Tasks and keys of the school stage of the All-Russian Olympiad for Schoolchildren in Mathematics
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school stage
4th grade
1. Rectangle area 91
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
5th grade
The maximum score for each task is 7 points
3. Cut the figure into three identical (coinciding when superimposed) figures:
4. Replace the letter A
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
6th grade
The maximum score for each task is 7 points
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
7th grade
The maximum score for each task is 7 points
1. - different numbers.
4. Replace the letters Y, E, A and R with numbers so that you get the correct equality:
YYYY ─ EEE ─ AA + R = 2017 .
5. There is something alive on the island th number of people, with her
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
8th grade
The maximum score for each task is 7 points
AVM, CLD and ADK respectively. Find∠ MKL .
6. Prove that if a, b, c and - whole numbers, then a fractionwill be an integer.
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
Grade 9
The maximum score for each task is 7 points
2. Numbers a and b are such that the equations and also has a solution.
6. At what natural x expression
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
Grade 10
The maximum score for each task is 7 points
4 – 5 – 7 – 11 – 19 = 22
3. In the equation
5. In triangle ABC held a bisector B.L. It turned out that . Prove that the triangle ABL - isosceles.
6. By definition,
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Tasks of the All-Russian Olympiad for Schoolchildren in Mathematics
school stage
Grade 11
The maximum score for each task is 7 points
1. The sum of two numbers is 1. Can their product be greater than 0.3?
2. Segments AM and BH ABC.
It is known that AH = 1 and . Find the length of a side BC.
3. a inequality true for all values X ?
Preview:
4th grade
1. Rectangle area 91. The length of one of its sides is 13 cm. What is the sum of all sides of the rectangle?
Answer. 40
Solution. The length of the unknown side of the rectangle is found from the area and the known side: 91:13 cm = 7 cm.
The sum of all sides of a rectangle is 13 + 7 + 13 + 7 = 40 cm.
2. Cut the figure into three identical (coinciding when superimposed) figures:
Solution.
3. Restore the addition example, where the digits of the terms are replaced by asterisks: *** + *** = 1997.
Answer. 999 + 998 = 1997.
4 . Four girls were eating candy. Anya ate more than Yulia, Ira - more than Sveta, but less than Yulia. Arrange the names of the girls in ascending order of the sweets eaten.
Answer. Sveta, Ira, Julia, Anya.
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Keys of the School Olympiad in Mathematics
5th grade
1. Without changing the order of the numbers 1 2 3 4 5, put signs of arithmetic operations and brackets between them so that the result is one. It is impossible to “glue” adjacent numbers into one number.
Solution. For example, ((1 + 2) : 3 + 4) : 5 = 1. Other solutions are possible.
2. Geese and piglets were walking in the barnyard. The boy counted the number of heads, there were 30, and then he counted the number of legs, there were 84. How many geese and how many pigs were in the school yard?
Answer. 12 piglets and 18 geese.
Solution.
1 step. Imagine that all the pigs raised two legs up.
2 step. There are 30 ∙ 2 = 60 legs left to stand on the ground.
3 step. Raised up 84 - 60 \u003d 24 legs.
4 step. Raised 24: 2 = 12 piglets.
5 step. 30 - 12 = 18 geese.
3. Cut the figure into three identical (coinciding when superimposed) figures:
Solution.
4. Replace the letter A to a non-zero digit to get the correct equality. It suffices to give one example.
Answer. A = 3.
Solution. It is easy to show that BUT = 3 is suitable, we prove that there are no other solutions. Reduce equality by BUT . We get .
If A ,
if A > 3, then .
5. Girls and boys went to the store on their way to school. Each student bought 5 thin notebooks. In addition, each girl bought 5 pens and 2 pencils, and each boy bought 3 pencils and 4 pens. How many notebooks were bought if the children bought 196 pieces of pens and pencils in total?
Answer. 140 notebooks.
Solution. Each student bought 7 pens and pencils. A total of 196 pens and pencils were purchased.
196: 7 = 28 students.
Each of the students bought 5 notebooks, which means that everything was bought
28
⋅ 5=140 notebooks.
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Keys of the School Olympiad in Mathematics
6th grade
1. There are 30 points on a straight line, the distance between any two adjacent points is 2 cm. What is the distance between the two extreme points?
