Prototypes of tasks No. 1 2015
1) Prototype of task 1 (No. 26616)
The cheese costs 7 rubles 20 kopecks. What is the largest number of cheesecakes you can buy for 60 rubles?
2) Prototype of task 1 (No. 26617)
The ship is designed for 750 passengers and 25 crew members. Each lifeboat can
accommodate 70 people. What is the minimum number of boats that should be on the ship so that, if necessary, they can accommodate all passengers and all crew members?
3) Prototype of task 1 (No. 26618)
A bottle of shampoo costs 160 rubles. What is the largest number of bottles you can buy for 1000 rubles?
during the sale when the discount is 25%?
4) Prototype of task 1 (No. 26619)
A ballpoint pen costs 40 rubles. What is the largest number of such pens that can be purchased for 900
rubles after a price increase of 10%?
5) Prototype of task 1 (No. 26620)
The notebook costs 40 rubles. What is the largest number of such notebooks that can be purchased for 750 rubles?
after a 10% price cut?
6) Prototype of task 1 (No. 26621)
The store purchases flower pots By wholesale price 120 rubles per piece and sells with a 20% markup.
What is the largest number of such pots that can be bought in this store for 1000 rubles?
7) Prototype of task 1 (No. 26622)
IN a pack of 500 sheets of A4 paper. The office uses 1,200 sheets per week. Which
What is the smallest number of packs of paper you need to buy for your office for 4 weeks?
8) Prototype of task 1 (No. 26623)
The cost of a ticket for a month is 580 rubles, and the cost of a ticket for one trip is
20 rubles. Anya bought a travel card and made 41 trips in a month. How many more rubles would she spend if she bought tickets for one trip?
9) Prototype of task 1 (No. 26624)
The patient is prescribed a medicine that needs to be taken 0.5 g 3 times a day for 21 days. One
a package of 10 tablets of medicine, 0.5 g each. What is the smallest number of packages that will be enough for the entire course of treatment?
10) Prototype of task 1 (No. 26625)
To prepare cucumber marinade, 1 liter of water requires 12 g citric acid. Lemon
acid is sold in 10 g bags. What is the smallest number of packs a housewife needs to buy to prepare 6 liters of marinade?
11) Prototype of task 1 (No. 26626)
Chocolate costs 35 rubles. On Sunday the supermarket has a special offer:
Having paid for two chocolates, the buyer receives three (one as a gift). How many chocolates can you get for 200 rubles on Sunday?
12) Prototype of task 1 (No. 26627)
The wholesale price of the textbook is 170 rubles. The retail price is 20% higher than the wholesale price. What is the largest number
Can such textbooks be purchased at a retail price of 7,000 rubles?
13) Prototype of task 1 (No. 26628)
A train ticket for an adult costs 720 rubles. Ticket price for a student
is 50% of the ticket price for an adult. The group consists of 15 schoolchildren and 2 adults. How many rubles are tickets for the whole group?
14) Prototype of task 1 (No. 26629)
Price for Electric kettle was increased by 16% and amounted to 3,480 rubles. How many rubles did it cost?
kettle before the price increase?
15) Prototype of task 1 (No. 26630)
The T-shirt cost 800 rubles. After the price was reduced, it began to cost 680 rubles. How long
the price of the T-shirt was reduced by one percent?
16) Prototype of task 1 (No. 26631)
IN City N has 200,000 inhabitants. Among them, 15% are children and adolescents. Among adult residents 45%
does not work (pensioners, students, housewives, etc.). How many adult residents work?
17) Prototype of task 1 (No. 26632)
The taxi driver drove 6000 km in a month. The cost of 1 liter of gasoline is 20 rubles. Average consumption gasoline for
100 km is 9 liters. How many rubles did the taxi driver spend on gasoline this month?
18) Prototype of task 1 (No. 26633)
The client took out a loan from the bank for 12,000 rubles for a year at 16%. He must repay the loan by depositing in the bank
monthly the same amount of money, in order to repay the entire amount borrowed, along with interest, in a year. How many rubles should he deposit into the bank monthly?
19) Prototype of task 1 (No. 26634)
IN Summer camp allows each participant 40 grams of sugar per day. There are 166 people in the camp. How many
kilogram packs of sugar will be needed for the entire camp for 5 days?
20) Prototype of task 1 (No. 26635)
IN summer camp there are 218 children and 26 teachers. The bus can accommodate no more than 45 passengers. How many
buses are required to transport everyone from the camp to the city?
21) Prototype of task 1 (No. 26636)
In summer, a kilogram of strawberries costs 80 rubles. Masha bought 1 kg 200 g of strawberries. How many rubles in change
should she get 500 rubles?
22) Prototype of task 1 (No. 26637)
On a birthday, people are supposed to give a bouquet of an odd number of flowers. Tulips cost 30 rubles per
piece. Vanya has 500 rubles. What is the largest number of tulips he can buy a bouquet for Masha for her birthday?
23) Prototype of task 1 (No. 26640)
Pavel Ivanovich bought an American car, the speedometer of which shows the speed in miles in
hour. An American mile is equal to 1609 m. What is the speed of a car in kilometers per hour if the speedometer shows 65 miles per hour? Round your answer to a whole number.
24) Prototype of task 1 (No. 26641)
IN the university library brought new textbooks on geometry for 1-3 courses, 360 pieces each
for each course. All books are the same size. IN bookcase 9 shelves, each shelf holds 25 textbooks. How many cabinets can be completely filled with new textbooks?
25) Prototype of task 1 (No. 26642)
To prepare cherry jam, 1 kg of cherries requires 1.5 kg of sugar. How many kilograms
packages of sugar do you need to buy to make jam from 27 kg of cherries?
26) Prototype of task 1 (No. 26643)
Income tax is 13% of wages. Ivan Kuzmich's salary is 12500
rubles How much will he receive after deducting income taxes? Give your answer in rubles.
27) Prototype of task 1 (No. 26644)
Income tax is 13% of wages. After income tax withholding Maria
Konstantinovna received 9,570 rubles. How many rubles is wage Maria Konstantinovna?
28) Prototype of task 1 (No. 26645)
The retail price of the textbook is 180 rubles, which is 20% higher than the wholesale price. What is the largest number of such
Can textbooks be purchased at a wholesale price of 10,000 rubles?
29) Prototype of task 1 (No. 77331)
On Mashiny's account mobile phone there were 53 rubles, and after talking with Lena there were 8 left
rubles How many minutes did the conversation with Lena last if one minute of conversation costs 2 rubles 50 kopecks?
30) Prototype of task 1 (No. 77332)
Graduates of 11 "A" buy bouquets of flowers for last call: from 3 roses to each teacher and from 7
roses to the class teacher and director. They are going to give bouquets to 15 teachers (including the director and class teacher), roses are purchased at a wholesale price of 35 rubles per piece. How many rubles are all the roses worth?
31) Prototype of task 1 (No. 77333)
The electricity meter reading on November 1 was 12,625 kilowatt-hours, and on December 1 – 12,802
kilowatt-hour. How much should you pay for electricity for November if 1 kilowatt-hour of electricity costs 1 ruble 80 kopecks? Give your answer in rubles.
32) Prototype of task 1 (No. 77334)
IN exchange office 1 hryvnia costs 3 rubles 70 kopecks. Vacationers exchanged rubles for hryvnia and
bought 3 kg of tomatoes at a price of 4 hryvnia per 1 kg. How many rubles did this purchase cost them? Round your answer to a whole number.
