Olympiad tasks for the academic year. What subjects are included in the Olympiad list? In which subjects can a student take part in Olympiads?

It is a whole system of Olympiads in subjects included in compulsory program educational institutions of the country. Participation in such an Olympiad is an honorable and responsible mission, because this is a student’s chance to show his accumulated knowledge, defend the honor of his educational institution, and in case of victory, also the opportunity to receive financial incentives and earn a privilege when entering the best universities in Russia.

The practice of holding subject Olympiads has existed in the country for more than a hundred years - back in 1886, representatives of educational authorities initiated competitions between young talents. During times Soviet Union this movement not only did not cease to exist, but also received an additional impetus for development. Since the 60s of the last century, intellectual competitions on an all-Union and then all-Russian scale began to be held in almost all major school disciplines.

What subjects are included in the Olympiad list?

In the 2017-2018 academic year, the country's schoolchildren will be able to compete for prizes in several categories of disciplines:

in the exact sciences, which include computer science and mathematics;
in the natural sciences, which include geography, biology, astronomy, physics, chemistry and ecology;
in the field of philology, including Olympiads in German, English, Chinese, French, Italian, as well as Russian language and literature;
in the field of humanities, consisting of history, social studies, law and economics;
in other disciplines, which include physical education, world art culture, technology and life safety.

In the Olympiad tasks for each of the listed disciplines, there are usually two blocks of tasks: a part that tests theoretical preparation, and a part aimed at identifying practical skills.

Main stages of the 2017-2018 Olympiad

Conducting the All-Russian school Olympiad includes the organization of four stages of competitions held at various levels. Final schedule intellectual battles between schoolchildren are determined by representatives of schools and regional educational authorities, however, you can focus on such periods of time.

Schoolchildren will have 4 stages of competitions of varying difficulty levels

Stage 1. School. Competitions between representatives of the same school will be held in September-October 2017. The Olympiad is held between parallel students, starting in the fifth grade. In this case, the development of tasks for conducting subject Olympiads is entrusted to members of the city-level methodological commission.
Stage 2. Municipal. The stage, where competitions take place between the winners of schools in the same city, representing grades 7-11, will be held from December 2017 to January 2018. The mission of compiling Olympiad tasks is entrusted to the organizers at the regional level, and local officials are responsible for issues related to the provision of a place and ensuring the procedure for the Olympiads.
Stage 3. Regional. The third level of the Olympiad, which will be held in January-February 2018. At this stage, schoolchildren who received prizes at the city Olympiad and those who won the regional selections last year take part in the competition.
Stage 4. All-Russian. The highest level of subject Olympiads will be organized by representatives of the Ministry of Education Russian Federation in March-April 2018. Regional winners and the guys who won last year are invited to attend. However, not every winner of the regional selection can become a participant this stage. The exception is schoolchildren who received 1st place in their region, but are behind in points from the winners at the level of other cities. Prize-winners All-Russian stage can then go to international competitions that take place in the summer.

Where can I find standard tasks for the Olympiad?

Of course, to perform well in this event, you need to have a high level of preparation. The All-Russian Olympiad is represented on the Internet by its own website - rosolymp.ru - on which students can familiarize themselves with tasks from previous years, check their level with the help of answers to them, find out specific dates and requirements for organizational issues.

Tasks and keys for the school stage of the All-Russian Olympiad for schoolchildren in mathematics

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School stage

4th grade

1. Area of rectangle 91

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Objectives 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 (matching when overlapping) figures:

4. Replace letter A

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Objectives 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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Objectives of the All-Russian Olympiad for Schoolchildren in Mathematics

School stage

7th grade

The maximum score for each task is 7 points

1. - various numbers.

4. Replace the letters Y, E, A and R with numbers so that you get the correct equation:

YYYY ─ EEE ─ AA + R = 2017 .

5. Something lives on the island number of people, including her

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Objectives 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 fractionswill be an integer.

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Objectives of the All-Russian Olympiad for Schoolchildren in Mathematics

School stage

9th grade

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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Objectives 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 Eq.

5. In triangle ABC drew a bisector BL. It turned out that . Prove that the triangle ABL – isosceles.

6. By definition,

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Objectives 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 side length B.C.

3. and inequality true for all values X ?

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4th grade

1. Area of rectangle 91. The length of one of its sides is 13 cm. What is the sum of all sides of the rectangle?

Answer. 40

Solution. We find the length of the unknown side of the rectangle from the area and the known side: 91:13 cm = 7 cm.

The sum of all sides of the rectangle is 13 + 7 + 13 + 7 = 40 cm.