Answer. 58 cm
Solution. 29 parts of 2 cm are placed between the extreme points.
2 cm * 29 = 58 cm.
2. Will the sum of the numbers 1 + 2 + 3 + ......+ 2005 + 2006 + 2007 be divisible by 2007? Justify the answer.
Answer. Will be.
Solution. We represent this sum in the form of the following terms:
(1 + 2006) + (2 + 2005) + …..+ (1003 + 1004) + 2007.
Since each term is divisible by 2007, the whole sum will be divisible by 2007.
3. Cut the figurine into 6 equal checkered figurines.
Solution. The figurine can only be cut
4. Nastya arranges the numbers 1, 3, 5, 7, 9 in the cells of a 3 by 3 square. She wants the sum of the numbers along all horizontals, verticals and diagonals to be divisible by 5. Give an example of such an arrangement, provided that Nastya is going to use each number no more than two times.
Solution. Below is one of the arrangements. There are other solutions as well.
5. Usually dad comes to pick up Pavlik after school by car. Once the lessons ended earlier than usual and Pavlik went home on foot. After 20 minutes, he met dad, got into the car and arrived home 10 minutes early. How many minutes early did class end that day?
Answer. 25 minutes early.
Solution. The car arrived home earlier, because it did not have to travel from the meeting point to the school and back, which means that the car travels twice this way in 10 minutes, and in one direction - in 5 minutes. So, the car met with Pavlik 5 minutes before the usual end of the lessons. By this time, Pavlik had already been walking for 20 minutes. Thus, the lessons ended 25 minutes early.
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Keys of the School Olympiad in Mathematics
7th grade
1. Find the solution to the numerical puzzle a,bb + bb,ab = 60 , where a and b - different numbers.
Answer. 4.55 + 55.45 = 60
2. After Natasha ate half of the peaches from the jar, the compote level dropped by one third. By what part (from the received level) will the compote level decrease if you eat half of the remaining peaches?
Answer. For one quarter.
Solution. It is clear from the condition that half of the peaches take up a third of the jar. So, after Natasha ate half of the peaches, the jar of peaches and compote remained equally (one third each). So half of the number of remaining peaches is a quarter of the total content
banks. If you eat this half of the remaining peaches, the compote level will drop by a quarter.
3. Cut the rectangle shown in the figure along the grid lines into five rectangles of different sizes.
Solution. For example, so
4. Replace the letters Y, E, A and R with numbers so that you get the correct equality: YYYY ─ EEE ─ AA + R = 2017.
Answer. With Y=2, E=1, A=9, R=5 we get 2222 ─ 111 ─ 99 + 5 = 2017.
5. There is something alive on the island th number of people, with yo m each of them is either a knight who always tells the truth, or a liar who always lies yo m. Once all the knights said: - "I am friends with only 1 liar", and all the liars: - "I am not friends with the knights." Who is more on the island, knights or knaves?
Answer. more knights
Solution. Every knave is friends with at least one knight. But since each knight is friends with exactly one knave, two knaves cannot have a common knight friend. Then each knave can be associated with his friend a knight, whence it turns out that there are at least as many knights as there are knaves. Since there are no inhabitants on the island yo number, then equality is impossible. So more knights.
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Keys of the School Olympiad in Mathematics
8th grade
1. There are 4 people in the family. If Masha's scholarship is doubled, the total income of the whole family will increase by 5%, if instead mom's salary is doubled - by 15%, if dad's salary is doubled - by 25%. By what percentage will the income of the whole family increase if grandfather's pension is doubled?
Answer. By 55%.
Solution . When Masha's scholarship is doubled, the total family income increases exactly by the amount of this scholarship, so it is 5% of income. Similarly, mom and dad's salaries are 15% and 25%. So, grandfather's pension is 100 - 5 - 15 - 25 = 55%, and if e yo doubled, the family income will increase by 55%.
2. On the sides AB, CD and AD of the square ABCD equilateral triangles are built outside AVM, CLD and ADK respectively. Find∠ MKL .
Answer. 90°.
Solution. Consider a triangle MAK : angle MAK equals 360° - 90° - 60° - 60° = 150°. MA=AK by condition, then a triangle MAC isosceles,∠AMK = ∠AKM = (180° - 150°) : 2 = 15°.
