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18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
 | Solution. With probability $\frac{1}{3}$, the pedestrian falls into the interval when the green light is on (event $G$), and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\bar{G}$). In this case, the waiting time is uniformly distributed over the interval from 0 to 2 minutes, so the conditional expected value of $T$ given $\bar{G}$ is one minute: $\mathrm{E}(T \mid G)=\frac{0+2}{2}=1$ (min).
We get:
$$
\mathrm{E}(T)=\mathrm{E}(T \mid G) \cdot \mathrm{P}(G)+\mathrm{E}(T \mid \bar{G}) \cdot \mathrm{P}(\bar{G})=\frac{1}{3} \cdot 0+\frac{2}{3} \cdot 1=\frac{2}{3}(\text { min })
$$
Answer: 40 seconds. | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It is known that trains run at approximately equal intervals in both directions, and that the train traveling on the northern route (through "Belorusskaya") takes 17 minutes to travel from "Kievskaya" to "Taganskaya" or vice versa, while the train on the southern route (through "Paveletskaya") takes 11 minutes.
Out of habit, the Scientist always calculates everything. Once, he calculated that over many years of observation:
- the train traveling counterclockwise arrives at "Kievskaya" on average 1 minute and 15 seconds after the train traveling clockwise arrives at the same station. The same is true for "Taganskaya";
- the average travel time from home to work is 1 minute less than the travel time from work to home.
Find the expected interval between trains traveling in the same direction.
 | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is
$$
11 p + 17(1-p) = 17 - 6p
$$
On the return trip from "Kievskaya" to "Taganskaya," the expected travel time is
$$
17 p + 11(1-p) = 11 + 6p
$$
According to the condition, $11 + 6p - (17 - 6p) = 1$, from which $p = \frac{7}{12}$. Let the interval between trains be $T$. Then $T(1-p) = Y$, where $Y$ is the time between the arrival of a train "clockwise" and the arrival of a train "counterclockwise" at the favorite stations. Then
$$
\mathrm{E} T = \frac{\mathrm{E} Y}{1-p} = \frac{5}{4} \cdot \frac{12}{5} = 3
$$
Answer: 3 minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It is known that trains run at approximately equal intervals in both directions, and that the train traveling on the northern route (through "Belorusskaya") takes 17 minutes to travel from "Kievskaya" to "Taganskaya" or vice versa, while the train on the southern route (through "Paveletskaya") takes 11 minutes.
Out of habit, the Scientist always calculates everything. Once, he calculated that over many years of observation:
- the train traveling counterclockwise arrives at "Kievskaya" on average 1 minute and 15 seconds after the train traveling clockwise arrives at the same station. The same is true for "Taganskaya";
- the average travel time from home to work is 1 minute less than the travel time from work to home.
Find the expected interval between trains traveling in the same direction.
 | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is
$$
11 p + 17(1-p) = 17 - 6p
$$
On the return trip from "Kievskaya" to "Taganskaya," the expected travel time is
$$
17 p + 11(1-p) = 11 + 6p
$$
According to the condition, $11 + 6p - (17 - 6p) = 1$, from which $p = \frac{7}{12}$. Let the interval between trains be $T$. Then $T(1-p) = Y$, where $Y$ is the time between the arrival of a train "clockwise" and the arrival of a train "counterclockwise" at the favorite stations. Then
$$
\mathrm{E} T = \frac{\mathrm{E} Y}{1-p} = \frac{5}{4} \cdot \frac{12}{5} = 3
$$
Answer: 3 minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
13. Ring Line (from 8th grade. 3 points). On weekdays, the Absent-Minded Scientist travels to work on the ring line of the Moscow metro from the station "Taganskaya" to the station "Kievskaya", and back in the evening (see the diagram).
Upon entering the station, the Scientist boards the first train that arrives. It is known that trains run at approximately equal intervals in both directions, and that the train traveling on the northern route (through "Belorusskaya") takes 17 minutes to travel from "Kievskaya" to "Taganskaya" or vice versa, while the train on the southern route (through "Paveletskaya") takes 11 minutes.
Out of habit, the Scientist always calculates everything. Once, he calculated that over many years of observation:
- the train traveling counterclockwise arrives at "Kievskaya" on average 1 minute and 15 seconds after the train traveling clockwise arrives at the same station. The same is true for "Taganskaya";
- the average travel time from home to work is 1 minute less than the travel time from work to home.
Find the expected interval between trains traveling in the same direction.
 | Solution. If the Scientist boarded trains of different directions with equal probabilities, the average travel time in one direction and the average travel time in the other would be the same. Therefore, the probabilities are not equal.