33) Prototype of task 1 (No. 77335)
Masha sent SMS messages with New Year's greetings to her 16 friends. Price
one SMS message costs 1 ruble 30 kopecks. Before sending the message, Masha had 30 rubles in her account. How many rubles will Masha have left after sending all the messages?
34) Prototype of task 1 (No. 77336)
The Novosibirsk-Krasnoyarsk train departs at 15:20 and arrives at 4:20 the next day (time
Moscow). How many hours does the train travel?
35) Prototype of task 1 (No. 77337)
IN The school has three-person tourist tents. What is the minimum number of tents you need to take in
a trip involving 20 people?
36) Prototype of task 1 (No. 77338)
IN The institute's dormitory can accommodate four people in each room. What is the smallest
How many rooms are needed to accommodate 83 out-of-town students?
37) Prototype of task 1 (No. 77339)
IN On average, 70 tea bags are consumed per day during a conference. The conference lasts 6 days. IN
a pack of tea contains 50 bags. What is the smallest number of packs of tea that will last for all days of the conference?
38) Prototype of task 1 (No. 77340)
IN school French studied by 124 students, which is 25% of all students
schools. How many students are there in the school?
39) Prototype of task 1 (No. 77341)
27 school graduates are going to study at technical universities. They make up 30% of the number
graduates. How many graduates does the school have?
40) Prototype of task 1 (No. 77342)
Tutu butter costs 60 rubles. The store gives a 5% discount to pensioners. How many rubles
Will a pensioner pay for a stick of butter?
41) Prototype of task 1 (No. 77343)
The notebook costs 24 rubles. How many rubles will the buyer pay for 60 notebooks, if more when purchasing
Does the store give 50 notebooks a 10% discount on the entire purchase price?
42) Prototype of task 1 (No. 77344)
48 students became winners of the city Olympiad in mathematics, which accounted for 12% of the number
participants. How many people took part in the Olympiad?
43) Prototype of task 1 (No. 77345)
Only 94% of the city's 27,500 graduates solved problem B1 correctly. How many people is correct?
solved problem B1?
44) Prototype of task 1 (No. 77346)
The mobile phone cost 3,500 rubles. After some time, the price for this model was reduced to 2800
rubles By what percentage was the price reduced?
45) Prototype of task 1 (No. 77347)
IN the school has 800 students, of which 30%− students primary school. Among middle and high school students
schools study 20% German. How many students study German in school if German is not taught in primary school?
46) Prototype of task 1 (No. 77348)
Among the city's 40,000 residents, 60% are not interested in football. Among football fans 80%
I watched the Champions League final on TV. How many residents of the city watched this match on TV?
47) Prototype of task 1 (No. 77349)
IN In September, 1 kg of grapes cost 60 rubles; in October, grapes rose in price by 25%, and in November by another
20%. How many rubles did 1 kg of grapes cost after the price increase in November?
48) Prototype of task 1 (No. 77350)
IN The house where Petya lives has one entrance. There are six apartments on each floor. Petya lives in
apartment 50. On what floor does Petya live?
49) Prototype of task 1 (No. 77351)
IN The house in which Masha lives has 9 floors and several entrances. On each floor there are 4
apartments. Masha lives in apartment No. 130. In which entrance does Masha live?
50) Prototype of task 1 (No. 77352)
When paying for services via payment terminal a 5% commission will be charged. The terminal accepts amounts
multiples of 10 rubles. Anya wants to deposit at least 300 rubles into her mobile phone account. Which minimum amount should she put it in the receiving device of this terminal?
51) Prototype of task 1 (No. 77353)
IN September 1 kg of plums cost 60 rubles. In October, plum prices increased by 25%. How many rubles did 1 kg cost?
plum after the rise in price in October?
52) Prototype of task 1 (No. 77354)
The store gives pensioners a discount of a certain percentage of the purchase price. Plastic bag
kefir costs 40 rubles in the store. The pensioner paid 38 rubles for a package of kefir. What percentage is the discount for pensioners?
53) Prototype of task 1 (No. 77355)
The student received his first fee in the amount of 700 rubles for the completed translation. He decided to
use all the money received to buy a bouquet of tulips for your teacher in English. Which greatest number Can a student buy tulips if the income tax withheld from him is 13% of the fee, tulips cost 60 rubles apiece and the bouquet must consist of an odd number of flowers?
54) Prototype of task 1 (No. 77356)
The car's speedometer shows speed in miles per hour. What speed (in miles per hour) does it show?
speedometer if the car is moving at a speed of 36 km per hour? (Consider 1 mile equal to 1.6 km.)
55) Prototype of task 1 (No. 77365)
Holders of a bookstore discount card receive a 5% discount upon purchase. The book costs 200
rubles How many rubles will the discount card holder pay for this book?
56) Prototype of task 1 (No. 282847)
At a gas station, the client gave the cashier 1,000 rubles and filled the tank with 28 liters of gasoline at a price of 28 rubles. 50
cop. per liter How much change should the client receive? Give your answer in rubles.
57) Prototype of task 1 (No. 282848)
At a gas station, the client gave the cashier 1,000 rubles and asked to fill up with gas until the tank was full. Price
gasoline 31 rub. 20 kopecks The client received 1 ruble in change. 60 kopecks How many liters of gasoline were poured into the tank?
58) Prototype of task 1 (No. 314867)
IN the apartment where Alexey lives has a flow meter installed cold water(counter).
On September 1, the meter showed a consumption of 103 cubic meters of water, and on October 1 - 114 cubic meters. What amount should Alexey pay for cold water in September, if the price of 1 cubic meter of cold water is 19 rubles. 20 kopecks? Give your answer in rubles.
59) Prototype of task 1 (No. 314968)
One tablet of the medicine weighs 20 mg and contains 5% active substance. Child under 6 years of age
months, the doctor prescribes 1.4 mg of the active substance per kilogram of weight per day. How many tablets of this medicine should be given to a child aged four months and weighing 5 kg during the day?
60) Prototype of task 1 (No. 318579)
The diagonal of the TV screen is 64 inches. Express the screen diagonal in centimeters if
one inch is 2.54 cm. Round the result to the nearest whole number of centimeters.
61) Prototype of task 1 (No. 318580)
The man is 6 feet 1 inch tall. Express his height in centimeters if 1 foot is equal to 0.305 m and 1 inch
equals 2.54 cm. Round the result to the nearest whole number of centimeters.
62) Prototype of task 1 (No. 318581)
A runner ran 50 m in 5 seconds. Find average speed runner at a distance. Give the answer in
kilometers per hour.
63) Prototype of task 1 (No. 318582)
IN Elena Molokhovets’ book “A Gift for Young Housewives” contains a recipe for prune pie. For
For a pie for 10 people, use 1/10 of a pound of prunes. How many grams of prunes should I use for a pie for 3 people? Consider that 1 pound is equal to 0.4 kg.
64) Prototype of task 1 (No. 318583)
The navigation system built into the back of the aircraft seat informs the passenger that
the flight takes place at an altitude of 37,000 feet. Express the flight altitude in meters. Consider that 1 foot is equal to
65) Prototype of task 1 (No. 323510)
To renovate an apartment, 63 rolls of wallpaper are required. How many packs of wallpaper paste should you buy if
Is one pack of glue for 6 rolls?