Similarly, we get that the angle DKL equals 15°. Then the required angle MKL is the sum of ∠MKA + ∠AKD + ∠DKL = 15° + 60° + 15° = 90°.
3. Nif-Nif, Naf-Naf and Nuf-Nuf shared three pieces of truffle with masses of 4 g, 7 g and 10 g. The wolf decided to help them. He can cut off and eat 1 g of truffle from any two pieces at the same time. Can the wolf leave the piglets equal pieces of truffle? If so, how?
Answer. Yes.
Solution. The wolf can first cut off 1 g three times from pieces of 4 g and 10 g. You will get one piece of 1 g and two pieces of 7 g. Now it remains to cut and eat 1 g six times from pieces of 7 g, then the piglets will get 1 g of truffle.
4. How many four-digit numbers are there that are divisible by 19 and end in 19?
Answer. 5 .
Solution. Let - such a number. Thenis also a multiple of 19. But
Since 100 and 19 are coprime, a two-digit number is divisible by 19. And there are only five of them: 19, 38, 57, 76 and 95.
It is easy to make sure that all numbers 1919, 3819, 5719, 7619 and 9519 suit us.
5. A team of Petit, Vasya and a single scooter is participating in the race. The distance is divided into sections of the same length, their number is 42, at the beginning of each there is a checkpoint. Petya runs the section in 9 minutes, Vasya - in 11 minutes, and on a scooter any of them passes the section in 3 minutes. They start at the same time, and at the finish line, the time of the one who came last is taken into account. The guys agreed that one of them rides the first part of the way on a scooter, the rest is running, and the other - vice versa (the scooter can be left at any checkpoint). How many sections does Petya have to ride on a scooter for the team to show the best time?
Answer. eighteen
Solution. If the time of one becomes less than the time of the other of the guys, then the time of the other will increase and, consequently, the time of the team. So, the time of the guys should coincide. Denoting the number of sections Petya passes through x and solving the equation, we get x = 18.
6. Prove that if a, b, c and - whole numbers, then a fractionwill be an integer.
Solution.
Consider , by the condition this number is an integer.
Then and will also be an integer as the difference N and double integer.
Preview:
Keys of the School Olympiad in Mathematics
Grade 9
1. Sasha and Yura are now together for 35 years. Sasha is now twice as old as Yura was when Sasha was as old as Yura is now. How old is Sasha now and how old is Yura?
Answer. Sasha is 20 years old, Yura is 15 years old.
Solution. Let Sasha now x years, then Yura and when Sasha wasyears, then Yura, according to the condition,. But the time for both Sasha and Yura has passed equally, so we get the equation
from which .
2. Numbers a and b are such that the equations and have solutions. Prove that the equationalso has a solution.
Solution. If the first equations have solutions, then their discriminants are nonnegative, whence and . Multiplying these inequalities, we get or , whence it follows that the discriminant of the last equation is also non-negative and the equation has a solution.
3. The fisherman caught a large number of fish weighing 3.5 kg. and 4.5 kg. His backpack can hold no more than 20 kg. What is the maximum weight of fish he can take with him? Justify the answer.
Answer. 19.5 kg.
Solution. The backpack can hold 0, 1, 2, 3 or 4 fish weighing 4.5 kg.
(no more because). For each of these options, the remaining capacity of the backpack is not divisible by 3.5 and at best it will be possible to pack kg. fish.
4. The shooter fired ten times at the standard target and hit 90 points.
How many hits were in the seven, eight and nine, if there were four ten, and there were no other hits and misses?
Answer. Seven - 1 hit, eight - 2 hits, nine - 3 hits.
Solution. Since the shooter hit only the seven, eight and nine in the remaining six shots, then for three shots (since the shooter hit the seven, eight and nine at least once) he will scorepoints. Then for the remaining 3 shots you need to score 26 points. What is possible with a single combination of 8 + 9 + 9 = 26. So, the shooter hit the seven 1 time, the eight - 2 times, the nine - 3 times.
5 . The midpoints of adjacent sides in a convex quadrilateral are connected by segments. Prove that the area of the resulting quadrilateral is half the area of the original.