Let $p$ be the probability that the Scientist boards a train going clockwise. Then the expected travel time from "Taganskaya" to "Kievskaya" is
$$
11 p + 17(1-p) = 17 - 6p
$$
On the return trip from "Kievskaya" to "Taganskaya," the expected travel time is
$$
17 p + 11(1-p) = 11 + 6p
$$
According to the condition, $11 + 6p - (17 - 6p) = 1$, from which $p = \frac{7}{12}$. Let the interval between trains be $T$. Then $T(1-p) = Y$, where $Y$ is the time between the arrival of a train "clockwise" and the arrival of a train "counterclockwise" at the favorite stations. Then
$$
\mathrm{E} T = \frac{\mathrm{E} Y}{1-p} = \frac{5}{4} \cdot \frac{12}{5} = 3
$$
Answer: 3 minutes. | 3 | Logic and Puzzles | math-word-problem | Yes | Yes | olympiads | false |
18. Traffic Light (from 10th grade. 2 points). A traffic light at a pedestrian crossing allows pedestrians to cross the street for one minute and prohibits crossing for two minutes. Find the average waiting time for the green light for a pedestrian who approaches the intersection.
 | Solution. With probability $\frac{1}{3}$, the pedestrian falls into the interval when the green light is on (event $G$), and therefore the conditional expected waiting time $T$ in this case is 0: $\mathrm{E}(T \mid G)=0$. With probability $\frac{2}{3}$, the pedestrian falls on the red signal and has to wait (event $\bar{G}$). In this case, the waiting time is uniformly distributed over the interval from 0 to 2 minutes, so the conditional expected value of $T$ given $\bar{G}$ is one minute: $\mathrm{E}(T \mid G)=\frac{0+2}{2}=1$ (min).
We get:
$$
\mathrm{E}(T)=\mathrm{E}(T \mid G) \cdot \mathrm{P}(G)+\mathrm{E}(T \mid \bar{G}) \cdot \mathrm{P}(\bar{G})=\frac{1}{3} \cdot 0+\frac{2}{3} \cdot 1=\frac{2}{3}(\text { min })
$$
Answer: 40 seconds. | 40 | Other | math-word-problem | Yes | Yes | olympiads | false |
4. Stem-and-leaf plot. (From 6th grade, 2 points). To represent whole numbers or decimal fractions, a special type of diagram called a "stem-and-leaf plot" is often used. Such diagrams are convenient for representing people's ages. Suppose that in the studied group, there are 5 people aged 19, 34, 37, 42, and 48. For this group, the diagram will look as shown in Fig. 2. The left column is the "stem," and to the right of it are the "leaves."
When studying a certain group of patients, on December 1, the doctor created a diagram of their ages (Fig. 3a). On Fig. 3b, a new diagram of their ages is shown, which was also created on December 1, several years later. Over these years, the composition of the group remained the same - all those who were there remained, and no one new joined the group. However, the numbers on the new diagram are not visible - they are replaced by asterisks. Determine how many years have passed and restore the diagram.
$$
\begin{array}{l|llllll}
0 & & & & & & \\
1 & 0 & 0 & 1 & 2 & 2 & 3 \\
2 & 1 & 5 & 6 & & & \\
3 & 0 & 2 & 4 & 6 & & \\
4 & 1 & 6 & & & &
\end{array}
$$
Fig. 3 a)
Fig. 3 b) | Solution. The digits from 0 to 5, representing decades of years, can be placed immediately (Fig. 4a). It is clear that less than 10 years have passed, otherwise there would be no digits in line "1".
If 7 or more years had passed, then the person who is 13 years old would have moved to line "2", and there would be fewer than six values left in line "1". But this did not happen, so less than 7 years have passed.
No one from line "1" moved to line "2", but two people, aged 25 and 26, moved from line "2" to line "3". However, the number of values in line "3" did not change, which means that those who were 34 and 36 years old moved to line "4". Therefore, at least 6 years have passed.
Thus, exactly 6 years have passed, and the diagram shown in Fig. 4b) was obtained. [^0]
Fig. 4 a)
Fig. 4 b)
Answer: 6 years, see Fig. 4 b). | 6 | Logic and Puzzles | 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 |
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. Errors are possible. For example, a low minivan may be classified as class 1 by the system, 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. is 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 class 2 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: approx. 190 cm. | 190 | Other | math-word-problem | Yes | Yes | olympiads | false |
12. Ivan the Tsarevich's Arrows (from 8th grade. 3 points). Ivan the Tsarevich is learning to shoot a bow. He put 14 arrows in his quiver and went to shoot at pine cones in the forest. He knocks down a pine cone with a probability of 0.1, and for each pine cone he knocks down, the Frog-Princess gives him 3 more arrows. Ivan the Tsarevich shoots until he runs out of arrows. Find the expected number of shots Ivan the Tsarevich will make. | Solution. First method. Let Ivan have $n$ arrows at the present moment. Let $X_{0}$ be the random variable "the number of shots needed to reduce the number of arrows by one." Ivan makes a shot. Consider the random variable - the indicator $I$ of a successful shot. $I=0$, if the shot is unsuccessful (probability of this is 0.9), or $I=1$, if the pine cone is knocked down (probability 0.1). Then
$$
X_{0}=(1-I) \cdot 1+I \cdot\left(1+X_{1}+X_{2}+X_{3}\right)
$$
where $X_{1}, X_{2}$, and $X_{3}$ are the same random variables as $X_{0}$. Each of them equals the number of shots needed to reduce the number of arrows in reserve by one after receiving three new arrows from the Princess.