66) Prototype of task 1 (No. 323511)
The cost of a six-month subscription to the magazine is 460 rubles, and the cost of one issue
magazine - 24 rubles. In six months, Anya bought 25 issues of the magazine. How many less rubles would she spend if she subscribed to the magazine?
67) Prototype of task 1 (No. 323513)
For painting 1 sq. m of ceiling requires 240 g of paint. The paint is sold in 2.5 kg cans. Which
The smallest number of cans of paint you need to buy to paint a ceiling with an area of 50 square meters. m?
68) Prototype of task 1 (No. 323514)
One roll of wallpaper is enough to cover a 1.6 m wide strip from floor to ceiling. How many rolls?
Do you need to buy wallpaper to cover a rectangular room measuring 2.3 m by 4.2 m?
69) Prototype of task 1 (No. 323515)
IN All furniture in the store is sold disassembled. The buyer can order furniture assembly at
home, the cost of which is 10% of the cost of purchased furniture. The wardrobe costs 3,300 rubles. How many rubles will it cost to purchase this cabinet along with assembly?
70) Prototype of task 1 (No. 323516)
At a gas station, one liter of gasoline costs 32 rubles. 60 kopecks The driver poured 30 liters of gasoline into the tank and
I took a bottle of water for 48 rubles. How many rubles will he receive in change from 1,500 rubles?
71) Prototype of task 1 (No. 323517)
Installation of two water meters (cold and hot) costs 3,300 rubles. Before installing meters
water was paid 800 rubles monthly. After installing the meters, the monthly payment for water began to be 300 rubles. In what minimum number of months will the savings on water bills exceed the costs of installing meters if water tariffs do not change?
Answers to prototypes No. 1
y-axis | – torque | ||||||||
N∙m. Speed | car | ||||||||
approximately | is expressed | formula |
|||||||
ν 0.036 n, where n is the engine speed
V minute. From what lowest speed the car must move so that the torque is at least 120
N∙m? Give your answer in kilometers per hour.
2. Prototype of task 2 (No. 26864)
The graph shows the dependence
torque of a car engine based on its number of revolutions per minute. The number of revolutions per minute is plotted on the abscissa axis. On the ordinate - torque in N∙ m. For the car to start moving, the torque must be at least 60 N∙ m. What is the minimum number of engine revolutions per minute that is enough for the car to start moving?
3. Prototype of task 2 (No. 26866)
The graph shows the heating process
engine passenger car. The abscissa axis shows the time in minutes that has elapsed since the engine was started, and the y-axis shows the engine temperature in degrees Celsius. Determine from the graph how many minutes the engine heated from a temperature of 60 to a temperature of 90.
4. Prototype of task 2 (No. 26868)
The picture shows
change in air temperature over three days. The date and time of day are indicated horizontally, and the temperature value in degrees Celsius is indicated vertically. Determine the highest air temperature on January 22 from the figure. Give your answer in degrees Celsius.
5. Prototype of task 2 (No. 26869)
The picture shows
change in air temperature over three days. The date and time of day are indicated horizontally, and the temperature value in degrees Celsius is indicated vertically. Determine the lowest air temperature on April 27 from the figure. Give your answer in degrees Celsius.
6. Prototype of task 2 (No. 26870)
The picture shows
change in air temperature over three days. The date and time are indicated horizontally, and the temperature value in degrees Celsius is indicated vertically. Determine the difference between the highest and lowest air temperatures on July 15 from the figure. Give your answer in degrees Celsius.
7. Prototype of task 2 (No. 26871)
In the figure, bold dots
shows the daily amount of precipitation that fell in Kazan from February 3 to February 15, 1909. The dates of the month are indicated horizontally, and the amount of precipitation that fell on the corresponding day in millimeters is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the picture on what date 5 millimeters of rain fell for the first time.
8. Prototype of task 2 (No. 26872)
In the figure, bold dots
The price of oil is shown at the close of exchange trading on all working days from August 17 to August 31, 2004. The dates of the month are indicated horizontally, and the price of a barrel of oil in US dollars is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the figure the lowest price of oil at the close of trading in the specified period (in US dollars per barrel).
9. Prototype of task 2 (No. 26873)
The figure shows bold dots
nickel price at the close of exchange trading on all working days from May 6 to May 20, 2009. The dates of the month are indicated horizontally, and the price of a ton of nickel in US dollars is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the picture highest price nickel at the close of trading for the specified period (in US dollars per ton).
10. Prototype of task 2 (No. 26874)
In the figure, bold dots
The price of gold is shown at the close of exchange trading on all working days from March 5 to March 28, 1996. The dates of the month are indicated horizontally, and the price of an ounce of gold in US dollars is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the figure what date the price of gold at the close of trading was the lowest for the given period.
11. Prototype of task 2 (No. 26875)
In the figure, bold dots
The price of tin is shown at the close of exchange trading on all working days from September 3 to September 18, 2007. The dates of the month are indicated horizontally, and the price of a ton of tin in US dollars is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the figure what date the price of tin at the close of trading was the highest for the given period.
12. Prototype of task 2 (No. 26876)
In the figure, bold dots show the daily amount of precipitation that fell in Tomsk from 8 to 24
January 2005. The dates of the month are indicated horizontally, and the amount of precipitation that fell on the corresponding day in millimeters is indicated vertically. For clarity, the bold points in the figure are connected by a line. Determine from the picture what the greatest amount of precipitation fell in the period from January 13 to January 20. Give your answer in millimeters.
Average general education
Line UMK G. K. Muravin. Algebra and beginnings mathematical analysis(10-11) (in-depth)
UMK Merzlyak line. Algebra and beginnings of analysis (10-11) (U)
Mathematics
The profile level examination lasts 3 hours 55 minutes (235 minutes).
Minimum threshold- 27 points.
The examination paper consists of two parts, which differ in content, complexity and number of tasks.
The defining feature of each part of the work is the form of the tasks:
Panova Svetlana Anatolevna, mathematic teacher highest category schools, work experience 20 years:
“In order to receive a school certificate, a graduate must pass two mandatory exams in the form of the Unified State Examination, one of which is mathematics. In accordance with the Concept of development of mathematics education in Russian Federation The Unified State Examination in mathematics is divided into two levels: basic and specialized. Today we will look at profile-level options.”
Task No. 1- tests the Unified State Exam participants’ ability to apply the skills acquired in the 5th to 9th grade course in elementary mathematics in practical activities. The participant must have computational skills, be able to work with rational numbers, be able to round decimals, be able to convert one unit of measurement to another.
Example 1. In the apartment where Peter lives, a cold water flow meter (meter) was installed. On May 1, the meter showed a consumption of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. m. What amount should Peter pay for cold water in May, if the price is 1 cubic meter? m of cold water is 34 rubles 17 kopecks? Give your answer in rubles.
Solution:
1) Find the amount of water spent per month:
177 - 172 = 5 (cubic m)
2) Let’s find how much money they will pay for wasted water:
34.17 5 = 170.85 (rub)
Answer: 170,85.
Task No. 2- is one of the simplest exam tasks. The majority of graduates successfully cope with it, which indicates knowledge of the definition of the concept of function. Type of task No. 2 according to the requirements codifier is a task on the use of acquired knowledge and skills in practical activities and Everyday life. Task No. 2 consists of describing, using functions, various real relationships between quantities and interpreting their graphs. Task No. 2 tests the ability to extract information presented in tables, diagrams, and graphs. Graduates need to be able to determine the value of a function by the value of its argument when in various ways specifying a function and describing the behavior and properties of the function based on its graph. You also need to be able to find the greatest or smallest value and build graphs of the studied functions. Errors made are random in reading the conditions of the problem, reading the diagram.
