rubles are deducted for a D. If the amount turns out to be negative, Kolya simply gets nothing. The math teacher gives a grade for the quarter by calculating the arithmetic mean and rounding according to rounding rules. What is the maximum amount of money Kolya could have received if it is known that his quarterly grade is two points, the quarter lasted exactly 2 months, there were 14 math lessons each month, and Kolya receives no more than one grade per lesson?
#
|
# Solution:
If Kolya received a final grade of 2 for the quarter, then for each 5 he received more than 5 2s, for each 4 - more than 3 2s, and for each 3 - more than 1 2. This means that the number of 2s was greater than the total number of all other grades combined, so Kolya could receive money in at most one of the months.
Thus, the maximum amount is achieved if in one of the months he receives only 2s, and in the other - only 4s and 5s, from which the maximum number of 2s received is 14. Kolya could not receive 3 or more 5s, as in this case he would have 15 or more 2s.
This leaves 3 cases:
a) Kolya has 2 fives, then he can get 1 four, the sum is 250
b) Kolya has 1 five, then he can get 2 fours, the sum is 200
c) Kolya has no fives, in which case he can get 4 fours, the sum is 200.
Thus, the maximum amount of money is 250 rubles.
Criteria:
Correct answer - 2 points
Method of obtaining the correct answer - 2 points
Explanation of why money cannot be received in both months - 2 points
Correct enumeration - 2 points
+1 bonus point for explicitly stating the advantage of receiving all 2s in the second month, as money loses value over time and it is more advantageous to receive it in the first month rather than the second.
|
250
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6. (10 points)
Vasily is planning to graduate from college in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles. When Vasily finished school, he could have chosen not to go to college and instead work as a real estate assistant, like his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in college? Note: Bachelor's degree education lasts 4 years.
|
# Solution:
In 4 years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points).
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300 * $(1 / 5 * 60000 + 1 / 10 * 80$ $000 + 1 / 20 * 25000 + (1 - 1 / 5 - 1 / 10 - 1 / 20) * 40000) + 30 / 300 * 25000 = 45025$ rubles (6 points).
|
45025
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1. (4 points)
In the run-up to the New Year, a fair is being held at the school where students exchange festive toys. As a result, the following exchange norms have been established:
1 Christmas tree ornament can be exchanged for 2 crackers, 5 sparklers can be exchanged for 2 garlands, and 4 Christmas tree ornaments can be exchanged for 1 garland.
a) How many crackers can be obtained for 10 sparklers? (1 point)
b) Which is more valuable: 5 Christmas tree ornaments and 1 cracker or 2 sparklers? (3 points)
#
|
# Solution:
a) 10 sparklers $=4$ garlands = 16 ornaments = $\mathbf{32}$ crackers. (1 point)
b) Convert everything to crackers. In the first case, we have $\mathbf{11}$ crackers. In the second case, 2 sparklers $=4 / 5$ garlands $=16 / 5$ ornaments $=32 / 5$ crackers $=\mathbf{6.4}$ crackers. Answer: 5 ornaments and 1 cracker are more expensive than 2 sparklers. (1 point for converting each set to a single unit + 1 point for the correct answer)
|
32
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. (8 points)
In the Sidorov family, there are 3 people: the father works as a programmer with an hourly rate of 1500 rubles. The mother works as a hairdresser at home and charges 1200 rubles per haircut, which takes her 1.5 hours. The son tutors in mathematics and earns 450 rubles per academic hour (45 minutes). Every day, the family needs to prepare food, walk the dog, and go to the store. The mother spends 2 hours preparing food, the father 1 hour, and the son 4 hours. Walking the dog takes 1 hour for any family member. A trip to the store takes 1 hour for the father, 2 hours for the mother, and 3 hours for the son. In addition, each family member sleeps for 8 hours and spends another 8 hours on rest and personal matters.
What is the maximum amount of money the family can earn in one day?
|
# Solution
In this problem, there are 2 possible interpretations, both of which were counted as correct.