The variable $I$ is only related to the nearest shot, while the variables $X_{k}$ for $k=1,2,3$ are only related to subsequent shots. Therefore, the random variables $I$ and $X_{k}$ are independent. We will transition to expectations in the obtained equality:
$$
\mathrm{E} X_{0}=(1-\mathrm{E} I)+\mathrm{E} I \cdot\left(1+\mathrm{E} X_{1}+\mathrm{E} X_{2}+\mathrm{E} X_{3}\right)
$$
Since all variables from $X_{0}$ to $X_{3}$ are identically distributed, their expectations are equal. Let's denote them by $a$. In addition, $\mathrm{E} I=0.1$. Therefore,
$$
a=(1-0.1)+0.1 \cdot(1+3 a), \text { from which } a=\frac{10}{7}
$$
Thus, to get rid of one arrow, Ivan needs on average 10/7 shots. To get rid of all 14 arrows that were initially present, he needs on average $14 \cdot 10 / 7=20$ shots.
Second method. Determine how many fewer arrows Ivan has as a result of the $k$-th shot (taking into account the new ones given by the Princess). Let this random variable be $X_{k}$ ( $k=1,2,3, \ldots$ ). From the condition, it is clear that $X_{k}=1$, if Ivan did not knock down the pine cone, or $X_{k}=-2$ (the number of arrows increased by 2), if Ivan knocked down the pine cone. That is,
$$
X_{k} \sim\left(\begin{array}{cc}
1 & -2 \\
0.9 & 0.1
\end{array}\right)
$$
independently of $k$. Let's find the expectation: $\mathrm{E} X=0.9-0.2=0.7$ : with each shot, Ivan has on average 0.7 fewer arrows.
Introduce indicators of individual shots $I_{k}$, where $k=1,2,3, \ldots$ :
$$
I_{k}=\left\{\begin{array}{l}
0, \text { if there were not enough arrows for the } k \text {-th shot, } \\
1, \text { if the } k \text {-th shot was made. }
\end{array}\right.
$$
The total number of shots is
$$
S=I_{1}+I_{2}+I_{3}+\ldots+I_{k}+\ldots
$$
and the total change (decrease) in the number of arrows is
$$
I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots
$$
By the time all the arrows are gone, this sum equals 14. We form the equation
$$
14=I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots
$$
and transition to the expectation in the right part (the left part is a constant):
$$
14=\mathrm{E}\left(I_{1} X_{1}+I_{2} X_{2}+\ldots+I_{k} X_{k}+\ldots\right)
$$
The random variable $I_{k}$ depends only on how many arrows are left after $k-1$ shots, so the variables $I_{k}$ and $X_{k}$ are independent. Therefore,
$$
\begin{gathered}
14=\mathrm{E} I_{1} \cdot \mathrm{E} X_{1}+\mathrm{E} I_{2} \cdot \mathrm{E} X_{2}+\ldots+\mathrm{E} I_{k} \cdot \mathrm{E} X_{k}+\ldots= \\
=0.7\left(\mathrm{E} I_{1}+\mathrm{E} I_{2}+\ldots+\mathrm{E} I_{k}+\ldots\right)=0.7 \mathrm{E}\left(I_{1}+I_{2}+I_{3}+\ldots+I_{k}+\ldots\right)=0.7 \mathrm{E} S
\end{gathered}
$$
Therefore, $\mathrm{E} S=14: 0.7=20$.
Answer: 20. | 20 | Logic and Puzzles | 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 diagram shows a track layout for karting. The start and finish are at point $A$, and the kart driver can make as many laps as they like, returning to point $A$.
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 |
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. 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 class 2 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 |
7. Solution. Suppose for clarity of reasoning that when a bite occurs, the Absent-Minded Scholar immediately pulls out and re-casts the fishing rod, and does so instantly. After this, he waits again. Consider a 6-minute time interval. During this time, on average, there are 3 bites on the first fishing rod and 2 bites on the second fishing rod. Therefore, in total, there are on average 5 bites on both fishing rods in these 6 minutes. Consequently, the average waiting time for the first bite is $6: 5=1.2$ minutes. | Answer: 1 minute 12 seconds.