Solution. Let's denote the quadrilateral by ABCD , and the midpoints of the sides AB , BC , CD , DA for P , Q , S , T respectively. Note that in the triangle ABC segment PQ is the median line, which means that it cuts off the triangle from it PBQ four times less area than area ABC. Likewise, . But triangles ABC and CDA add up to the whole quadrilateral ABCD means Similarly, we get thatThen the total area of these four triangles is half the area of the quadrilateral ABCD and the area of the remaining quadrilateral PQST is also half the area ABCD.
6. At what natural x expression is the square of a natural number?
Answer. For x = 5.
Solution. Let . Note that is also the square of some integer, less than t . We get that . Numbers and - natural and the first is greater than the second. Means, a . Solving this system, we get, , what gives .
Preview:
Keys of the School Olympiad in Mathematics
Grade 10
1. Arrange the signs of the module so that the correct equality is obtained
4 – 5 – 7 – 11 – 19 = 22
Solution. For example,
2. When Winnie the Pooh came to visit the Rabbit, he ate 3 plates of honey, 4 plates of condensed milk and 2 plates of jam, and after that he could not go outside because he was very fat from such food. But it is known that if he ate 2 plates of honey, 3 plates of condensed milk and 4 plates of jam or 4 plates of honey, 2 plates of condensed milk and 3 plates of jam, he could easily leave the hole of the hospitable Rabbit. What makes them fatter more: from jam or from condensed milk?
Answer. From condensed milk.
Solution. Let us denote through M - the nutritional value of honey, through C - the nutritional value of condensed milk, through B - the nutritional value of jam.
By condition 3M + 4C + 2B > 2M + 3C + 4B, whence M + C > 2B. (*)
By condition, 3M + 4C + 2B > 4M + 2C + 3B, whence 2C > M + B (**).
Adding inequality (**) with inequality (*), we obtain M + 3C > M + 3B, whence C > B.
3. In the equation one of the numbers is replaced by dots. Find this number if one of the roots is known to be 2.
Answer. 2.
Solution. Since 2 is the root of the equation, we have:
whence we get that, which means that the number 2 was written instead of the ellipsis.
4. Marya Ivanovna came out of the town into the village, and Katerina Mikhailovna simultaneously came out to meet her from the village into the town. Find the distance between the village and the city, if it is known that the distance between the pedestrians was 2 km twice: first, when Marya Ivanovna walked half the way to the village, and then, when Katerina Mikhailovna walked a third of the way to the city.
Answer. 6 km.
Solution. Let us denote the distance between the village and the city as S km, the speeds of Marya Ivanovna and Katerina Mikhailovna as x and y , and calculate the time spent by pedestrians in the first and second cases. We get in the first case
In the second. Hence, excluding x and y , we have
, whence S = 6 km.
5. In triangle ABC held a bisector B.L. It turned out that . Prove that the triangle ABL - isosceles.
Solution. By the property of the bisector, we have BC:AB = CL:AL. Multiplying this equation by, we get , whence BC:CL = AC:BC . The last equality implies similarity of triangles ABC and BLC by angle C and adjacent sides. From the equality of the corresponding angles in similar triangles, we obtain, from where to
triangle ABL vertex angles A and B are equal, i.e. he is equilateral: AL=BL.
6. By definition, . Which factor should be removed from the productso that the remaining product becomes the square of some natural number?
Answer. ten!
Solution. notice, that
x = 0.5 and is 0.25.2. Segments AM and BH are the median and height of the triangle, respectively ABC.
It is known that AH = 1 and . Find the length of a side BC.
Answer. 2 cm
Solution. Let's spend a segment MN, it will be the median of a right triangle BHC drawn to the hypotenuse BC and equal to half of it. Thenisosceles, therefore, so, therefore, AH = HM = MC = 1 and BC = 2MC = 2 cm.
3. At what values of the numerical parameter and inequality true for all values X ?
Answer . .
Solution . When we have , which is not true.
At 1 reduce the inequality by, keeping the sign:
This inequality is true for all x only for .
At reduce inequality by, changing the sign to the opposite:. But the square of a number is never negative.