In one case, it is assumed that 8 hours are spent on work on average over the month, in the other that no more than 8 hours are spent on work each day.
## First Case:
1) Determine the hourly wage for each family member. (2 points)
a) Father = 1500 rubles per hour
b) Mother = 1200 / 1.5 = 800 rubles per hour
c) Son = 450 / 0.75 = 600 rubles per hour
2) Calculate how much money the family will lose for each mandatory task depending on who does it. (1 point for each sub-item, total 3 points)
a) Cooking: father - 1500 rubles, mother - 1600 rubles, son - 2400 rubles.
Therefore, the father cooks and spends 1 hour on it.
b) Walking the dog: father - 1500 rubles, mother - 800 rubles, son - 600 rubles.
Therefore, the son walks the dog and spends 1 hour on it.
c) Going to the store: father - 1500 rubles, mother - 1600 rubles, son - 1800 rubles.
Therefore, the father goes to the store and spends 1 hour on it.
3) Calculate the final earnings of each family member: (3 points)
a) Father: 1500 * (24 - 8 - 8 - 1 - 1) = 9000 rubles.
b) Mother: 800 * (24 - 8 - 8) = 6400 rubles.
c) Son: 600 * (24 - 8 - 8 - 1) = 4200 rubles.
d) Total: 9000 + 6400 + 4200 = 19600 rubles.
Answer: 19600 rubles
## Second Case:
1) The number of hours each family member can spend on work or household chores per day: 24 - 8 - 8 = 8 hours.
2) Estimate the maximum income of each family member assuming they do not do household chores (2 points):
a) Father: 1500 * 8 = 12000 rubles.
b) Mother: 1200 * 5 = 6000 rubles + 30 minutes of free time.
c) Son: 450 * 10 = 4500 rubles + 30 minutes of free time.
3) Calculate the lost income from each type of activity (3 points, 1 point for each type of work):
a) Cooking: father - 1500, mother - 1200 (if she didn't go to the store) or -2400 (if she did go to the store), son - 2250.
b) Going to the store: father - 1500, mother - 1200 (if she didn't cook) or -2400 (if she did cook), son - 1800.
c) Walking the dog: father - 1500, mother - 1200, son - 450.
4) Choose the smallest losses: the father and mother cook and go to the store (it doesn't matter who does what, but each does one thing), the son walks the dog (2 points for correctly determining the tasks for each family member).
5) Thus, the maximum earnings: 12000 - 1500 + 6000 - 1200 + 4500 - 450 = 19350 rubles (1 point)
#
|
19600
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. (10 points)
On December 31 at 16:35, Misha realized he had no New Year's gifts for his entire family. He wants to give different gifts to his mother, father, brother, and sister. Each of the gifts is available in 4 stores: Romashka, Odynachik, Nezabudka, and Lysichka, which close at 20:00. The journey from home to each store and between any two stores takes 30 minutes.
The table below shows the cost of the gifts in all four stores and the time Misha will need to spend shopping in each store. What is the minimum amount of money Misha can spend if he must definitely manage to buy all 4 gifts?
| | mother | father | brother | sister | Time spent in the store (min.) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Romashka | 1000 | 750 | 930 | 850 | 35 |
| Odynachik | 1050 | 790 | 910 | 800 | 30 |
| Nezabudka | 980 | 810 | 925 | 815 | 40 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Lysichka | 1100 | 755 | 900 | 820 | 25 |
|
# Solution:
Notice that in each of the stores, there is a "unique" gift with the lowest price.
If Misha managed to visit all 4 stores, he would spend the minimum amount of $980+750+900+800=3430$ rubles. However, visiting any three stores would take Misha at least $30 * 3+25+30+35=180$ minutes. Considering the additional 30 minutes to travel to the fourth store, he would not make it. On the other hand, Misha can manage to visit any three out of the four stores—along with the travel time, he would arrive at the third store no later than $3 * 30+35+40=165$ minutes, i.e., at $19:10$.