Evaluation Criteria
| Correct and justified solution | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| The solution is incorrect or missing (in particular, only the answer is given) | 0 points | | 1 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 8. Solution.
a) Suppose there are 9 numbers in the set. Then five of them do not exceed the median, which is the number 2. Another four numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed
$$
5 \cdot 2 + 4 \cdot 13 = 62
$$
Since the arithmetic mean is 7, the sum of the numbers in the set is $7 \cdot 9 = 63$. This is a contradiction. The set cannot consist of 9 numbers.
b) Let the number of numbers in the set be $2n + 1$ ($n$ is a natural number). In the set, there will be exactly $n + 1$ numbers, each of which does not exceed the median, which is the number 2. The remaining $n$ numbers do not exceed the number 13. Therefore, the sum of all numbers in the set does not exceed
$$
13n + 2(n + 1) = 15n + 2
$$
On the other hand, this sum is equal to $7(2n + 1) = 14n + 7$. From the inequality $14n + 7 \leq 15n + 2$, we get that $n \geq 5$. Therefore, the set contains no fewer than $2 \cdot 5 + 1 = 11$ numbers. To show that this is possible, consider the following example. The set
$$
2; 2; 2; 2; 2; 2; 13; 13; 13; 13; 13
$$
consists of 11 numbers and satisfies conditions 1-4. | Answer: a) no; b) 11.
Scoring criteria
| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| The correct estimate of the number of numbers in part (b) is found, but no example is given | 2 points |
| Part (a) is solved correctly | 1 point |
| The solution is incorrect or missing (in particular, only the answer is given) | 0 points | | 11 | Other | math-word-problem | Yes | Yes | olympiads | false |
7. Solution. For clarity, let's assume that when a bite occurs, the Absent-Minded Scholar immediately reels in and casts the fishing rod again, and does so instantly. After this, he waits again. Consider a 5-minute time interval. During this time, on average, there are 5 bites on the first fishing rod and 1 bite on the second. Therefore, in total, there are on average 6 bites on both fishing rods in these 5 minutes. Consequently, the average waiting time for the first bite is $\frac{5}{6}$ minutes. | Answer: 50 seconds.
## Grading Criteria
| Solution is correct and well-reasoned | 3 points |
| :--- | :---: |
| It is shown that on average there are 5 bites in 6 minutes, or an equivalent statement is proven | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points | | 50 | Other | math-word-problem | Yes | Yes | olympiads | false |
# 8. Solution.
a) Suppose the set contains 7 numbers. Then four of them are not less than the median, which is the number 10. Another three numbers are not less than one. Then the sum of all numbers in the set is not less than
$$
3+4 \cdot 10=43
$$
Since the arithmetic mean is 6, the sum of the numbers in the set is $6 \cdot 7=42$. Contradiction. The set cannot consist of 7 numbers.
b) Let the number of numbers in the set be $2 n+1$ (where $n$ is a natural number). In the set, there will be exactly $n+1$ numbers, each of which is not less than the median, which is the number 10. The remaining $n$ numbers are not less than the number 1. Then the sum of all numbers in the set is not less than
$$
n+10(n+1)=11 n+10
$$
On the other hand, this sum is equal to $6(2 n+1)=12 n+6$. From the inequality $12 n+6 \geq 11 n+10$ we get that $n \geq 4$. Therefore, the set contains no fewer than $2 \cdot 4+1=9$ numbers. Let's provide an example to show that this is possible. The set
$$
\text { 1; 1; 1; 1; 10; 10; 10; 10; } 10
$$
consists of 9 numbers and satisfies conditions 1-4. | Answer: a) no; b) 9.
## Grading Criteria
| Both parts solved correctly or only part (b) | 3 points |
| :--- | :---: |
| Correct estimate of the number of numbers in part (b), but no example | 2 points |
| Part (a) solved correctly | 1 point |
| Solution is incorrect or missing (including only the answer) | 0 points | | 9 | Other | 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 |
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-profile UAZ passenger car may be incorrectly classified as class 2, and its driver will not be happy. He can file a claim, and the operating company will have to refund him 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 | Algebra | 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 |
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. Errors are possible. For example, a low minivan may be classified as class 1 by the system, and its driver will be pleased. A low SUV may be incorrectly classified as class 2, and its driver will not be happy. He can file a claim, and the operating company will have to refund him 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 class 2 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: approx. 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 |
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 |
Problem 3. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did the winner score? | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise, $a_{1}+a_{2}+a_{3} \leq 6+7+8=21$).