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Example 2. The figure shows the change in the exchange value of one share of a mining company in the first half of April 2017. On April 7, the businessman purchased 1,000 shares of this company. On April 10, he sold three-quarters of the shares he purchased, and on April 13, he sold all the remaining shares. How much did the businessman lose as a result of these operations?
Solution:
2) 1000 · 3/4 = 750 (shares) - constitute 3/4 of all shares purchased.
6) 247500 + 77500 = 325000 (rub) - the businessman received 1000 shares after selling.
7) 340,000 – 325,000 = 15,000 (rub) - the businessman lost as a result of all operations.
Answer: 15000.
Task No. 3- is a task basic level first part, tests the ability to perform actions with geometric shapes on the content of the course “Planimetry”. Task 3 tests the ability to calculate the area of a figure on checkered paper, the ability to calculate degree measures of angles, calculate perimeters, etc.
Example 3. Find the area of a rectangle drawn on checkered paper with a cell size of 1 cm by 1 cm (see figure). Give your answer in square centimeters.
Solution: To calculate the area of a given figure, you can use the Peak formula:
To calculate the area of a given rectangle, we use Peak’s formula:
|
S= B + |
G | |
| 2 |
|
S = 18 + |
6 | |
| 2 |
Example 4. There are 5 red and 1 blue dots marked on the circle. Determine which polygons are larger: those with all the vertices red, or those with one of the vertices blue. In your answer, indicate how many there are more of some than others.
Solution: 1) Let's use the formula for the number of combinations of n elements by k:
whose vertices are all red.
3) One pentagon with all vertices red.
4) 10 + 5 + 1 = 16 polygons with all red vertices.
which have red tops or with one blue top.
which have red tops or with one blue top.
8) One hexagon with red vertices and one blue vertex.
9) 20 + 15 + 6 + 1 = 42 polygons with all red vertices or one blue vertex.
10) 42 – 16 = 26 polygons using the blue dot.
11) 26 – 16 = 10 polygons – how many more polygons in which one of the vertices is a blue dot are there than polygons in which all the vertices are only red.
Answer: 10.
Task No. 5- the basic level of the first part tests the ability to solve simple equations (irrational, exponential, trigonometric, logarithmic).
Example 5. Solve equation 2 3 + x= 0.4 5 3 + x .
Solution. Divide both sides of this equation by 5 3 + X≠ 0, we get
| 2 3 + x | = 0.4 or | 2 | 3 + X | = | 2 | , | ||
| 5 3 + X | 5 | 5 |
whence it follows that 3 + x = 1, x = –2.
Answer: –2.
Task No. 6 in planimetry to find geometric quantities (lengths, angles, areas), modeling real situations in the language of geometry. Study of constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary theorems of planimetry.
Area of a triangle ABC equals 129. DE– midline parallel to the side AB. Find the area of the trapezoid ABED.
Solution. Triangle CDE similar to a triangle CAB at two angles, since the angle at the vertex C general, angle СDE equal to angle CAB as the corresponding angles at DE || AB secant A.C.. Because DE– the middle line of the triangle by condition, then by property midline | DE = (1/2)AB. This means that the similarity coefficient is 0.5. The areas of similar figures are related as the square of the similarity coefficient, therefore
Hence, S ABED = S Δ ABC – S Δ CDE = 129 – 32,25 = 96,75.
Task No. 7- checks the application of the derivative to the study of a function. Successful implementation requires meaningful, non-formal knowledge of the concept of derivative.
Example 7. To the graph of the function y = f(x) at the abscissa point x 0 a tangent is drawn that is perpendicular to the line passing through the points (4; 3) and (3; –1) of this graph. Find f′( x 0).
Solution. 1) Let’s use the equation of a line passing through two given points and find the equation of a line passing through points (4; 3) and (3; –1).
(y – y 1)(x 2 – x 1) = (x – x 1)(y 2 – y 1)
(y – 3)(3 – 4) = (x – 4)(–1 – 3)
(y – 3)(–1) = (x – 4)(–4)
–y + 3 = –4x+ 16| · (-1)
y – 3 = 4x – 16
y = 4x– 13, where k 1 = 4.
2) Find the slope of the tangent k 2, which is perpendicular to the line y = 4x– 13, where k 1 = 4, according to the formula:
3) Slope factor tangent – derivative of the function at the point of tangency. Means, f′( x 0) = k 2 = –0,25.
Answer: –0,25.
Task No. 8- tests the exam participants’ knowledge of elementary stereometry, the ability to apply formulas for finding surface areas and volumes of figures, dihedral angles, compare the volumes of similar figures, be able to perform actions with geometric figures, coordinates and vectors, etc.
The volume of a cube circumscribed around a sphere is 216. Find the radius of the sphere.
Solution. 1) V cube = a 3 (where A– length of the edge of the cube), therefore
A 3 = 216
A = 3 √216
2) Since the sphere is inscribed in a cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, therefore d = a, d = 6, d = 2R, R = 6: 2 = 3.
Task No. 9- requires the graduate to have the skills to transform and simplify algebraic expressions. Task No. 9 of an increased level of difficulty with a short answer. The tasks from the “Calculations and Transformations” section in the Unified State Exam are divided into several types:
transformation of numerical rational expressions;
converting algebraic expressions and fractions;
conversion of numeric/letter irrational expressions;
actions with degrees;
transformation logarithmic expressions;
Example 9. Calculate tanα if it is known that cos2α = 0.6 and
| 3π | < α < π. |
| 4 |
Solution. 1) Let’s use the double argument formula: cos2α = 2 cos 2 α – 1 and find
| tan 2 α = | 1 | – 1 = | 1 | – 1 = | 10 | – 1 = | 5 | – 1 = 1 | 1 | – 1 = | 1 | = 0,25. |
| cos 2 α | 0,8 | 8 | 4 | 4 | 4 |
This means tan 2 α = ± 0.5.
3) By condition
| 3π | < α < π, |
| 4 |
this means α is the angle of the second quarter and tgα< 0, поэтому tgα = –0,5.
Answer: –0,5.
#ADVERTISING_INSERT# Task No. 10- tests students’ ability to use acquired early knowledge and skills in practical activities and everyday life. We can say that these are problems in physics, and not in mathematics, but all the necessary formulas and quantities are given in the condition. The problems are reduced to solving linear or quadratic equation, or linear or quadratic inequality. Therefore, it is necessary to be able to solve such equations and inequalities and determine the answer. The answer must be given as a whole number or a finite decimal fraction.
Two bodies of mass m= 2 kg each, moving at the same speed v= 10 m/s at an angle of 2α to each other. The energy (in joules) released during their absolutely inelastic collision is determined by the expression Q = mv 2 sin 2 α. At what smallest angle 2α (in degrees) must the bodies move so that at least 50 joules are released as a result of the collision?
Solution. To solve the problem, we need to solve the inequality Q ≥ 50, on the interval 2α ∈ (0°; 180°).
mv 2 sin 2 α ≥ 50
2 10 2 sin 2 α ≥ 50
200 sin 2 α ≥ 50
Since α ∈ (0°; 90°), we will only solve
Let us represent the solution to the inequality graphically:
Since by condition α ∈ (0°; 90°), it means 30° ≤ α< 90°. Получили, что smallest angleα is 30°, then the smallest angle is 2α = 60°.