4. There is one kilogram of 20% saline solution. The laboratory assistant placed the flask with this solution into an apparatus in which water is evaporated from the solution and at the same time a 30% solution of the same salt is poured into it at a constant rate of 300 g/h. The evaporation rate is also constant at 200 g/h. The process stops as soon as a 40% solution is in the flask. What will be the mass of the resulting solution?
Answer. 1.4 kilograms.
Solution. Let t be the time during which the apparatus worked. Then, at the end of the work in the flask, it turned out 1 + (0.3 - 0.2)t = 1 + 0.1t kg. solution. In this case, the mass of salt in this solution is 1 0.2 + 0.3 0.3 t = 0.2 + 0.09t. Since the resulting solution contains 40% salt, we get
0.2 + 0.09t = 0.4(1 + 0.1t), that is, 0.2 + 0.09t = 0.4 + 0.04t, hence t = 4 h. Therefore, the mass of the resulting solution is 1 + 0.1 4 = 1.4 kg.
5. In how many ways can 13 different numbers be chosen among all natural numbers from 1 to 25 so that the sum of any two chosen numbers does not equal 25 or 26?
Answer. The only one.
Solution. Let's write all our numbers in the following order: 25,1,24,2,23,3,…,14,12,13. It is clear that any two of them add up to 25 or 26 if and only if they are adjacent in this sequence. Thus, among the thirteen numbers we have chosen, there should not be neighboring ones, from which we immediately get that these must be all members of this sequence with odd numbers - the only choice.
6. Let k be a natural number. It is known that among 29 consecutive numbers 30k+1, 30k+2, ..., 30k+29 there are 7 primes. Prove that the first and last of them are simple.
Solution. Let's cross out the numbers that are multiples of 2, 3 or 5 from this row. There will be 8 numbers left: 30k+1, 30k+7, 30k+11, 30k+13, 30k+17, 30k+19, 30k+23, 30k+29. Let's assume that among them there is a composite number. Let us prove that this number is a multiple of 7. The first seven of these numbers give different remainders when divided by 7, since the numbers 1, 7, 11, 13, 17, 19, 23 give different remainders when divided by 7. Hence, one of of these numbers is a multiple of 7. Note that the number 30k+1 is not a multiple of 7, otherwise 30k+29 will also be a multiple of 7, and the composite number must be exactly one. Hence the numbers 30k+1 and 30k+29 are prime.
All-Russian Olympiads for schoolchildren are held under the auspices of the Russian Ministry of Education and Science after the official confirmation of the calendar of their dates. Such events cover almost all disciplines and subjects included in the compulsory curriculum of general education schools.
When participating in such competitions, students are given the opportunity to gain experience in answering questions of intellectual competitions, as well as to expand and demonstrate their knowledge. Students begin to respond calmly to various forms of knowledge testing, are responsible for representing and protecting the level of their school or region, which develops a sense of duty and discipline. In addition, a good result can bring a well-deserved cash bonus or benefits during admission to the country's leading universities.
The Olympiads for schoolchildren of the 2017-2018 academic year are held in 4 stages, subdivided according to the territorial aspect. These stages in all cities and regions are held within the general calendar dates established by the regional leadership of educational municipal departments.
Schoolchildren participating in competitions go through four levels of competition in stages:
- Level 1 (school). In September-October 2017, competitions will be held within each individual school. Independently of each other, all parallels of students are tested, starting from the 5th grade and ending with graduates. Tasks for this level are prepared by the methodological commissions of the city level, they also provide tasks for district and rural secondary schools.
- Tier 2 (regional). In December 2017 - January 2018, the next level will be held, in which the winners of the city and district - students of grades 7-11 will take part. Tests and assignments at this stage are developed by the organizers of the regional (third) stage, and all questions on preparation and locations for conducting are assigned to the local authorities.
- Tier 3 (regional). The time frame is from January to February 2018. Participants are the winners of the Olympiads of the current and completed year of study.
- Level 4 (All-Russian). Organized by the Ministry of Education and takes place from March to April 2018. Prize-winners of regional stages and winners of the last year participate in it. However, not all winners of the current year can take part in the All-Russian Olympiads. The exception is children who took 1st place in the region, but are significantly behind other winners in points.
Winners of the All-Russian level, if they wish, can take part in international competitions taking place during the summer holidays.