Therefore, between $a_{3}$ and $a_{n-2}$, there can only be the numbers $10,11,12,13,14$. Their sum is 60, but it should be 47. Therefore, all these numbers are present except for the number 13.
This results in the sequence $a_{1}, a_{2}, 9,10,11,12,14,15, a_{n-1}, a_{n}$, where $a_{1}+a_{2}=14$ and $a_{n-1}+a_{n}=34$. Thus, $n=10$, $a_{1}=6$, $a_{2}=8$, $a_{n-1}=16$, and $a_{n}=18$.
Note that it is not enough to simply specify the required set. A proof of uniqueness is needed.
Answer: A) 10 students; B) 18 points.
Answer to variant 212: A) 10 participants; B) 18 points. | 10 | Number Theory | math-word-problem | Yes | Yes | olympiads | false |
Problem 4. One mole of an ideal gas undergoes a closed cycle, in which:
$1-2$ - isobaric process, during which the volume increases by 4 times;
$2-3$ - isothermal process, during which the pressure increases;
$3-1$ - a process in which the gas is compressed according to the law $T=\gamma V^{2}$.
Find how many times the volume in state 3 exceeds the initial volume in state 1. | Solution. Let the initial volume and pressure be denoted as $\left(V_{0} ; P_{0}\right)$. Then $V_{2}=4 V_{0}$.
From the Mendeleev-Clapeyron law, we have three relationships:
$$
P_{0} V_{0}=R T_{1}, P_{0} V_{2}=R T, P_{3} V_{3}=R T
$$
Dividing the third relationship by the second, we get: $\frac{P_{3}}{P_{0}}=\frac{V_{2}}{V_{3}}$.
From the dependence of temperature on volume in the process $3-1$, there is a linear relationship between pressure and volume: $P=\gamma V$. Then we can write the relationship between the parameters corresponding to states 1 and $3: \frac{P_{3}}{P_{0}}=\frac{V_{3}}{V_{0}}$.
From the two obtained relationships, it follows that:
$$
\frac{V_{2}}{V_{3}}=\frac{V_{3}}{V_{0}} \Rightarrow V_{3}^{2}=V_{2} V_{0} \Rightarrow V_{3}=2 V_{0}
$$
Answer: 2.
Answer to variant 212: 2. | 2 | Other | math-word-problem | Yes | Yes | olympiads | false |
Problem 3. All students in the class scored a different number of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did the winner score? | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise, $a_{1}+a_{2}+a_{3} \leq 6+7+8=21$).
Therefore, between $a_{3}$ and $a_{n-2}$, there can only be the numbers $10,11,12,13,14$. Their sum is 60, but it should be 47. Therefore, all these numbers are present except for the number 13.
This results in the sequence $a_{1}, a_{2}, 9,10,11,12,14,15, a_{n-1}, a_{n}$, where $a_{1}+a_{2}=14$ and $a_{n-1}+a_{n}=34$. Thus, $n=10$, $a_{1}=6$, $a_{2}=8$, $a_{n-1}=16$, and $a_{n}=18$.
Note that it is not enough to simply specify the required set. A proof of uniqueness is needed.
Answer: A) 10 students; B) 18 points.
Answer to variant 212: A) 10 participants; B) 18 points. | 10 | Number Theory | 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 |
4. A ten-liter bucket was filled to the brim with currants. Gavrila immediately said that there were 10 kg of currants in the bucket. Glafira thought about it and estimated the weight of the berries in the bucket more accurately. How can this be done if the density of the currant can be approximately considered equal to the density of water? | Solution. In approximate calculations, the sizes of the berries can be considered the same and much smaller than the size of the bucket. If the berries are laid out in one layer, then in the densest packing, each berry will have 6 neighbors: the centers of the berries will be at the vertices of equilateral triangles with sides equal to the diameter of the berries. When pouring the next layer, the berries will be located in the depressions between the berries of the previous layer. In such a packing, each berry will have 12 neighbors, and the centers of the berries will be at the vertices of regular tetrahedrons.
To calculate the fraction of the volume occupied by the berries, one can mentally "cut out" a parallelepiped from the lattice formed by the centers of the berries with edges of length $2 N R$, where $R$ is the radius of the berries, and $N$ is the number of berries that fit along one side, and in two opposite vertices of the edge, they meet at an angle of $60^{\circ}$. The volume of such a parallelepiped is $4 \sqrt{2} N^{3} R^{3}$, and it contains approximately $N^{3}$ berries, each with a volume of $\frac{4}{3} \pi R^{3}$ (berries near the walls are not fully placed, but their number is of the order of $N^{2}$, so this can be neglected for large $N$). Therefore, the ratio of the volume of the berries to the volume of the bucket is approximately $\frac{\pi \sqrt{2}}{6} \approx 0.74$, so in a 10-liter bucket, with the densest packing, there are about 7.4 kg of berries. In reality, there are somewhat fewer due to the lack of density.
The ratio of the volume of the berries to the volume of the container can also be calculated differently. For example, consider a rectangular parallelepiped in which the berries are laid out in layers, the distance between which is equal to the height of the tetrahedron mentioned above $\frac{2 \sqrt{2} R}{\sqrt{3}}$. In each layer, the distance between rows is equal to the height of the equilateral triangle $R \sqrt{3}$. Thus, $N^{3}$ berries will occupy a parallelepiped with a length of $2 R N$, a width of $\sqrt{3} R N$, and a height of $\frac{2 \sqrt{2} R N}{\sqrt{3}}$. The ratio of volumes is $\frac{N^{3} \cdot \frac{4}{3} \pi R^{3}}{2 R N \cdot \sqrt{3} R N \cdot \frac{2 \sqrt{2} R N}{\sqrt{3}}}=\frac{\pi}{3 \sqrt{2}}=\frac{\pi \sqrt{2}}{6}$.
Answer: about 7 kg.
Criteria: 20 points - correct (not necessarily the same as above) solution and the correct answer (from 7 to 7.4 kg); 15 points - generally ideologically correct solution, with errors that did not affect the answer; 10 points - ideologically correct solution with significant errors; also evaluated are solutions in which the packing density coefficient $\frac{\pi \sqrt{2}}{6}$ (or 74%) is used without proof; 5 points - the packing in the layer is taken into account, but the packing between layers is not; **0** points - everything else. | 7 | Logic and Puzzles | 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 |
4. Hot oil at a temperature of $100^{\circ} \mathrm{C}$ in a volume of two liters is mixed with one liter of cold oil at a temperature of $20^{\circ} \mathrm{C}$. What volume will the mixture have when thermal equilibrium is established in the mixture? Heat losses to the external environment can be neglected. The coefficient of volumetric expansion of the oil is $2 \cdot 10^{-3} 1 /{ }^{\circ} \mathrm{C}$. | Answer: 3
Let $V_{1}=2$ L be the volume of hot oil, and $V_{2}=1$ L be the volume of cold oil. Then we can write $V_{1}=U_{1}\left(1+\beta t_{1}\right), V_{2}=U_{2}\left(1+\beta t_{2}\right)$, where $U_{1}, U_{2}$ are the volumes of the respective portions of oil at zero temperature; $t_{1}=100^{\circ} \mathrm{C}, t_{2}=20^{\circ} \mathrm{C} ; \beta=2 \cdot 10^{-3} 1 /{ }^{\circ} \mathrm{C}$ is the coefficient of volumetric expansion.
From the heat balance equation: $c m_{1}\left(t-t_{1}\right)+c m_{2}\left(t-t_{2}\right)=0 \Leftrightarrow t=\frac{m_{1} t_{1}+m_{2} t_{2}}{m_{1}+m_{2}}$, where $m_{1}, m_{2}$ are the masses of the hot and cold portions of oil, respectively; $t$ is the temperature of the mixture. Note that $m_{1}=U_{1} \rho, m_{2}=U_{2} \rho$, where $\rho$ is the density of the oil at zero temperature.
Then $W_{1}=U_{1}(1+\beta t), W_{2}=U_{2}(1+\beta t)$. Here $W_{1}, W_{2}$ are the volumes of the initial portions of oil at temperature $t$.
The new total volume of the oil will be equal to
$$
\begin{aligned}
& W_{1}+W_{2}=\left(U_{1}+U_{2}\right)(1+\beta t)=\left(\frac{m_{1}}{\rho}+\frac{m_{2}}{\rho}\right)(1+\beta t)=\frac{1}{\rho}\left(m_{1}+m_{2}+\left(m_{1}+m_{2}\right) \beta t\right)= \\
& \frac{1}{\rho}\left(m_{1}+m_{2}+m_{1} \beta t_{1}+m_{2} \beta t_{2}\right)=\frac{1}{\rho}\left(m_{1}+m_{1} \beta t_{1}\right)+\frac{1}{\rho}\left(m_{2}+m_{2} \beta t_{2}\right)=\frac{m_{1}}{\rho}\left(1+\beta t_{1}\right)+\frac{m_{2}}{\rho}(1+
\end{aligned}
$$
$\left.\beta t_{2}\right)$
That is, $W_{1}+W_{2}=U_{1}\left(1+\beta t_{1}\right)+U_{2}\left(1+\beta t_{2}\right)=V_{1}+V_{2}$
Thus, the volume will not change. | 3 | 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 2. Grandma baked 19 pancakes. The grandchildren came from school and started eating them. While the younger grandson eats 1 pancake, the older grandson eats 3 pancakes, and during this time, grandma manages to cook 2 more pancakes. When they finished, there were 11 pancakes left on the plate. How many pancakes did the older grandson eat? | Solution. In one "cycle", the grandsons eat $1+3=4$ pancakes, and the grandmother bakes 2 pancakes, which means the number of pancakes decreases by 2. There will be ( $19-11$ ) $/ 2=4$ such cycles. This means, in these 4 cycles, the younger grandson ate 4 pancakes, the older grandson ate 12 pancakes, and the grandmother baked 8 pancakes during this time. Indeed, $19+8-4-12=11$.
Answer: 12. | 12 | Logic and Puzzles | 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 4. All students in the class scored different numbers of points (positive integers) on the test, with no duplicate scores. In total, they scored 119 points. The sum of the three lowest scores is 23 points, and the sum of the three highest scores is 49 points. How many students took the test? How many points did the winner score? | Solution. Let's denote all the results in ascending order $a_{1}, a_{2}, \ldots, a_{n}$, where $n$ is the number of students. Since $a_{1}+a_{2}+a_{3}=23$ and $a_{n-2}+a_{n-1}+a_{n}=49$, the sum of the numbers between $a_{3}$ and $a_{n-2}$ is $119-23-49=47$.
Since $a_{1}+a_{2}+a_{3}=23$, then $a_{3} \geq 9$ (otherwise, $a_{1}+a_{2}+a_{3} \leq 6+7+8=21$).
Therefore, between $a_{3}$ and $a_{n-2}$, there can only be the numbers $10,11,12,13,14$. Their sum is 60, but it should be 47. Therefore, all these numbers are present except for the number 13.
Thus, we get the sequence $a_{1}, a_{2}, 9,10,11,12,14,15, a_{n-1}, a_{n}$, where $a_{1}+a_{2}=14$ and $a_{n-1}+a_{n}=34$. Therefore, $n=10$, $a_{1}=6$, $a_{2}=8$, $a_{n-1}=16$, and $a_{n}=18$.
Answer: A) 10 students; B) 18 points. | 10 | Number Theory | 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 |
3.1. The friends who came to visit Gavrila occupied all the three-legged stools and four-legged chairs in the room, but there was no place left for Gavrila himself. Gavrila counted that there were 45 legs in the room, including the "legs" of the stools and chairs, the legs of the visiting guests (two for each!), and Gavrila's own two legs. How many people were in the room? | Answer: 9. Solution. If there were $n$ stools and $m$ chairs, then the number of legs in the room is $3 n+4 m+2 \cdot(n+m)+2$, from which we get $5 n+6 m=43$. This equation in integers has the solution $n=5-6 p, m=3+5 p$. The values of $n$ and $m$ are positive only when $p=0$. Therefore, there were 5 stools and 3 chairs. Thus, the number of people in the room is: $5+3+$ Gavril. | 9 | Logic and Puzzles | 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 |
5.1. A grenade lying on the ground explodes into a multitude of small identical fragments, which scatter in a radius of $L=90$ m. Determine the time interval (in seconds) between the moments of impact on the ground of the first and the last fragment, if such a grenade explodes in the air at a height of $H=10 \mathrm{m}$. The acceleration due to gravity is considered to be $g=10$ m/s ${ }^{2}$. Air resistance is negligible. | Answer: 6. Solution. From the motion law for a body thrown from ground level at an angle $\alpha$ to the horizontal, the range of flight is determined by the relation $L=\frac{V_{0}^{2}}{g} \sin 2 \alpha$. Therefore, the maximum range of flight is achieved at $\alpha=45^{\circ}$ and is equal to $L=\frac{V_{0}^{2}}{g}$. This means $V_{0}=\sqrt{g L}$.
The fragment that falls to the ground first is the one that flies vertically downward, and the last one is the one that flies vertically upward. After time $\tau=\frac{2 V_{0}}{g}$, the last fragment will return to the explosion site, which means it will be in the same situation as the first fragment. Therefore, the required time interval is $\tau=\frac{2 V_{0}}{g}$ (independent of height $H$). Thus, $\tau=\frac{2 \sqrt{g L}}{g}=2 \sqrt{\frac{L}{g}}=6$ seconds. | 6 | 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. Gavriil and Glafira took a glass filled to the brim with water and poured a little water into three ice cube trays, then placed them in the freezer. When the ice froze, they put the three resulting ice cubes back into the glass. Gavriil predicted that some water would spill out of the glass because ice expands in volume when it freezes. Glafira, however, claimed that the water level would be below the rim of the glass because part of the floating ice would protrude above the water surface. Who is right and why? | Solution. Let $V$ be the volume of water in the molds. Then the volume $W$ of ice in the molds can be determined from the law of conservation of mass $V \cdot \rho_{\text {water }}=W \cdot \rho_{\text {ice }}$. When ice of volume $W$ is floating, the submerged part of this volume $U$ can be determined from the condition of floating bodies $U \cdot \rho_{\text {water }}=W \cdot \rho_{\text {ice }}$. It is therefore obvious that $V=U$.