Task No. 11- is typical, but turns out to be difficult for students. The main source of difficulty is the construction of a mathematical model (drawing up an equation). Task No. 11 tests the ability to solve word problems.
Example 11. During spring break, 11th-grader Vasya had to solve 560 practice problems to prepare for the Unified State Exam. On March 18, on the last day of school, Vasya solved 5 problems. Then every day he solved the same number of problems more than the previous day. Determine how many problems Vasya solved on April 2, the last day of the holidays.
Solution: Let's denote a 1 = 5 – the number of problems that Vasya solved on March 18, d– daily number of tasks solved by Vasya, n= 16 – number of days from March 18 to April 2 inclusive, S 16 = 560 – total number of tasks, a 16 – the number of problems that Vasya solved on April 2. Knowing that every day Vasya solved the same number of problems more compared to the previous day, we can use formulas for finding the sum of an arithmetic progression:560 = (5 + a 16) 8,
5 + a 16 = 560: 8,
5 + a 16 = 70,
a 16 = 70 – 5
a 16 = 65.
Answer: 65.
Task No. 12- they test students’ ability to perform operations with functions, and to be able to apply the derivative to the study of a function.
Find the maximum point of the function y= 10ln( x + 9) – 10x + 1.
Solution: 1) Find the domain of definition of the function: x + 9 > 0, x> –9, that is, x ∈ (–9; ∞).
2) Find the derivative of the function:
4) The found point belongs to the interval (–9; ∞). Let's determine the signs of the derivative of the function and depict the behavior of the function in the figure:
The desired maximum point x = –8.
Download for free the working program in mathematics for the line of teaching materials G.K. Muravina, K.S. Muravina, O.V. Muravina 10-11 Download free teaching aids on algebraa) Solve the equation 2log 3 2 (2cos x) – 5log 3 (2cos x) + 2 = 0
b) Find all the roots of this equation that belong to the segment.
Solution: a) Let log 3 (2cos x) = t, then 2 t 2 – 5t + 2 = 0,
|
|
log 3(2cos x) = | 2 | ⇔ |
|
2cos x = 9 | ⇔ |
|
cos x = | 4,5 | ⇔ because |cos x| ≤ 1, |
| log 3(2cos x) = | 1 | 2cos x = √3 | cos x = | √3 | ||||||
| 2 | 2 |
| then cos x = | √3 |
| 2 |
|
|
x = | π | + 2π k |
| 6 | |||
| x = – | π | + 2π k, k ∈ Z | |
| 6 |
b) Find the roots lying on the segment .
The figure shows that the roots of the given segment belong to
| 11π | And | 13π | . |
| 6 | 6 |
| Answer: A) | π | + 2π k; – | π | + 2π k, k ∈ Z; b) | 11π | ; | 13π | . |
| 6 | 6 | 6 | 6 |
The diameter of the circle of the base of the cylinder is 20, the generatrix of the cylinder is 28. The plane intersects its base along chords of length 12 and 16. The distance between the chords is 2√197.
a) Prove that the centers of the bases of the cylinder lie on one side of this plane.
b) Find the angle between this plane and the plane of the base of the cylinder.
Solution: a) A chord of length 12 is at a distance = 8 from the center of the base circle, and a chord of length 16, similarly, is at a distance of 6. Therefore, the distance between their projections onto a plane parallel to the bases of the cylinders is either 8 + 6 = 14, or 8 − 6 = 2.
Then the distance between the chords is either
= = √980 = = 2√245
= = √788 = = 2√197.
According to the condition, the second case was realized, in which the projections of the chords lie on one side of the cylinder axis. This means that the axis does not intersect this plane within the cylinder, that is, the bases lie on one side of it. What needed to be proven.
b) Let us denote the centers of the bases as O 1 and O 2. Let us draw from the center of the base with a chord of length 12 a perpendicular bisector to this chord (it has length 8, as already noted) and from the center of the other base to the other chord. They lie in the same plane β, perpendicular to these chords. Let's call the midpoint of the smaller chord B, the larger chord A and the projection of A onto the second base - H (H ∈ β). Then AB,AH ∈ β and therefore AB,AH are perpendicular to the chord, that is, the straight line of intersection of the base with the given plane.
This means that the required angle is equal to
| ∠ABH = arctan | A.H. | = arctan | 28 | = arctg14. |
| B.H. | 8 – 6 |
Task No. 15- increased level of complexity with a detailed answer, tests the ability to solve inequalities, which is most successfully solved among tasks with a detailed answer of an increased level of complexity.
Example 15. Solve inequality | x 2 – 3x| log 2 ( x + 1) ≤ 3x – x 2 .
Solution: The domain of definition of this inequality is the interval (–1; +∞). Consider three cases separately:
1) Let x 2 – 3x= 0, i.e. X= 0 or X= 3. In this case, this inequality becomes true, therefore, these values are included in the solution.
2) Let now x 2 – 3x> 0, i.e. x∈ (–1; 0) ∪ (3; +∞). Moreover, this inequality can be rewritten as ( x 2 – 3x) log 2 ( x + 1) ≤ 3x – x 2 and divide by a positive expression x 2 – 3x. We get log 2 ( x + 1) ≤ –1, x + 1 ≤ 2 –1 , x≤ 0.5 –1 or x≤ –0.5. Taking into account the domain of definition, we have x ∈ (–1; –0,5].
3) Finally, let's consider x 2 – 3x < 0, при этом x∈ (0; 3). In this case, the original inequality will be rewritten in the form (3 x – x 2) log 2 ( x + 1) ≤ 3x – x 2. After dividing by positive 3 x – x 2 , we get log 2 ( x + 1) ≤ 1, x + 1 ≤ 2, x≤ 1. Taking into account the region, we have x ∈ (0; 1].
Combining the solutions obtained, we obtain x ∈ (–1; –0.5] ∪ ∪ {3}.
Answer: (–1; –0.5] ∪ ∪ {3}.
Task No. 16- advanced level refers to tasks in the second part with a detailed answer. The task tests the ability to perform actions with geometric shapes, coordinates and vectors. The task contains two points. In the first point, the task must be proven, and in the second point, calculated.
In an isosceles triangle ABC with an angle of 120°, the bisector BD is drawn at vertex A. Rectangle DEFH is inscribed in triangle ABC so that side FH lies on segment BC, and vertex E lies on segment AB. a) Prove that FH = 2DH. b) Find the area of rectangle DEFH if AB = 4.
Solution: A)
1) ΔBEF – rectangular, EF⊥BC, ∠B = (180° – 120°): 2 = 30°, then EF = BE by the property of the leg lying opposite the angle of 30°.
2) Let EF = DH = x, then BE = 2 x, BF = x√3 according to the Pythagorean theorem.
3) Since ΔABC is isosceles, it means ∠B = ∠C = 30˚.
BD is the bisector of ∠B, which means ∠ABD = ∠DBC = 15˚.
4) Consider ΔDBH – rectangular, because DH⊥BC.
| 2x | = | 4 – 2x |
| 2x(√3 + 1) | 4 |
| 1 | = | 2 – x |
| √3 + 1 | 2 |
√3 – 1 = 2 – x
x = 3 – √3
EF = 3 – √3
2) S DEFH = ED EF = (3 – √3 ) 2(3 – √3 )
S DEFH = 24 – 12√3.