Then, in one of the stores, he needs to choose 2 gifts to minimize the extra cost. Let's create a table of the extra costs:
| | mom | dad | brother | sister |
| :--- | :--- | :--- | :--- | :--- |
| Daisy | 20 | 0 | 30 | 50 |
| Dandelion | 70 | 40 | 10 | 0 |
| Forget-me-not | 0 | 60 | 25 | 15 |
| Lily of the Valley | 120 | 5 | 0 | 20 |
From the table, it is easy to see that the minimum extra cost is 5 rubles at the "Lily of the Valley" store for the gift for dad.
Therefore, Misha goes to Dandelion, Forget-me-not, and Lily of the Valley, and the total cost for the gifts will be $3430+5=3435$ rubles.
Answer: 3435 rubles.
Criteria:
Correct answer - 2 points
Reasoning that he cannot visit all 4 stores - 2 points
Reasoning that he can visit any 3 stores - 2 points
Explanation that the minimum extra cost is 5 rubles - 4 points
#
|
3435
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make the purchase.
## Maximum 30 points
## Problem 5
Schoolboy Vanechka informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use an additional card issued to her bank account. Vanya needed to buy 100 large chocolate bars for organizing and conducting a quiz at the New Year's party. However, the card has a protection system that analyzes the last 5 purchases, determines their average value, and compares it with the planned purchase amount. If the planned purchase exceeds the average by three times, the bank blocks the transaction and requires additional verification (for example, a call from his mother to the call center). Over the last month, his mother made purchases in the following amounts: 785 rubles, 2033 rubles, 88 rubles, 3742 rubles, 1058 rubles.
Calculate whether Vanechka will be able to buy the chocolate bars for the New Year's party in one transaction without calling his mother to the call center if the cost of one chocolate bar is 55 rubles?
|
# Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an acceptable purchase would be no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, one can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct answer.
7 points - correct calculation of the average value of purchases
3 points - partial solution
## Appendix to Problem No. 1
Insurance is a system of relationships between insurers (insurance company) and insured individuals (for example, any citizen), which allows for the reduction of property risks by insuring, for example, property against possible adverse events...
Property insurance is one of the types of insurance according to the Civil Code of the Russian Federation. Real estate is one type of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
84
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. A stationery store is running a promotion: there is a sticker on each notebook, and for every 5 stickers, a customer can get another notebook (also with a sticker). Fifth-grader Katya thinks she needs to buy as many notebooks as possible before the new semester. Each notebook costs 4 rubles, and Katya has 150 rubles. How many notebooks will Katya get?
|
Answer: 46.
1) Katya buys 37 notebooks for 148 rubles.
2) For 35 stickers, Katya receives 7 more notebooks, after which she has notebooks and 9 stickers.
3) For 5 stickers, Katya receives a notebook, after which she has 45 notebooks and 5 stickers.
4) For 5 stickers, Katya receives the last 46th notebook.
|
46
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give the answer in rubles, rounding to the nearest whole number (the answer is a whole number).
|
Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997$. Since rounding was required, the answer is 24.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/1.html
|
24
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
|
Answer: 15.
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let's find the number of pairs of different colored tulips in the bouquet: $600: 81=7$ (remainder 33). This means Vanya needs a bouquet with $7 * 2+1=15$ tulips.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 12. (6 points)
Victor received a large sum of money as a birthday gift in the amount of 45 thousand rubles. The young man decided to save this part of his savings in dollars on a currency deposit. The term of the deposit agreement was 2 years, with an interest rate of 4.7% per annum, compounded quarterly. On the day the deposit was opened, the commercial bank bought dollars at a rate of 59.60 rubles per 1 US dollar and sold them at a rate of 56.60 rubles per 1 US dollar. What amount in US dollars will be on Victor's account at the end of the term of the deposit agreement (rounding to the nearest whole number)?
|
Answer: 873 USD.