2. Cut the figure into three identical (matching when overlapping) figures:

Solution.

3. Recreate the example for addition, 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 girls' names in ascending order of the candies eaten.

Answer. Sveta, Ira, Julia, Anya.

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Keys for the school mathematics Olympiad

5th grade

1. Without changing the order of the numbers 1 2 3 4 5, place arithmetic signs and parentheses between them so that the result is one. You cannot “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 piglets were there in the schoolyard?

Answer. 12 piglets and 18 geese.

Solution.

1 step. Imagine that all the piglets raised two legs up.

Step 2. There are 30 ∙ 2 = 60 legs left standing on the ground.

Step 3. Raised up 84 - 60 = 24 legs.

Step 4 Raised 24: 2 = 12 piglets.

Step 5 30 - 12 = 18 geese.

3. Cut the figure into three identical (matching when overlapping) figures:

Solution.

4. Replace letter A by a non-zero number to obtain a true equality. It is enough to give one example.

Answer. A = 3.

Solution. It is easy to show that A = 3 is suitable, let us prove that there are no other solutions. Let's reduce the equality by A . We'll get it.
If A ,
if A > 3, then .

5. Girls and boys went into a 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 purchased if the children bought 196 pens and pencils in total?

Answer. 140 notebooks.

Solution. Each of the students bought 7 pens and pencils. A total of 196 pens and pencils were purchased.

196: 7 = 28 students.

Each student bought 5 notebooks, which means they bought a total
28 ⋅ 5=140 notebooks.

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Keys for the school mathematics Olympiad

6th grade

1. There are 30 points on a straight line, the distance between any two adjacent ones is 2 cm. What is the distance between the two extreme points?

Answer. 58 cm.

Solution. Between the extreme points there are 29 pieces of 2 cm each.

2 cm * 29 = 58 cm.

2. Will the sum of the numbers 1 + 2 + 3 + ......+ 2005 + 2006 + 2007 be divisible by 2007? Justify your answer.

Answer. Will.

Solution. Let's imagine this amount in the form of the following terms:
(1 + 2006) + (2 + 2005) + …..+ (1003 + 1004) + 2007.

Since each term is divisible by 2007, the entire sum will be divisible by 2007.

3. Cut the figure into 6 equal checkered figures.

Solution. This is the only way to cut a figurine

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.

5. Usually dad comes to pick Pavlik up after school by car. One day, classes ended earlier than usual and Pavlik walked home. 20 minutes later he met his dad, got into the car and arrived home 10 minutes early. How many minutes earlier did classes end that day?

Answer. 25 minutes earlier.

Solution. The car arrived home earlier because it didn’t have to drive from the meeting place to school and back, which means the car covers twice this distance in 10 minutes, and one way in 5 minutes. So, the car met Pavlik 5 minutes before the usual end of classes. By this time, Pavlik had already been walking for 20 minutes. Thus, classes ended 25 minutes early.

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Keys for the school mathematics Olympiad

7th grade

1. Find the solution to a number puzzle a,bb + bb,ab = 60, where a and b - various numbers.

Answer. 4.55 + 55.45 = 60

2. After Natasha ate half of the peaches from the jar, the level of the compote dropped by one third. By what part (from the obtained level) will the compote level decrease if you eat half of the remaining peaches?

Answer. One quarter.

Solution. From the condition it is clear that half of the peaches occupy a third of the jar. This means that after Natasha ate half of the peaches, there were equal amounts of peaches and compote left in the jar (one third each). This means that half of the number of remaining peaches is a quarter of the total volume of contents

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 varying sizes.

Solution. For example, like this

4. Replace the letters Y, E, A and R with numbers so that you get the correct equation: YYYY ─ EEE ─ AA + R = 2017.

Answer. With Y=2, E=1, A=9, R=5 we get 2222 ─ 111 ─ 99 + 5 = 2017.

5. Something lives on the island number of people, including e m each of them is either a knight who always tells the truth, or a liar who always lies e t. Once all the knights said: “I am friends with only 1 liar,” and all the liars: “I am not friends with knights.” Who is more on the island, knights or knaves?

Answer. There are more knights

Solution. Every liar is friends with at least one knight. But since each knight is friends with exactly one liar, two liars cannot have a common knight friend. Then each liar can be matched with his knight friend, which means that there are at least as many knights as there are liars. Since the total number of inhabitants on the island e number, then equality is impossible. This means there are more knights.

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Keys for the school mathematics Olympiad

8th grade

1. There are 4 people in the family. If Masha's scholarship is doubled, the total income of the entire 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 entire family increase if grandfather’s pension is doubled?