Answer: 24 – 12√3.
Example 17. A deposit of 20 million rubles is planned to be opened for four years. At the end of each year, the bank increases the deposit by 10% compared to its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the investor annually replenishes the deposit by X million rubles, where X - whole number. Find highest value X, in which the bank will accrue less than 17 million rubles to the deposit over four years.
Solution: At the end of the first year, the contribution will be 20 + 20 · 0.1 = 22 million rubles, and at the end of the second - 22 + 22 · 0.1 = 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24.2 + X), and at the end - (24.2 + X) + (24,2 + X)· 0.1 = (26.62 + 1.1 X). At the beginning of the fourth year the contribution will be (26.62 + 2.1 X), and at the end - (26.62 + 2.1 X) + (26,62 + 2,1X) · 0.1 = (29.282 + 2.31 X). By condition, you need to find the largest integer x for which the inequality holds
(29,282 + 2,31x) – 20 – 2x < 17
29,282 + 2,31x – 20 – 2x < 17
0,31x < 17 + 20 – 29,282
0,31x < 7,718
| x < | 7718 |
| 310 |
| x < | 3859 |
| 155 |
| x < 24 | 139 |
| 155 |
The largest integer solution to this inequality is the number 24.
Answer: 24.
At what a system of inequalities
| x 2 + y 2 ≤ 2ay – a 2 + 1 | |
| y + a ≤ |x| – a |
has exactly two solutions?
Solution: This system can be rewritten in the form
| x 2 + (y– a) 2 ≤ 1 | |
| y ≤ |x| – a |
If we draw on the plane the set of solutions to the first inequality, we get the interior of a circle (with a boundary) of radius 1 with center at point (0, A). The set of solutions to the second inequality is the part of the plane lying under the graph of the function y = |
x| –
a,
and the latter is the graph of the function
y = |
x|
, shifted down by A. The solution to this system is the intersection of the sets of solutions to each of the inequalities.
Consequently, this system will have two solutions only in the case shown in Fig. 1.
The points of contact of the circle with the lines will be the two solutions of the system. Each of the straight lines is inclined to the axes at an angle of 45°. So it's a triangle PQR– rectangular isosceles. Dot Q has coordinates (0, A), and the point R– coordinates (0, – A). In addition, the segments PR And PQ equal to the radius of the circle equal to 1. This means
| Qr= 2a = √2, a = | √2 | . |
| 2 |
| Answer: a = | √2 | . |
| 2 |
Let Sn sum P terms of an arithmetic progression ( a p). It is known that S n + 1 = 2n 2 – 21n – 23.
a) Provide the formula P th term of this progression.
b) Find the smallest absolute sum S n.
c) Find the smallest P, at which S n will be the square of an integer.
Solution: a) It is obvious that a n = S n – S n- 1 . Using this formula, we get:
S n = S (n – 1) + 1 = 2(n – 1) 2 – 21(n – 1) – 23 = 2n 2 – 25n,
S n – 1 = S (n – 2) + 1 = 2(n – 1) 2 – 21(n – 2) – 23 = 2n 2 – 25n+ 27
Means, a n = 2n 2 – 25n – (2n 2 – 29n + 27) = 4n – 27.
B) Since S n = 2n 2 – 25n, then consider the function S(x) = | 2x 2 – 25x|. Its graph can be seen in the figure.
Obviously, the smallest value is achieved at the integer points located closest to the zeros of the function. Obviously these are points X= 1, X= 12 and X= 13. Since, S(1) = |S 1 | = |2 – 25| = 23, S(12) = |S 12 | = |2 · 144 – 25 · 12| = 12, S(13) = |S 13 | = |2 · 169 – 25 · 13| = 13, then the smallest value is 12.
c) From the previous paragraph it follows that Sn positive, starting from n= 13. Since S n = 2n 2 – 25n = n(2n– 25), then the obvious case, when this expression is a perfect square, is realized when n = 2n– 25, that is, at P= 25.
It remains to check the values from 13 to 25:
S 13 = 13 1, S 14 = 14 3, S 15 = 15 5, S 16 = 16 7, S 17 = 17 9, S 18 = 18 11, S 19 = 19 13, S 20 = 20 13, S 21 = 21 17, S 22 = 22 19, S 23 = 23 21, S 24 = 24 23.
It turns out that for smaller values P perfect square is not achieved.
Answer: A) a n = 4n– 27; b) 12; c) 25.
________________
*Since May 2017, the united publishing group "DROFA-VENTANA" has been part of the Russian Textbook corporation. The corporation also includes the Astrel publishing house and the LECTA digital educational platform. General Director Alexander Brychkin, graduate of the Financial Academy under the Government of the Russian Federation, candidate economic sciences, head of innovative projects of the DROFA publishing house in the field of digital education (electronic forms of textbooks, Russian Electronic School, digital educational platform LECTA). Before joining the DROFA publishing house, he held the position of vice president for strategic development and investments of the publishing holding EKSMO-AST. Today, the publishing corporation "Russian Textbook" has the largest portfolio of textbooks included in the Federal List - 485 titles (approximately 40%, excluding textbooks for special schools). The corporation's publishing houses own the most popular Russian schools sets of textbooks on physics, drawing, biology, chemistry, technology, geography, astronomy - areas of knowledge that are needed for the development of the country's production potential. The corporation's portfolio includes textbooks and teaching aids for primary schools, awarded the Presidential Prize in the field of education. These are textbooks and manuals on subject areas, which are necessary for the development of the scientific, technical and production potential of Russia.
Problem No. 5922.
The owner agreed with the workers that they would dig a well under the following conditions: for the first meter he would pay them 3,500 rubles, and for each subsequent meter - 1,600 rubles more than for the previous one. How much money will the owner have to pay the workers if they dig a well 9 meters deep?
Since the payment for each next meter differs from the payment for the previous one by the same number, we have before us.
In this progression - the payment for the first meter, - the difference in payment for each subsequent meter, - the number of working days.
The sum of the terms of an arithmetic progression is found by the formula:
Let's substitute these problems into this formula.
Answer: 89100.
Problem No. 5943.
At the exchange office you can perform one of two operations:
· for 2 gold coins you get 3 silver and one copper;
· for 5 silver coins you get 3 gold and one copper.
Nicholas only had silver coins. After several visits to the exchange office, his silver coins became smaller, no gold coins appeared, but 100 copper coins appeared. How much did Nicholas's number of silver coins decrease??
Problem No. 5960.
The grasshopper jumps along the coordinate line in any direction for a unit segment per jump. How many different points are there on the coordinate line at which the grasshopper can end up after making exactly 5 jumps, starting from the origin?
If the grasshopper makes five jumps in one direction (right or left), then it will end up at points with coordinates 5 or -5:
Note that the grasshopper can jump both to the right and to the left. If he makes 1 jump to the right and 4 jumps to the left (5 jumps in total), he will end up at the point with coordinate -3. Similarly, if the grasshopper makes 1 jump to the left and 4 jumps to the right (5 jumps in total), it will end up at the point with coordinate 3:
If the grasshopper makes 2 jumps to the right and 3 jumps to the left (5 jumps in total), it will end up at the point with coordinate -1. Similarly, if the grasshopper makes 2 jumps to the left and 3 jumps to the right (5 jumps in total), it will end up at point with coordinate 1:
Note that if the total number of jumps is odd, then the grasshopper will not return to the origin of coordinates, that is, it can only get to points with odd coordinates:
There are only 6 of these points.