## Comment:
45000 RUB / 56.60 RUB $\times(1+4.7\% / 4 \text { quarters })^{2 \text { years } \times 4 \text { quarters }}=873$ USD.
|
873
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 17-19. (2 points per Task)
Alena opened a multi-currency deposit at "Severny" Bank for 3 years. The deposit involves the deposit of funds in three currencies: euros, dollars, and rubles. At the beginning of the deposit agreement, Alena's account contained 3000 euros, 4000 dollars, and 240000 rubles. The interest rate on the ruble deposit is 7.9% per annum, and on the dollar and euro deposits, it is 2.1% per annum. Interest is accrued annually, and the interest is capitalized. After one year, Alena decided to change the structure of her deposit, sold 1000 euros, and used the proceeds to increase the share of funds in dollars. At the time of conversion, the selling rate for dollars was 58.90 rubles, the buying rate for dollars was 56.90 rubles, the selling rate for euros was 61.15 rubles, and the buying rate for euros was 60.10 rubles. Another year later, Alena reviewed the structure of her deposit again, exchanged 2000 dollars, and bought euros with the proceeds. At the time of conversion, the selling rate for dollars was 60.10 rubles, the buying rate for dollars was 58.50 rubles, the selling rate for euros was 63.20 rubles, and the buying rate for euros was 61.20 rubles.
Question 17. Determine the balance on Alena's dollar account at the end of the deposit agreement (round to the nearest whole number).
Question 18. Determine the balance on Alena's euro account at the end of the deposit agreement (round to the nearest whole number).
Question 19. Determine the balance on Alena's ruble account at the end of the deposit agreement (round to the nearest whole number).
|
Answer 17: $3280;
Answer 18: 4040 euros,
Answer 19: 301492 rubles.
## Comment:
1 year
Euros: 3000 euros $\times(1+2.1 \%)=3063$ euros.
Dollars: 4000 dollars $\times(1+2.1 \%)=4084$ dollars.
Rubles: 240000 rubles $\times(1+7.9 \%)=258960$ rubles.
2 year
Euros: (3063 euros - 1000 euros $) \times(1+2.1 \%)=2106$ euros.
Dollars: (4084 dollars + 1000 euros × 60.10 rubles / 58.90 rubles) × (1 + 2.1 \%) = 5212 dollars.
Rubles: 258960 rubles $\times(1+7.9 \%)=279418$ rubles.
3 year
Euros: (2106 euros + 2000 dollars $\times$ 58.50 rubles / 63.20 rubles $) \times(1+2.1 \%)=4040$ euros.
Dollars: (5212 dollars - 2000 dollars) $\times(1+2.1 \%)=3279.45$ dollars.
Rubles: 279418 rubles $\times(1+7.9 \%)=301492$ rubles.
|
3280
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. (4 points)
Ivan bought a used car from 2010 for 90,000 rubles with an engine power of 150 hp and registered it on January 29. On August 21 of the same year, the citizen sold his car and a month later bought a horse and a cart for 7,500 rubles. The transport tax rate is set at 20 rubles per 1 hp. What amount of transport tax should the citizen pay? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 2000.
## Comment:
|
Solution: transport tax $=150 \times 20 \times 8 / 12=2000$ rubles. A horse and a cart are not subject to transport tax.
|
2000
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. (6 points)
Sergei, being a student, worked part-time in a student cafe after classes for a year. Sergei's salary was 9000 rubles per month. In the same year, Sergei paid for his medical treatment at a healthcare facility in the amount of 100000 rubles and purchased medications on a doctor's prescription for 20000 rubles (eligible for deduction). The following year, Sergei decided to apply for a social tax deduction. What amount of the paid personal income tax is subject to refund to Sergei from the budget? (Provide the answer as a whole number, without spaces and units of measurement.)