List of disciplines
In the 2017-2018 academic season, Russian schoolchildren can test their strength in the following areas:
- exact sciences - analytical and physical and mathematical direction;
- natural sciences - biology, ecology, geography, chemistry, etc.;
- philological sector - various foreign languages, native language and literature;
- humanitarian direction - economics, law, historical sciences, etc.;
- other items - art and, BZD.
This year, the Ministry of Education officially announced the holding of 97 Olympiads, which will be held in all regions of Russia from 2017 to 2018 (9 more than last year).
Benefits for winners and runners up
Each Olympiad has its own level: I, II or III. Level I is the most difficult, but it gives its diplomats and prize-winners the most advantages when entering many prestigious universities in the country.
Benefits for winners and prize-winners are of two categories:
- enrollment without exams in the selected university;
- awarding the highest USE score in the discipline in which the student received a prize.
The most famous level I state competitions include the following Olympiads:
- St. Petersburg Astronomical;
- "Lomonosov";
- St. Petersburg State Institute;
- "Young talents";
- Moscow school;
- "The highest standard";
- "Information Technology";
- "Culture and Art", etc.
Level II Olympiad 2017-2018:
- Herzenovskaya;
- Moscow;
- "Eurasian linguistic";
- "Teacher of the school of the future";
- Tournament named after Lomonosov;
- "TechnoCup", etc.
The 2017-2018 level III competitions include the following:
- "Star";
- "Young talents";
- Competition of scientific works "Junior";
- "Hope of Energy";
- "Step into the Future";
- "Ocean of Knowledge", etc.
According to the Order “On Amendments to the Procedure for Admission to Universities”, winners or prize-winners of the final stage have the right to enter any university without entrance examinations for the direction corresponding to the profile of the Olympiad. At the same time, the correlation between the direction of training and the profile of the Olympiad is determined by the university itself and publishes this information on its official website without fail.
The right to use the benefit is retained by the winner for 4 years, after which it is canceled and admission occurs on a general basis.
Preparation for the Olympics
The standard structure of the Olympiad tasks is divided into 2 types:
- verification of theoretical knowledge;
- the ability to translate theory into practice or demonstrate practical skills.
A decent level of preparation can be achieved with the help of the official website of the Russian state Olympiads, which contains the tasks of the past rounds. They can be used both to test your knowledge and to identify problem areas in training. There you can also check the dates of the tours and get acquainted with the official results on the website.
Video: assignments for the All-Russian Olympiad for schoolchildren appeared online
2019-2020 academic year
ORDER No. 336 of 06/05/2019 "On holding the school stage of the All-Russian Olympiad for schoolchildren in the 2019-2020 academic year".
Parental Consent(legal representatives) for the processing of personal data (form).
Analytical report template.
ATTENTION!!! Protocols on the results of the VSS 4-11 classes are accepted ONLY in the program Excel(archived documents in programs ZIP and RAR, except 7z).
Data for the 2019-2020 academic year
- Guidelines for the school stage of the 2018-2019 academic year in subjects you can download on the website.
- Presentation meetings on the All-Russian Olympiad for schoolchildren 2019-2020 academic year.
- Presentation "Peculiarities of organizing and conducting the school stage of the Higher School of Education for students with disabilities"
- Presentation "Regional Center for Gifted Children".
- Diploma winner / prize-winner of the school stage of the Higher School of Education.
- Regulations fulfillment of the Olympiad tasks of the school stage of the All-Russian Olympiad for schoolchildren.
- Schedule holding the school stage of the All-Russian Olympiad for schoolchildren in the 2018-2019 academic year.
Clarifications on the procedure for holding the All-Russian Olympiad for schoolchildren - the school stage for grade 4
According to the order of the Ministry of Education and Science of the Russian Federation dated December 17, 2015 No. 1488, the All-Russian Olympiad for schoolchildren has been held since September 2016 for 4th grade students only in Russian and mathematics. According to the schedule 09/21/2018 - in Russian; 09/26/2018 - in mathematics. A detailed schedule for the school stage of the Higher School of Education for all parallels of students is posted in the plan of the MBU "Center for Educational Innovations" for September 2018.
Time to complete the work in the Russian language 60 minutes, in mathematics - 9 0 minutes.
To the attention of those responsible for holding the Olympiads
in educational institutions!