Answer. By 55%.

Solution . When Masha's scholarship doubles, the total family income increases exactly by the amount of this scholarship, so it is 5% of income. Likewise, mom and dad's salaries are 15% and 25%. This means that grandfather’s pension is 100 – 5 – 15 - 25 = 55%, and if e double, then family income will increase by 55%.

2. On sides AB, CD and AD of square ABCD equilateral triangles are constructed on the 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 according to the condition, it means a triangle MAK isosceles,∠ AMK = ∠ AKM = (180° - 150°) : 2 = 15°.

Similarly we find that the angle DKL equal to 15°. Then the required angle MKL is equal to the sum of ∠ MKA + ∠ AKD + ∠ DKL = 15° + 60° + 15° = 90°.

3. Nif-Nif, Naf-Naf and Nuf-Nuf were sharing three pieces of truffle weighing 4 g, 7 g and 10 g. The wolf decided to help them. He can cut off any two pieces at the same time and eat 1 g of truffle each. Will the wolf be able to leave equal pieces of truffle for the piglets? If so, how?

Answer. Yes.

Solution. The wolf can first cut 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 six times and eat 1 g each from pieces of 7 g, then the piglets you 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 relatively prime, 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 verify that all the numbers 1919, 3819, 5719, 7619 and 9519 are suitable for us.

5. A team of Petya, Vasya and a single-seater scooter is participating in the race. The distance is divided into sections of equal 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, either of them covers 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 would ride the first part of the journey on a scooter, then run the rest, and the other would do the opposite (the scooter can be left at any checkpoint). How many sections must Petya cover on a scooter for the team to show the best time?

Answer. 18

Solution. If the time of one becomes less than the time of another of the guys, then the time of the other and, consequently, the time of the team will increase. This means that the guys’ time must coincide. Having indicated 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 fractionswill be an integer.

Solution.

Let's consider , by convention this is an integer.

Then will also be an integer as the difference N and double the integer.

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Keys for the school mathematics Olympiad

9th grade

1. Sasha and Yura have now been together for 35 years. Sasha is now twice as old as Yura was then, 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 passed equally for both Sasha and Yura, 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 non-negative, whence And . Multiplying these inequalities, we get or , from which 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 holds no more than 20 kg. Which Weight Limit can he take fish with him? Justify your 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 backpack capacity is not divisible by 3.5, and in the best case it will be possible to pack kg. fish.

4. The shooter fired ten times at a standard target and scored 90 points.

How many hits were there on the seven, eight and nine, if there were four tens, and there were no other hits or misses?

Answer. Seven – 1 hit, eight – 2 hits, nine – 3 hits.

Solution. Since the shooter hit only seven, eight and nine in the remaining six shots, then in three shots (since the shooter hit seven, eight and nine at least once each) he will scorepoints Then for the remaining 3 shots you need to score 26 points. What is possible with the only combination 8 + 9 + 9 = 26. So, the shooter hit the seven once, the eight - 2 times, and 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 one.

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 midline, which means it cuts off the triangle from it PBQ four times less area than area ABC. Likewise, . But triangles ABC and CDA in total they make up the entire 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 equal to half the area ABCD.

6. At what natural x expression is the square of a natural number?

Answer. At x = 5.

Solution. Let . Note that – 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 .

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Keys for the school mathematics Olympiad

Grade 10

1. Arrange the modulus signs so that you get the correct equality

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 had become 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 you fatter: jam or condensed milk?

Answer. From condensed milk.

Solution. Let us denote by M the nutritional value of honey, by C the nutritional value of condensed milk, and by B the nutritional value of jam.

By condition, 3M + 4C + 2B > 2M + 3C + 4B, whence M + C > 2B. (*)

According to the 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 Eq. one of the numbers is replaced with dots. Find this number if it is known that one of the roots is 2.

Answer. 2.

Solution. Since 2 is the root of the equation, we have:

where do we get that, which means the number 2 was written instead of an ellipsis.

4. Marya Ivanovna came out from the city into the village, and Katerina Mikhailovna came out to meet her from the village into the city at the same time. Find the distance between the village and the city if it is known that the distance between 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. In the first case we get

In the second. Hence, excluding x and y, we have
, from where S = 6 km.

5. In triangle ABC drew a bisector BL. It turned out that . Prove that the triangle ABL – isosceles.

Solution. By the bisector property we have BC:AB = CL:AL. Multiplying this equality by, we get , from where BC:CL = AC:BC . The last equality implies the similarity of triangles ABC and BLC at 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. it is isosceles: AL = BL.