## Answer: 14040.
## Comment:
|
Solution: the amount of the social tax deduction for medical treatment will be: $100000+20000=$ 120000 rubles. The possible tax amount eligible for refund under this deduction will be $120000 \times 13\% = 15600$ rubles. However, in the past year, Sergey paid income tax (NDFL) in the amount of $13\% \times (9000 \times 12) = 14040$ rubles. Therefore, the amount of personal income tax paid, eligible for refund to Sergey from the budget, will be 14040 rubles.
|
14040
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. (2 points)
After graduating from a technical university, Oleg started his own business producing water heaters. This year, Oleg plans to sell 5000 units of water heaters. Variable costs for production and sales of one water heater amount to 800 rubles, and total fixed costs are 1000 thousand rubles. Oleg wants his income to exceed expenses by 1500 thousand rubles. At what price should Oleg sell the water heaters? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 1300.
## Comment:
|
Solution: the price of one kettle $=((1000000+0.8 \times 5000)+1500$ 000) / $(1000000$ + $0.8 \times 5000)=1300$ rub.
|
1300
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 18. (4 points)
By producing and selling 4000 items at a price of 6250 rubles each, a budding businessman earned 2 million rubles in profit. Variable costs for one item amounted to 3750 rubles. By what percentage should the businessman reduce the production volume to make his revenue equal to the cost? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 20.
## Comment:
|
Solution: fixed costs $=-2$ million, gap $+4000 \times 6250-3750 \times 4000$ million $=8$ million.
$6.25 \times Q=3750 Q+8 \text{ million}$.
$Q=3200$ units.
Can be taken out of production $=4000-3200=800$ units, that is, $20\%$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make the purchase.
## Maximum 30 points
## Problem 5
Schoolboy Vanechka informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use an additional card issued to her bank account. Vanya needed to buy 100 large chocolate bars for organizing and conducting a quiz at the New Year's party. However, the card has a protection system that analyzes the last 5 purchases, calculates their average value, and compares it with the planned purchase amount. If the planned purchase exceeds the average by three times, the bank blocks the transaction and requires additional verification (for example, a call from his mother to the call center). Over the last month, his mother made purchases in the following amounts: 785 rubles, 2033 rubles, 88 rubles, 3742 rubles, 1058 rubles.
Calculate whether Vanechka will be able to buy the chocolate bars for the New Year's party in one transaction without calling his mother to the call center if the cost of one chocolate bar is 55 rubles?
|
# Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an allowable purchase is no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, you can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct answer.
7 points - correct calculation of the average value of purchases
3 points - partial solution
## Appendix to Problem No. 1
Insurance is a system of relationships between insurers (insurance company) and insured individuals (for example, any citizen), which allows reducing property risks by insuring, for example, property against possible adverse events...
Property insurance is one of the types of insurance according to the Civil Code of the Russian Federation. Real estate is one type of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
84
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give your answer in rubles, rounding to the nearest whole number.
|
Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997 \approx 24$.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/23.html
|
24
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Alexey plans to buy one of two car brands: "A" for 900 thousand rubles or "B" for 600 thousand rubles. On average, Alexey drives 15 thousand km per year. The cost of gasoline is 40 rubles per liter. The cars consume the same type of gasoline. The car is planned to be used for 5 years, after which Alexey will be able to sell the car of brand "A" for 500 thousand rubles, and the car of brand "B" for 350 thousand rubles.
| Car Brand | Fuel Consumption (l/100km) | Annual Insurance Cost (rubles) | Average Annual Maintenance Cost (rubles) |
| :--- | :--- | :--- | :--- |
| "A" | 9 | 35000 | 25000 |
| "B" | 10 | 32000 | 20000 |
Using the data in the table, answer the question: how much more expensive will the purchase and ownership of the more expensive car be for Alexey?
|
Answer: 160000.