Tasks for the school stage of the All-Russian Olympiad for schoolchildren 2018-2019 ac. year. for grades 4-11 will be sent to educational organizations by e-mail, starting from September 10, 2018. Please send all changes and clarifications related to e-mail addresses to e-mail: [email protected], no later than 09/06/2018
Olympiad assignments (at 08.00) and solutions (at 15.00) will be sent to the school's email address. And also the answers will be duplicated the next day on the website www.site
If you have not received the tasks of the school stage, please kindly look at them in the "spam" folder from the mail [email protected]
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Evaluation criteria for an ESSAY on a creative project.
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Answers of the school stage according to the MHC 11 cells.
Answers of the school stage in social studies for grade 8.
Answers of the school stage in social studies for grade 9.
Answers of the school stage in social studies for 10 cells.
Answers of the school stage in social studies for grade 11.
Answers of the school stage on ecology for 7-8 cells.
Answers of the school stage in ecology for grade 9.
Answers of the school stage on ecology for 10-11 cells.
Answers of the school stage in physics.
Answers of the school stage in the history of 7th grade.
Answers of the school stage in the history of 8th grade.
Answers of the school stage in the history of grade 9.
Answers of the school stage in the history of 10-11 cells.
Answers of the school stage in physical culture (grades 7-8).
Answers of the school stage in physical culture (grades 9-11).
Answers of the school stage in German 7-8 cells.
It has become a good tradition to hold the All-Russian School Olympiad. Its main task is to identify gifted children, motivate schoolchildren to study subjects in depth, develop creative abilities and non-standard thinking in children.
The Olympic movement is gaining more and more popularity among schoolchildren. And there are reasons for this:
- winners of the All-Russian round are accepted to universities without competition, if the profile subject is an olympiad subject (diplomas of the winners are valid for 4 years);
- participants and prize-winners receive additional chances for admission to educational institutions (if the subject is not in the profile of the university, the winner receives an additional 100 points upon admission);
- significant monetary reward for prizes (60 thousand, 30 thousand rubles;
- and, of course, fame throughout the country.
Before becoming a winner, you must go through all the stages of the All-Russian Olympiad:
- The initial school stage, at which worthy representatives are determined for the next level, is held in September-October 2017. The organization and conduct of the school stage is carried out by specialists of the methodological office.
- The municipal stage is held between the schools of the city or district. It takes place at the end of December 2017. – early January 2018
- The third round is more difficult. Talented students from all over the region take part in it. The regional stage takes place in January-February 2018.
- The final stage determines the winners of the All-Russian Olympiad. In March-April, the best children of the country compete: the winners of the regional stage and the winners of last year's Olympiad.
The organizers of the final round are representatives of the Ministry of Education and Science of Russia, they also sum up the results.
You can show your knowledge in any subject: mathematics, physics, geography, even physical education and technology. You can compete in erudition in several subjects at once. There are 24 disciplines in total.
Olympiad subjects are divided into areas:
| № | Direction | Items |
| 1 | Exact disciplines | mathematics, computer science |
| 2 | Natural sciences | geography, biology, physics, chemistry, ecology, astronomy |
| 3 | Philological disciplines | literature, Russian language, foreign languages |
| 4 | Humanities | economics, social studies, history, law |
| 5 | Other | art, technology, physical culture, basics of life safety |
The peculiarity of the final stage of the Olympiad consists in two types of tasks: theoretical and practical. For example, in order to get good results in geography, students must complete 6 theoretical tasks, 8 practical tasks, and also answer 30 test questions.
The first stage of the Olympiad begins in September, which means that those wishing to take part in the intellectual marathon should prepare in advance. But first of all, they must have a good base at the school level, which must be constantly replenished with additional knowledge that goes beyond the school curriculum.
The official website of the Olympiad www.rosolymp.ru places tasks from previous years. These materials can be used in preparation for an intellectual marathon. And of course, you can’t do without the help of teachers: additional classes after school, classes with tutors.
The winners of the final stage will take part in international olympiads. They form the national team of Russia, which will be trained at training camps in 8 subjects.
To provide methodological assistance on the site, orientation webinars are held, the Central Organizing Committee of the Olympiad, subject-methodical commissions have been formed.