6. By definition, . Which factor should be deleted from the product?, so that the remaining product becomes the square of some natural number?

Answer. 10!

Solution. notice, that

x = 0.5 and is 0.25.

2. Segments AM and BH - the median and altitude of the triangle, respectively ABC.

It is known that AH = 1 and . Find the side length B.C.

Answer. 2 cm.

Solution. Let's draw a segment MN, it will be the median of the right triangle B.H.C. , drawn to the hypotenuse B.C. and is equal to half of it. Then– isosceles, therefore, 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 incorrect.

At 1 reduce the inequality by, keeping the sign:

This inequality is true for everyone x only at .

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 in an apparatus in which water is evaporated from the solution and at the same time a 30% solution of the same salt is added to it at a constant rate of 300 g/hour. The evaporation rate is also constant and amounts to 200 g/h. The process stops as soon as there is a 40% solution 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 device worked. Then, at the end of the work, the result in the flask was 1 + (0.3 – 0.2)t = 1 + 0.1t kg. solution. In this case, the mass of salt in this solution is equal to 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 hours. Therefore, the mass of the resulting solution is 1 + 0.1 · 4 = 1.4 kg.

5. In how many ways can you choose 13 different numbers from 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 are equal in sum to 25 or 26 if and only if they are adjacent in this sequence. Thus, among the thirteen numbers we have chosen there should be no neighboring ones, from which we immediately obtain that these must be all members of this sequence with odd numbers - there is only one choice.

6. Let k be a natural number. It is known that among the 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 numbers that are multiples of 2, 3 or 5 from this series. There will be 8 numbers left: 30k+1, 30k+7, 30k+11, 30k+13, 30k+17, 30k+19, 30k+23, 30k+29. Let us 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. This means that one of these numbers are 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. This means that the numbers 30k+1 and 30k+29 are prime numbers.

In a few months a new one begins academic year, which means subject Olympiads are just around the corner. The school stage of the All-Russian Subject Olympiad will be held in September-October 2017. Any student who wishes (not counting students) primary school) can take part in it. What else do you need to know about this large-scale event?

Useful facts about the All-Russian Subject Olympiad for Schoolchildren 2017-2018

A student who takes part in the VOS receives advantages upon admission;
The list of subjects for the title of best has not been changed and consists of both humanitarian and technical subjects;
You can take part in all subjects at once, participation in them does not affect each other;
The structure of the competition and materials for preparing for the Olympiad will remain unchanged.

The All-Russian Olympiad is held in all subjects of the compulsory school curriculum.

By taking part, the student has the opportunity to test his knowledge and receive an incentive in the form of a cash prize, which will help in purchasing the necessary items for in-depth study; protect the honor of your home school or city, even gain some benefits when entering outstanding higher institutions countries.

That is why each participant tries to approach preparation and participation in the competition as responsibly and seriously as possible.

Competitions between the smartest talents have been held for quite a long period of time, more than a century: the first Olympiads date back to the year one thousand eight hundred and eighty-six.

Video review of the All-Russian School Olympiad

In which subjects can a student take part in Olympiads?

All items can be divided into several types:

mathematical sciences: computer science and branches of mathematics;
natural sciences such as physics, chemistry, geography, astronomy, biology;
sciences that study literature (Russian Olympiads, as well as foreign languages, literature);
humanities: history, social studies, economics, law;
remaining items: Physical Culture, life safety, technology, art.

The organizers of the Olympiads test both the student’s theoretical knowledge and the ability to apply this knowledge in practice.

Stages of the All-Russian Olympiad 2017-2018

Determining the smartest schoolchildren goes through several stages, there are four in total:

Olympiad between school students. Middle and high school students can participate. This stage falls on September-October 2017. Responsibility for the organization rests with the members of the city Education Committee, in accordance with the taught program in school textbooks.
Competition for the winners of the city’s intra-school Olympiads at the municipal level. The honor of representing the school falls to students in grades seven through eleven. Municipal Olympiads are held from December to January 2017-2018. Organizers at the regional level are responsible for compiling assignments.
Continuation of the competition at the regional level between the winners of the previous stage of the Olympiad and the winners of last year. High school students (grades nine to eleven) in January - February 2018, winners of the municipal stage, receive a ticket to the regional stage.
The final stage. It is held among the winners of the regional stage among all of Russia who have collected a sufficient number of points for regional stage, and last year's winners. Period: March-April 2018. The event is managed by representatives of the Ministry of Education of the Russian Federation.