Use of car brand "A":
$900000+(15000 / 100) * 9 * 5 * 40+35000 * 5+25000 * 5-500000=970000$
Use of car brand "B":
$600000+(15000 / 100) * 10 * 5 * 40+32000 * 5+20000 * 5-350000=810000$
Difference: $970000-810000=160000$
|
160000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A family of 4, consisting of a mom, a dad, and two children, has arrived in city $\mathrm{N}$ for 5 days. They plan to make 10 trips on the subway each day. What is the minimum amount they will have to spend on tickets, given the following tariffs in city $\mathrm{N}$?
| Adult ticket for one trip | 40 rubles |
| :--- | :--- |
| Child ticket for one trip | 20 rubles |
| Unlimited daily pass for one person | 350 rubles |
| Unlimited daily pass for a group of up to 5 people | 1500 rubles |
| Unlimited 3-day pass for one person | 900 rubles |
| Unlimited 3-day pass for a group of up to 5 people | 3500 rubles |
|
Answer: 5200. The family will spend this amount if the parents buy a three-day pass for themselves, and for the remaining two days, they will buy a one-day pass. For this, they will spend ($900 + 350 * 2$) * $2 = 3200$ rubles.
For the children, it is most cost-effective to buy single-trip tickets for all 5 days, spending $2 * 10 * 20 * 5 = 2000$ rubles.
Thus, the total amount will be $3200 + 2000 = 5200$ rubles.
|
5200
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. (12 points)
Andrey lives near the market, and during the summer holidays, he often helped one of the traders lay out fruits on the counter early in the morning. For this, the trader provided Andrey with a $10 \%$ discount on his favorite apples. But autumn came, and the price of apples increased by $10 \%$. Despite the fact that Andrey started school and stopped helping the trader, the $10 \%$ discount for him remained. What will Andrey's monthly expenses on apples be now, considering that he buys 2 kilograms monthly? Before the price increase, apples at the market cost 50 rubles per kg for all customers. (Provide the answer as a whole number, without spaces or units of measurement.)
#
|
# Answer: 99.
## Comment
Solution: the new price of apples at the market is 55 rubles per kg, with a discount of $10 \%$ applied to this price. Thus, the price for 1 kg for Andrei will be 49.5 rubles, and for 2 kg Andrei will pay 99 rubles monthly.
|
99
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 10. (12 points)
The Ivanov family owns an apartment with necessary property worth 3 million rubles, a car that is currently valued at 900 thousand rubles on the market, and savings, part of which, amounting to 300 thousand rubles, is placed in a bank deposit, part is invested in securities worth 200 thousand rubles, and part, amounting to 100 thousand rubles, is in liquid form on a bank card and in cash. In addition, there are outstanding loans. The remaining balance on the mortgage is 1.5 million rubles, the remaining balance on the car loan is 500 thousand rubles. It is also necessary to return a debt to relatives in the amount of 200 thousand rubles. What net worth (net wealth) does the Ivanov family possess? (Provide the answer as a whole number, without spaces or units of measurement.)
|
# Answer: 2300000
## Comment
Solution: equity (net worth) = value of assets - value of liabilities. Value of assets $=3000000+900000+300000+200000+100000=$ 4500000 rubles. Value of liabilities $=1500000+500000+200000=2200000$ rubles. Net worth $=4500000-2200000=2300000$ rubles
## MOSCOW FINANCIAL LITERACY OLYMPIAD 5-7 GRADE Variant 2
|
2300000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. (8 points)
Konstantin's mother opened a foreign currency deposit in the "Western" Bank for an amount of 10 thousand dollars for a term of 1 year. Literally 4 months later, the Bank of Russia revoked the license of the "Western" Bank. The exchange rate on the date of the license revocation was 58 rubles 15 kopecks per 1 dollar. Konstantin's mother was not too upset, as the deposits in the bank were insured under the deposit insurance system. What amount of the deposit should be returned to Konstantin's mother according to the legislation (do not consider interest payments in the calculations)?
a. 10 thousand dollars;
b. 581,500 rubles;
c. 1,400 thousand rubles;
d. 10 thousand rubles.
|
# Answer: b.
## Comment
$10000 \times 58.15$ RUB $=581500$ RUB.
|
581500
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
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