Answer: No one is right. The water will fill the glass exactly to its edges.
Grading criteria: 20 points - correct (not necessarily the same as above) solution and correct answer; 10 points - correct answer, but insufficiently clear justification. | 1 | Logic and Puzzles | 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 |
2. What is the greatest whole number of liters of water that can be heated to boiling temperature using the amount of heat obtained from the combustion of solid fuel, if in the first 5 minutes of combustion, 480 kJ is obtained from the fuel, and for each subsequent five-minute period, 25% less than the previous one. The initial temperature of the water is $20^{\circ} \mathrm{C}$, the boiling temperature is $100^{\circ} \mathrm{C}$, and the specific heat capacity of water is 4.2 kJ. | Answer: 5 liters
Solution: The amount of heat required to heat a mass $m$ of water under the conditions of the problem is determined by the relation $Q=4200(100-20) m=336 m$ kJ. On the other hand, if the amount of heat received in the first 5 minutes is $Q_{0}=480$ kJ. Then the total (indeed over an infinite time) amount of heat received will be $Q=4 Q_{0}$. Then the mass $m$ will be determined from the relation $m=\frac{4 Q_{0}}{336} \approx 5.7$
That is, the whole number of liters of water heated to 100 degrees is 5. | 5 | 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 |
1. A new model car travels $4 \frac{1}{6}$ kilometers more on one liter of gasoline compared to an old model car. At the same time, its fuel consumption per 100 km is 2 liters less. How many liters of gasoline does the new car consume per 100 km? | Answer: 6 liters.
Instructions. The fuel consumption of the new car is $x$ liters, and the consumption of the old car is $x+2$
liters. Equation: $\frac{100}{x}-\frac{100}{x+2}=\frac{25}{6} \Leftrightarrow \frac{4(x+2-x)}{x(x+2)}=\frac{1}{6} \Leftrightarrow x^{2}+2 x-48=0 \Leftrightarrow x=-8 ; x=6$. Therefore, $x=6$ liters. | 6 | 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 |
8. Solve the equation $\sqrt{15 x^{2}-52 x+45} \cdot(3-\sqrt{5 x-9}-\sqrt{3 x-5})=1$. | Solution. Rewrite our equation in the form
$$
\sqrt{3 x-5} \cdot \sqrt{5 x-9} \cdot(3-\sqrt{5 x-9}-\sqrt{3 x-5})=1
$$
Such a transformation is possible because the solution to the original equation exists only for $x>\frac{9}{5}$. Let $\sqrt{3 x-5}=a>0, \sqrt{5 x-9}=b>0$. We have
$$
a+b+\frac{1}{a b}=3
$$
Apply the Cauchy inequality to the left side
$$
a+b+\frac{1}{a b} \geq 3 \sqrt[3]{a \cdot b \cdot \frac{1}{a b}}=3
$$
Thus, for the solution of our equation, it is necessary that the Cauchy inequality holds as an equality and $a=b=\frac{1}{a b}$.
Answer: 2. | 2 | Algebra | math-word-problem | Yes | Yes | olympiads | false |
8. Solve the equation $\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}\left(21 x^{2}-76 x+64\right)\right)=1$. | Solution. Rewrite our equation in the form
$$
\log _{5}(3 x-4) \cdot \log _{5}(7 x-16) \cdot\left(3-\log _{5}(3 x-4)-\log _{5}(7 x-16)\right)=1
$$
Such a transformation is possible because the solution to the original equation exists only for $x>\frac{16}{7}$. Let $\log _{5}(3 x-4)=a>0, \log _{5}(7 x-16)=b>0$. Note that both logarithms must be positive. Indeed, the positivity of the first logarithm follows from the domain of existence. Suppose that $\log _{5}(7 x-16)<0 \Leftrightarrow \frac{16}{7}<x<\frac{17}{7}$. In this case, two factors in the left-hand side are positive, and one is negative, which is impossible. Thus, we have
$$
a+b+\frac{1}{a b}=3
$$
Apply the Cauchy inequality to the left-hand side
$$
a+b+\frac{1}{a b} \geq 3 \sqrt[3]{a \cdot b \cdot \frac{1}{a b}}=3
$$
Thus, for the solution of our equation, it is necessary that the Cauchy inequality holds as an equality and $a=b=\frac{1}{a b}$.
## Answer: 3. | 3 | Algebra | 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 |
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