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2. When insuring property, the insurance amount cannot exceed its actual value (insurance value) at the time of concluding the insurance contract.
Insurance tariff - the rate of the insurance premium or the insurance premium (insurance premium) expressed in rubles, payable per unit of the insurance amount, which is usually 100 rubles. In simpler terms, it is the percentage of the insurance value of the property established by the insurance company within the framework of laws concerning insurance activities.
An underwriter of an insurance company is a person authorized by the insurance company to analyze, accept for insurance (reinsurance), and reject various types of risks, as well as classify the selected risks to obtain the optimal insurance premium for them.
Title insurance (title insurance) is protection against the risk of losing the right of ownership of real estate as a result of the transaction for the acquisition of real estate being declared invalid or illegal. A transaction can be declared invalid by a court in the following situations:
- the presence of unaccounted property interests of minors, heirs, persons serving time in places of deprivation of liberty;
- the discovery of errors or fraud in the current or previous transactions with the real estate;
- in the case of the incapacity of one of the parties to the transaction, etc.
The appraised value of the property is the final result of the appraisal study, calculated by a licensed appraiser.
The cadastral value of housing is the value of the real estate object established through state valuation, primarily for the calculation of taxes.
## Instruction:
In the event of a discrepancy between the insurance amount and the actual value of the property, the following consequences are established:
a) if the insurance amount is set below the actual value of the property in the Insurance Contract, the Insurer, in the event of an insurance case, compensates the Insured (Beneficiary) for the damage incurred within the limits of the insurance amount established by the Insurance Contract without applying the ratio of the insurance amount and the actual value of the property (first risk payment condition).
b) if the insurance amount specified in the Insurance Contract exceeds the actual value of the property, the Contract is void to the extent that the insurance amount exceeds the actual value of the property.
Gross insurance tariffs (title insurance)
| Group of insured objects | Number of previous transactions with the real estate object | Annual tariff |
| :--- | :--- | :---: |
| Apartments and parts of apartments, consisting of one or several isolated rooms, non-residential buildings | Primary market (one transaction) | 0.18 |
| Secondary market (no more than 3 transactions) | One transaction preceding the current transaction | 0.20 |
| No more than 3 transactions*** | | |
**** in the case of more than 3 transactions over the past 3 years preceding the current transaction, the tariff is agreed individually.
## List of corrective coefficients (title insurance)
| Factors | Coefficient |
| :--- | :---: |
| The presence in the history of the real estate object of one or more of the following facts: - a time gap of less than one year between two preceding (or between the current and the preceding) transfers of ownership of the real estate object; - the presence of lease transactions; - transfer of rights as a result of inheritance or gift; - transactions carried out by persons on the basis of a power of attorney; - more than 2 transfers of ownership of the real estate object. | |
| - the last transaction preceding the current one was more than 3 years ago | 1.2 |
| Absence of one of the necessary documents | 0.8 |
## MOSCOW OLYMPIAD OF SCHOOL STUDENTS IN FINANCIAL LITERACY FINAL STAGE $10-11$ GRADES 2nd variant Answers and solutions
## Problem 1
You are an underwriter of an insurance company. According to the instructions for title insurance (Appendix to this problem), based on the data provided below, you need to set the tariff and determine the amount of the insurance premium. Be sure to keep the glossary in front of you.
Data:
Insured Ostrozhnov Konstantin Petrovich is purchasing an apartment on the secondary market with the participation of credit funds. The loan amount secured by real estate is 20 million rubles. Konstantin Petrovich has a good credit history and easily obtained approval for such an amount. The appraised value of the apartment is 14,500,000 rubles, the cadastral value is 15,000,000 rubles. The bank-lender requires Ostrozhnov to purchase a title insurance policy to protect its financial interests.
Seller - Ivanov G.L., born in 1952, the sole owner of the apartment, purchased it more than 5 years ago under a construction investment agreement. Married, has adult children, no one is registered in the apartment. At the time of purchasing the apartment, he was married. Provided all necessary documents except for certificates from the psychiatric and narcological dispensaries. Does not drive, does not have a driver's license, and is not subject to military service.
|
# Solution:
In accordance with the instruction, the base rate is $0.2\%$ of the insurance amount, apply a reducing factor for the absence of a change in ownership over the past 3 years $(0.8)$ and an increasing factor for the absence of certificates from the PND and ND $(1.3)$.
In total: $0.2 * 0.8 * 1.3=0.208\%$
The loan amount exceeds both the appraised and the cadastral value. Therefore, we choose the maximum value to minimize the difference between the loan amount (bank requirement) and the allowable amount by law - 15000000 and take it as the insurance amount. $15000000 * 0.00208=31200$
Answer: rate $0.208\%$, SP=31,200 rubles.
Maximum 20 points
Evaluation criteria:
2 points for each correctly found coefficient and 4 points in total for the correct explanation of the use of these coefficients (if there is a correct explanation of only one coefficient, only 1 point out of 4 is given, and if there is a correct explanation of only two coefficients, only 3 points out of 4 are given). In total, 10 points for this part of the correct solution.
2 points for the correct rate.
A total of 8 points for the correct insurance premium amount.
- Participants receive only 2 points if the insurance premium is calculated without arithmetic errors based on an incorrect appraised value of the apartment.
- Participants receive only 4 points if the insurance premium is correctly determined but lacks a complete correct justification for its application.
## Task 2
The following information is available for JSC «Total Trade».
The annual revenue of JSC «Total Trade» is 2,500,000 rubles (after VAT payment). It is known that the ratio of operating expenses to revenue is 3/5 annually. The authorized capital of JSC «Total Trade» consists of the nominal value of the company's shares acquired by shareholders and is eight times the minimum authorized capital of a public company established under Russian law. The authorized capital is made up of 1600 ordinary shares, $35\%$ of which belong to the general director.
|
31200
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Maria Ivanovna decided to use the services of an online clothing store and purchase summer clothing: trousers, a skirt, a jacket, and a blouse. Being a regular customer of this store, Maria Ivanovna received information about two ongoing promotions. The first promotion allows the customer to use an electronic coupon worth 1000 rubles, which can cover no more than $15 \%$ of the cost of the ordered items. The second promotion allows the customer to get every third item, which has the lowest price, for free when making a purchase. The online store offers the option to pick up the items from any pickup point located in the city where Maria Ivanovna lives, or to have them delivered to any address within the city. Delivery from a pickup point costs the customer 100 rubles, while delivery to a specified address costs 350 rubles if the order amount does not exceed 5000 rubles. If the order amount exceeds 5000 rubles, delivery to the specified address is free. Maria Ivanovna has selected the following set of clothing:
| Trousers | 2800 rubles |
| :--- | :--- |
| Skirt | 1300 rubles |
| Jacket | 2600 rubles |
| Blouse | 900 rubles |
a) What is the smallest amount of money Maria Ivanovna will have to pay for her purchases, including delivery, if the promotions of the online store cannot be combined (used simultaneously)?
b) Explain why Maria Ivanovna's actual expenses for this purchase in this online store may be higher than the monetary amount found in the previous question. (20 points)
|
Solution:
(a) Maria Ivanovna can make one purchase, using only one of the promotions, or she can "split" the selected items into two purchases, using both promotions in this case.
Let's consider all possible options:
1) One purchase. In this case, Maria Ivanovna can save either 900 rubles by using the "third item free" promotion, or $((2800+1300+2600+900) * 0.85=1140>1000) \quad 1000$ rubles by using the coupon. Delivery will be free in both of these cases. Thus, if Maria Ivanovna buys all the items in one purchase, the maximum savings will be 1000 rubles, and the cost of the purchase will be $2800+1300+2600+900-1000=6600$ rubles.
2) Two purchases. The most expensive of the cheapest third items in the purchase can be either the skirt or the blouse. If the free item is the skirt, then the coupon should be used for the blouse and it should be ordered as a separate purchase. If the free item is the blouse, then the coupon is most profitable to use for the most expensive item - the trousers.
Let's consider the cost of each such purchase.
1st option:
Cost of purchase 1 (trousers, jacket, skirt) = 2800 + 2600 = 5400 rubles. Delivery of this purchase will be free.
Cost of purchase 2 (blouse) $= 900 * 0.85 = 765$. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Total cost of the option: $5400 + 765 + 100 = 6265$
2nd option:
Cost of purchase 1 (jacket, skirt, blouse) = 2600 + 1300 = 3900 rubles. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Cost of purchase 2 (trousers) $= 2800 * 0.85 = 2380$. Delivery of this purchase will cost a minimum of 100 rubles for self-pickup.
Total cost of the option: $3900 + 2380 + 200 = 6480$
Thus, Maria Ivanovna will spend the least amount if she orders two purchases from the online store, one of which will include trousers, jacket, and skirt, and the other - blouse. The cost of such a purchase, including delivery, will be 6265 rubles.
(b) Attention should be paid to at least two main points that can increase the actual cost of the purchase for Maria Ivanovna.
1) The conditions of self-pickup imply that Maria Ivanovna (or the person she entrusts to pick up her order) will spend time and, possibly, money to get to the pickup point. Despite the low cost of this delivery method, the pickup point may be far from the customer. Considering this fact, many people choose delivery to the address specified by the customer, and some may even prefer (in this case) not to split the purchase into two, but to buy all the selected items in one purchase.
2) When placing two purchases instead of one, the customer will also spend more time, the cost of which for him may exceed the benefit of "double" order placement.
Grading criteria (total 20 points):
(a) (total 15 points)
The idea of dividing orders into 1 or 2 purchases is present - 2 points.
Explicit monetary expenses for 1 purchase are correctly calculated - 3 points.
Explicit expenses for 2 purchases are correctly calculated - 4 points for each option.
Comparison is made and the correct answer about the lowest cost of orders is given - 2 points.
A penalty for each arithmetic error is 1 point. However, in the case of a conceptually correct approach to the solution, the penalty for arithmetic errors, if their number exceeds 1 error in each critical section, cannot exceed 50% of the maximum score assigned to it.
(b) (total 5 points)
The idea that implicit (opportunity) costs of purchasing items in an online store are not taken into account is present - 2 points.
At least 2 examples of implicit costs are provided - 3 points.
|
6265
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11. (16 points)
The Dorokhov family plans to purchase a vacation package to Crimea. The family plans to travel with the mother, father, and their eldest daughter Polina, who is 5 years old. They carefully studied all the offers and chose the "Bristol" hotel. The head of the family approached two travel agencies, "Globus" and "Around the World," to calculate the cost of the tour from July 10 to July 18, 2021.
The first agency offered the tour on the following terms: 11,200 rubles per person under 5 years old and 25,400 rubles per person over 5 years old. In addition, the "Globus" agency provides the Dorokhov family with a discount of 2% of the total cost of the tour as a loyal customer.
The terms of purchasing the tour from the "Around the World" agency are as follows: 11,400 rubles per person under 6 years old and 23,500 rubles per person over 6 years old. When purchasing a tour from the "Around the World" agency, a commission of 1% of the total cost of the tour is charged.
Determine which agency is more cost-effective to purchase the tour from. In your answer, indicate the minimum expenses of the Dorokhov family for the vacation in Crimea.
In your answer, provide only the number without units of measurement!
|
# Solution:
Cost of the tour with the company "Globus"
$(3 * 25400) *(1-0.02)=74676$ rubles.
Cost of the tour with the company "Around the World"
$(11400+2 * 23500) * 1.01=58984$ rubles.
Answer: 58984
|
58984
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 12. (16 points)
The Vasilievs' family budget consists of the following income items:
- parents' salary after income tax deduction - 71000 rubles;
- income from renting out property - 11000 rubles;
- daughter's scholarship - 2600 rubles
The average monthly expenses of the family include:
- utility payments - 8400 rubles;
- food - 18000 rubles;
- transportation expenses - 3200 rubles;
- tutor services - 2200 rubles;
- other expenses - 18000 rubles.
10 percent of the remaining amount is transferred to a deposit for the formation of a financial safety cushion. The father wants to buy a car on credit. Determine the maximum amount the Vasilievs family can pay monthly for the car loan.
In your answer, provide only the number without units of measurement!
|
# Solution:
family income
$71000+11000+2600=84600$ rubles
average monthly expenses
$8400+18000+3200+2200+18000=49800$ rubles
expenses for forming a financial safety cushion
$(84600-49800) * 0.1=3480$ rubles
the amount the Petrovs can save monthly for the upcoming vacation
$84600-49800-3480=31320$ rubles
## Answer: 31320
|
31320
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 13. (8 points)
Natalia Petrovna has returned from her vacation, which she spent traveling through countries in North America. She has a certain amount of money left in foreign currency.
Natalia Petrovna familiarized herself with the exchange rates at the nearest banks: "Rebirth" and "Garnet." She decided to take advantage of the most favorable offer. What amount will she receive in rubles for exchanging 120 US dollars, 80 Canadian dollars, and 10 Mexican pesos at one of the two banks?
| Type of Currency | Exchange Rate | |
| :--- | :---: | :---: |
| | Rebirth | Garnet |
| US Dollar | 74.9 rub. | 74.5 rub. |
| Canadian Dollar | 59.3 rub. | 60.1 rub. |
| Mexican Peso | 3.7 rub. | 3.6 rub. |
In your answer, provide only the number without units of measurement!
|
# Solution:
1) cost of currency at Bank "Vozrozhdenie":
$$
120 * 74.9 + 80 * 59.3 + 10 * 3.7 = 13769 \text{ RUB}
$$
2) cost of currency at Bank "Garant":
$$
120 * 74.5 + 80 * 60.1 + 10 * 3.6 = 13784 \text{ RUB}
$$
Answer: 13784
|
13784
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 14. (8 points)
To attend the section, Mikhail needs to purchase a tennis racket and a set of tennis balls. Official store websites have product catalogs. Mikhail studied the offers and compiled a list of stores where the items of interest are available:
| Item | Store | |
| :--- | :---: | :---: |
| | Higher League | Sport-Guru |
| Tennis racket | 5600 rub. | 5700 rub. |
| Set of tennis balls (3 pcs.) | 254 rub. | 200 rub. |
Mikhail plans to use delivery for the items. The delivery cost from the store "Higher League" is 550 rub., and from the store "Sport-Guru" it is 400 rub. If the purchase amount exceeds 5000 rub., delivery from the store "Higher League" is free.
Mikhail has a discount card from the store "Sport-Guru," which provides a 5% discount on the purchase amount.
Determine the total expenses for purchasing the sports equipment, including delivery costs, assuming that Mikhail chooses the cheapest option.
In your answer, provide only the number without units of measurement!
|
# Solution:
1) cost of purchase in the store "Higher League":
$$
\text { 5600+254=5854 rub. }
$$
1) cost of purchase in the store "Sport-guru": $(2700+200)^{*} 0.95+400=6005$ rub.
## Answer: 5854
|
5854
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 15. (8 points)
The fast-food network "Wings and Legs" offers summer jobs to schoolchildren. The salary is 25000 rubles per month. Those who work well receive a monthly bonus of 5000 rubles.
How much will a schoolchild who works well at "Wings and Legs" earn per month (receive after tax) after the income tax is deducted?
In your answer, provide only the number without units of measurement!
|
Solution:
The total earnings will be 25000 rubles + 5000 rubles $=30000$ rubles
Income tax $13 \%-3900$ rubles
The net payment will be 30000 rubles - 3900 rubles $=26100$ rubles
## Correct answer: 26100
## 2nd Option
|
26100
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 11. (16 points)
One way to save on utility bills is to use the night tariff (from 23:00 to 07:00). To apply this tariff, a multi-tariff meter needs to be installed.
The Romanov family is considering purchasing a multi-tariff meter to reduce their utility bills. The cost of the meter is 3500 rubles. The installation cost of the meter is 1100 rubles. On average, electricity consumption is 300 kWh per month, with 230 kWh used from 23:00 to 07:00.