If the number of jumps were even, then the grasshopper would be able to return to the origin of coordinates and all points on the coordinate line that he could hit would have even coordinates.
Answer: 6
Problem No. 5990
A snail climbs up a tree 2 m in a day, and slides down 1 m in a night. The height of the tree is 9 m. How many days will it take the snail to crawl to the top of the tree?
Note that in this problem we should distinguish between the concept of “day” and the concept of “day”.
The problem asks exactly how long days the snail will crawl to the top of the tree.
In one day the snail rises to 2 m, and in one day the snail rises to 1 m (it rises by 2 m during the day, and then descends by 1 m during the night).
In 7 days the snail rises 7 meters. That is, on the morning of the 8th day she will have to crawl 2 m to the top. And on the eighth day she will cover this distance.
Answer: 8 days.
Problem No. 6010.
In all entrances of the house same number floors, and each floor has the same number of apartments. In this case, the number of floors in the house is greater than the number of apartments on the floor, the number of apartments on the floor is greater than the number of entrances, and the number of entrances is more than one. How many floors are in the building if there are 105 apartments in total?
To find the number of apartments in a house, you need to multiply the number of apartments on the floor ( ) by the number of floors ( ) and multiply by the number of entrances ( ).
That is, we need to find ( ), based on following conditions:
(1)
The last inequality reflects the condition “the number of floors in a building is greater than the number of apartments on a floor, the number of apartments on a floor is greater than the number of entrances, and the number of entrances is more than one.”
That is, ( ) is the most larger number.
Let's factor 105 into prime factors:
Taking into account condition (1), .
Answer: 7.
Problem No. 6036.
There are 30 mushrooms in the basket: saffron milk caps and milk mushrooms. It is known that among any 12 mushrooms there is at least one saffron milk cap, and among any 20 mushrooms there is at least one milk mushroom. How many saffron milk caps are in the basket?
Because among any 12 mushrooms there is at least one camelina(or more) the number of milk mushrooms must be less than or equal to.
It follows that the number of saffron milk caps is greater than or equal to .
Because among any 20 mushrooms at least one mushroom(or more), the number of saffron milk caps must be less than or equal to
Then we found that, on the one hand, the number of saffron milk caps is greater than or equal to 19 , and on the other hand - less than or equal to 19 .
Therefore, the number of saffron milk caps equals 19.
Answer: 19.
Problem No. 6047.
Sasha invited Petya to visit, saying that he lived in the seventh entrance in apartment No. 333, but forgot to say the floor. Approaching the house, Petya discovered that the house was nine stories high. What floor does Sasha live on? (On each floor the number of apartments is the same; apartment numbers in the building begin with one.)
Let there be apartments on each floor.
Then the number of apartments in the first six entrances is equal to
Let's find the maximum natural value that satisfies the inequality ( - the number of the last apartment in the sixth entrance, and it is less than 333.)
From here
The number of the last apartment in the sixth entrance is
The seventh entrance starts from apartment 325.
Therefore, apartment 333 is on the second floor.
Answer: 2
Problem No. 6060.
On the surface of the globe, 17 parallels and 24 meridians were drawn with a felt-tip pen. How many parts do the drawn lines divide the surface of the globe into? Meridian is an arc of a circle connecting the North and South pole. parallel is a circle lying in a plane parallel to the plane of the equator.
Let's imagine a watermelon that we cut into pieces.
By making two cuts from the top to the bottom (drawing two meridians), we will cut the watermelon into two slices. Therefore, by making 24 cuts (24 meridians), we will cut the watermelon into 24 slices.
Now we will cut each slice.
If we make 1 transverse cut (parallel), then we will cut one slice into 2 parts.
If we make 2 transverse cuts (parallels), we will cut one slice into 3 parts.
This means that by making 17 cuts we will cut one slice into 18 parts.
So, we cut 24 slices into 18 pieces and got a piece.
Consequently, 17 parallels and 24 meridians divide the surface of the globe into 432 parts.
Answer: 432.
Problem No. 6069
The stick is marked with transverse lines of red, yellow and Green colour. If you cut a stick along the red lines, you will get 5 pieces, if along the yellow lines, 7 pieces, and if along the green lines, 11 pieces. How many pieces will you get if you cut a stick along the lines of all three colors?
If you make 1 cut, you will get 2 pieces.
If you make 2 cuts, you will get 3 pieces.
In general: if you make cuts, you will get a piece.
Back: to get pieces, you need to make a cut.
Let's find the total number of lines along which the stick was cut.
If you cut a stick along the red lines, you get 5 pieces - therefore, there were 4 red lines;
if on yellow – 7 pieces - therefore, there were 6 yellow lines;
and if on the green ones - 11 pieces - therefore, there were 10 green lines.
Hence the total number of lines is equal to . If you cut a stick along all the lines, you will get 21 pieces.
Answer: 21.
Problem No. 9626.
There are four gas stations on the ring road: A, B, B, and D. The distance between A and B is 50 km, between A and B is 40 km, between C and D is 25 km, between G and A is 35 km (all distances are measured along the ring road in the shortest direction). Find the distance between B and C.
Let's see how gas stations can be located. Let's try to arrange them like this:
With this arrangement, the distance between G and A cannot be equal to 35 km.
Let's try this:
With this arrangement, the distance between A and B cannot be 40 km.
Let's consider this option:
This option satisfies the conditions of the problem.
Answer: 10.
Problem No. 10041.
The list of quiz tasks consisted of 25 questions. For each correct answer, the student received 7 points, for an incorrect answer, 9 points were deducted from him, and for no answer, 0 points were given. How many correct answers did a student give who scored 56 points, if it is known that he was wrong at least once?
Let the student give correct and incorrect answers ( ). Since there were possibly other questions that he answered, we get the inequality:
Moreover, according to the condition,
Since the correct answer adds 7 points and the wrong answer subtracts 9, and the student ends up with 56 points, the equation is:
This equation must be solved in whole numbers.
Since 9 is not divisible by 7, it must be divisible by 7.
Let it be then.
In this case, all conditions are met.
Problem No. 10056.
The rectangle is divided into four small rectangles by two straight cuts. The areas of three of them, starting from the top left and then clockwise, are 15, 18, 24. Find the area of the fourth rectangle.
The area of a rectangle is equal to the product of its sides.
The yellow and blue rectangles have a common side, so the ratio of the areas of these rectangles is equal to the ratio of the lengths of the other sides (not equal to each other).
The white and green rectangles also have a common side, so the ratio of their areas is equal to the ratio of the other sides (not equal to each other), that is, the same ratio:
By the property of proportion we get
From here.
Problem No. 10071.
The rectangle is divided into four small rectangles by two straight cuts. The perimeters of three of them, starting from the top left and then clockwise, are 17, 12, 13. Find the perimeter of the fourth rectangle.
Perimeter of a rectangle equal to the sum the lengths of all its sides.
Let us designate the sides of the rectangles as indicated in the figure and express the perimeters of the rectangles through the indicated variables. We get:
Now we need to find what the value of the expression is.
Let's subtract the second from the third equation and add the third. We get:
Simplifying the right and left sides, we get:
So, .
Answer: 18.
Problem No. 10086.
The table has three columns and several rows. A natural number was placed in each cell of the table so that the sum of all numbers in the first column is 72, in the second – 81, in the third – 91, and the sum of the numbers in each row is more than 13, but less than 16. How many rows are there in the table?