The cost of electricity when using a multi-tariff meter: from 07:00 to 23:00 - 5.2 rubles per kWh, from 23:00 to 07:00 - 3.4 rubles per kWh.
The cost of electricity when using a standard meter is 4.6 rubles per kWh.
Determine how much the Romanov family will save by using a multi-tariff meter over three years.
In your answer, provide only the number without units of measurement!
|
# Solution:
2) use of a multi-tariff meter:
$$
3500+1100+(230 * 3.4+(300-230) * 5.2) * 12 * 3=45856 \text { rub. }
$$
3) use of a typical meter
$$
300 * 4.6 * 12 * 3=49680 \text { rub. } \quad \text {. } \quad \text {. }
$$
the savings will be
$49680-45856=3824$ rub.
## Answer: 3824
|
3824
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 12. (16 points)
The budget of the Petrovs consists of the following income items:
- parents' salary after income tax deduction - 56000 rubles;
- grandmother's pension - 14300 rubles;
- son's scholarship - 2500 rubles
Average monthly expenses of the family include:
- utility payments - 9800 rubles;
- food - 21000 rubles;
- transportation expenses - 3200 rubles;
- leisure - 5200 rubles;
- other expenses - 15000 rubles
10 percent of the remaining amount is transferred to a deposit for the formation of a financial safety cushion. Determine the amount that the Petrovs can save monthly for an upcoming vacation.
In the answer, indicate only the number without units of measurement!
|
# Solution:
family income
$56000+14300+2500=72800$ rubles.
average monthly expenses
$9800+21000+3200+5200+15000=54200$ rubles.
expenses for forming a financial safety cushion
$(72800-54200) * 0.1=1860$ rubles.
the amount the Petrovs can save monthly for the upcoming vacation
$72800-54200-1860=16740$ rubles.
Answer: 16740.
|
16740
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 13. (8 points)
Maxim Viktorovich returned from a trip to Asian countries. He has a certain amount of money in foreign currency left.
Maxim Viktorovich familiarized himself with the exchange rates at the nearest banks: "Voskhod" and "Elfa". He decided to take advantage of the most favorable offer. What amount will he receive as a result of exchanging 110 Chinese yuan, 80 Japanese yen, and 50 Hong Kong dollars into rubles at one of the two banks?
| Type of Currency | Exchange Rate | |
| :--- | :---: | :---: |
| | Voskhod | Elfa |
| Chinese yuan | 11.7 rub. | 11.6 rub. |
| Japanese yen | 72.1 rub. | 71.9 rub. |
| Hong Kong dollar | 9.7 rub. | 10.1 rub. |
In your answer, indicate only the number without units of measurement!
|
# Solution:
1) cost of currency at "Voskhod" bank:
$110 * 11.7 + 80 * 72.1 + 50 * 9.7 = 7540$ rubles.
2) cost of currency at "Alpha" bank: $110 * 11.6 + 80 * 71.9 + 50 * 10.1 = 7533$ rubles.
Answer: 7540.
|
7540
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 14. (8 points)
Elena decided to get a pet - a budgerigar. She faced the question of where to buy a cage and a bath more cost-effectively.
On the official websites of the stores, product catalogs are posted. Elena studied the offers and compiled a list of stores where the items she is interested in are available:
| Item | Store | |
| :--- | :---: | :---: |
| | ZooWorld | ZooIdea |
| Cage | 4500 rub. | 3700 rub. |
| Bath | 510 rub. | 680 rub. |
Elena plans to use delivery. The delivery cost from the store "ZooWorld" is 500 rub., and from the store "ZooIdea" it is 400 rub. If the purchase amount exceeds 5000 rub., delivery from the store "ZooWorld" is free.
Elena has a discount card from the store "ZooIdea," which provides a 5% discount on the purchase amount.
Determine the minimum total cost for purchasing a cage and a bath for the budgerigar, including delivery costs.
In your answer, provide only the number without units of measurement!
|
# Solution:
2) cost of purchase in the "Zoimir" store: $4500+510=5010$ rubles
3) cost of purchase in the "Zooidea" store: $(3700+680) * 0.95+400=4561$ rubles
## Answer: 4561
|
4561
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 15. (8 points)
Announcement: "Have free time and want to earn money? Write now and earn up to 2500 rubles a day working as a courier with the service 'Food.There-Here!'. Delivery of food from stores, cafes, and restaurants.
How much will a school student working as a courier with the service 'Food.There-Here!' earn in a month (4 weeks), working 4 days a week with a minimum load for 1250 rubles a day after the income tax deduction?
In your answer, provide only the number without units of measurement! #
|
# Solution:
The total earnings will be (1250 rubles * 4 days) * 4 weeks = 20000 rubles
Income tax 13% - 2600 rubles
The amount of earnings (net pay) will be 20000 rubles - 2600 rubles = 17400 rubles
Correct answer: 17400
|
17400
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. There are several technologies for paying with bank cards: chip, magnetic stripe, paypass, cvc. Arrange the actions performed with a bank card in the order corresponding to the payment technologies.
1 - tap
2 - pay online
3 - swipe
4 - insert into terminal
|
Answer in the form of an integer, for example 1234.
Answer: 4312
|
4312
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 1. (4 points)
The price of a new 3D printer is 625000 rubles. Under normal operating conditions, its resale value decreases by $20 \%$ in the first year, and then by $8 \%$ each subsequent year. After how many years will the resale value of the printer be less than 400000 rubles?
|
Solution:
Let's calculate the cost of the printer year by year:
1 year $=625000 * 0.8=500000$ rubles
2 year $=500000 * 0.92=460000$ rubles (1 point)
3 year $=460000 * 0.92=423200$ rubles (1 point)
4 year $=423200 * 0.92=397694$ rubles. (1 point)
Answer: in 4 years. ( $\mathbf{1}$ point)
|
4
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. (8 points)
Kolya's parents give him pocket money once a month, calculating it as follows: 100 rubles for each A in math, 50 rubles for a B, 50 rubles are deducted for a C, and 200 rubles are deducted for a D. If the amount turns out to be negative, Kolya simply gets nothing. The math teacher gives a grade for the quarter by calculating the arithmetic mean and rounding according to rounding rules. What is the maximum amount of money Kolya could have received if it is known that his quarterly grade is two points, the quarter lasted exactly 2 months, there were 14 math lessons each month, and Kolya receives no more than one grade per lesson?
#
|
# Solution:
If Kolya received a final grade of 2 for the quarter, then for each 5 he received more than 5 2s, for each 4 - more than 3 2s, and for each 3 - more than 1 2. This means that the number of 2s was greater than the total number of all other grades combined, so Kolya could receive money in at most one of the months.
Thus, the maximum amount is achieved if in one of the months he receives only 2s, and in the other - only 4s and 5s, from which the maximum number of 2s received is 14. Kolya could not receive 3 or more 5s, as in this case he would have 15 or more 2s.
This leaves 3 cases:
a) Kolya has 2 fives, then he can get 1 four, the sum is 250
b) Kolya has 1 five, then he can get 2 fours, the sum is 200
c) Kolya has no fives, in which case he can get 4 fours, the sum is 200.
Thus, the maximum amount of money is 250 rubles.
Criteria:
Correct answer - 2 points
Method of obtaining the correct answer - 2 points
Explanation of why money cannot be received in both months - 2 points
Correct enumeration - 2 points
+1 bonus point for explicitly stating the advantage of receiving all 2s in the second month, as money loses value over time and it is more advantageous to receive it in the first month rather than the second.
|
250
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6. (10 points)
Vasily is planning to graduate from college in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles. When Vasily finished school, he could have chosen not to go to college and instead work as a real estate assistant, like his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in college? Note: Bachelor's degree education lasts 4 years.
|
# Solution:
In 4 years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points).
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300 * $(1 / 5 * 60000 + 1 / 10 * 80$ $000 + 1 / 20 * 25000 + (1 - 1 / 5 - 1 / 10 - 1 / 20) * 40000) + 30 / 300 * 25000 = 45025$ rubles (6 points).
|
45025
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 1. (4 points)
In the run-up to the New Year, a fair is being held at the school where students exchange festive toys. As a result, the following exchange norms have been established:
1 Christmas tree ornament can be exchanged for 2 crackers, 5 sparklers can be exchanged for 2 garlands, and 4 Christmas tree ornaments can be exchanged for 1 garland.
a) How many crackers can be obtained for 10 sparklers? (1 point)
b) Which is more valuable: 5 Christmas tree ornaments and 1 cracker or 2 sparklers? (3 points)
#
|
# Solution:
a) 10 sparklers $=4$ garlands = 16 ornaments = $\mathbf{32}$ crackers. (1 point)
b) Convert everything to crackers. In the first case, we have $\mathbf{11}$ crackers. In the second case, 2 sparklers $=4 / 5$ garlands $=16 / 5$ ornaments $=32 / 5$ crackers $=\mathbf{6.4}$ crackers. Answer: 5 ornaments and 1 cracker are more expensive than 2 sparklers. (1 point for converting each set to a single unit + 1 point for the correct answer)
|
32
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 3. (8 points)
In the Sidorov family, there are 3 people: the father works as a programmer with an hourly rate of 1500 rubles. The mother works as a hairdresser at home and charges 1200 rubles per haircut, which takes her 1.5 hours. The son tutors in mathematics and earns 450 rubles per academic hour (45 minutes). Every day, the family needs to prepare food, walk the dog, and go to the store. The mother spends 2 hours preparing food, the father 1 hour, and the son 4 hours. Walking the dog takes 1 hour for any family member. A trip to the store takes 1 hour for the father, 2 hours for the mother, and 3 hours for the son. In addition, each family member sleeps for 8 hours and spends another 8 hours on rest and personal matters.
What is the maximum amount of money the family can earn in one day?
|
# Solution
In this problem, there are 2 possible interpretations, both of which were counted as correct.
In one case, it is assumed that 8 hours are spent on work on average over the month, in the other that no more than 8 hours are spent on work each day.
## First Case:
1) Determine the hourly wage for each family member. (2 points)
a) Father = 1500 rubles per hour
b) Mother = 1200 / 1.5 = 800 rubles per hour
c) Son = 450 / 0.75 = 600 rubles per hour
2) Calculate how much money the family will lose for each mandatory task depending on who does it. (1 point for each sub-item, total 3 points)
a) Cooking: father - 1500 rubles, mother - 1600 rubles, son - 2400 rubles.
Therefore, the father cooks and spends 1 hour on it.
b) Walking the dog: father - 1500 rubles, mother - 800 rubles, son - 600 rubles.
Therefore, the son walks the dog and spends 1 hour on it.
c) Going to the store: father - 1500 rubles, mother - 1600 rubles, son - 1800 rubles.
Therefore, the father goes to the store and spends 1 hour on it.
3) Calculate the final earnings of each family member: (3 points)
a) Father: 1500 * (24 - 8 - 8 - 1 - 1) = 9000 rubles.
b) Mother: 800 * (24 - 8 - 8) = 6400 rubles.
c) Son: 600 * (24 - 8 - 8 - 1) = 4200 rubles.
d) Total: 9000 + 6400 + 4200 = 19600 rubles.
Answer: 19600 rubles
## Second Case:
1) The number of hours each family member can spend on work or household chores per day: 24 - 8 - 8 = 8 hours.
2) Estimate the maximum income of each family member assuming they do not do household chores (2 points):
a) Father: 1500 * 8 = 12000 rubles.
b) Mother: 1200 * 5 = 6000 rubles + 30 minutes of free time.
c) Son: 450 * 10 = 4500 rubles + 30 minutes of free time.
3) Calculate the lost income from each type of activity (3 points, 1 point for each type of work):
a) Cooking: father - 1500, mother - 1200 (if she didn't go to the store) or -2400 (if she did go to the store), son - 2250.
b) Going to the store: father - 1500, mother - 1200 (if she didn't cook) or -2400 (if she did cook), son - 1800.
c) Walking the dog: father - 1500, mother - 1200, son - 450.
4) Choose the smallest losses: the father and mother cook and go to the store (it doesn't matter who does what, but each does one thing), the son walks the dog (2 points for correctly determining the tasks for each family member).
5) Thus, the maximum earnings: 12000 - 1500 + 6000 - 1200 + 4500 - 450 = 19350 rubles (1 point)
#
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19600
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Other
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math-word-problem
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Yes
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Yes
|
olympiads
| false
|
# Problem 4. (10 points)
On December 31 at 16:35, Misha realized he had no New Year's gifts for his entire family. He wants to give different gifts to his mother, father, brother, and sister. Each of the gifts is available in 4 stores: Romashka, Odynachik, Nezabudka, and Lysichka, which close at 20:00. The journey from home to each store and between any two stores takes 30 minutes.
The table below shows the cost of the gifts in all four stores and the time Misha will need to spend shopping in each store. What is the minimum amount of money Misha can spend if he must definitely manage to buy all 4 gifts?
| | mother | father | brother | sister | Time spent in the store (min.) |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Romashka | 1000 | 750 | 930 | 850 | 35 |
| Odynachik | 1050 | 790 | 910 | 800 | 30 |
| Nezabudka | 980 | 810 | 925 | 815 | 40 |
| :--- | :--- | :--- | :--- | :--- | :--- |
| Lysichka | 1100 | 755 | 900 | 820 | 25 |
|
# Solution:
Notice that in each of the stores, there is a "unique" gift with the lowest price.
If Misha managed to visit all 4 stores, he would spend the minimum amount of $980+750+900+800=3430$ rubles. However, visiting any three stores would take Misha at least $30 * 3+25+30+35=180$ minutes. Considering the additional 30 minutes to travel to the fourth store, he would not make it. On the other hand, Misha can manage to visit any three out of the four stores—along with the travel time, he would arrive at the third store no later than $3 * 30+35+40=165$ minutes, i.e., at $19:10$.
Then, in one of the stores, he needs to choose 2 gifts to minimize the extra cost. Let's create a table of the extra costs:
| | mom | dad | brother | sister |
| :--- | :--- | :--- | :--- | :--- |
| Daisy | 20 | 0 | 30 | 50 |
| Dandelion | 70 | 40 | 10 | 0 |
| Forget-me-not | 0 | 60 | 25 | 15 |
| Lily of the Valley | 120 | 5 | 0 | 20 |
From the table, it is easy to see that the minimum extra cost is 5 rubles at the "Lily of the Valley" store for the gift for dad.
Therefore, Misha goes to Dandelion, Forget-me-not, and Lily of the Valley, and the total cost for the gifts will be $3430+5=3435$ rubles.
Answer: 3435 rubles.
Criteria:
Correct answer - 2 points
Reasoning that he cannot visit all 4 stores - 2 points
Reasoning that he can visit any 3 stores - 2 points
Explanation that the minimum extra cost is 5 rubles - 4 points
#
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3435
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make the purchase.
## Maximum 30 points
## Problem 5
Schoolboy Vanechka informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use an additional card issued to her bank account. Vanya needed to buy 100 large chocolate bars for organizing and conducting a quiz at the New Year's party. However, the card has a protection system that analyzes the last 5 purchases, determines their average value, and compares it with the planned purchase amount. If the planned purchase exceeds the average by three times, the bank blocks the transaction and requires additional verification (for example, a call from his mother to the call center). Over the last month, his mother made purchases in the following amounts: 785 rubles, 2033 rubles, 88 rubles, 3742 rubles, 1058 rubles.
Calculate whether Vanechka will be able to buy the chocolate bars for the New Year's party in one transaction without calling his mother to the call center if the cost of one chocolate bar is 55 rubles?
|
# Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an acceptable purchase would be no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, one can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct answer.
7 points - correct calculation of the average value of purchases
3 points - partial solution
## Appendix to Problem No. 1
Insurance is a system of relationships between insurers (insurance company) and insured individuals (for example, any citizen), which allows for the reduction of property risks by insuring, for example, property against possible adverse events...
Property insurance is one of the types of insurance according to the Civil Code of the Russian Federation. Real estate is one type of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
84
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Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. What was NOT used as money?
1) gold
2) stones
3) horses
4) dried fish
5) mollusk scales
6) all of the above were used
|
Answer: 6. All of the above were used as money.
|
6
|
Logic and Puzzles
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
2. A stationery store is running a promotion: there is a sticker on each notebook, and for every 5 stickers, a customer can get another notebook (also with a sticker). Fifth-grader Katya thinks she needs to buy as many notebooks as possible before the new semester. Each notebook costs 4 rubles, and Katya has 150 rubles. How many notebooks will Katya get?
|
Answer: 46.
1) Katya buys 37 notebooks for 148 rubles.
2) For 35 stickers, Katya receives 7 more notebooks, after which she has notebooks and 9 stickers.
3) For 5 stickers, Katya receives a notebook, after which she has 45 notebooks and 5 stickers.
4) For 5 stickers, Katya receives the last 46th notebook.
|
46
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give the answer in rubles, rounding to the nearest whole number (the answer is a whole number).
|
Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997$. Since rounding was required, the answer is 24.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/1.html
|
24
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Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
|
Answer: 15.
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let's find the number of pairs of different colored tulips in the bouquet: $600: 81=7$ (remainder 33). This means Vanya needs a bouquet with $7 * 2+1=15$ tulips.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. What is a sign of a financial pyramid?
1) an offer of income significantly above average
2) incomplete information about the company
3) aggressive advertising
4) all of the above
|
Answer: 4. All of the above are signs of a financial pyramid.
|
4
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Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Problem 12. (6 points)
Victor received a large sum of money as a birthday gift in the amount of 45 thousand rubles. The young man decided to save this part of his savings in dollars on a currency deposit. The term of the deposit agreement was 2 years, with an interest rate of 4.7% per annum, compounded quarterly. On the day the deposit was opened, the commercial bank bought dollars at a rate of 59.60 rubles per 1 US dollar and sold them at a rate of 56.60 rubles per 1 US dollar. What amount in US dollars will be on Victor's account at the end of the term of the deposit agreement (rounding to the nearest whole number)?
|
Answer: 873 USD.
## Comment:
45000 RUB / 56.60 RUB $\times(1+4.7\% / 4 \text { quarters })^{2 \text { years } \times 4 \text { quarters }}=873$ USD.
|
873
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 17-19. (2 points per Task)
Alena opened a multi-currency deposit at "Severny" Bank for 3 years. The deposit involves the deposit of funds in three currencies: euros, dollars, and rubles. At the beginning of the deposit agreement, Alena's account contained 3000 euros, 4000 dollars, and 240000 rubles. The interest rate on the ruble deposit is 7.9% per annum, and on the dollar and euro deposits, it is 2.1% per annum. Interest is accrued annually, and the interest is capitalized. After one year, Alena decided to change the structure of her deposit, sold 1000 euros, and used the proceeds to increase the share of funds in dollars. At the time of conversion, the selling rate for dollars was 58.90 rubles, the buying rate for dollars was 56.90 rubles, the selling rate for euros was 61.15 rubles, and the buying rate for euros was 60.10 rubles. Another year later, Alena reviewed the structure of her deposit again, exchanged 2000 dollars, and bought euros with the proceeds. At the time of conversion, the selling rate for dollars was 60.10 rubles, the buying rate for dollars was 58.50 rubles, the selling rate for euros was 63.20 rubles, and the buying rate for euros was 61.20 rubles.
Question 17. Determine the balance on Alena's dollar account at the end of the deposit agreement (round to the nearest whole number).
Question 18. Determine the balance on Alena's euro account at the end of the deposit agreement (round to the nearest whole number).
Question 19. Determine the balance on Alena's ruble account at the end of the deposit agreement (round to the nearest whole number).
|
Answer 17: $3280;
Answer 18: 4040 euros,
Answer 19: 301492 rubles.
## Comment:
1 year
Euros: 3000 euros $\times(1+2.1 \%)=3063$ euros.
Dollars: 4000 dollars $\times(1+2.1 \%)=4084$ dollars.
Rubles: 240000 rubles $\times(1+7.9 \%)=258960$ rubles.
2 year
Euros: (3063 euros - 1000 euros $) \times(1+2.1 \%)=2106$ euros.
Dollars: (4084 dollars + 1000 euros × 60.10 rubles / 58.90 rubles) × (1 + 2.1 \%) = 5212 dollars.
Rubles: 258960 rubles $\times(1+7.9 \%)=279418$ rubles.
3 year
Euros: (2106 euros + 2000 dollars $\times$ 58.50 rubles / 63.20 rubles $) \times(1+2.1 \%)=4040$ euros.
Dollars: (5212 dollars - 2000 dollars) $\times(1+2.1 \%)=3279.45$ dollars.
Rubles: 279418 rubles $\times(1+7.9 \%)=301492$ rubles.
|
3280
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. (4 points)
Ivan bought a used car from 2010 for 90,000 rubles with an engine power of 150 hp and registered it on January 29. On August 21 of the same year, the citizen sold his car and a month later bought a horse and a cart for 7,500 rubles. The transport tax rate is set at 20 rubles per 1 hp. What amount of transport tax should the citizen pay? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 2000.
## Comment:
|
Solution: transport tax $=150 \times 20 \times 8 / 12=2000$ rubles. A horse and a cart are not subject to transport tax.
|
2000
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. (6 points)
Sergei, being a student, worked part-time in a student cafe after classes for a year. Sergei's salary was 9000 rubles per month. In the same year, Sergei paid for his medical treatment at a healthcare facility in the amount of 100000 rubles and purchased medications on a doctor's prescription for 20000 rubles (eligible for deduction). The following year, Sergei decided to apply for a social tax deduction. What amount of the paid personal income tax is subject to refund to Sergei from the budget? (Provide the answer as a whole number, without spaces and units of measurement.)
## Answer: 14040.
## Comment:
|
Solution: the amount of the social tax deduction for medical treatment will be: $100000+20000=$ 120000 rubles. The possible tax amount eligible for refund under this deduction will be $120000 \times 13\% = 15600$ rubles. However, in the past year, Sergey paid income tax (NDFL) in the amount of $13\% \times (9000 \times 12) = 14040$ rubles. Therefore, the amount of personal income tax paid, eligible for refund to Sergey from the budget, will be 14040 rubles.
|
14040
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 8. (2 points)
After graduating from a technical university, Oleg started his own business producing water heaters. This year, Oleg plans to sell 5000 units of water heaters. Variable costs for production and sales of one water heater amount to 800 rubles, and total fixed costs are 1000 thousand rubles. Oleg wants his income to exceed expenses by 1500 thousand rubles. At what price should Oleg sell the water heaters? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 1300.
## Comment:
|
Solution: the price of one kettle $=((1000000+0.8 \times 5000)+1500$ 000) / $(1000000$ + $0.8 \times 5000)=1300$ rub.
|
1300
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 10. (4 points)
To buy new headphones costing 275 rubles, Katya decided to save money on sports activities. Until now, she has been buying a single-visit ticket to the swimming pool, including a visit to the sauna for 250 rubles, to warm up. However, summer has arrived, and the need to visit the sauna has disappeared. Visiting only the swimming pool costs 200 rubles more than visiting the sauna. How many times does Katya need to visit the swimming pool without the sauna to save enough to buy the headphones? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 11.
## Comment:
|
Solution: one visit to the sauna costs 25 rubles, the price of one visit to the swimming pool is 225 rubles. Katya needs to visit the swimming pool 11 times without going to the sauna in order to save up for buying headphones.
|
11
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 18. (4 points)
By producing and selling 4000 items at a price of 6250 rubles each, a budding businessman earned 2 million rubles in profit. Variable costs for one item amounted to 3750 rubles. By what percentage should the businessman reduce the production volume to make his revenue equal to the cost? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 20.
## Comment:
|
Solution: fixed costs $=-2$ million, gap $+4000 \times 6250-3750 \times 4000$ million $=8$ million.
$6.25 \times Q=3750 Q+8 \text{ million}$.
$Q=3200$ units.
Can be taken out of production $=4000-3200=800$ units, that is, $20\%$.
|
20
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive than a, which means the highest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend $4800 + 1500 - 4800 * 0.2 = 5340$ rubles. This is the most profitable way to make the purchase.
## Maximum 30 points
## Problem 5
Schoolboy Vanechka informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use an additional card issued to her bank account. Vanya needed to buy 100 large chocolate bars for organizing and conducting a quiz at the New Year's party. However, the card has a protection system that analyzes the last 5 purchases, calculates their average value, and compares it with the planned purchase amount. If the planned purchase exceeds the average by three times, the bank blocks the transaction and requires additional verification (for example, a call from his mother to the call center). Over the last month, his mother made purchases in the following amounts: 785 rubles, 2033 rubles, 88 rubles, 3742 rubles, 1058 rubles.
Calculate whether Vanechka will be able to buy the chocolate bars for the New Year's party in one transaction without calling his mother to the call center if the cost of one chocolate bar is 55 rubles?
|
# Solution:
The average value of the last purchases is $(785+2033+88+3742+1058) / 5 = 1541.2$ rubles. Therefore, an allowable purchase is no more than $1541.2 * 3 = 4623.6$ rubles. With this amount, you can buy $4623.6 / 55 \approx 84$ chocolates.
## Maximum 20 points
20 points - fully detailed solution and correct answer.
7 points - correct calculation of the average value of purchases
3 points - partial solution
## Appendix to Problem No. 1
Insurance is a system of relationships between insurers (insurance company) and insured individuals (for example, any citizen), which allows reducing property risks by insuring, for example, property against possible adverse events...
Property insurance is one of the types of insurance according to the Civil Code of the Russian Federation. Real estate is one type of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
84
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Leshа has 10 million rubles. Into what minimum number of banks should he deposit them to receive the full amount through ACB insurance payouts in case the banks cease operations?
|
Answer: 8. The maximum insurance payout is 1,400,000, which means no more than this amount should be deposited in each bank.
|
8
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How much did the US dollar exchange rate change over the 2014 year (from January 1, 2014 to December 31, 2014)? Give your answer in rubles, rounding to the nearest whole number.
|
Answer: 24. On January 1, 2014, the dollar was worth 32.6587, and on December 31, it was 56.2584.
$56.2584-32.6587=23.5997 \approx 24$.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/23.html
|
24
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
7. Alexey plans to buy one of two car brands: "A" for 900 thousand rubles or "B" for 600 thousand rubles. On average, Alexey drives 15 thousand km per year. The cost of gasoline is 40 rubles per liter. The cars consume the same type of gasoline. The car is planned to be used for 5 years, after which Alexey will be able to sell the car of brand "A" for 500 thousand rubles, and the car of brand "B" for 350 thousand rubles.
| Car Brand | Fuel Consumption (l/100km) | Annual Insurance Cost (rubles) | Average Annual Maintenance Cost (rubles) |
| :--- | :--- | :--- | :--- |
| "A" | 9 | 35000 | 25000 |
| "B" | 10 | 32000 | 20000 |
Using the data in the table, answer the question: how much more expensive will the purchase and ownership of the more expensive car be for Alexey?
|
Answer: 160000.
Use of car brand "A":
$900000+(15000 / 100) * 9 * 5 * 40+35000 * 5+25000 * 5-500000=970000$
Use of car brand "B":
$600000+(15000 / 100) * 10 * 5 * 40+32000 * 5+20000 * 5-350000=810000$
Difference: $970000-810000=160000$
|
160000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. A family of 4, consisting of a mom, a dad, and two children, has arrived in city $\mathrm{N}$ for 5 days. They plan to make 10 trips on the subway each day. What is the minimum amount they will have to spend on tickets, given the following tariffs in city $\mathrm{N}$?
| Adult ticket for one trip | 40 rubles |
| :--- | :--- |
| Child ticket for one trip | 20 rubles |
| Unlimited daily pass for one person | 350 rubles |
| Unlimited daily pass for a group of up to 5 people | 1500 rubles |
| Unlimited 3-day pass for one person | 900 rubles |
| Unlimited 3-day pass for a group of up to 5 people | 3500 rubles |
|
Answer: 5200. The family will spend this amount if the parents buy a three-day pass for themselves, and for the remaining two days, they will buy a one-day pass. For this, they will spend ($900 + 350 * 2$) * $2 = 3200$ rubles.
For the children, it is most cost-effective to buy single-trip tickets for all 5 days, spending $2 * 10 * 20 * 5 = 2000$ rubles.
Thus, the total amount will be $3200 + 2000 = 5200$ rubles.
|
5200
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. (12 points)
Andrey lives near the market, and during the summer holidays, he often helped one of the traders lay out fruits on the counter early in the morning. For this, the trader provided Andrey with a $10 \%$ discount on his favorite apples. But autumn came, and the price of apples increased by $10 \%$. Despite the fact that Andrey started school and stopped helping the trader, the $10 \%$ discount for him remained. What will Andrey's monthly expenses on apples be now, considering that he buys 2 kilograms monthly? Before the price increase, apples at the market cost 50 rubles per kg for all customers. (Provide the answer as a whole number, without spaces or units of measurement.)
#
|
# Answer: 99.
## Comment
Solution: the new price of apples at the market is 55 rubles per kg, with a discount of $10 \%$ applied to this price. Thus, the price for 1 kg for Andrei will be 49.5 rubles, and for 2 kg Andrei will pay 99 rubles monthly.
|
99
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 10. (12 points)
The Ivanov family owns an apartment with necessary property worth 3 million rubles, a car that is currently valued at 900 thousand rubles on the market, and savings, part of which, amounting to 300 thousand rubles, is placed in a bank deposit, part is invested in securities worth 200 thousand rubles, and part, amounting to 100 thousand rubles, is in liquid form on a bank card and in cash. In addition, there are outstanding loans. The remaining balance on the mortgage is 1.5 million rubles, the remaining balance on the car loan is 500 thousand rubles. It is also necessary to return a debt to relatives in the amount of 200 thousand rubles. What net worth (net wealth) does the Ivanov family possess? (Provide the answer as a whole number, without spaces or units of measurement.)
|
# Answer: 2300000
## Comment
Solution: equity (net worth) = value of assets - value of liabilities. Value of assets $=3000000+900000+300000+200000+100000=$ 4500000 rubles. Value of liabilities $=1500000+500000+200000=2200000$ rubles. Net worth $=4500000-2200000=2300000$ rubles
## MOSCOW FINANCIAL LITERACY OLYMPIAD 5-7 GRADE Variant 2
|
2300000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. (8 points)
Konstantin's mother opened a foreign currency deposit in the "Western" Bank for an amount of 10 thousand dollars for a term of 1 year. Literally 4 months later, the Bank of Russia revoked the license of the "Western" Bank. The exchange rate on the date of the license revocation was 58 rubles 15 kopecks per 1 dollar. Konstantin's mother was not too upset, as the deposits in the bank were insured under the deposit insurance system. What amount of the deposit should be returned to Konstantin's mother according to the legislation (do not consider interest payments in the calculations)?
a. 10 thousand dollars;
b. 581,500 rubles;
c. 1,400 thousand rubles;
d. 10 thousand rubles.
|
# Answer: b.
## Comment
$10000 \times 58.15$ RUB $=581500$ RUB.
|
581500
|
Other
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Problem 9. (12 points)
Ivan, a full-time student, started working at his long-desired job on March 1 of the current year. In the first two months, during the probationary period, Ivan's salary was 20000 rubles. After the probationary period, the salary increased to 25000 rubles per month. In December, for exceeding the plan, Ivan was awarded a bonus of 10000 rubles. In addition, while studying in a full-time budget-funded graduate program, Ivan received a scholarship of 2800 rubles per month throughout the year. What amount of personal income tax should Ivan pay to the budget? (Provide the answer as a whole number, without spaces and units of measurement)
|
Answer: 32500.
## Comment
Solution: Personal Income Tax from salary $=(20000 \times 2+25000 \times 8+10000) \times 13\% = 32500$ rubles. The scholarship is not subject to Personal Income Tax.
|
32500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 4. (8 points)
Lena receives 50,000 rubles per month and spends 45,000 rubles per month. She gets her salary on the 6th of each month. Lena has: a deposit in the bank $\mathrm{X}$ with a $1\%$ monthly interest rate with monthly capitalization and the possibility of topping up, but no withdrawals allowed, with the calculation day being the 6th of each month; a debit card, on which 6% of the minimum balance for the month is credited on the 6th of each month at an annual rate of $8\%$; a credit card with a payment due date of the 7th of each month and a limit of 100,000 rubles, with a $3\%$ fee for cash withdrawals or transfers.
Lena can pay all her expenses using her bank card. Describe the optimal strategy for Lena and calculate the annual income she can earn from her cooperation with the bank (in rubles).
|
Solution:
1) Of all the bank's products, the deposit brings Lena the highest income. She can use $50-45=5$ thousand rubles for savings, depositing them immediately after receiving her salary. Over a year, this will bring Lena $5000 * \frac{1.01 * (1.01^{12}-1)}{1.01-1} - 5000 * 12 \approx 4000$.
2) The remaining 45 thousand rubles per month cannot be deposited by Lena, as she will spend them during the month after receiving her salary, and the deposit does not allow withdrawals. Therefore, Lena needs to use the second most profitable instrument - a debit card with interest on the balance.
3) If Lena simply deposits the money into a debit card and uses it for payments, she will not gain any additional benefit, because by the end of the month, after receiving her salary, the balance on the card will be zero, and interest is calculated on the minimum balance.
4) To maximize her income, it is profitable for Lena to pay with a credit card and keep her salary on a debit card. This way, the minimum balance on the debit card will always be 45 thousand rubles, so over a year, this will bring Lena $45000 * 0.08 = 860$ rubles.
Let's describe Lena's strategy in detail: On the 6th of each month, after receiving her salary, she should deposit 5000 rubles into the deposit and 45000 rubles into the debit card. Then, on the 7th, pay off the credit card debt. All expenses during the month should be made using the credit card. Thus, Lena's annual income from cooperation with the bank will be $4000 + 860 = 4860$ rubles.
Criteria: description of the correct strategy - 4 points, calculation of income from the debit card - 2 points, correct formula for deposit income - 2 points.
|
4860
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Problem 6. (8 points)
Vasily is planning to graduate from the institute in a year. Only 270 out of 300 third-year students successfully pass their exams and complete their bachelor's degree. If Vasily ends up among the 30 expelled students, he will have to work with a monthly salary of 25,000 rubles. It is also known that every fifth graduate gets a job with a salary of 60,000 rubles per month, every tenth graduate earns 80,000 rubles per month, every twentieth graduate cannot find a job in their field and has an average salary of 25,000 rubles per month, while the salary of all others is 40,000 rubles.
When Vasily finished school, he could have chosen not to go to the institute but to work as a real estate assistant, as his friend Fyodor did. Fyodor's salary increases by 3,000 rubles each year. What is Vasily's expected salary? Whose salary will be higher in a year and by how much - Vasily's expected salary or Fyodor's actual salary, if Fyodor started working with a salary of 25,000 rubles at the same time Vasily enrolled in the institute?
Note: Bachelor's education lasts 4 years.
|
Solution:
Four years after graduating from school, Fedor will earn $25000 + 3000 * 4 = 37000$ rubles (2 points)
The expected salary of Vasily is the expected value of the salary Vasily can earn under all possible scenarios (2 points). It will be 270/300*(1/5*60 000 + 1/10*80 000 + 1/20*25 000 + (1 - 1/5 - 1/10 - 1/20) * 40000) + 30/300 * 25000 = 45025 rubles (4 points).
|
45025
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. How much did the euro exchange rate change over the 2012 year (from January 1, 2012 to December 31, 2012)? Provide the answer in rubles, rounded to the nearest whole number.
|
Answer: 1 or -1. On January 1, 2012, the euro was worth 41.6714, and on December 31, it was 40.2286. $40.2286-41.6714=-1.4428 \approx-1$.
Note: This problem could have been solved using the internet. For example, the website https://news.yandex.ru/quotes/region/23.html
|
-1
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. The Petrovs family has decided to renovate their apartment. They can hire a company for a "turnkey renovation" for 50,000 or buy materials for 20,000 and do the renovation themselves, but for that, they will have to take unpaid leave. The husband earns 2000 per day, and the wife earns 1500. How many working days can they spend on the renovation so that it turns out to be more cost-effective than hiring workers?
|
Answer: 8. The combined daily salary of the husband and wife is $2000+1500=3500$ rubles. The difference between the cost of a turnkey repair and buying materials is $50000-20000=30000$.
$30000: 3500 \approx 8.57$, so the family can spend no more than 8 days on the repair.
|
8
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In the store "Third is Not Excessive," there is a promotion: if a customer presents three items at the cash register, the cheapest of them is free. Ivan wants to buy 11 items costing $100, 200, 300, \ldots, 1100$ rubles. For what minimum amount of money can he buy these items?
|
Answer: 4800. It is clear that items should be listed in descending order of price, then the cost of the purchase will be $1100+1000+800+700+500+400+200+100=4800$ rubles.
|
4800
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. A supermarket discount card that gives a $3 \%$ discount costs 100 rubles. Masha bought 4 cakes for 500 rubles each and fruits for 1600 rubles for her birthday. The seller at the cash register offered her to buy the discount card before the purchase. Should Masha agree?
1) no, they offer these cards to everyone
2) yes, of course, she should agree
3) it will not affect the cost of the purchase
|
Answer: 2. The cost of Masha's purchase is $4 * 500 + 1600 = 3600$. If Masha buys a discount card, she will spend $100 + 3600 * 0.97 = 3592$.
|
3592
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
7. Vanya decided to give Masha a bouquet of an odd number of flowers for her birthday, consisting of yellow and red tulips, so that the number of flowers of one color differs from the number of flowers of the other by exactly one. Yellow tulips cost 50 rubles each, and red ones cost 31 rubles. What is the largest number of tulips he can buy for Masha's birthday, spending no more than 600 rubles?
|
Answer: 15.
A bouquet with one more red tulip than yellow ones is cheaper than a bouquet with the same total number of flowers but one more yellow tulip. Therefore, Vanya should buy a bouquet with one more red tulip. The remaining flowers can be paired into red and yellow tulips, with each pair costing 81 rubles. Let's find the number of pairs of different colored tulips in the bouquet: $600: 81=7$ (remainder 33). Therefore, Vanya needs a bouquet with $7 * 2+1=$ 15 tulips.
|
15
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 4. (8 points)
Irina Mikhailovna opened a foreign currency deposit in the "Western" Bank for an amount of $23,904 for a term of 1 year. The interest rate on the deposit was 5% per annum. Exactly 3 months later, the Bank of Russia revoked the license of the "Western" Bank. The official exchange rate on the date of the license revocation was 58 rubles and 15 kopecks per 1 dollar. All deposits of customers in the "Western" Bank were insured. What amount of the deposit should be returned to Irina Mikhailovna, taking into account the period during which the money was stored in the account?
a. 1,390,017 rubles and 60 kopecks;
b. 1,400,000 rubles;
c. 1,407,393 rubles;
d. $24,075.67.
|
Answer: $b$.
## Comment:
23904 USD $\times 58.15$ RUB $\times(1+5\% / 4)=1407393$ RUB. Since the Deposit Insurance Agency compensates deposits up to 1400000 RUB, this amount will be paid to Irina Mikhailovna.
|
1400000
|
Algebra
|
MCQ
|
Yes
|
Yes
|
olympiads
| false
|
Problem 13. (8 points)
Dar'ya received a New Year's bonus of 60 thousand rubles, which she decided to save for a summer vacation. To prevent the money from losing value, the girl chose between two options for saving the money - to deposit the money at an interest rate of $8.6 \%$ annually for 6 months or to buy dollars and deposit them in a foreign currency deposit at an interest rate of $1.5 \%$ annually, also for 6 months. The bank's selling rate for dollars was 59.65 rubles, and the bank's buying rate for dollars was 56.65 rubles. Dar'ya chose the second option. After 6 months, Dar'ya withdrew the dollars from the foreign currency deposit, exchanged them for rubles, and went to a travel agency to purchase a tour package. The exchange rate on the day of the exchange operation was: the bank's selling rate for dollars was 58.95 rubles, and the bank's buying rate for dollars was 55.95 rubles. Determine the final financial result (profit or loss) from the operations performed by Dar'ya (rounding to the nearest whole number).
|
Answer: the loss incurred from the second option for placing funds is (rounded to the nearest whole number) $\underline{3300 \text{ RUB}}$
|
3300
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 6. (8 points)
Anna Ivanovna bought a car from her neighbor last November for 300,000 rubles with an engine power of 250 hp, and in May she purchased a used rowing catamaran for 6 rubles. The transport tax rate is set at 75 rubles per 1 hp. How much transport tax should Anna Ivanovna pay? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 3125.
## Comment:
|
Solution: transport tax $=250 \times 75 \times 2 / 12=3125$ rubles. A rowing catamaran is not a taxable object.
|
3125
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry competition, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
|
Solution: Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles.
|
39540
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
# Task 14. (1 point)
Calculate the amount of personal income tax (NDFL) paid by Sergey for the past year, if he is a Russian resident and during this period had a stable income of 30000 rub./month and a one-time vacation bonus of 20000 rub. In the past year, Sergey sold a car he inherited two years ago for 250000 rub. and bought a land plot for building a residential house for 300000 rub. Sergey claimed all the tax deductions he was entitled to. (Provide the answer without spaces and units of measurement.)
|
Answer: 10400.
## Comment:
Solution: tax base $=30000 \times 12+20000+250000=630000$ rubles. The amount of the tax deduction $=250000+300000=550000$ rubles. The amount of personal income tax $=13 \% \times(630000-$ $550000)=10400$ rubles.
|
10400
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 20. (6 points)
Ivan Sergeyevich decided to raise quails. In a year, he sold 100 kg of poultry meat at a price of 500 rubles per kg, and also 20000 eggs at a price of 50 rubles per dozen. The expenses for the year amounted to 100000 rubles. What profit did Ivan Sergeyevich receive for this year? (Provide the answer as a whole number, without spaces and units of measurement.)
|
Answer: 50000.
Comment:
Solution: revenue $=100 \times 500 + 50 \times 20000 / 10 = 150000$ rubles. Profit $=$ revenue costs $=150000-100000=50000$ rubles.
|
50000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 21. (8 points)
Dmitry's parents decided to buy him a laptop for his birthday. They calculated that they could save the required amount in two ways. In the first case, they need to save one-tenth of their salary for six months. In the second case, they need to save half of their salary for one month, and then deposit it in the bank for ten months at $3\%$ per month (calculated using simple interest). In the first case, the money will be just enough for the laptop, while in the second case, after buying the laptop, there will be some money left, which will be enough to buy a computer desk for 2875 rubles. What is the mother's salary, if the father's salary is 30% higher? (Provide the answer as a whole number, without spaces or units of measurement.)
|
Answer: 25000.
## Comment:
Solution:
Mom's salary is $x$, then dad's salary is $1.3x$. We set up the equation:
$(x + 1.3x) / 10 \times 6 = (x + 1.3x) / 2 \times (1 + 0.03 \times 10) - 2875$
$1.38x = 1.495x - 2875$
$x = 25000$
|
25000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. When insuring property, the insurance amount cannot exceed its actual value (insurance value) at the time of concluding the insurance contract.
Insurance tariff - the rate of the insurance premium or the insurance premium (insurance premium) expressed in rubles, payable per unit of the insurance amount, which is usually 100 rubles. In simpler terms, it is the percentage of the insurance value of the property established by the insurance company within the framework of laws concerning insurance activities.
An underwriter of an insurance company is a person authorized by the insurance company to analyze, accept for insurance (reinsurance), and reject various types of risks, as well as classify the selected risks to obtain the optimal insurance premium for them.
Title insurance (title insurance) is protection against the risk of losing the right of ownership of real estate as a result of the transaction for the acquisition of real estate being declared invalid or illegal. A transaction can be declared invalid by a court in the following situations:
- the presence of unaccounted property interests of minors, heirs, persons serving time in places of deprivation of liberty;
- the discovery of errors or fraud in the current or previous transactions with the real estate;
- in the case of the incapacity of one of the parties to the transaction, etc.
The appraised value of the property is the final result of the appraisal study, calculated by a licensed appraiser.
The cadastral value of housing is the value of the real estate object established through state valuation, primarily for the calculation of taxes.
## Instruction:
In the event of a discrepancy between the insurance amount and the actual value of the property, the following consequences are established:
a) if the insurance amount is set below the actual value of the property in the Insurance Contract, the Insurer, in the event of an insurance case, compensates the Insured (Beneficiary) for the damage incurred within the limits of the insurance amount established by the Insurance Contract without applying the ratio of the insurance amount and the actual value of the property (first risk payment condition).
b) if the insurance amount specified in the Insurance Contract exceeds the actual value of the property, the Contract is void to the extent that the insurance amount exceeds the actual value of the property.
Gross insurance tariffs (title insurance)
| Group of insured objects | Number of previous transactions with the real estate object | Annual tariff |
| :--- | :--- | :---: |
| Apartments and parts of apartments, consisting of one or several isolated rooms, non-residential buildings | Primary market (one transaction) | 0.18 |
| Secondary market (no more than 3 transactions) | One transaction preceding the current transaction | 0.20 |
| No more than 3 transactions*** | | |
**** in the case of more than 3 transactions over the past 3 years preceding the current transaction, the tariff is agreed individually.
## List of corrective coefficients (title insurance)
| Factors | Coefficient |
| :--- | :---: |
| The presence in the history of the real estate object of one or more of the following facts: - a time gap of less than one year between two preceding (or between the current and the preceding) transfers of ownership of the real estate object; - the presence of lease transactions; - transfer of rights as a result of inheritance or gift; - transactions carried out by persons on the basis of a power of attorney; - more than 2 transfers of ownership of the real estate object. | |
| - the last transaction preceding the current one was more than 3 years ago | 1.2 |
| Absence of one of the necessary documents | 0.8 |
## MOSCOW OLYMPIAD OF SCHOOL STUDENTS IN FINANCIAL LITERACY FINAL STAGE $10-11$ GRADES 2nd variant Answers and solutions
## Problem 1
You are an underwriter of an insurance company. According to the instructions for title insurance (Appendix to this problem), based on the data provided below, you need to set the tariff and determine the amount of the insurance premium. Be sure to keep the glossary in front of you.
Data:
Insured Ostrozhnov Konstantin Petrovich is purchasing an apartment on the secondary market with the participation of credit funds. The loan amount secured by real estate is 20 million rubles. Konstantin Petrovich has a good credit history and easily obtained approval for such an amount. The appraised value of the apartment is 14,500,000 rubles, and the cadastral value is 15,000,000 rubles. The bank-lender requires Ostrozhnov to purchase a title insurance policy to protect its financial interests.
The seller - Ivanov G.L., born in 1952, is the sole owner of the apartment, which he purchased more than 5 years ago under a construction investment agreement. He is married, has adult children, and no one is registered in the apartment. At the time of purchasing the apartment, he was married. He provided all necessary documents except for certificates from the psychiatric and narcological dispensaries. He does not drive, does not have a driver's license, and is not subject to military service.
|
# Solution:
In accordance with the instruction, the base rate is $0.2\%$ of the insurance amount, apply a reducing factor for the absence of a change in ownership over the past 3 years $(0.8)$ and an increasing factor for the absence of certificates from the PND and ND $(1.3)$.
In total: $0.2 * 0.8 * 1.3=0.208\%$
The loan amount exceeds both the appraised and the cadastral value. Therefore, we choose the maximum value to minimize the difference between the loan amount (bank requirement) and the allowable amount by law - 15000000 and take it as the insurance amount. $15000000 * 0.00208=31200$
Answer: rate $0.208\%$, SP=31,200 rubles.
Maximum 20 points
Evaluation criteria:
2 points for each correctly found coefficient and 4 points in total for the correct explanation of the use of these coefficients (if there is a correct explanation of only one coefficient, only 1 point out of 4 is given, and if there is a correct explanation of only two coefficients, only 3 points out of 4 are given). In total, 10 points for this part of the correct solution.
2 points for the correct rate.
A total of 8 points for the correct insurance premium amount.
- Participants receive only 2 points if the insurance premium is calculated without arithmetic errors based on an incorrect appraised value of the apartment.
- Participants receive only 4 points if the insurance premium is correctly determined but lacks a complete correct justification for its application.
## Task 2
The following information is available for JSC «Total Trade».
The annual revenue of JSC «Total Trade» is 2,500,000 rubles (after VAT payment). It is known that the ratio of operating expenses to revenue is 3/5 annually. The authorized capital of JSC «Total Trade» consists of the nominal value of the company's shares acquired by shareholders and is eight times the minimum authorized capital of a public company established under Russian law. The authorized capital is made up of 1600 ordinary shares, $35\%$ of which belong to the general director.
|
31200
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. (15 points) Purchase a meat grinder at "Technomarket" first, as it is more expensive, which means the largest bonuses can be earned on it, and then purchase a blender using the accumulated bonuses. In this case, she will spend
$$
\text { 4800+1500-4800*0.2=5340 rubles. }
$$
This is the most cost-effective way to make the purchases.
## Maximum 30 points
## Problem 5
Student Alexei informed his parents that he is already an adult and can manage his finances independently. His mother suggested he use a duplicate of the bank card linked to her account. To participate in a charitable New Year's program, Alexei wants to buy 40 "Joy" chocolate bars and donate them to a children's home. However, the bank, whose clients are Alexei's parents, has implemented a new system to protect against unauthorized card payments. The protection system analyzes the root mean square (RMS) value of the last 3 purchases (S) using the formula $S=\sqrt{\frac{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}}{3}}$, where $\mathrm{x}_{1}, \mathrm{x}_{2}$, and $\mathrm{x}_{3}$ are the amounts of the last 3 purchases, and compares the value of $S$ with the amount of the current purchase. If the current payment exceeds the value of $S$ by three times, the bank blocks the transaction and requires additional verification (e.g., a call from his mother to the call center). Over the past month, only payments for mobile phone service have been made, each for 300 rubles. How many minimum receipts should Alexei split the purchase into to buy all 40 "Joy" chocolate bars at a cost of 50 rubles each?
|
# Solution:
The root mean square value of the last purchases is $\sqrt{(300 * 300+300 * 300+300 * 300) / 3}=300$ rubles. Therefore, the permissible first purchase is no more than $300 * 3=900$ rubles, with which 18 chocolates can be bought. It remains to buy 22 chocolates for a total of $22 * 50=1100$ rubles.
For the second purchase, the average value considering the first purchase will be $\sqrt{(300 * 300+300 * 300+900 * 900) / 3}=\sqrt{330000}$, which is more than 500 rubles. Therefore, the second purchase can be for more than $500 * 3=1500$ rubles, which is sufficient to buy the remaining chocolates. The minimum number of receipts is 2.
## Maximum 15 points
15 points - a fully detailed correct solution and the correct answer.
10 points - correct calculation of the permissible amount of the first purchase, the correct number of chocolates in the first purchase, and the correct calculation of the amount for the second purchase.
5 points - correct calculation of the permissible amount of the first purchase.
## 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 of the types of property. Houses, apartments, and land plots are considered real estate.
## To insure a risk:
|
2
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In July, Volodya and Dima decided to start their own business producing non-carbonated bottled mineral water called "Dream," investing 1,500,000 rubles, and used these funds to purchase equipment for 500,000 rubles. The technical passport for this equipment indicates that the maximum production capacity is 100,000 bottles.
At the end of July, Dima and Volodya decided to launch a trial batch of water production and received 200 bottles, 5 of which were not full. In August, the equipment started operating at full capacity, and 15,000 bottles of water were produced. In the 20th of September, the equipment broke down and was idle for several days, but 12,300 bottles of water were produced over 20 days of this month.
In October, the friends decided to stop producing water, as it would not be in demand during the winter season, and decided to sell the equipment.
a) Determine the total depreciation of the equipment.
b) Determine the residual value of the equipment at the time of sale.
c) For what amount should the equipment be sold to achieve a profit of 10,000 rubles? (20 points)
|
# Answer:
a) The norm for 1 bottle of water = initial cost / maximum quantity: $500000 / 100000 = 5$ rubles;
- depreciation in July $5 \times 200 = 1000$ rubles;
- depreciation in March $15000 \times 5 = 75000$ rubles;
- depreciation in September $12300 \times 5 = 61500$ rubles, Total depreciation 137500 (6 points).
b) Residual value $-500000 - 137500 = 362500$ (4 points)
c) Sale amount - $362500 + 10000 = 372500$ (10 points)
- calculation errors were made, an incorrect numerical answer was given, a correct conclusion was made (2 points)
|
372500
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Newlyweds Alexander and Natalia successfully got jobs at an advertising company in April. With their earnings, they want to buy new phones next month: phone "A" for Alexander, which costs 57,000 rubles, and phone "B" for Natalia, costing 37,000 rubles. Will they be able to do this, given the following data?
- Alexander's salary is 125,000 rubles.
- Natalia's salary is 61,000 rubles.
- They spend 17,000 rubles per month on utilities, transportation, and daily needs.
- They spend 15,000 rubles per month on loan repayment and maintenance.
- Cultural entertainment expenses per month (1 theater visit - 5,000 rubles, and 1 movie visit - 1,000 rubles - ticket price per person).
- Savings for a trip to Crimea - they save 20,000 rubles per month, planning to go in August.
- Dining out expenses for both on weekdays - 1,500 rubles, and on weekends - 3,000 rubles (20 weekdays and 10 weekends in a month).
a) Determine the total expenses.
b) Determine the net income.
c) Will the couple be able to buy the phones? Provide a detailed answer. (20 points)
|
Answer:
a) Total expenses: $17000+15000+12000+20000+30000+30000=$ $=124000$ (4 points);
b) Net income: $(125000+61000) \times 13 \% = 24180.186000 - 24180=$ $=161820$ (6 points);
c) Remaining funds: $161820-124000=37820$. The phone can only be bought for Natalia, and to buy a phone for Alexander, it is necessary to forego the trip to Crimea or postpone the trip to September and catch the off-peak season (10 points).
- calculation errors were made, an incorrect numerical answer was given, a correct conclusion was drawn (2 points)
|
37820
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let's say you have two bank cards for making purchases: a debit card and a credit card. Today, you decided to buy airline tickets for 20,000 rubles. If you pay for the purchase with a credit card (the credit limit allows it), you will have to return the money to the bank in $\mathrm{N}$ days to avoid going beyond the grace period, during which you can repay the credit for free. In this case, the bank will pay a cashback of $0.5 \%$ of the purchase amount after 1 month. If you pay for the purchase with a debit card (there is more than enough money on the card), you will receive a cashback of $1 \%$ of the purchase amount after 1 month. It is known that the annual interest rate on the average monthly balance of funds on the debit card is $6 \%$ per year (for simplicity, assume that each month has 30 days, interest on the card is paid at the end of each month, and interest earned on the balance is not capitalized). Determine the smallest number of days $\mathrm{N}$, under equal conditions, for which it is more profitable to pay for this purchase of airline tickets with a credit card. (15 points)
|
# Solution:
When paying by credit card, the amount of 20,000 rubles will be on your debit card for $\mathrm{N}$ days, which will earn you $\frac{6 \mathrm{~N}}{100 \cdot 12 \cdot 30} \cdot 20000$ rubles in interest on the remaining funds.
You will also receive $20000 \times 0.005 = 100$ rubles in cashback.
When paying by debit card, you will receive a cashback of 200 rubles after 1 month.
For it to be more profitable to pay for this purchase with a debit card, the inequality $\frac{6 N}{100 \cdot 12 \cdot 30} \cdot 20000 + 100 > 200$ must be satisfied.
This inequality holds if $\mathrm{N} > 30$. Therefore, the minimum number of days in the grace period, during which it is more profitable to pay for this purchase with a credit card, is 31 days.
Moscow Schoolchildren's Olympiad in Financial Literacy. 2017-2018 academic year. Final stage. $10-11$ grades. Variant 1
## Criteria:
Correctly accounted for the fact that interest is accrued on the average monthly balance of funds - up to 4 points.
Correctly calculated the cashback when paying with a credit card - up to 3 points.
Correctly calculated the cashback when paying with a debit card - up to 3 points.
Correctly formulated the inequality to find the required number of days - up to 3 points.
Obtained a correct and justified answer - up to 2 points
|
31
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Two friends, Arthur and Timur, with the support of their parents, decided to open several massage salons in Moscow. For this purpose, a business plan was drawn up, the economic indicators of which are presented below.
- Form of ownership - LLC
- Number of employees - no more than 50 people
- Planned revenues for the year:
- with their own funds, they purchased property and contributed it as a capital contribution, its market value is 7,000,000 rubles;
- planned revenue 120,000,000 rubles;
- advances received from a large company that will sign a contract for massage services for its employees 30,000,000 rubles.
- Planned monthly expenses:
- rental of massage rooms including utilities 770,000 rubles;
- expenses for the purchase of various oils 1,450,000 rubles;
- average salary for the entire staff 4,600,000 rubles;
- social insurance contributions paid on salaries 1,380,000 rubles;
- accounting services for all salons 340,000 rubles;
- advertising campaign 1,800,000 rubles;
- expenses for independent qualification assessment and retraining of personnel 800,000 rubles;
- other expenses (KKT maintenance, publication of reports in the media, office supplies) 650,000 rubles.
Which tax object would you recommend to the friends under the USN regime?
a) Revenues at a rate of $6 \%$;
b) Revenues minus expenses, at a rate of $15 \%$, or a minimum tax of $1 \%$ of revenues
c) Make the calculation, fill in the tables, and justify your answer. (20 points)
|
# Solution:
Reference information: Chapter 26.2, Part 2 of the Tax Code of the Russian Federation.
- Criteria applicable under the simplified tax system (STS) - Article 346.13;
- Taxable object - Article 346.14;
- Determination of income - Article 346.15;
- Determination of expenses - Article 346.16;
- Recognition of income and expenses - Article 346.17;
- Tax base and minimum tax - Article 346.18;
- Procedure for calculating the tax - Article 346.21
a) Income at a rate of 6%:
$120000000 + 30000000 = 150000000$ (2 points);
$150000000 \times 6\% = 9000000$ (2 points);
Reduction of the amount by 50% of the paid insurance contributions:
$1380000 \times 12$ months / $2 = 8280000$
To be paid to the budget at the end of the period:
$9000000 - 8280000 = 720000$ (4 points).
b) Income minus expenses at 15% or minimum tax at 1%:
Income 150000000
Expenses (770000 + 1450000 + 4600000 + 1380000 + 340000 + 1800000 + 800000 + 650000) $\times 12$ months = 141480000 (4 points);
Tax base: 150000000 - 141480000 = 8520000 (2 points);
Tax amount 8520000 * 0.15 = 1278000; (1 point);
Minimum tax amount - 150000000 * 1% = 1500000; (1 point);
Tax amount to be paid - 1500000 - 1278000 = 222000 (2 points).
c) Based on these indicators, it is more advantageous to use the income minus expenses method, as it is necessary to compare the calculated tax amount under options a) and b), using the cash method of determining the tax base (Article 346.17) (2 points for a justified answer).
| Indicators | Calculation for 6% |
| :--- | :--- |
| Income | 150000000 |
| Expenses | |
| Special calculation | 8280000 |
| Tax base | 150000000 |
| Tax amount | 9000000 |
| Tax amount to be paid to the budget | 720000 |
| Indicators | Calculation for 15% |
| :--- | :--- |
| Income | 150000000 |
| Expenses | 141480000 |
| Special calculation | |
| Tax base | 8520000 |
| Tax amount | 1278000 |
| Tax amount to be paid to the budget | 222000 |
| Indicators | Calculation for 1% |
| Income | 150000000 |
| Expenses | |
| Special calculation | |
| Tax base | 150000000 |
| Tax amount | 1500000 |
| Tax amount to be paid to the budget | |
|
222000
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Let's say you have two bank cards for making purchases: a debit card and a credit card. Today, at the beginning of the month, you decided to buy airline tickets for 12,000 rubles. If you pay for the purchase with a credit card (the credit limit allows it), you will have to return the money to the bank in $\mathrm{N}$ days to avoid going beyond the grace period, during which you can repay the credit for free. In this case, the bank will pay a cashback of $1 \%$ of the purchase amount after 1 month. If you pay for the purchase with a debit card (there is more than enough money on the card), you will receive a cashback of $2 \%$ of the purchase amount after 1 month. It is known that the annual interest rate on the average monthly balance of funds on the debit card is $6 \%$ per year (for simplicity, assume that each month has 30 days, interest on the card is paid at the end of each month, and interest earned on the balance is not capitalized). Determine the maximum number of days $\mathrm{N}$, under equal conditions, for which it is more profitable to pay for this purchase of airline tickets with a debit card. (15 points)
|
# Solution:
When paying by credit card, the amount of 12,000 rubles will be on your debit card for $\mathrm{N}$ days, which will earn you $\frac{6 \mathrm{~N}}{100 \cdot 12 \cdot 30} \cdot 12000$ rubles in interest on the remaining funds. Additionally, you will receive $12000 \times 0.01 = 120$ rubles in cashback.
When paying by debit card, you will receive a cashback of $12000 \times 0.02 = 240$ rubles after 1 month.
For it to be more profitable to pay for this purchase with a debit card, the inequality $\frac{6 N}{100 \times 2 \times 0} \geqslant 2000 + 120 < 240$ must be satisfied. This is true if N is 60. Therefore, the maximum number of days in the grace period, during which it is more profitable to pay for this purchase with a debit card, is 59 days.
Moscow Schoolchildren's Olympiad in Financial Literacy. 2017-2018 academic year.
Final stage. 10-11 grades. Variant 2
## Grading criteria:
Correctly accounted for the fact that interest is accrued on the average monthly balance of funds - up to 4 points.
Correctly calculated the cashback when paying with a credit card - up to 3 points.
Correctly calculated the cashback when paying with a debit card - up to 3 points.
Correctly formulated the inequality to find the required number of days - up to 3 points.
Obtained a correct and justified answer - up to 2 points
|
59
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Twin brothers, Anton Sergeyevich, a civil aviation pilot by profession, and Mikhail Sergeyevich, a neurologist, born on 05.06.1977, decided to go on vacation together and purchase a life and health insurance policy for 2,000,000 rubles. Anton Sergeyevich and Mikhail Sergeyevich had the same height - 187 cm and weight - 98 kg. Mikhail Sergeyevich decided to preliminarily calculate the cost of the policy on the insurance company's website. He looked at the rates and, without consulting and filling out the application, concluded that the cost of the insurance was the same for both brothers. Find the difference in the cost of the brothers' policies using the data in the appendix, given that insurers apply increasing coefficients for the insured person's belonging to a high-risk profession and excess weight (increased risk of cardiovascular diseases). (24 points)
Appendix: Body Mass Index (BMI) formula: BMI = weight (kg) / height^2 (m) (weight in kilograms must be divided by the square of the height, expressed in meters)
Increasing coefficients for BMI:
| BMI Range | 25-29 | 30-35 | 36-40 |
| :--- | :---: | :---: | :---: |
| Increasing Coefficient | 1 | 1.2 | 1.7 |
Increasing coefficient for occupation class:
pilot of civil aviation 1.5; neurologist 1.02
Basic rates:
| | basic rate age \ gender | | | | basic rate age \ gender | |
| :---: | :---: | :---: | :---: | :---: | :---: | :---: |
| Age | male | female | | Age | male | female |
| **31** | 0.21 | 0.12 | | **41** | 0.33 | 0.22 |
| **32** | 0.22 | 0.12 | | **42** | 0.35 | 0.24 |
| **33** | 0.22 | 0.14 | | **43** | 0.38 | 0.25 |
| **34** | 0.22 | 0.14 | | **44** | 0.41 | 0.25 |
| **35** | 0.24 | 0.15 | | **45** | 0.42 | 0.28 |
| **36** | 0.25 | 0.17 | | **46** | 0.5 | 0.32 |
| **37** | 0.26 | 0.18 | | **47** | 0.55 | 0.32 |
| **38** | 0.28 | 0.18 | | **48** | 0.61 | 0.4 |
| **39** | 0.3 | 0.2 | | **49** | 0.66 | 0.4 |
| **40** | 0.32 | 0.21 | | **50** | 0.74 | 0.43 |
|
# Solution:
Let's calculate the age. Both men are 40 years old. The base rate is $0.32\%$. By occupation class: Mikhail (doctor) $0.32\% \times 1.02 = 0.3264\%$.
Anton (pilot) $0.32\% \times 1.5 = 0.48\%$.
Calculate the BMI $= 98 / 1.87^2 = 98 / 3.4969 = $ (approximately) 28.025.
Find the increasing coefficient from the table. It is 1 (i.e., there is no increasing coefficient).
Apply the final rate: $2000000 \times 0.48\% = 9600$ (Anton) and $2000000 \times 0.3264\% = 6528$.
Answer: The difference in the cost of the insurance policies is 3072 rubles.
## Grading Criteria:
1) All calculation steps are correct, and the correct answer is obtained - 24 points
2) The logic of the calculation is correct (the increasing coefficients are correctly applied), but there is an arithmetic error in the final summation or percentage calculation - up to 20 points
3) The logic of the calculation is correct (the increasing coefficients are correctly applied), but the cost of the policy was calculated without considering that the base rate is expressed as a percentage - up to 12 points
4) The BMI scale is incorrectly determined, but all other calculations are correct - up to 6 points
5) The logic of applying the rates is violated, but the increasing coefficients are correctly found - up to 3 points
6) The logic of applying the rates is violated, and the data from the appendix is incorrectly found and other issues - **0** points
|
3072
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Anna and Ekaterina have opened a cosmetic salon in New Moscow. The enterprise applies the general taxation system. Ekaterina attended a seminar on taxation and learned about the Simplified System of Taxation (USNO). To avoid changing the document flow and control over financial and economic operations, the friends decided to compare the two taxation systems and choose the best one for the new year.
Financial indicators of the preceding year:
- Annual income of the cosmetic salon - 4,500,000 rubles;
- Monthly expenses of the cosmetic salon:
- Rent of premises - 60,000 rubles;
- Purchase of cosmetic products - 40,000 rubles;
- Wages of all employees - 120,000 rubles;
- Social insurance contributions on wages - 36,000 rubles;
- Advertising expenses - 15,000 rubles;
- Expenses for independent qualification assessment and retraining of personnel - 12,000 rubles;
- Other expenses (KKT maintenance, publication of reports in the media, expenses for office supplies) - 20,000 rubles;
- On average, no more than 45% of all expenses were paid in the tax period.
Help the girls with the following calculations:
a) Tax amount under the general taxation system (accrual method) - rate 20%;
b) Tax amount under the USNO - Income with a rate of 6%;
c) Income minus expenses with a rate of 15%, or a minimum tax of 1% of income;
d) Fill in the tables.
What recommendations would you give the girls? (20 points)
#
|
# Solution:
Reference information: Chapter 26.2, Chapter 25, Part 2 of the Tax Code of the Russian Federation.
- Criteria applicable under the simplified tax system (STS) - Article 346.13;
- Taxable object - Article 346.14;
- Determination of income - Article 346.15;
- Determination of expenses - Article 346.16;
- Recognition of income and expenses - Article 346.17;
- Tax base and minimum tax - Article 346.18;
- Calculation of tax - Article 346.21;
- Recognition of income under the accrual method - Article 271;
- Recognition of expenses under the accrual method - Article 272;
- Recognition of income and expenses under the cash method - Article 273.
a) Income at a rate of 6%:
$4500000 \times 6\% = 270000$; (2 points)
Reduction of the amount by 50% of the paid insurance contributions:
$36000 \times 12$ months / $2 = 216000$; paid - $45\% - 97200$;
By the end of the period to be paid to the budget:
$270000 - 97200 = 172800$; (4 points).
b) Income minus expenses 15% or minimum tax 1%:
Income 4500000;
Expenses $(60000 + 40000 + 120000 + 36000 + 15000 + 12000 + 20000) \times 12$ months = $3636000$; paid $45\% - 1636200$ (4 points);
Tax base: $4500000 - 1636200 = 2863800$ (2 points);
Tax amount: $2863800 \times 0.15 = 429570$; (2 points)
Minimum tax amount: $4500000 \times 1\% = 45000$; (1 point)
c) General tax system (GTS), accrual method, rate 20%:
Income - 4500000
Expenses - $(60000 + 40000 + 120000 + 36000 + 15000 + 12000 + 20000) \times 12$ months = $3636000$
Tax base: $4500000 - 3636000 = 864000$; (2 points)
Tax amount: $864000 \times 0.2 = 172800$ (1 point);
d) Comparison of tax amounts:
$6\% - 172800$
$15\% - 429570$
$20\% - 172800$.
Conclusion: It is possible to remain on the GTS or switch to the STS at 6%, if the revenue does not increase and the insurance contributions are fully paid. (2 points for a justified conclusion)
| Indicators | Calculation for 6% |
| :--- | :--- |
| Income | 4500000 |
| Expenses | |
| Special calculation | 97200 |
| Tax base | 4500000 |
| Tax amount | 270000 |
| Tax amount to be paid to the budget | 172800 |
| Indicators | Calculation for 15% |
| :--- | :--- |
| Income | 4500000 |
| Expenses | 3636000 |
| Special calculation | 163600 |
| Tax base | 2863800 |
| Tax amount | 429570 |
| Tax amount to be paid to the budget | 429570 |
Moscow School Olympiad in Financial Literacy. 2017-2018 academic year.
Final stage. 10-11 grades. Variant 2
| Indicators | Calculation for 1% |
| :--- | :--- |
| Income | 4500000 |
| Expenses | |
| Special calculation | |
| Tax base | 4500000 |
| Tax amount | 45000 |
| Tax amount to be paid to the budget | |
| Indicators | Calculation for 20% |
| :--- | :--- |
| Income | 4500000 |
| Expenses | 3636000 |
| Special calculation | |
| Tax base | 864000 |
| Tax amount | |
| Tax amount to be paid to the budget | 172800 |
|
172800
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Problem 7. (8 points)
The earned salary of the citizen was 23,000 rubles per month from January to June inclusive, and 25,000 rubles from July to December. In August, the citizen, participating in a poetry contest, won a prize and was awarded an e-book worth 10,000 rubles. What amount of personal income tax needs to be paid to the budget? (Provide the answer as a whole number, without spaces or units of measurement.)
Answer: 39540.
## Comment:
|
Solution: Personal Income Tax from salary $=(23000 \times 6+25000 \times 6) \times 13\%=37440$ rubles.
Personal Income Tax from winnings $=(10000-4000) \times 35\%=2100$ rubles.
Total Personal Income Tax = 37440 rubles +2100 rubles$=39540$ rubles.
|
39540
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
Task 4.
Every day after lunch, 7 half-eaten pieces of white bread are left on the desks of the second-grade students. If these pieces are put together, they make up half a loaf of bread. How many loaves of bread will the second-grade students save in 20 days if they do not leave these pieces? How much money will the school save on this in 20 days and 60 days if a loaf of bread costs 35 rubles? (Answer)
Suggest two topics for extracurricular activities for this class that involve fostering concepts of economic actions and thrift among students. (Answer)
|
Solution: 1. 0.5 (1/2) * 20 = 10 (loaves); 10 * 35 = 350 rubles; 2. 0.5 (1/2) * 60 = 30 (loaves); 30 * 35 = 1050 rubles.
Themes for extracurricular activities: "Young Economist," "Bread is the Head of Everything," "Saving and Frugality in Our School Canteen," "Journey to the School of the Frugal," "Frugality - the Main Source of Wealth," etc.
## Criteria:
20 points - for the correct answer with justifications, correct calculations, and 2 proposed themes for extracurricular activities
15 points - correct answer with justifications, 1 proposed theme for an extracurricular activity or general thematic areas.
10 points - for the correct answer without justifications, with calculations and one proposed extracurricular activity.
5 points - for the correct answer, but with calculation errors and one proposed extracurricular activity
3 points - incorrect answer and calculation, 2 proposed extracurricular activities
2 points - incorrect answer and calculation, 1 proposed extracurricular activity.
|
350
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
|
Solution. Let the first crow eat $x$ grams of cheese. Then the fox received $100-x$ grams of cheese from the first crow. The second crow ate $\frac{x}{2}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40$. The fox ate 240 grams.
|
240
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
|
Solution. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will also meet for the first time at the 53rd minute.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
53
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many solutions in natural numbers does the equation $(a+1)(b+1)(c+1)=2 a b c$ have?
|
Solution. Rewrite the equation as $(1+1 / a)(1+1 / b)(1+1 / c)=2$. Due to symmetry, it is sufficient to find all solutions with $a \leqslant b \leqslant c$. Then $(1+1 / a)^{3} \geqslant 2$, which means $a \leqslant(\sqrt[3]{2}-1)^{-1}<4$ and $a \in\{1,2,3\}$. In the case $a=1$, the inequality $2(1+1 / b)^{2} \geqslant 2$ holds, so there are no solutions. If $a=2$, then $\frac{3}{2}(1+1 / b)^{2} \geqslant 2$, which means $2 \leqslant b \leqslant\left(\frac{2 \sqrt{3}}{3}-1\right)^{-1}<7$. In this case, there are 3 solutions $(a, b, c)=(2,4,15),(2,5,9),(2,6,7)$ (for $b=2$ and $b=3$, the equation for $c$ has no solutions in natural numbers). Finally, if $a=3$, then $\frac{4}{3}(1+1 / b)^{2} \geqslant 2$, which means $3 \leqslant b \leqslant\left(\sqrt{\frac{3}{2}}-1\right)^{-1}<5$. This gives 2 more solutions $(a, b, c)=(3,3,8),(3,4,5)$. Taking into account permutations, there are a total of 27 solutions.
Criteria. If only part of the solutions is found, no more than 2 points are given.
|
27
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. The park is a $10 \times 10$ grid of cells. A lamp can be placed in any cell (but no more than one lamp per cell).
a) The park is called illuminated if, no matter which cell a visitor is in, there is a $3 \times 3$ square of 9 cells that contains both the visitor and at least one lamp. What is the minimum number of lamps in an illuminated park?
b) The park is called reliably illuminated if it remains illuminated even after the failure of any one lamp. What is the minimum number of lamps in a reliably illuminated park?
|
Solution. a) 4. Divide the park into 4 quarters (squares $5 \times 5$), then there must be at least one lamp in each quarter (to illuminate, for example, the corner cells). By placing one lamp in the center of each quarter, we get an example.
b) 10.
Estimate. In each corner square $3 \times 3$ there must be at least two lamps (to illuminate the corner cell). Temporarily leave only these 8 lamps. Each of them illuminates only within its own quarter, and if the lamp in the center of the quarter breaks (or if it is absent), then a five-cell strip inside this quarter, adjacent to another quarter, will definitely not be illuminated. Note that the union of two such strips for opposite quarters cannot be illuminated by one lamp in any case, so at least two more lamps are needed.
Example:
Criteria. In part a) 1 point is given for the estimate and 1 point for the example. In part b) 3 points are given for the estimate (1 point if it is proven that 8 lamps are insufficient), and 2 points for the example.
International Mathematical Olympiad
«Formula of Unity» / «The Third Millennium»
Year 2022/2023. Qualifying round
## Problems for grade R9
Each task is assessed at 7 points. Some problems have their own criteria (printed in gray).
|
4
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Marina needs to buy a notebook, a pen, a ruler, and a pencil to participate in the Olympiad. If she buys a notebook, a pencil, and a ruler, she will spend 47 tugriks. If she buys a notebook, a ruler, and a pen, she will spend 58 tugriks. If she buys only a pen and a pencil, she will spend 15 tugriks. How much money will she need for the entire set?
|
Solution. If Marina buys all three sets from the condition at once, she will spend $47+58+$ $15=120$ tugriks, and she will buy each item twice, so the full set of school supplies costs $120 / 2=60$ tugriks.
Criteria. Only the answer without explanation - 1 point. If in the solution they try to determine the cost of the pen and pencil (although it is not stated in the condition that the cost must be an integer) - 0 points.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the perimeter of the original sheet.
|
Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus, the perimeter is: $(28+18) \cdot 2=92$.
Criteria. An answer without justification or found by trial and error - 0 points.
|
92
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 25?
|
Solution. The number must be divisible by 25, so “$\lambda$A” equals 25, 50, or 75 (00 cannot be, as the letters are different). If “LA” equals 50, then for the other letters (“G”, “V”, “T”, “E”, “M”) there are $A_{8}^{5}$ options; otherwise, for the other letters there are $7 \cdot A_{7}^{4}$ options. In total, $8!/ 6 + 2 \cdot 7 \cdot 7!/ 6 = 18480$ ways.
Criteria. It is explicitly stated that the option 00 is not suitable for the letters “LA”, and all 3 cases are demonstrated - 3 points. If the 2 variants are further analyzed without errors, then another 4 points. If the answer is given as an expression not fully calculated - do not deduct points.
|
18480
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
|
Solution. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will also meet for the first time at the 53rd minute.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution uses that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
International Mathematical Olympiad
«Formula of Unity» / «The Third Millennium»
Year 2022/2023. Qualifying round
## Problems for grade R6
Each task is assessed at 7 points. Some problems have their own criteria (printed in gray).
|
53
|
Other
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. A few years ago, in the computer game "Minecraft," there were 11 different pictures (see the figure): one horizontal with dimensions $2 \times 1$, and two each with dimensions $1 \times 1$, $1 \times 2$ (vertical), $2 \times 2$, $4 \times 3$ (horizontal), and $4 \times 4$. In how many ways can all 11 pictures be placed on a rectangular wall that is 12 blocks long and 6 blocks high? The pictures should not overlap; they cannot be rotated.
(P. D. Mulyenko)
|
Answer: 16896.
Solution. We will say that two pictures are in different columns if no block of the first picture is in the same column as any block of the second. It is clear that the $4 \times 4$ pictures are in different columns from each other and from the $4 \times 3$ pictures in any arrangement. Thus, the $4 \times 3$ pictures will necessarily be strictly one below the other. Both $4 \times 4$ pictures are pressed against the floor or ceiling, as at least 6 columns must have 2 free adjacent cells for the remaining pictures of height 2. There are $3 \cdot 2^{4}=48$ ways to place the 4-wide pictures (3 ways to choose the column with the $3 \times 3$ picture, $2^{2}$ ways to choose "floor/ceiling" in the other columns, 2 ways to permute the $4 \times 4$ pictures, and another 2 ways to permute the $4 \times 3$ pictures). Out of these, in 16 cases, a $8 \times 2$ empty area remains (there are 4 "degrees of freedom" with 2 options each, as shown by the colored arrows in the middle picture). In the remaining 32 cases, two separate $4 \times 2$ areas remain.
Note that in any case, the $1 \times 1$ pictures together with the $2 \times 1$ picture will form a $2 \times 2$ square, so they can be replaced by one glued picture (and the answer for the new set of pictures should be multiplied by 4 ways to unglue it).
If an $8 \times 2$ area remains, it needs to be filled with five vertical blocks in some order, for which there are $5!=120$ ways.
Consider the case where two $4 \times 2$ free areas remain. Then one area must be divided into two $2 \times 2$ pictures, and the other into one $2 \times 2$ picture and two $1 \times 2$ pictures. There are two ways to choose which area gets which, then 3 ways to choose which $2 \times 2$ picture (including the composite one) will be on the second area; after that, $2!$ ways to order the blocks for the first area and $3!$ for the second. In total, $2 \cdot 3 \cdot 6 \cdot 2=72$ ways.
In total, $4 \cdot(32 \cdot 72+16 \cdot 120)=16896$ variants.
Criteria. It is obtained that the three smallest pictures (two $1 \times 1$ and one $1 \times 2$) can only be placed together as a $2 \times 2$ square - 1 point.
It is obtained that the 4-wide pictures can be arranged in 48 ways - another 2 points (with an error in the calculation, ±1 point).
It is obtained that in the case of two "windows" $4 \times 2$, there are 72 ways to arrange the small pictures - another 2 points (with an error in the calculation, ±1 point).
It is obtained that in the case of a single $8 \times 2$ space, there are 120 ways to arrange the small pictures - another 2 points (with an error in the calculation, ±1 point).
|
16896
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. In the Thrice-Tenth Kingdom, there are 17 islands, each inhabited by 119 people. The inhabitants of the kingdom are divided into two castes: knights, who always tell the truth, and liars, who always lie. During the census, each person was first asked: "Excluding you, are there an equal number of knights and liars on your island?". It turned out that on 7 islands, everyone answered "Yes", while on the others, everyone answered "No". Then each person was asked: "Is it true that, including you, there are fewer people of your caste than half the island's population?". This time, on some 7 islands, everyone answered "No", while on the others, everyone answered "Yes". How many liars are there in the kingdom?
(P. D. Mulyenko)
|
Answer: 1013.
## Solution.
1) Consider the first question. A "yes" answer would be given by either a knight on an island with exactly 60 knights, or a liar if the number of knights is different. A "no" answer would be given by either a liar on an island with 59 knights, or a knight if the number of knights is different. Therefore, on the 7 islands where everyone answered "yes" to the first question, the number of knights is either 60 or 0; on the other 10 islands, the number of knights is either 59 or 119.
2) For the second question, regardless of who answers, a "yes" answer indicates that the number of knights is less than half, and a "no" answer indicates that the number of liars is less than half. Therefore, on the 7 islands where everyone answered "no" to the second question, the number of knights is at least 60 (either 60 or 119); on the other 10 islands, the number of knights is no more than 59 (either 59 or 0).
3) Let $x$ be the number of islands with 60 knights, and $y$ be the number of islands with 59 knights; then (see point 1) on 7 - $x$ islands there are 0 knights, and on 10 - $y$ islands there are 119 knights. From point 2, we get: $x + (10 - y) = 7, y + (7 - x) = 10$. Both equations are equivalent to the equation $y - x = 3$.
4) Calculate the total number of knights:
$60 \cdot x + 59 \cdot y + (7 - x) \cdot 0 + (10 - y) \cdot 119 = 60x + 59(x + 3) + 119(7 - x) = 59 \cdot 3 + 119 \cdot 7 = 1010$.
Therefore, the number of liars is 1013.
Criteria. Correctly listed all possible numbers of knights and liars on each type of island - 2 points.
The problem is solved for the case where the 7 islands in the first and second questions are the same - 2 points. The problem is solved correctly, but the answer indicates the number of knights, not liars - 6 points.
|
1013
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Find all real solutions to the system of equations
$$
\left\{\begin{array}{l}
\sqrt{x-997}+\sqrt{y-932}+\sqrt{z-796}=100 \\
\sqrt{x-1237}+\sqrt{y-1121}+\sqrt{3045-z}=90 \\
\sqrt{x-1621}+\sqrt{2805-y}+\sqrt{z-997}=80 \\
\sqrt{2102-x}+\sqrt{y-1237}+\sqrt{z-932}=70
\end{array}\right.
$$
(L. S. Korechkova, A. A. Tessler)
|
Answer: $x=y=z=2021$.
Solution. First, we prove that the solution is unique if it exists. Let $\left(x_{1}, y_{1}, z_{1}\right)$ and $\left(x_{2}, y_{2}, z_{2}\right)$ be two different solutions and, without loss of generality, $x_{1} \leqslant x_{2}$. Then there are four possible cases: $y_{1} \leqslant y_{2}$ and $z_{1} \leqslant z_{2}$ (with at least one of the three inequalities being strict); $y_{1} \leqslant y_{2}$ and $z_{1}>z_{2} ; y_{1}>y_{2}$ and $z_{1} \leqslant z_{2} ; y_{1}>y_{2}$ and $z_{1}>z_{2}$. Each case contradicts the corresponding equation due to monotonicity.
The solution itself can be found by assuming it is an integer and all roots are extracted as integers. For example, let $x=1621+a^{2}=1237+b^{2}$ for integers $a, b \geqslant 0$, then $(b+a)(b-a)=384$; if we try all possible values of $a$, it becomes clear that only $a=20$ allows the other roots to be extracted. Additionally, one can notice that some roots look similar (e.g., $\sqrt{x-997}$ and $\sqrt{z-997}$), so it is convenient to look for a solution where $x=y=z$. After this, the numbers 1121, 1621, 2102 can help guess the answer.
Criteria. 2 points for finding the solution (verifying the calculations that it fits or somehow motivating its discovery is not required); 5 points for proving uniqueness.
|
2021
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In each cell of a $10 \times 10$ table, a natural number was written. Then, each cell was shaded if the number written in it was less than one of its neighbors but greater than another neighbor. (Two numbers are called neighbors if they are in cells sharing a common side.) As a result, only two cells remained unshaded. What is the minimum possible sum of the numbers in these two cells?
|
Solution. Answer: 20. This value is achieved if the unshaded cells are in opposite corners and contain the numbers 1 and 19.
Evaluation.
1) The cells that contain the minimum and maximum numbers are definitely not shaded. This means that the minimum and maximum each appear exactly once, and they are in the unshaded cells. Let their values be $m$ and $M$ respectively.
2) If a cell contains the number $n$, then the length of the path from it to the minimum cell is no more than $n-1$. Indeed, from a cell, you can move to an adjacent cell with a number at least 1 less, and no later than after $n-1$ steps, we will definitely reach the cell with the minimum number.
Similarly, this is true "from the other side": if the number in a cell is less than the maximum by $k$, then the distance from that cell to the maximum is no more than $k$ steps.
3) We will prove that there is a corner cell for which the sum of the distances to the two unshaded cells is at least 18. (We will call the distance between cells the length of the shortest path between them.)
Indeed, let $(a, b)$ and $(c, d)$ be the coordinates of the unshaded cells. The sum of the distances from the first unshaded cell to the four corner cells is $(b-1+a-1)+(b-1+10-a)+(10-b+a-1)+(10-b+10-a)=36$. The same is true for the second unshaded cell, so the sum of all 8 distances from the corner cells to the unshaded cells is 72. Therefore, the sum of the distances from the unshaded cells to some one corner is at least 18.
4) Let the number in the corner cell found in point 3 be $x$. Then $M-m=(M-x)+(x-m)$ is no less than the sum of the lengths of the paths from the corner cell to the unshaded cells, which is at least 18. Since $m \geqslant 1$, then $M \geqslant 19$, from which $M+m \geqslant 20$.
|
20
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. How many five-digit numbers are divisible by their last digit?
将上面的文本翻译成英文，请保留源文本的换行和格式，直接输出翻译结果。
|
2. The total number of five-digit numbers is $-99999-9999=90000$, and among them, there are an equal number of numbers ending in $0,1, \ldots, 9$, that is, 9000 numbers of each type.
Let $n_{i}$, where $i=0,1, \ldots, 9$, be the number of numbers ending in $i$ that are divisible by $i$. Then
$n_{0}=0$ (a number cannot be divisible by 0);
$n_{1}=9000$ (all numbers are divisible by 1);
$n_{2}=9000$ and $n_{5}=9000$ by the divisibility rules for 2 and 5.
Let's find $n_{3}$. The number $\overline{a b c d 3} \vdots 3$ if $\overline{a b c d 0} \vdots 3$, i.e., $10 \cdot \overline{a b c d}: 3$, which is equivalent to $\overline{a b c d}: 3$ (since 3 and 10 are coprime). Thus, we need to find the number of four-digit numbers divisible by 3. Since there are 9000 four-digit numbers and among any three consecutive numbers, exactly one is divisible by 3, then $n_{3}=9000: 3=3000$.
Similarly, $n_{9}$ is the number of four-digit numbers divisible by 9, i.e., $n_{9}=9000: 9=1000$. Similarly, $n_{7}$ is the number of four-digit numbers divisible by 7. The smallest of them is $7 \cdot 143=1001$, and the largest is $7 \cdot 1428=9996$, so $n_{7}=1428-143+1=1286$.
Now let's find $n_{4}$. The number $\overline{a b c d 4} \vdots 4 \Leftrightarrow 4+10 \cdot \overline{a b c d} \vdots 4 \Leftrightarrow 10 \cdot \overline{a b c d} \vdots 4 \Leftrightarrow 5 \cdot \overline{a b c d} \vdots 2 \Leftrightarrow \overline{a b c d} \vdots 2$. Therefore, $n_{4}$ is the number of even four-digit numbers, which is half of the total number of four-digit numbers. $n_{4}=9000 / 2=4500$.
Similarly, $\overline{a b c d 6}: 6 \Leftrightarrow \overline{a b c d 0} \vdots 6 \Leftrightarrow 10 \cdot \overline{a b c d}: 6 \Leftrightarrow 5 \cdot \overline{a b c d} \vdots 3 \Leftrightarrow \overline{a b c d} \vdots 3$, so $n_{6}=9000$ : $3=3000$.
$\overline{a b c d 8}: 8 \Leftrightarrow \overline{a b c d 0}: 8 \Leftrightarrow 10 \cdot \overline{a b c d}: 8 \Leftrightarrow 5 \cdot \overline{a b c d}: 4 \Leftrightarrow \overline{a b c d}: 4$, and $n_{8}=9000: 4=2250$.
The total number of the numbers of interest is
$n_{0}+n_{1}+n_{2}+\ldots+n_{9}=0+9000+9000+3000+4500+9000+3000+1286+2250+1000=42036$.
|
42036
|
Number Theory
|
proof
|
Yes
|
Yes
|
olympiads
| false
|
1. In one move, you can either add one of its digits to the number or subtract one of its digits from the number (for example, from the number 142 you can get $142+2=144, 142-4=138$ and several other numbers).
a) Can you get the number 2021 from the number 2020 in several moves?
b) Can you get the number 2021 from the number 1000 in several moves?
|
Solution. a) Yes, for example, like this: $20 \mathbf{2 0} \rightarrow 20 \mathbf{1 8} \rightarrow \mathbf{2 0 1 9} \rightarrow 2021$.
b) Yes. For example, by adding the first digit (one), we can reach the number 2000; by adding the first digit (two), we can reach 2020; then see part a.
Criteria. Part a) 3 points, b) 4 points. In part (b), 1 point is given for progressing to 1999 (for example, for a "solution" of the form "from 1000, we add one at a time and there is the result 2021").
|
2021
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. In a rectangular grid 303 cells long and 202 cells wide, two diagonals were drawn and all cells through which they passed were painted. How many cells were painted?
(O. A. Pyayve, A. A. Tseler)
|
Solution. Mentally divide the large rectangle into $2 \times 3$ rectangles. (The central part of the rectangle is shown in the figure.)
Notice that each diagonal intersects 101 such rectangles (passing through their vertices), and in each of them, it passes through 4 cells. Thus, the two diagonals, it seems, pass through $404 \cdot 2 = 808$ cells. However, the central $2 \times 3$ rectangle is common to both diagonals, and in it, only 6 cells are shaded (not 8 as our calculation suggested).
Answer: 806 cells.
Criteria. The idea of dividing the rectangle into $2 \times 3$ rectangles or indicating that the diagonal passes through their nodes - at least 1 point (but for a "solution" like "in a $2 \times 3$ rectangle, the diagonals intersect 6 cells, so in a $202 \times 303$ rectangle, there will be 101 times more" - 0 points, as there is no idea of dividing into rectangles).
Missing the correction related to the center (i.e., the answer 808) - 3 points; the center is accounted for, but incorrectly - 4 points.
|
806
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. There are 28 students in the class. On March 8th, each boy gave each girl one flower - a tulip, a rose, or a daffodil. How many roses were given, if it is known that there were 4 times as many roses as daffodils, but 3 times fewer than tulips?
(A. A. Tesler)
|
Solution. Let the number of narcissus be $x$, then the number of roses is $4x$, and the number of tulips is $12x$, so the total number of flowers is $17x$. The number of flowers is the product of the number of boys and the number of girls. Since 17 is a prime number, one of these quantities must be divisible by 17, meaning it is 17 and 11. Therefore, $17x = 17 \cdot 11$, so $x = 11$, and the number of roses is $4x = 44$.
Answer: 44 roses.
Criteria. An incomplete enumerative solution - no more than 4 points. Only the answer - 1 point.
|
44
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Once Valera left home, walked to the cottage, painted 11 fence boards there, and returned home 2 hours after leaving. Another time, Valera and Olga went to the cottage together, painted 9 fence boards (without helping or hindering each other), and returned home together 3 hours after leaving. How many boards will Olga be able to paint alone if she needs to return home 1 hour after leaving? The physical abilities, diligence, and working conditions of Valera and Olga remain unchanged.
(V. P. Fedorov)
|
Solution. The strange result (working together for a longer time, the characters managed to do less work) is explained by the different times spent walking, since the speed of "joint" walking is equal to the lower of the two walkers' speeds. The second time, Valery's working time decreased, which means the travel time increased by more than an hour; hence, Olga spends more than an hour on the trip to the cottage and back. Therefore, in an hour, she wouldn't even have time to get there and return.
Answer: 0 boards
Criteria. Only the answer - 0 points. Understanding that two people walk at the speed of the slower walker (i.e., Olga) - 2 points.
|
0
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. Once, a god sent a little cheese to two ravens. The first raven received 100 g, from which a part was taken by a fox. The piece of the second raven turned out to be twice as large as that of the first, but she managed to eat only half as much as the first raven. The portion of cheese that the fox got from the second raven turned out to be three times larger than from the first. How much cheese did the fox get in total?
|
Solution. Let the first crow eat $x$ grams of cheese. Then the fox got $100-x$ grams of cheese from the first crow. The second crow ate ${ }_{2}^{x}$ grams of cheese. From the second crow, the fox received $200-\frac{x}{2}$ grams of cheese. This was three times more, so: $200-\frac{x}{2}=3(100-x)$. Solution: $x=40$. The fox ate 240 grams.
|
240
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. A natural number $n>5$ is called new if there exists a number that is not divisible by $n$, but is divisible by all natural numbers less than $n$. What is the maximum number of consecutive numbers that can be new?
|
Solution. Answer: 3.
Example: the number 7 is new (60 is divisible by the numbers from 1 to 6, but not by 7);
the number 8 is new (420 is divisible by the numbers from 1 to 7, but not by 8);
the number 9 is new (840 is divisible by the numbers from 1 to 8, but not by 9).
Evaluation: every fourth number has the form $n=4k+2=2(2k+1)$; if some number is divisible by 2 and $2k+1$, then it is also divisible by $2(2k+1)$, therefore such an $n$ cannot be new.
|
3
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will $C$ and $D$ meet for the first time?
|
Solution. Since $A$ with $C$ and $B$ with $D$ meet every 7 minutes, their approach speeds are equal: $V_{A}+V_{C}=V_{B}+V_{D}$. In other words, the speeds of separation of $A, B$ and $C, D$ are equal: $V_{A}-V_{B}=V_{D}-V_{C}$. Therefore, since $A$ and $B$ meet for the first time at the 53rd minute, $C$ and $D$ will also meet for the first time at the 53rd minute.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
53
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
8. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 30?
|
Solution. The letter A must equal 0. The remaining 6 letters are non-zero digits with a sum that is a multiple of 3. Note that each remainder when divided by 3 appears three times. By enumeration, we find all sets of remainders whose sum is a multiple of three: 000111, 000222, 111222, 001122. We count the 6-element subsets of digits: the first three types have one each, and the last type has \(3^3 = 27\). Each of these can be permuted in \(6!\) ways. In total: \(30 \cdot 6! = 21600\).
Criteria. It is shown that “A” equals 0 - 1 point. Combinations of the remaining letters are found by enumeration - do not deduct, but if cases are lost - no more than 3 points. Error in calculations -1 point. The answer is given as an expression, not fully calculated - also -1 point.
International Mathematical Olympiad
«Formula of Unity» / «The Third Millennium»
Year 2022/2023. Qualifying round
## Problems for grade R8
Each task is assessed at 7 points. Some problems have their own criteria (printed in gray).
|
21600
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
2. The pond has a square shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by $19 \%$. How long will it take for the pond to freeze completely?
|
Solution. It is not hard to understand that a pond of $200 \times 200$ fits, for which the answer is - in 10 days (since each day the side decreases by 20 meters). There are no other options, as the larger the side of the pond, the smaller the percentage that will freeze on the first day.
More rigorously: let the side of the pond be $x$ meters, then the initial area is $x^{2}$ m $^{2}$; then after the first day, $0.81 x^{2}=(0.9 x)^{2}$ remains, that is, the side of the pond after the first day is $0.9 x$. Therefore, the side decreased by $0.1 x$ on the first day. At the same time, it decreased by 20 m, hence $0.1 x=20, x=200$.
|
10
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. How many ways are there to cut a $10 \times 10$ square into several rectangles along the grid lines such that the sum of their perimeters is 398? Ways that can be matched by rotation or flipping are considered different.
|
Solution: 180 ways.
If the entire square is cut into 100 unit squares, the sum of the perimeters will be $4 \times 100=400$. Therefore, we need to reduce this sum by 2, which is achieved by keeping one internal partition intact (in other words, the square is cut into 98 squares and 1 domino). There are a total of 180 internal partitions - 9 in each of the 10 rows and 9 in each of the 10 columns.
|
180
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. A rectangle $11 \times 12$ is cut into several strips $1 \times 6$ and $1 \times 7$. What is the minimum total number of strips?
|
Solution. Answer: 20. The example is shown in the figure.
Evaluation: we will paint every seventh diagonal so that 20 cells are shaded (see figure). Each strip contains no more than one cell, so there are no fewer than 20 strips.
|
20
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In several packages, there are 20 candies, and there are no two packages with the same number of candies and no empty packages. Some packages may be inside other packages (then it is considered that a candy lying in the inner package also lies in the outer package). However, it is forbidden to place a package inside another package that already contains a package. What is the maximum possible number of packages?
|
Solution. 8. Example: ((6)(2)) ((3)(4)) ((1)4) (there are other examples).
There cannot be more than 8 packages. Indeed, then the sum of the number of candies in the packages (or rather, the number of incidences of candies to packages) is not less than $1+2+\ldots+9=45$. But there are 20 candies, so at least one of them lies in at least three packages, which is not allowed.
|
8
|
Combinatorics
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. There is a rectangular sheet, white on one side and gray on the other. It was folded as shown in the picture. The perimeter of the first rectangle is 20 more than the perimeter of the second rectangle. And the perimeter of the second rectangle is 16 more than the perimeter of the third rectangle. Find the area of the original sheet.
|
Solution. From the figure, it can be seen that when folding, the perimeter of the rectangle decreases by twice the short side, so the short side of rectangle-1 is $20 / 2=10$, the short side of rectangle-2 is $16 / 2=8$. Therefore, the long side of rectangle-1 is 18, and the long side of the original sheet is 28. Thus, the area is: $28 \cdot 18=504$.
Criteria. If the perimeter is given instead of the area - 3 points. Any trial and error in selecting lengths (or the lengths of the numbered rectangles are simply stated without justification) - 0 points.
|
504
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Egor wrote a number on the board and encrypted it according to the rules of letter puzzles (different letters correspond to different digits, the same letters correspond to the same digits). The result was the word "GUATEMALA". How many different numbers could Egor have initially written if his number was divisible by 8?
|
Solution. For a number to be divisible by 8, "АЛА" must be divisible by 8, with "А" -
the expression in parentheses is clearly divisible by 8, so it is sufficient to require that ("А" + 2 * "Л") : 8. By enumeration, we find 11 options: (0,4), (0,8), (2,3), (2,7), (4,2), (4,6), (6,1), (6,5), (6,9), (8,0), (8,4). In three of them, where there is a zero, for the remaining five letters ("Г", "В", "Т", "Е", "М") there are \(A_{8}^{5} = 8!/3!\) options; in the other eight \(7 \cdot A_{7}^{4} = 7 \cdot 7!/3!\). In total: \(3 \cdot 8!/3! + 8 \cdot 7 \cdot 7!/3! = 67200\).
Criteria. The divisibility rule for 8 is formulated and it is explicitly stated that "А" is an even digit - 1 point. It is proven that "А" + 2 "Л" or 5 "А" + 10 "Л" is divisible by 8 - 3 points. For each
|
67200
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. Four cars $A, B, C$, and $D$ start simultaneously from the same point on a circular track. $A$ and $B$ drive clockwise, while $C$ and $D$ drive counterclockwise. All cars move at constant (but pairwise different) speeds. Exactly 7 minutes after the start of the race, $A$ meets $C$ for the first time, and at the same moment, $B$ meets $D$ for the first time. After another 46 minutes, $A$ and $B$ meet for the first time. After how much time from the start of the race will all the cars meet for the first time?
|
Solution. $A$ and $C$ meet every 7 minutes, while $A$ and $B$ meet every 53 minutes. Therefore, all three will meet at a time that is a multiple of both 7 and 53, which is $7 \cdot 53 = 371$ minutes. At the same time, $B$ and $D$ also meet every 7 minutes, so on the 371st minute, car $D$ will be at the same point as the other three cars.
Criteria. If 53 minutes is replaced with 46 - no more than 3 points. If the solution assumes that any of the speeds are equal - 1 point. Only the answer without explanation - 0 points.
|
371
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
1. The pond has a rectangular shape. On the first frosty day, the part of the pond within 10 meters of the nearest shore froze. On the second day, the part within 20 meters froze, on the third day, the part within 30 meters, and so on. On the first day, the area of open water decreased by 20.2%, and on the second day, it decreased by 18.6% of the original area. On which day will the pond be completely frozen?
|
Solution. First method. Let the sides of the pond be $a$ and $b$ meters, then
$(a-20)(b-20)=(1-0.202) a b, (a-40)(b-40)=(1-0.388) a b$,
from which $20(a+b)-400=0.202 a b, 40(a+b)-1600=0.388 a b$, that is, $800=0.016 a b, a b=5000$ and further $a+b=525$. It turns out that the sides are 400 and 125 meters.
Answer: on the seventh day
Second method. Notice that each day 800 m $^{2}$ less freezes than the previous day. This is evident from the diagram, which shows that the "outer frame" consists of pieces equal to the corresponding pieces of the "inner frame," plus eight 10 x 10 m squares. Therefore, the percentage of the frozen part also decreases by the same amount each day. That is, on the first day, 20.2% of the area freezes, on the second day 18.6%, on the third day 17.0%, and so on. Note that the sum of the first six terms of this progression is less than 100%, while the sum of the first seven terms is already more than 100%. Therefore, the pond will freeze on the seventh day.
|
7
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
3. Here is a problem from S. A. Rachinsky's problem book (late 19th century): "How many boards 6 arshins long and 6 vershoks wide are needed to cover the floor of a square room with a side of 12 arshins?" The answer to the problem is: 64 boards. Determine from these data how many vershoks are in an arshin.
|
Solution. The area of the room is $12 \cdot 12=144$ square arshins. Therefore, the area of each board is $144 / 64=2.25$ square arshins. Since the length of the board is 6 arshins, its width should be $2.25 / 6=3 / 8=6 / 16$ arshins. Thus, 6 vershoks make up $6 / 16$ arshins, meaning 1 vershok is $1 / 16$ arshin.
Another solution. Note that the total area of the boards covering the room does not depend on their arrangement. Therefore, we can assume that they lie in two rows, with 32 boards in each row. The length of each row is equal to the side of the room (12 arshins), and at the same time, it is 32 times greater than the width of the board (6 vershoks). Thus, 12 arshins are 32 times greater than 6 vershoks. Therefore, 6 arshins are 16 times greater than 6 vershoks, meaning an arshin is 16 times greater than a vershok.
Answer: there are 16 vershoks in an arshin.
|
16
|
Geometry
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
4. On a glade, two firs, each 30 meters tall, grow 20 meters apart from each other. The branches of the firs grow very densely, and among them are some that are directed straight towards each other, and the length of each branch is half the distance from it to the top. A spider can crawl up or down the trunk (strictly vertically), along the branches (strictly horizontally), or descend vertically down a web from one branch to another. What is the shortest distance the spider will have to crawl to get from the top of one fir to the top of the other?
|
Solution. From the diagram, it can be seen that the branches of the firs intersect at a height of no more than 10 meters from the ground. Indeed, at this height, the distance to the treetop is 20 meters, so the length of each branch is $20 / 2 = 10$ meters, and the total length of the branches of the two firs is equal to the distance between them. Therefore, the spider must descend to a height of 10 meters and then climb up. This means it has to overcome at least 40 meters vertically. Additionally, it needs to travel at least 20 meters horizontally. In total, its path will be at least 60 meters. Note that there are many paths of 60 meters (one of them is shown in black in the diagram).
There is another interpretation of the problem, where the path the spider traveled downward along the web is not counted (it falls rather than crawls). In this case, the answer is 40 meters (20 meters horizontally and 20 meters vertically).
|
60
|
Logic and Puzzles
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. Nikita has a magic jar. If you put $n$ candies in the jar and close it for an hour, the number of candies inside will increase by the sum of the digits of $n$. For example, if there were 137 candies, it would become $137+1+3+7=148$. What is the maximum number of candies Nikita can get in 20 hours and 16 minutes, if he starts with one candy?
|
Solution. We need to strive to have as many candies as possible at the end of each hour. However, this does not mean that we should always put all the candies in the jar. The greatest sum of digits (i.e., the greatest increase in the number of candies) is achieved with a number where all digits (except the first) are nines.
1 hour: $1+1=2$
2 hour: $2+2=4$
3 hour: $4+4=8$
4 hour: $8+8=16$
5 hour: (put 9 candies) $16+9=25$
6 hour: (put 19 candies) $25+10=35$
7 hour: (put 29 candies) $35+11=46$
8 hour: (put 39 candies) $46+12=58$
9 hour: (put 49 candies) $58+13=71$
10 hour: (put 69 candies) $71+15=86$
11 hour: (put 79 candies) $86+16=102$
12 hour: (put 99 candies) $102+18=120$
13-17 hours: put 99 candies each hour, the number of candies increases by 18 each hour, resulting in $138,156,174,192,210$ candies.
18-20 hours: put 199 candies each hour, the number of candies increases by 19 each hour, resulting in $229,248,267$ candies.
In the remaining 16 minutes, it is impossible to gain anything.
Answer: 267 candies.
|
267
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
6. How many numbers from 1 to 999 without the digit "0" are written in the Roman numeral system exactly one symbol longer than in the decimal system?
(P. D. Mulyenko)
Reference. To write a number in Roman numerals, you need to break it down into place value addends, write each place value addend according to the table, and then write them sequentially from the largest to the smallest. For example, let's write the number 899. According to the table, $800 = \text{DCCC}$, $90 = \text{XC}$, $9 = \text{IX}$, we get DCCCXCIX.
| 1 I | $10 \text{X}$ | $100 \text{C}$ | $1000 \text{M}$ |
| :--- | :--- | :--- | :--- |
| $2 \text{II}$ | $20 \text{XX}$ | $200 \text{CC}$ | $2000 \text{MM}$ |
| 3 III | $30 \text{XXX}$ | $300 \text{CCC}$ | $3000 \text{MMM}$ |
| $4 \text{IV}$ | $40 \text{XL}$ | $400 \text{CD}$ | |
| 5 V | $50 \text{L}$ | $500 \text{D}$ | |
| $6 \text{VI}$ | $60 \text{LX}$ | $600 \text{DC}$ | |
| $7 \text{VII}$ | $70 \text{LXX}$ | $700 \text{DCC}$ | |
| $8 \text{VIII}$ | $80 \text{LXXX}$ | $800 \text{DCCC}$ | |
| 9 IX | $90 \text{XC}$ | $900 \text{CM}$ | |
|
Solution. Note that, regardless of the digit place and other digits of the number, a decimal digit $a$ is written as:
- one Roman numeral when $a=1$ and $a=5$,
- two Roman numerals when $a$ is $2,4,6,9$,
- three Roman numerals when $a=3$ and $a=7$,
- four Roman numerals when $a=8$.
Thus, in suitable numbers, only the digits 1, 5, and exactly one of the digits $2,4,6,9$ are used (otherwise, the total length of the record will be longer by more than one character). This means that 4 single-digit numbers, $4 \cdot 2 + 2 \cdot 4 = 16$ two-digit numbers, and $4 \cdot 2 \cdot 2 + 2 \cdot 4 \cdot 2 + 2 \cdot 2 \cdot 4 = 48$ three-digit numbers are suitable. In total, 68 numbers.
Criteria. Correct answer without justification - 2 points.
One of the above cases is missing or extra cases are considered - 0 points.
The condition about the absence of zero is incorrectly interpreted (for example, as "excluding zero," meaning that the length of the Roman record of the number should be one more than the number of non-zero digits - no more than 3 points.
|
68
|
Number Theory
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
5. In a spring math camp, between 50 and 70 children arrived. On Pi Day (March 14), they decided to give each other squares if they were just acquaintances, and circles if they were friends. Andrey calculated that each boy received 3 circles and 8 squares, while each girl received 2 squares and 9 circles. And Katya noticed that the total number of circles and squares given was the same. How many children arrived at the camp?
(P. D. Mulyenko)
|
Answer: 60.
Solution. Let the number of boys be $m$, and the number of girls be $-d$. Then $3 m + 9 d = 8 m + 2 d$ (the number of circles equals the number of squares). Transforming, we get $5 m = 7 d$, which means the number of boys and girls are in the ratio $7: 5$. Therefore, the total number of children is divisible by 12. Between 50 and 70, only the number 60 fits.
Criteria. Incorrectly understood condition (for example, that an equal number of gifts were given to boys and girls) 0 points.
Correct answer without explanation -2 points. Correct answer with verification -3 points.
|
60
|
Algebra
|
math-word-problem
|
Yes
|
Yes
|
olympiads
| false
|
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