Let's find the sum of all the numbers in the table: .
Let the number of rows in the table be .
According to the problem, the sum of the numbers in each line more than 13 but less than 16.
Since the sum of numbers is a natural number, only two natural numbers satisfy this double inequality: 14 and 15.
If we assume that the sum of the numbers in each row is 14, then the sum of all the numbers in the table is equal to , and this sum satisfies the inequality.
If we assume that the sum of the numbers in each row is 15, then the sum of all the numbers in the table is equal to , and this number satisfies the inequality.
So, a natural number must satisfy the system of inequalities:
The only natural that satisfies this system is
Answer: 17.
It is known about the natural numbers A, B and C that each of them is greater than 4 but less than 8. They guessed a natural number, then multiplied it by A, then added it to the resulting product B and subtracted C. The result was 165. What number was guessed?
Integers A, B and C can be equal to the numbers 5, 6 or 7.
Let the unknown natural number be equal to .
We get: ;
Let's consider various options.
Let A=5. Then B=6 and C=7, or B=7 and C=6, or B=7 and C=7, or B=6 and C=6.
Let's check: ; (1)
165 is divisible by 5.
The difference between numbers B and C is either equal to or equal to 0 if these numbers are equal. If the difference is equal to , then equality (1) is impossible. Therefore, the difference is 0 and
Let A=6. Then B=5 and C=7, or B=7 and C=5, or B=7 and C=7, or B=5 and C=5.
Let's check: ; (2)
The difference between numbers B and C is either equal to or equal to 0 if these numbers are equal. If the difference is equal to or 0, then equality (2) is impossible, since it is an even number, and the sum (165 + an even number) cannot be an even number.
Let A=7. Then B=5 and C=6, or B=6 and C=5, or B=6 and C=6, or B=5 and C=5.
Let's check: ; (3)
The difference between numbers B and C is either equal to or equal to 0 if these numbers are equal. The number 165 when divided by 7 leaves a remainder of 4. Consequently, it is also not divisible by 7, and equality (3) is impossible.
Answer: 33
Several consecutive sheets fell out of the book. The number of the last page before the dropped sheets is 352, the number of the first page after the dropped sheets is written with the same numbers, but in a different order. How many sheets fell out?
Obviously, the number of the first page after the dropped sheets is greater than 352, which means it can be either 532 or 523.
Each dropped sheet contains 2 pages. Therefore, there is an even number of pages. 352 is an even number. If we add an even number to an even number, we get an even number. Therefore, the number of the last dropped page is an even number, and the number of the first page after the dropped sheets must be odd, that is, 523. Therefore, the number of the last dropped page is 522. Then the result is 
sheets.
Answer: 85
Masha and the Bear ate 160 cookies and a jar of jam, starting and finishing at the same time. At first Masha ate jam, and Bear ate cookies, but at some point they switched. The bear eats both three times faster than Masha. How many cookies did the Bear eat if they ate the jam equally?
If Masha and the Bear ate jam equally, and the bear ate three times as much jam per unit time, then he ate jam in three times less time than Masha. In other words, Masha ate jam three times longer than Bear. But while Masha was eating jam, the bear was eating cookies. Consequently, the bear ate cookies three times longer than Masha. But the Bear, moreover, ate three times more cookies per unit of time than Masha, therefore, in the end he ate 9 times more cookies than Masha.
Now it's easy to create an equation. Let Masha eat the cookies, then the Bear ate the cookies. Together they ate the cookies. we get the equation:
Answer: 144
On the counter flower shop There are 3 vases with roses: orange, white and blue. There are 15 roses to the left of the orange vase, and 12 roses to the right of the blue vase. There are a total of 22 roses in the vases. how many roses are there in an orange vase?
Since 15+12=27, and 27>22, therefore, the number of flowers in one vase was counted twice. And this is a white vase, because it should be the vase that stands to the right of the blue one and to the left of the orange one. So, the vases are in this order: 
From here we get the system:
Subtracting the first from the third equation, we get O = 7.
Answer: 7
Ten pillars are connected to each other by wires so that exactly 8 wires come from each pillar. How many wires are there between these ten poles?
Solution
Let's simulate the situation. Let us have two pillars, and they are connected to each other by wires so that exactly 1 wire comes from each pillar. Then it turns out that there are 2 wires coming from the poles. But we have this situation:
That is, even though there are 2 wires coming from the poles, only one wire will be stretched between the poles. This means that the number of extended wires is two times less than the number of outgoing ones.
We get: - the number of outgoing wires.
Number of wires pulled.
Answer: 40
Of the ten countries, seven signed a friendship treaty with exactly three other countries, and each of the remaining three signed a friendship treaty with exactly seven. How many contracts were signed?
This task is similar to the previous one: two countries sign one general treaty. Each agreement has two signatures. That is, the number of signed agreements is half as large as the number of signatures.
Let's find the number of signatures:
Let's find the number of signed contracts:
Answer: 21
Three rays emanating from one point divide the plane into three different angles, measured in an integer number of degrees. The largest angle is 3 times the smallest. How many values can the average angle take?
Let the smallest angle be equal to , then the largest angle is equal to . Since the sum of all angles is equal, the value of the average angle is equal.
The average angle must be greater than the smallest and less than the largest angle.
We obtain a system of inequalities:
Therefore, it takes values in the range from 52 to 71 degrees, that is, all possible values.
Answer: 20
Misha, Kolya and Lesha are playing table tennis: the player who lost the game gives way to the player who did not participate in it. In the end, it turned out that Misha played 12 games, and Kolya - 25. How many games did Lesha play?
Solution
It should be explained how the tournament is structured: the tournament consists of a fixed number of games; the loser in a given game gives way to a player who did not participate in this game. At the end of the next game, the player who did not take part in it takes the place of the loser. Consequently, each player takes part in at least one of two consecutive games.
Let's find how many games there were in total.
Since Kolya played 25 games, therefore, at least 25 games were played in the tournament.
Misha played 12 games. Since he definitely took part in every second game, therefore, no more than games were played. That is, the tournament consisted of 25 games.
If Misha played 12 games, then Lesha played the remaining 13.
Answer: 13
At the end of the quarter, Petya wrote down all his grades in a row for one of the subjects, there were 5 of them, and put multiplication signs between some of them. The product of the resulting numbers turned out to be equal to 3495. What mark does Petya get in a quarter in this subject if the teacher gives only marks 2, 3, 4 or 5 and the final mark in a quarter is the arithmetic mean of all current marks, rounded according to the rounding rules? (For example, 3.2 is rounded to 3; 4.5 - to 5; 2.8 - to 3)
Let's factor 3495 into prime factors. The last digit of the number is 5, therefore the number is divisible by 5; The sum of the digits is divisible by 3, therefore the number is divisible by 3.
Got that
Therefore, Petit’s estimates are 3, 5, 2, 3, 3. Let’s find the arithmetic mean:
Answer: 3
The arithmetic mean of 6 different natural numbers is 8. By how much should the largest of these numbers be increased so that their arithmetic mean becomes 1 larger?
The arithmetic mean is equal to the sum of all numbers divided by their number. Let the sum of all numbers be equal. According to the conditions of the problem, therefore.
The arithmetic mean became 1 more, that is, it became equal to 9. If one of the numbers was increased by , then the sum increased by and became equal to .
The number of numbers has not changed and is equal to 6.
We get equality:
