Single state exam The baseline mathematics consists of 20 tasks. In the task 20, the skills of solving logical tasks are checked. A schoolboy should be able to apply his knowledge to solve problems in practice, including an arithmetic and geometric progression. Here you can learn how to solve the task of 20 EGE on the mathematics of the base level, as well as study the examples and methods of solving on the basis of detailed disassembled tasks.
All tasks EEG base All tasks (263) EE Base Task 1 (5) EE Base Task 2 (6) EE Base Task 3 (45) EE Base Task 4 (33) EE Base Task 5 (2) EE Base Task 6 (44 ) EE Base Task 7 (1) EE Base Task 8 (12) EE Basebook Task 10 (22) EE Base Task 12 (5) EE Base Task 13 (20) EE Base Task 15 (13) EE Base Task 19 (23) EE Base Task 20 (32)
On the ribbon from the middle of the middle there are two transverse stripes
On the ribbon from different sides of the middle there are two transverse stripes: blue and red. If you cut the tape on the blue strip, then one part will be longer than another on a see. If you cut over red, then one part will be longer than another on b. See Distance from red to the blue stripes.
The task about the ribbon is part of the Mathematics of the base level for the 11th grade under the number 20.
Biologists opened a variety of amoe
Biologists opened the variety of amos, each of which is smoothly divided into two minutes after a minute. The biologist is putting Ameba to the test tube, and exactly after N clock turns out to be completely filled with amos. How many minutes will need all the test tube to be filled with amos, if you are not alone in it, and K Ameb?
When demonstrating summer clothes outfits of each mannequer
When demonstrating summer clothes, each of each mannequer differ at least one of the three elements: blouse, skirt and shoes. In total, the fashion designer prepared for the demonstration of a species of blouses, B like skirts and c like shoes. How many different outfits will be shown on this demonstration?
The task of the outfits is part of the Mathematics of the basic level for the 11th grade at number 20.
A group of tourists overcame a mountain pass
A group of tourists overcame a mountain pass. The first kilometer of the lifting they were overcame for k minutes, and each next kilometer passed on L minutes longer than the previous one. The last kilometer before the vertex was passed for M minutes. After rest, n minutes on top of the tourists began the descent, which was more gentle. The first kilometer after the top was passed per p minutes, and each next next on R minutes faster than the previous one. How many hours the group spent on the entire route if the last kilometer of the descent was passed for s minutes.
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
The doctor prescribed a patient to take a medicine for such a scheme.
The doctor prescribed a patient to take a medicine according to such a scheme: on the first day he must take K drops, and on every next day - on N drops more than in the previous one. How many bubbles medication need to buy a patient for the entire course of reception, if each contain M drops?
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
According to the empirical law of Moore, the average number of transistors on chips
According to the empirical law of Moore, the average number of transistors on chips each year increases in n times. It is known that in 2005 the average number of transistors on the microcircuit was killi. Determine how many average millions of transistors were on the chip in 2003.
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
Oil Company Bujects a well for oil production
The oil company will bury the well for oil production, which lies, according to geological exploration, at a depth of N km. During the working day, the drillers pass L meters to the depth, but for the night of the well, "stuck" again, that is, filled with soil at k meters. For how many working days, oil workers will try the well to the depth of oil occurrence?
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
In the household appliances store, the volume of sales of refrigerators is seasonal.
In the home appliance store, the volume of sales of refrigerators is seasonal. In January, K refrigerators were sold, and in the three subsequent months they were sold on L refrigerators. Since May, sales increased by M units compared with the previous month. Since September, sales began to decline on N refrigerators every month regarding the previous month. How many refrigerators sold the store for the year?
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
The coach advised Andrey on the first day of classes on the treadmill
The coach advised Andrey on the first day of classes to spend on a running track L minutes, and at each next lesson to increase the time spent on the treadmill, on M minutes. How many classes are Andrei will spend on a treadmill a total of n hours of k minutes if followed by the coach's advice?
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
Every second, bacteria is divided into two new bacteria
Every second, bacteria is divided into two new bacteria. It is known that the entire volume of one glass of the bacteria is filled out for N hours. How many seconds a glass will be filled with bacteria on 1 / k part?
The task is included in the exam in the mathematics of the basic level for the 11th grade at number 20.
Four benzocolones are located on the ring road: A, B, B and G
On the ring road there are four benzocolones: A, B, B and G. Distance between A and B - K km, between A and B - L km, between B and M km, between g and a - n km (all distances Measured along the ring road through the shortest arc). Find the distance (in kilometers) between B and V.
The task about the benzocolones is part of the EGE on the mathematics of the base level for the 11th grade at number 20.
Sasha invited Petya to visit, saying that he lives
Sasha invited Petya to visit, saying that he lives in a k entrance in the apartment number M, and the floor to say forgot. Going to the house, Petya discovered that the house is N-storey. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
The task about the apartment and houses is part of the exam in the mathematics of the basic level for the 11th grade at number 20.
Task number 5922.
The owner agreed with the workers that they dig a well on the following conditions: for the first meter he will pay 3,500 rubles them, and for each next meter - for 1600 rubles more than the previous one. How much money the owner will have to pay the worker if they dig up a depth of 9 meters?
Since the payment of each next meter differs from the payment of the previous one and the same number before us.
In this progression, the fee for the first meter is the difference in payment of each subsequent meter - the number of working days.
The sum of the members of the arithmetic progression is by the formula:
We substitute these tasks in this formula.
Answer: 89100.
Task number 5943.
In the exchange station, you can make one of two operations:
· For 2 gold coins get 3 silver and one copper;
· For 5 silver coins get 3 gold and one copper.
Nicholas had only silver coins. After several visits to the exchange office of silver coins, it became less, the gold did not appear, but it appeared 100 copper. How much the amount of silver coins in Nicholas has decreased?
Task number 5960.
The grasshopper jumps along the coordinate direct in any direction on a single segment for the jump. How many different points are on the coordinate direct, in which the grasshopper may be, making exactly 5 jumps, starting to jump from the beginning of the coordinates?
If the grasshopper makes five jumps in one direction (right or left), it will turn out to be at points with 5 or -5 coordinates:
Note that the grasshopper can jump and right and left. If he does 1 jump right and 4 jump left (in the amount of 5 jumps), it will turn out to be at the point with the coordinate -3. Similarly, if the grasshopper will make 1 leap left and 4 jumping to the right (in the amount of 5 jumps), it will be at the point with the coordinate 3:
If the grasshopper does 2 jump right and 3 jump left (in the amount of 5 jumps), it will turn out to be at the point with the coordinate -1. Similarly, if the grasshopper does 2 jump left and 3 jump to the right (in the amount of 5 jumps), it will turn out to be at the point with the coordinate 1:
Note that if the total number of jumps is odd, then the coordinate of the grasshopper will not return, that is, he will be able to get only to points with odd coordinates:
These points are only 6.
If the number of jumps was even, the grasshopper would be able to return to the beginning of the coordinates and all the points on the coordinate direct, in which he could have come would have even coordinates.
Answer: 6.
Task number 5990.
Snail per day climbs up a tree on 2 m, and overnight slides on 1 m. The height of the tree is 9 m. For how many days the snail will crash before the top of the tree?
Note that in this task it is necessary to distinguish the concept of "day" and the concept of "day".
In the task, it is asked for how much dayssnail will crash to the top of the tree.
In one day, the snail rises to 2 m, and for one day snail rises to 1 m (for a day rises by 2 m, and then goes off for 1 m).
For 7 days, the snail rises by 7 meters. That is, in the morning of the 8th day, she will remain up to the top of 2 m. And over the eighth day, it will overcome this distance.
Answer: 8 days.
Task number 6010.
In all entrances at home same number Floors, and on each floor there is the same number of apartments. At the same time, 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 greater than one. How many floors in the house, if there are 105 apartments in it?
To find the number of apartments in the house, you need the number of apartments on the floor () multiply by the number of floors () and multiply by the number of entrances ().
That is, we need to find (), based on the following conditions:
(1)
The last inequality reflects the condition "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 greater than one."
That is () - the longer number.
Spread 105 to simple multipliers:
Taking into account the condition (1) ,.
Answer: 7.
Task # 6036.
In the basket there are 30 mushrooms: Ryzhik and freight. It is known that among any 12 mushrooms there are at least one redhead, and among any 20 mushrooms at least one germ. How many redheads in the basket?
Because among any 12 mushrooms have at least one redhead(or more) The number of Gruse should be less than or equal to what.
It follows that the number of fries is greater than or equal to what.
Because among any 20 mushrooms at least one germ(or more), the number of rhymes should be less than or equal to
Then they got that on the one hand, the number of rhymes is greater than or equal to 19 and on the other - less or equal than 19 .
Therefore, the number of rhymes equally 19.
Answer: 19.
Task # 6047.
Sasha invited Petya to visit, saying that he lives in the seventh entrance in Apartment number 333, and forgot the floor. Going to the house, Petya discovered that the house was nine-story. What floor does Sasha live on? (On each floor the number of apartments is the same, the rooms of the apartments in the house begin with the unit.)
Let the apartments on each floor.
Then the number of apartments in the first six entrances is equal
We 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 -
The seventh entrance begins with the 325th apartment.
Consequently, 333 Apartment is located on the second floor.
Answer: 2.
Task number 6060.
On the surface of the globe, the felt-tip pen was carried out 17 parallels and 24 meridian. How many parts conducted lines divide the surface of the globe? Meridian is an arc of a circle connecting the North and South Poles. Parallel is a circle lying in the plane parallel to the plane of the equator.
Imagine watermelon, which we cut into pieces.
By making two cuts from the top point to the bottom (by spending two meridian), we will cut the watermelon for two slices. Consequently, after spending 24 cuts (24 meridian) We will cut watermelon at 24 slices.
Now we will cut every slicker.
If we make 1 transverse section (parallel), then we cut one slicing on 2 parts.
If we make 2 transverse cuts (parallels), then we cut one slicing on 3 parts.
So, making 17 cuts, we cut one slicker on 18 parts.
So, we cut 24 slices on 18 parts, and got a piece.
Consequently, 17 parallels and 24 meridian share the surface of the globe on 432 parts.
Answer: 432.
Task # 6069.
On the stick marked the transverse lines of red, yellow and green. If you cut off the stick over the red lines, it will work out 5 pieces, if on yellow - 7 pieces, and if on green - 11 pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors?
If you make 1 incision, then you will get 2 pieces.
If you make 2 cuts, then it turns out 3 pieces.
In general, if you make cuts, you will get a piece.
Back: To get pieces, you need to make a cut.
We find the total number of lines for which the stick was cut.
If you cut a stick on the red lines, it will work out 5 pieces -consequently, the red lines were 4;
if on yellow - 7 pieces -consequently, yellow lines were 6;
and if on green - 11 pieces -consequently, green lines were 10.
Hence the total number of lines is equal. If you cut a stick over all lines, then there are 21 pieces.
Answer: 21.
Task number 9626.
On the ring road there are four benzocolones: A, B, B, and the distance between A and B - 50 km between A and 40 km, between B and G - 25 km, between G and A - 35 km (all Distances are measured along the annular road as soon as possible). Find the distance between B and V.
Let's see how the benzokolones can be located. Let's try to arrange them like this:
With this location, the distance between r and not can be equal to 35 km.
Let's try this:
With this location, the distance between A and B cannot be 40 km.
Consider this option:
This option satisfies the condition of the problem.
Answer: 10.
Task number 10041.
The list of tasks of the quiz consisted of 25 questions. For each correct answer, the student received 7 points, 9 points were written off for the wrong answer from it, and in the absence of a response, 0 points were given. How many faithful answers gave a student who scored 56 points, if it is known that at least once he was wrong?
Let the student gave the correct answers and wrong (). Since it was possible to have more questions to which he answered, we get inequality:
In addition, by condition
Since the correct answer adds 7 points, and incorrectly delivers 9, and ultimately the student scored 56 points, we obtain the equation:
This equation must be solved in integers.
Since 9 to 7 is not divided, must share by 7.
Let, then.
In this case, all conditions are performed.
Task number 10056.
The rectangle is divided into four small rectangles with two straightforward cuts. The area of \u200b\u200bthree of them, starting from the left upper and then clockwise is equal to 15, 18, 24. Locate the fourth rectangle area.
The area of \u200b\u200bthe rectangle is equal to the product of its sides.
Yellow and blue rectangles have a common side, therefore the attitude of the areas of these rectangles is equal to the ratio of lengths of other sides (not equal to each other).
White and green rectangles also have a common side, so the ratio of their areas is equal to the attitude of other parties (not equal to each other), that is, the same relation:
By property of the proportion, we get
From here.
Task number 10071.
The rectangle is divided into four small rectangles with two straightforward cuts. The perimeters of three of them, starting with the left upper and then the hourly arrow are equal to 17, 12, 13. Locate the fourth rectangle perimeter.
The perimeter of the rectangle is equal to the sum of the lengths of all of its sides.
Denote the side of the rectangles as indicated in the figure and express through the specified variable perimeters of rectangles. We get:
Now we need to find what is the value of the expression.
The second equation will be subtracted from the third equation and add the third. We get:
We simplify the right and left parts, we get:
So, .
Answer: 18.
Task number 10086.
The table has three columns and several lines. In each cell table set natural Number So that the sum of all numbers in the first column is 72, in the second - 81, in the third - 91, and the amount of numbers in each line is greater than 13, but less than 16. How many rows in the table?
We will find the sum of all numbers in the table :.
Let the number of rows in the table equals.
Under the condition of the task, the amount of numbers in each row more than 13, but less than 16.
Since the amount of numbers is a natural number, only two natural numbers are satisfied with this dual inequality: 14 and 15.
If we assume that the sum of numbers in each row is 14, then the sum of all numbers in the table is equal, and this amount satisfies inequality.
If we assume that the sum of numbers in each row is 15, then the sum of all numbers in the table is equal, and this number satisfies inequality.
So, the natural number should satisfy the system of inequalities:
The only natural satisfying this system is
Answer: 17.
About natural numbers A, in and with it is known that each of them is more than 4, but less than 8. The natural number was made, then it was multiplied by and then added to the product obtained in and deductible C. It turned out 165. What number was forgotten?
Integers A, B and Cmay be equal to numbers 5, 6 or 7.
Let an unknown natural number equal to.
We get:;
Consider various options.
Let A \u003d 5. Then b \u003d 6 and c \u003d 7, or b \u003d 7 and c \u003d 6, or b \u003d 7 and c \u003d 7, or b \u003d 6 and c \u003d 6.
Check:; (one)
165 is divided into 5.
The difference between numbers in and c is equal is or equal to 0 if these numbers are equal. If the difference is equal, then equality (1) is impossible. Consequently, the difference is 0 and
Let A \u003d 6. Then B \u003d 5 and C \u003d 7, or B \u003d 7 and C \u003d 5, or B \u003d 7 and C \u003d 7, or B \u003d 5 and C \u003d 5.
Check:; (2)
The difference between numbers in and c is equal is or equal to 0 if these numbers are equal. If the difference is or 0, then equality (2) is not possible, since it is an even number, and the sum (165 + even number) cannot be even numbers.
Let A \u003d 7. Then B \u003d 5 and C \u003d 6, or B \u003d 6 and C \u003d 5, or B \u003d 6 and C \u003d 6, or B \u003d 5 and C \u003d 5.
Check:; (3)
The difference between numbers in and c is equal is or equal to 0 if these numbers are equal. The number 165 during division by 7 gives the residue 4. Therefore, it is also not divided into 7, and equality (3) is impossible.
Answer: 33.
From the book there was a few running in a row. The last page number in front of the dropped sheets is 352, the first page number after the dropped sheets is written by the same numbers, but in a different order. How many sheets fell?
It is obvious that the number of the first page after the sheets dropped more than 352, it means that it can be either 532 or 523.
Each dropped sheet contains 2 pages. Hence the even number of pages fell. 352 - even number. If we add to the even number even, we get an even number. Consequently, the number of the last pages dropped - 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 page 522, then fell 
Sheets.
Answer: 85.
Masha and the Bear ate 160 cookies and jam jar, starting and finished simultaneously. First, Masha ate jam, and a bear - cookies, but at some point they changed. Bear and the other eats three times faster than Masha. How many cookies ate a bear if the jam they ate equally?
If Masha and the Bear have eaten the jam row, and the bear one of the time eaten three times more jam, which means he ate jam three times less than Masha. Other words, Masha ate jam three times longer than the bear. But while Masha ate jam, the bear ate cookies. Consequently, the bear eating cookies three times longer than Masha. But the Bear, besides, a unit of time eaten three times more cookies than Masha, therefore, as a result, he ate 9 times more cookies than Masha.
Now it is easy to make an equation. Let Masha ate cookies, then the bear ate cookies. Together they ate cookies. We get the equation:
Answer: 144.
On the counter of the flower shop there are 3 vases with roses: orange, white and blue. To the left of the orange vase of 15 roses, to the right of the blue vase of 12 roses. Total in vases 22 roses. How many roses in an orange vase?
Since 15 + 12 \u003d 27, and 27\u003e 22, therefore, the number of colors of one vase was considered twice. And this is a white vase, since this is due to be a vase, which stands to the right of blue and to the left of Orange. So, vases are in this order: 
From here we get the system:
Extinguished from the third equation first, we get o \u003d 7.
Answer: 7.
Ten posts are interconnected by wires so that exactly 8 wires are departed from each post. How many wires are stretched between these ten pillars?
Decision
We simulate the situation. Let we have two columns, and they are connected to the wires in each other so that the exact 1 wire is moving away from each post. Then it turns out that 2 wires leave from the poles. But we have such a situation:
That is, despite the fact that 2 wires leave from the poles, it is extended between the columns only one wire. So, the number of extended wires is two times less than the number of outgoing.
We get: - the number of outgoing wires.
Number of extended wires.
Answer: 40.
Of the ten countries, seven signed an agreement on friendship smoothly with three other countries, and each of the remaining three - exactly with the family. How many contracts were signed?
This task is similar to the previous one: two countries signed one general contract. There are two signatures on each contract. That is, the number of signed contracts is twice as smaller than the number of signatures.
We find the number of signatures:
Find the number of signed contracts:
Answer: 21.
Three rays emerging from one point break the plane into three different angle, measured by an integer degree. The largest angle is 3 times the smallest. How many values \u200b\u200bcan take the magnitude of the middle angle?
Let the smallest angle equal, then the greatest angle is equal. Since the sum of all angles is equal, the magnitude of the middle angle is equal.
The average angle should be greater than the smallest and less than the highest corner.
We receive the system of inequalities:
Consequently, it takes values \u200b\u200bin the range from 52 to 71 degrees, that is, the entire possible values.
Answer: 20.
Misha, Kolya and Lesha play table tennis: a player who losted by the party is inferior to the player who did not participate in it. As a result, it turned out that Misha played 12 parties, and Kolya - 25. How many parties played Lesha?
Decision
Explained how the tournament is arranged: the tournament consists of a fixed number of parties; The player's loser in this party is inferior to a player who did not participate in this party. According to the results of the next party, a player who did not participate in it, stands in the place of the loser. Consequently, each player takes part in at least one of two consecutive parties.
We find how many parties were.
Since Kohl played 25 parties, therefore, there were no less than 25 parties in the tournament.
Misha played 12 parties. Since he accurately participated in every second party, therefore, no more than parties were conducted. That is, the tournament consisted of 25 parties.
If Misha played 12 parties, then Lesha played the remaining 13.
Answer: 13.
At the end of the quarter, Petya discharged all his marks according to one of the items, they turned out to be 5, and put the signs of multiplication between some of them. The product of the resulting numbers turned out to be equal to 3495. What a mark comes out in Petit at a quarter to this subject if the teacher puts only the 2nd, 3, 4 or 5 marks and a final stage in a quarter is the average arithmetic all current marks, rounded according to the rounding rules? (For example, 3.2 rounded up to 3; 4.5 - to 5; 2.8 - to 3)
Spread 3495 to simple multipliers. The last figure of Numbers 5, therefore, the number is divided into 5; The amount of numbers is divided into 3, therefore the number is divided by 3.
Received that
Consequently, the estimates of Petit 3, 5, 2, 3, 3. We will find the arithmetic average:
Answer: 3.
The arithmetic average of 6 different natural numbers is 8. How much do you need to increase the greatest of these numbers so that their arithmetic average is 1 more?
The arithmetic average equal to the sum of all numbers divided by their number. Let the sum of all numbers equal to. Under the condition of the problem, therefore.
The arithmetic average was 1 more, that is, it became equal to 9. If one of the numbers increased by, then the amount increased and became equal to.
The number of numbers has not changed and equal to 6.
We get equality:
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Mathematics
Preparation for the exam in mathematics (profile level): tasks, solutions and explanations
We disassemble tasks and solve examples with the teacherExamination paper The profile level lasts 3 hours 55 minutes (235 minutes).
Minimum threshold - 27 points.
The examination work consists of two parts that differ in content, complexity and number of tasks.
The defining feature of each part of the work is the form of tasks:
- part 1 contains 8 tasks (tasks 1-8) with a brief response in the form of an integer or final decimal fraction;
- part 2 contains 4 tasks (tasks 9-12) with a brief response as an integer or final decimal fraction and 7 tasks (Tasks 13-19) with a detailed answer (full record of the solution with the justification of the actions performed).
Panova Svetlana Anatolyevna, teacher of mathematics of the highest category of school, work experience 20 years:
"In order to get a school certificate, a graduate must be handed over two required exam In the form of the exam, one of which mathematics. In accordance with the concept of the development of mathematical education in Russian Federation Ege in mathematics is divided into two levels: basic and profile. Today we will look at the options for the profile level. "
Task number 1. - checks from the participants of the exam skills to apply the skills obtained in the course of 5 - 9 classes on elementary mathematics in practical activity. The participant must own computing skills, be able to work with rational numbers, be able to round down decimal fractions, be able to translate some units of measurement to others.
Example 1. In the apartment where Peter lives, installed a consumption accounting device cold water (counter). On May 1, the counter showed the flow rate of 172 cubic meters. m of water, and on the first of June - 177 cubic meters. What the amount should pay Peter for cold water for May if the price is 1 cube. C cold water is 34 rubles 17 kopecks? Give the answer in rubles.
Decision:
1) We find the amount of water spent per month:
177 - 172 \u003d 5 (cubic meters)
2) We will find how much money will be paid for the spent water:
34.17 · 5 \u003d 170.85 (rub)
Answer: 170,85.
Task number 2.- There is one of the simplest tasks of the exam. It successfully copes the majority of graduates, which indicates the ownership of the concept of function. Type of task number 2 for the requirements codifier is a task to use acquired knowledge and skills in practical activity and everyday life. Task number 2 consists of a description using the functions of various actual dependencies between the values \u200b\u200band the interpretation of their graphs. Task number 2 checks the ability to extract information presented in tables in charts, charts. Graduates need to be able to determine the value of the function by the value of the argument when various methods Setting function and describe the behavior and properties of the function according to its schedule. It is also necessary to be able to find the most or smallest value on the schedule and build graphs of learned functions. Allowed errors are random in reading the terms of the task, reading the chart.
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Example 2. The figure shows a change in the stock exchange value of one promotion of the mining company in the first half of April 2017. On April 7, a businessman acquired 1000 shares of this company. On April 10, he sold three quarters of purchased shares, and on April 13 sold all the remaining. How many businessman lost as a result of these operations?
Decision:
2) 1000 · 3/4 \u003d 750 (shares) - are 3/4 of all purchased shares.
6) 247500 + 77500 \u003d 325000 (rub) - the businessman received after the sale of 1000 shares.
7) 340000 - 325000 \u003d 15000 (rub) - Lost businessman as a result of all operations.
Answer: 15000.
Task number 3.- is the task of the basic level of the first part, checks the skills to perform actions with geometric figures According to the content of the "Planimetry" course. In the task 3, the ability to calculate the area of \u200b\u200bthe shape on the checkered paper is checked, the ability to calculate degree of angles, calculate the perimeters, etc.
Example 3. Find the area of \u200b\u200bthe rectangle depicted on the checkered paper with a cell size of 1 cm per 1 cm (see Fig.). Give the answer in square centimeters.
Decision: To calculate the area of \u200b\u200bthis figure, you can use the peak formula:
To calculate the area of \u200b\u200bthis rectangle, we use the peak formula:
|
S. \u003d B +. |
G. | |
| 2 |
|
S. = 18 + |
6 | |
| 2 |
READ ALSO: EGE in physics: solving problems of fluctuations
Task number 4. - The task of the course "Theory of Probability and Statistics". Checked the ability to calculate the likelihood of an event in the simplest situation.
Example 4. 5 red and 1 blue dot are marked on the circle. Identify which polygons are more: those who have all the tops are red, or those that have one of the tops of blue. In response, specify how much more than others.
Decision: 1) We use the formula of the number of combinations from n. Elements in k.:
who have all the tops are red.
3) One pentagon, who has all the peaks is red.
4) 10 + 5 + 1 \u003d 16 polygons, who have all the peaks are red.
in which the peaks are red or with one blue vertex.
in which the peaks are red or with one blue vertex.
8) One hexagon, whose peaks are red with one blue vertex.
9) 20 + 15 + 6 + 1 \u003d 42 polygongs who have all the tops are red or with one blue vertex.
10) 42 - 16 \u003d 26 polygons in which a blue point is used.
11) 26 - 16 \u003d 10 polygons - as far as polygons, which have one of the vertices - a blue point, more than polygons, who have all the tops are only red.
Answer: 10.
Task number 5. - The basic level of the first part checks the ability to solve the simplest equations (irrational, indicative, trigonometric, logarithmic).
Example 5. Decide equation 2 3 + x. \u003d 0.4 · 5 3 + x. .
Decision. We divide both parts of this equation by 5 3 + h. ≠ 0, we get
| 2 3 + x. | \u003d 0.4 or | 2 | 3 + h. | = | 2 | , | ||
| 5 3 + h. | 5 | 5 |
where it follows that 3 + x. = 1, x. = –2.
Answer: –2.
Task number 6. By planimetry to find geometric values \u200b\u200b(lengths, corners, squares), modeling of real situations in geometry language. Study of constructed models using geometric concepts and theorems. The source of difficulties is, as a rule, ignorance or incorrect application of the necessary planimeters theorems.
Area of \u200b\u200ba triangle ABC equal to 129. DE. - middle line parallel to the side AB. Find the square of the trapezium Abed..
Decision. Triangle CDE Like a triangle Cab. on two corners, since the angle at the top C. General, corner SDE equal to the corner Cab. as corresponding angles with DE. || AB Sale AC. Because DE. - the middle line of the triangle under the condition, then by the medium line properties | DE. = (1/2)AB. Hence, the likeness factor is 0.5. Square of such figures include as the square of the likeness ratio, therefore
Hence, S Abed. = S. Δ ABC – S. Δ CDE = 129 – 32,25 = 96,75.
Task number 7.- Checks the use of a derivative to study the function. For a successful implementation, it is necessary to be meaningful, not formal ownership of the concept of derivative.
Example 7. To graph chart y. = f.(x.) at the point with the abscissa x. 0 was carried out by a tangent, which is perpendicular to the direct passing through points (4; 3) and (3; -1) of this schedule. Find f.′( x. 0).
Decision. 1) We use the equation of a straight line passing through two setpoints and find the equation of direct 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 | · (-one)
y. – 3 = 4x. – 16
y. = 4x. - 13, where k. 1 = 4.
2) find an angular coefficient of tangential k. 2, which is perpendicular to the direct y. = 4x. - 13, where k. 1 \u003d 4, according to the formula:
3) The angular coefficient of tangent - the derivative function at the touch point. It means f.′( x. 0) = k. 2 = –0,25.
Answer: –0,25.
Task number 8.- Checks the participants in the participants of the knowledge exam in elementary stereometry, the ability to apply the formulas for finding the surfaces of the surfaces and volumes of figures, dugrani angles, compare the volumes of such figures, be able to perform actions with geometric figures, coordinates and vectors, etc.
The volume of the cube described near the sphere is 216. Find the radius of the sphere.
Decision. 1) V. Cuba \u003d. a. 3 (where but - Cuba's edge length), so
but 3 = 216
but = 3 √216
2) as the sphere is written in the cube, it means that the length of the diameter of the sphere is equal to the length of the edge of the cube, so d. = a., d. = 6, d. = 2R., R. = 6: 2 = 3.
Task number 9. - requires a graduate the skills of transformation and simplifying algebraic expressions. Task number 9 of an increased level of difficulty with a brief answer. Tasks from the section "Calculations and Transformations" in the EE are divided into several types:
- transformation of numerical / letter trigonometric expressions.
transforming numerical rational expressions;
transformations of algebraic expressions and fractions;
transforming numeric / letter irrational expressions;
action with degrees;
transformation of logarithmic expressions;
Example 9. Calculate TGα, if it is known that cos2α \u003d 0.6 and
| 3π. | < α < π. |
| 4 |
Decision. 1) We use the dual argument formula: COS2α \u003d 2 COS 2 α - 1 and find
| TG 2 α \u003d | 1 | – 1 = | 1 | – 1 = | 10 | – 1 = | 5 | – 1 = 1 | 1 | – 1 = | 1 | = 0,25. |
| COS 2 α. | 0,8 | 8 | 4 | 4 | 4 |
So, TG 2 α \u003d ± 0.5.
3) by condition
| 3π. | < α < π, |
| 4 |
so, α is the angle II of the Queen and TGα< 0, поэтому tgα = –0,5.
Answer: –0,5.
# Advertising_insert # Task number 10.- Checks students the ability to use acquired early knowledge and skills in practical activity and everyday life. It can be said that these are tasks in physics, and not in mathematics, but all the necessary formulas and values \u200b\u200bare given in the condition. The tasks are reduced to solving a linear or square equation or linear or square inequality. Therefore, it is necessary to be able to solve such equations and inequalities, and determine the answer. The answer should be in the form of an integer or finite decimal fraction.
Two bodies mass m. \u003d 2 kg each, moving at the same speed v. \u003d 10 m / s at an angle 2α to each other. Energy (in joules), released in their absolutely inelastic collision is determined by the expression Q. = mV 2 SIN 2 α. By what lower angle 2α (in degrees) should be the body to move so that at least 50 joules seen as a result of the collision?
Decision. 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
I will depict the inequality decision graphically:
Since, according to the condition α ∈ (0 °; 90 °), it means 30 ° ≤ α< 90°. Получили, что наименьший угол α равен 30°, тогда наименьший угол 2α = 60°.
Task number 11. - It is typical, but it turns out to be difficult for students. The main source of difficulties is to build a mathematical model (drawing up equation). Task number 11 checks the ability to solve text tasks.
Example 11. On the spring holidays, the 11-grader Vasya had to solve 560 training tasks to prepare for the exam. March 18, on the last school day, Vasya decided 5 tasks. Next daily, he solved the same amount of tasks more compared to the previous day. Determine how many tasks Vasya decided on April 2 on the last day of the vacation.
Decision: Denote a. 1 \u003d 5 - the number of tasks that Vasya decided on March 18, d. - Daily number of tasks solved Vasya, n. \u003d 16 - the number of days from March 18 to April 2 inclusive, S. 16 \u003d 560 - the total number of tasks, a. 16 - the number of tasks that Vasya decided on April 2. Knowing that daily Vasya solved the same number of tasks more than the previous day, then the formulas for finding the sum of arithmetic progression can be used:560 = (5 + a. 16) · 8,
5 + a. 16 = 560: 8,
5 + a. 16 = 70,
a. 16 = 70 – 5
a. 16 = 65.
Answer: 65.
Task number 12.- Check in students the ability to perform actions with functions, be able to apply a derivative to the study of the function.
Find a maximum point y. \u003d 10LN ( x. + 9) – 10x. + 1.
Decision: 1) Find the function definition area: x. + 9 > 0, x. \u003e -9, that is, x ∈ (-9; ∞).
2) Find a derivative function:
4) The found point belongs to the gap (-9; ∞). We define the signs of the derived function and depicting the behavior of the function:
The desired point of maximum x. = –8.
Download free Working Program on Mathematics to Line UMK G.K. Muravina, K.S. Maravina, O.V. Charging 10-11 Download free algebra manualsTask number 13.- Asshole difficulty level with a detailed response that checks the ability to solve the equations most successfully solved among the tasks with the detailed response of an increased level of complexity.
a) decide the 2Log 3 2 equation (2cos x.) - 5log 3 (2cos x.) + 2 = 0
b) Find all the roots of this equation, segment belonging.
Decision: 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 | ⇔ T.K. | 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) We will find the roots lying on the segment.
From the figure it is clear that the specified segment belongs to the roots
| 11π. | and | 13π. | . |
| 6 | 6 |
| Answer: but) | π | + 2π. k.; – | π | + 2π. k., k. ∈ Z.; b) | 11π. | ; | 13π. | . |
| 6 | 6 | 6 | 6 |
The diameter of the base of the base of the cylinder is 20, forming the cylinder is 28. The plane crosses its bases along the chords of length 12 and 16. The distance between the chords is 2√197.
a) Prove that the base centers of the cylinder lie on one side of this plane.
b) Find the angle between this plane and the base plane of the cylinder.
Decision: a) chord length 12 is at a distance \u003d 8 from the center of the base of the base, and the chord length 16, similarly, is at a distance of 6. Therefore, the distance between their projections on the plane, parallel to the bases of the cylinders, is either 8 + 6 \u003d 14, or 8 - 6 \u003d 2.
Then the distance between chords is either
= = √980 = = 2√245
= = √788 = = 2√197.
By condition, the second case was realized, in it the projection of the chords lie on one side of the axis of the cylinder. It means that the axis does not cross this plane Within the cylinder, that is, the bases lie on one side of it. What was required to prove.
b) We denote the centers of the grounds for 1 and 2. We will spend from the center of the base with the chord length 12 of the middle perpendicular to this chord (it has a length of 8, as already noted) and from the center of another base - to another chord. They lie in the same plane β, perpendicular to these chordam. Let's call the middle of the smaller chord b, greater A and the projection A on the second base - h (h ∈ β). Then AB, ah ∈ β and then AB, AH perpendicular to the chord, that is, the direct intersection of the base with this plane.
So the desired angle is equal
| ∠abh \u003d Arctg. | AH. | \u003d Arctg. | 28 | \u003d arctg14. |
| Bh. | 8 – 6 |
Task number 15. - Increased difficulty with a detailed answer, checks the ability to solve inequalities, the most successfully solved among the tasks with the detailed response of an increased level of complexity.
Example 15. Decide inequality | x. 2 – 3x.| · Log 2 ( x. + 1) ≤ 3x. – x. 2 .
Decision: The area of \u200b\u200bdefinition of this inequality is the interval (-1; + ∞). Consider separately three cases:
1) Let x. 2 – 3x. \u003d 0, i.e. h.\u003d 0 or h. \u003d 3. In this case, this inequality turns into faithful, therefore, these values \u200b\u200bare included in the solution.
2) Let now x. 2 – 3x. \u003e 0, i.e. x. ∈ (-1; 0) ∪ (3; + ∞). In this case, this inequality can be rewritten in the form ( x. 2 – 3x.) · Log 2 ( x. + 1) ≤ 3x. – x. 2 and divided into 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. Considering the field of definition, we have x. ∈ (–1; –0,5].
3) Finally, consider x. 2 – 3x. < 0, при этом x. ∈ (0; 3). In this case, the initial inequality will rewrite in the form of (3 x. – x. 2) · Log 2 ( x. + 1) ≤ 3x. – x. 2. After dividing a positive expression 3 x. – x. 2, we get Log 2 ( x. + 1) ≤ 1, x. + 1 ≤ 2, x. ≤ 1. Considering the area we have x. ∈ (0; 1].
Combining the solutions received, get x. ∈ (–1; –0.5] ∪ ∪ {3}.
Answer: (–1; –0.5] ∪ ∪ {3}.
Task number 16.- Increased level refers to the tasks of the second part with a detailed answer. The task checks the skills to perform actions with geometric figures, coordinates and vectors. The task contains two points. In the first point, the task must be proved, and in the second paragraph to calculate.
In an equally chained triangle ABC with an angle of 120 °, with a vertex A, bisector BD was carried out. The DEFH rectangle is inscribed in the ABC triangle so that the FH side lies on the BC segment, and the vertex E is on the ab. a) Prove that fh \u003d 2dh. b) Find the DEFH rectangle area if AB \u003d 4.
Decision: but)
1) ΔBef is rectangular, EF⊥BC, ∠B \u003d (180 ° - 120 °): 2 \u003d 30 °, then EF \u003d BE according to the property of the category of the corner of 30 °.
2) Let EF \u003d DH \u003d x., then be \u003d 2 x., Bf \u003d. x.√3 on the Pythagorean theorem.
3) Since ΔABC isceived, it means, ∠b \u003d ∠c \u003d 30˚.
BD - bisector ∠B, it means ∠abd \u003d ∠dbc \u003d 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 \u003d 3 - √3
2) S. DEFH \u003d ED · EF \u003d (3 - √3) · 2 (3 - √3)
S. DEFH \u003d 24 - 12√3.
Answer: 24 – 12√3.
Task number 17. - Task with a detailed answer, this task checks the use of knowledge and skills in practical activity and everyday life, the ability to build and explore mathematical models. This task - text task with economic content.
Example 17. Deposit in the amount of 20 million rubles is planned to open for four years. At the end of each year, the Bank increases the contribution by 10% compared with its size at the beginning of the year. In addition, at the beginning of the third and fourth years, the depositor annually replenishes the contribution to h. million rubles, where h. - whole number. Find the greatest value h.In which the bank will be charged for less than 17 million rubles for four years.
Decision: At the end of the first year, the contribution will be 20 + 20 · 0.1 \u003d 22 million rubles, and at the end of the second - 22 + 22 · 0.1 \u003d 24.2 million rubles. At the beginning of the third year, the contribution (in million rubles) will be (24,2 + h.), and at the end - (24,2 + x) + (24,2 + x) · 0.1 \u003d (26,62 + 1.1 h.). At the beginning of the fourth year, the contribution will be (26.62 + 2.1 x), and at the end - (26.62 + 2.1 h.) + (26,62 + 2,1h.) · 0.1 \u003d (29,282 + 2.31 h.). Under the condition, it is necessary to find the greatest integer for which inequality
(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 greatest solution of this inequality is the number 24.
Answer: 24.
Task number 18. - A task of an increased level of difficulty with a detailed answer. This task is intended for competitive selection In universities with increased demands on mathematical preparation of applicants. The task high level Difficulties are a task not to apply one solution method, but to a combination of various methods. For the successful implementation of the task 18, except for durable mathematical knowledge, also a high level of mathematical culture.
With any a. The system of inequalities
| x. 2 + y. 2 ≤ 2aY. – a. 2 + 1 | |
| y. + a. ≤ |x.| – a. |
has exactly two solutions?
Decision: This system can be rewritten as
| x. 2 + (y.– a.) 2 ≤ 1 | |
| y. ≤ |x.| – a. |
If you draw a multitude of solutions of the first inequality on the plane, it turns out the inside of the circle (with the boundary) of the radius 1 with the center at the point (0, but). Many solutions of the second inequality - part of the plane lying under the graph of the function y. = |
x.| –
a.,
and the latter is a function schedule
y. = |
x.|
shifted but. The solution of this system is the intersection of sets of solutions of each of the inequalities.
Consequently, two solutions this system will only have in the case depicted in Fig. one.
Circle touch points with straight and will be two solutions of the system. Each of the straight lines is tilted to the axes at an angle of 45 °. So triangle Pqr. - Rectangular isceived. Dot Q. has coordinates (0, but) and point R. - coordinates (0, - but). In addition, segments Pr. and PQ. equal to the circle radius equal to 1. So
| QR= 2a. = √2, a. = | √2 | . |
| 2 |
| Answer: a. = | √2 | . |
| 2 |
Task number 19.- A task of an increased level of difficulty with a detailed answer. This task is intended for competitive selection to universities with increased requirements for mathematical preparation of applicants. The task of a high level of complexity is a task not to apply one decision method, but to a combination of various methods. To successfully fulfill the task 19, you must be able to search for solutions, choosing various approaches from among those known by modifying the studied methods.
Let SN. sum p members of arithmetic progression ( a P.). It is known that S N. + 1 = 2n. 2 – 21n. – 23.
a) specify the formula p- Member of this progression.
b) Find the smallest sum in the module S N..
c) find the smallest pin which S N. It will be a square of an integer.
Decision: a) obviously a N. = S N. – S N. - one . 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
it 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 picture.
Obviously, the smallest value is achieved in the integer points located closest to zero function. Obviously, these are points h.= 1, h.\u003d 12 I. h.\u003d 13. Since S.(1) = |S. 1 | = |2 – 25| = 23, S.(12) = |S. 12 | \u003d | 2 · 144 - 25 · 12 | \u003d 12, S.(13) = |S. 13 | \u003d | 2 · 169 - 25 · 13 | \u003d 13, then the smallest value is 12.
c) from the previous point implies that SN. Positive, starting with n. \u003d 13. Since S N. = 2n. 2 – 25n. = n.(2n. - 25), then the obvious case, when this expression is a complete square, is implemented at n. = 2n. - 25, that is, when p= 25.
It remains to check the values \u200b\u200bfrom 13 to 25:
S. 13 \u003d 13 · 1, S. 14 \u003d 14 · 3, S. 15 \u003d 15 · 5, S. 16 \u003d 16 · 7, S. 17 \u003d 17 · 9, S. 18 \u003d 18 · 11, S. 19 \u003d 19 · 13, S. 20 \u003d 20 · 13, S. 21 \u003d 21 · 17, S. 22 \u003d 22 · 19, S. 23 \u003d 23 · 21, S. 24 \u003d 24 · 23.
It turns out that with smaller values p full square Not achieving.
Answer: but) a N. = 4n. - 27; b) 12; c) 25.
________________
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Task 20 Basic eGE level
1) Snail per day cras up a tree on a 4 m tree, and overnight slides on 1 m. The height of the tree is 13 m. For how many days the snail will finish the top of the tree for the first time? (4-1 \u003d 3, Morning 4 days will be at an altitude of 9m, and the day will crack up 4m.Answer: 4. )
2) Snail per day cras up a tree on a 4 m tree, and overnight slides on 3 m. The height of the tree is 10 m. For how many days the snail will finish the top of the tree for the first time? Answer: 7.
3) Snail in the day climbs up the tree on a 3 m, and for the night it descends on 2 m. The height of the tree is 10 m. For how many days the snail will rise to the top of the tree? Answer: 8.
4) On the stick marked transverse lines of red, yellow and green. If you cut off the stick over the red lines, it will turn out 15 pieces if it is yellow - 5 pieces, and if the green is 7 pieces. How many pieces will turn out if you cut a stick along the lines of all three colors ? (If you cut a stick along the red lines, it will turn out 15 pieces, therefore, lines - 14. If you cut the stick on yellow - 5 pieces, therefore, lines - 4. If you cut into green - 7 pieces, lines - 6. Total lines: 14 + 4 + 6 \u003d 24 lines. Answer:25 )
5) The paste lines of red, yellow and green color are marked on the stick. If you cut off the stick over the red lines, it will work out 5 pieces, if on yellow - 7 pieces, and if on green - 11 pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors? Answer : 21
6) The sticky lines of red, yellow and green colors are marked on the stick. If you cut the stick over the red lines, it will work out 10 pieces, if on yellow - 8 pieces, if the green is 8 pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors? Answer : 24
7) In the exchange station, you can make one of two operations:
For 2 gold coins get 3 silver and one copper;
For 5 silver coins get 3 gold and one copper.
Nicholas had only silver coins. After several visits to the exchange office of silver coins, it became less, the gold did not appear, but 50 copper appeared. How much did the amount of silver coins decreased in Nicholas? Answer: 10.
8) In the exchange station, you can make one of the two operations:
· For 2 gold coins get 3 silver and one copper;
· For 5 silver coins get 3 gold and one copper.
Nicholas had only silver coins. After several visits to the exchange office of silver coins, it became less, the gold did not appear, but it appeared 100 copper. How much the amount of silver coins in Nicholas has decreased? Answer: 20.
9) In the exchange station, you can make one of two operations:
1) for 3 gold coins get 4 silver and one copper;
2) For 6 silver coins get 4 gold and one copper.
Nicola had only silver coins. After visits to the exchange office of silver coins, it became less, the gold did not appear, but 35 copper appeared. How much did the amount of silver coins decreased by Nikola? Answer: 10.
10) In exchange item, you can make one of two operations:
1) for 3 gold coins get 4 silver and one copper;
2) For 7 silver coins get 4 gold and one copper.
Nicola had only silver coins. After visits to the exchange office of silver coins, it became less, the gold did not appear, but 42 copper appeared. How much did the amount of silver coins decreased by Nikola? Answer: 30.
11) In exchange item, one of two operations can be made:
1) for 4 gold coins get 5 silver and one copper;
2) For 8 silver coins get 5 gold and one copper.
Nikolai had only silver coins. After several visits to the exchange office of silver coins, it became less, the gold did not appear, but 45 copper appeared. How much did the amount of silver coins decreased in Nicholas? Answer: 35.
12) There are 50 mushrooms in the basket: Ryzhik and freight. It is known that among any 28 mushrooms there are at least one redhead, and among any 24 mushrooms at least one germ. How many cans in the basket? ( (50-28)+1=23 - There must be rhymes. (50-24)+1=27 - There must be a weak. Answer: Gruss in the basket 27 .)
13) In the basket lies 40 mushrooms: Ryzhiki and freight. It is known that among any 17 mushrooms there are at least one rhyger, and among any 25 mushrooms at least one germ. How many redheads in the basket? ( According to the condition of the problem: (40-17)+1=24 - There must be rhymes. (40-25)+1=16 24 .)
14) The basket lies 30 mushrooms: Ryzhik and freight. It is known that among any 12 mushrooms there are at least one redhead, and among any 20 mushrooms at least one germ. How many redheads in the basket? (According to the condition of the problem: (30-12)+1=19 - There must be rhymes. (30-20)+1=11 - There must be a weak. Answer: Ryzhikov in the basket 19 .)
15) The basket lies 45 mushrooms: Ryzhiki and freight. It is known that among any 23 mushrooms there are at least one rhyger, and among any 24 mushrooms at least one germ. How many redheads in the basket? ( According to the condition of the problem: (45-23)+1=23 - There must be rhymes. (45-24)+1=22 - There must be a weak. Answer: Ryzhikov in the basket 23 .)
16) The basket lies 25 mushrooms: Ryzhiki and freight. It is known that among any 11 mushrooms there are at least one redhead, and among any 16 mushrooms at least one germ. How many redheads in the basket? ( Since among any 11 mushrooms, at least one - a rhyger, then we are no more than 10. Since among any 16 mushrooms, at least one - gently, then the rhymes are not more than 15. And since everything is in a cart 25 mushrooms, then Gruzno 10, And Ryzhikov RivneAnswer: 15.
17) The owner agreed with the workers that they dig a well on the following conditions: for the first meter he will pay them 4,200 rubles, and for each next meter, it is 1,300 rubles more than the previous one. How much money the owner will have to pay the worker if they dig a well depth 11 meters ? (Answer: 117700)
18) The owner agreed with workers that they dig a well on the following conditions: for the first meter he will pay 3,700 rubles them, and for each next meter - by 1700 rubles more than the previous one. How much money the owner will have to pay the worker if they dig up a depth of 8 meters? ( 77200 )
19) The owner agreed with the workers that they dig a well on the following conditions: for the first meter he will pay them 3,500 rubles, and for each next meter - for 1600 rubles more than the previous one. How much money the owner will have to pay the worker if they dig up a depth of 9 meters? ( 89100 )
20) The owner agreed with the workers that they dig a well on the following conditions: for the first meter he will pay them 3900 rubles, and for each next meter it will pay for 1200 rubles more than the previous one. How many rubles the owner will have to pay the workers if they dig a well of 6 meters deep? (41400)
21) The coach advised Andrei on the first day of occupations to spend 15 minutes on the treadmill, and at each next lesson increase the time spent on the treadmill, for 7 minutes. For how many lessons, Andrei will spend on a treadmill a total of 2 hours of 25 minutes, if you follow the advice of the coach? ( 5 )
22) The coach advised Andrei on the first day of occupations to spend on the treadmill 22 minutes, and at each next lesson to increase the time spent on the treadmill, for 4 minutes until it reaches 60 minutes, and then continue to train for 60 minutes every day. For how many lessons, starting from the first, Andrei will spend on a treadmill in the amount of 4 hours 48 minutes? ( 8 )
23) In the first row of the cinema of 24 places, and in each next 2 more than in the previous one. How many places in the eighth row? ( 38 )
24) The doctor prescribed a patient to take a medicine according to such a scheme: on the first day, it should take 3 drops, and on each next day - by 3 drops more than in the previous one. Having received 30 drops, it drinks 30 drops of drugs for another 3 days, and then daily reduces the reception by 3 drops. How many bubbles medication need to buy a patient for the entire course of reception, if each contain 20 ml of medication (which is 250 drops)? (2) the amount of arithmetic progression with the first term of 3, a difference equal to 3 and the last term of 30; 165 + 90 + 135 \u003d 390 drops; 3+ 3 (n.-1)=30; n.\u003d 10 and 27-3 (n.-1)=3; n.=9
25) The doctor prescribed a patient to take a medicine according to such a scheme: on the first day he must take 20 drops, and on each next day - 3 drops more than in the previous one. After 15 days of reception, the patient takes a break in 3 days and continues to take a medicine on the conference scheme: in the 19th day he takes the same drops as in the 15th day, and then reduces the dose daily by 3 drops, until the dosage becomes Less than 3 drops per day. How many bubbles medication need to buy a patient for the entire course of reception if 200 drops contain 200 drops? ( 7 ) drills 615 + 615 + 55 \u003d 1285; 1285: 200 \u003d 6.4
26) In the home appliance store, the volume of sales of refrigerators is seasonal. In January, 10 refrigerators were sold, and in the three subsequent months they were sold at 10 refrigerators. Since May, sales increased by 15 units compared with the previous month. Since September, sales began to decrease by 15 refrigerators every month regarding the previous month. How many refrigerators sold the store for the year? (360) (5*10+2*25+2*40+2*55+70=360
27) On the surface of the globe, 12 parallels and 22 meridian were held on the surface of the globe. How many parts conducted lines divided the surface of the globe?
Meridian is an arc of a circle connecting the Northern and South Poles. Parallel is a circle lying in the plane parallel to the plane of the equator. (13 · 22 \u003d286)
28) On the surface of the globe, the felt-tip pen was carried out 17 parallels and 24 meridian. How many parts conducted lines divided the surface of the globe? Meridian is an arc of a circle connecting the Northern and South Poles. Parallel is a circle lying in the plane parallel to the plane of the equator. (18 · 24 \u003d432)
29) What is the smallest number of people going in a row to take their work to be divided into 7? (2) If the condition of the task sounded like this: "What the smallest number of people going in a row should be taken to make their work guaranteed share on 7? " It would be necessary to take seven consecutive numbers.
30) What is the smallest number of in a row in a row of numbers need to take their work to share on 9? (2)
31) The work of ten walking in a row was divided into 7. What can be equal to the residue? (0) Among the 10 consecutive number, one of them will definitely be divided into 7, so the product of these numbers is multiple of seven. Consequently, the balance of division by 7 is zero.
32) The grasshopper jumps along the coordinate direct in any direction on a single segment for jumping. How many different points are on the coordinate direct, in which the grasshopper may be, making exactly 6 jumps, starting to jump from the beginning of the coordinates? ( The grasshopper may be at points: -6, -4, -2, 0, 2, 4 and 6; Total 7 points.)
33) The grasshopper jumps along the coordinate direct in any direction on a single segment for jumping. How many different points are on the coordinate direct, in which the grasshopper may be, making exactly 12 jumps, starting to jump from the beginning of the coordinates? ( the grasshopper may be at points: -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, 10 and 12; Total 13 points.)
34) The grasshopper jumps along the coordinate line in any direction on a single segment for the jump. How many different points are on the coordinate direct, in which the grasshopper may be, making exactly 11 jumps, starting to jump from the beginning of the coordinates? (It may be at points: -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9 and 11; only 12 points.)
35) The grasshopper jumps along the coordinate direct in any direction on a single segment for jumping. How many different points are on the coordinate direct, in which the grasshopper may turn out to be, making exactly 8 jumps, starting jumping from the start of coordinates?
Note that the grasshopper may be only at points with well-known coordinates, since the number of jumps that he does is clearly. The most for the grasshopper may be at points whose module does not exceed eight. Thus, the grasshopper may be at points: -8, -6,-2 ; -4, 0.2, 4, 6, 8 only 9 points.
Consider such a task plan. We have the following conditions:
Total amount:N.
From a thing at least 1 of another type, and from in the pieces of at least 1 of the first type
Then: (A-1) - the minimum number of the first species, and (B-1) - the second.
After doing check: (A-1) + (B-1) \u003dN..
EXAMPLE
AT
DECISION
So: Total fish we have 35 (perch and roach)
Consider the conditions: among any 21 fish there are at least one branch, which means the minimum of 1 the roach is there in this condition, therefore (21-1) \u003d 20 is at least the perch. Among any 16 fish are at least one perch, arguing similarly, (16-1) \u003d 15 is a minimum of roach. Now we do the check: 20 + 15 \u003d 35, that is, we received the total number of fish, which means 20 perch and 15 roaching.
Answer: 15 roach
Quiz and the number of correct answers
The list of tasks of the quiz consisted of a question. For each correct answer, the student received a glasses, for the wrong answer to it was written offb. Points, and in the absence of a response, they gave 0 points. How many faithful answers gave a student who scoredN. glasses, if you know that at least once he was wrong?
We know how many points he earned, know the price of the correct and incorrect response. Based on that, at least one incorrect answer was given, the number of points for the correct answers must exceed the number of penalty points onN. Points. Let it be given to the correct answers and in the wrong, then:
but*x.= N.+ b.* y.
x \u003d (N.+ b.* y.)/but
from this equality it can be seen that the number in brackets should be multiple a. With this in mind, we can appreciate the (it is also an integer). It should be noted that the number of correct and not correct answers should not exceed the total number of issues.
EXAMPLE
DECISION:
introduce the designation (for convenience) x - the correct, y - incorrect, then
5 * x \u003d 75 + 11 *
X \u003d (75 + 11 * y) / 5
Since 75 shares aimed at five, then 11 * y also should share a focus on five. Therefore, it can take the values \u200b\u200bof a multiple of five (5, 10, 15, etc.). We take the first value of y \u003d 5 if x \u003d (75 + 11 * 5) / 5 \u003d 26 all questions 26 + 5 \u003d 31
Y \u003d 10 x \u003d (75 + 11 * 10) \u003d 37 Total answers 37 + 10 \u003d 47 (more than questions) is not suitable.
It means everything was: 26 faithful and 5 wrong answers.
Answer: 26 faithful answers
On what floor?
Sasha invited Petya to visit, saying that he lives in and the entrance in the apartment numberN., and the floor forgot to say. Going to the house, Petya discovered that the housey-storey. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
DECISION
By the condition of the task, we know the apartment number, the entrance and the number of floors in the house. Based on this data, it is possible to assess the number of apartments on the floor. Let X be the number of apartments on the floor, then the following condition should be performed:
A * u * x should be greater than or equalN.
From this inequality, we estimate x
We pre-start the minimum integer value x, let it be with, and make a check: (A-1) * y * with lessN., and * y * with more or equalN..
By choosing the importance of X, we can easily calculate the floor (B): B \u003d (N.-( a.-1)* c.)/ c., moreover, in - an integer and getting a fractional value, we take the nearest whole (in the most part)
EXAMPLE
DECISION
We estimate the number of apartments on the floor: 7 * 7 * x larger than or equal to 462, hence Crabish or equal to 462 / (7 * 7) \u003d 9.42 means the minimum x \u003d 10. We make a check: 6 * 7 * 10 \u003d 420 and 7 * 7 * 10 \u003d 490 As a result, we got that the apartment hits the number in this diapass. Now we find the floor: (462-6 * 7 * 10) / 10 \u003d 4.2 So the boy lives on the fifth floor.
Answer: 5th floor
Apartments, floors, entrances
In all entrances of the house the same number of floors, and the PA of all floors the same number of apartments. At the same time, 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 greater than one. How many floors in the house, if there are still x apartments?
This type of task is based on the following condition: if in the house of e - floors, P - entrances and to - apartments on the floor, then the total number of apartments in the house should be equal to e * n * k \u003d x. So we need to represent in the form of a product of three numbers are not equal to 1 (under the condition of the problem). To do this, we will decompose the number of numbers on simple factors. By deciding and taking into account the terms of the task, we make a sample of the compliance of the numbers and the conditions that are specified in the tasks.
EXAMPLE
DECISION
Imagine the number 105 as a product of ordinary multipliers
105 \u003d 5 * 7 * 3, now we will return to the condition of the problem: since the number of floors is the largest, then it is 7, the number of apartments on the floor 5, and the entrances are 3.
Answer: Entrances - 7, apartments on the floor - 5, entrances - 3.
Exchange
AT
For and gold coins get silver and copper;
For x silver coins get in gold and 1 copper.
Nicholas had only silver coins. After the metabolic point of silver coins, it became less, the gold did not appear, but it appeared with copper. How much did the amount of silver coins decreased in Nicholas?
In the exchange of Punukte there are two exchange schemes:
EXAMPLE
AT exchange can be made one of two operations:
DECISION
5 gold \u003d 4 silver + 1 copper
10 silver \u003d 7 gold + 1 copper
since gold coins did not appear, we need a scheme of exchange without gold coins. Therefore, the amount of gold coins should be equal in both cases. We need to find the smallest common multiple numbers 5 and 7, and bring our gold in both cases to it:
35 gold \u003d 28 silver + 7 copper
50 silver \u003d 35 gold + 5 copper
as a result, we get
50 silver \u003d 28 silver + 12 copper
We found the exchange scheme by passing gold coins, now we need, knowing the number of copper coins, to find how many times such an operation was performed
N.=60/12=5
As a result, we get
250 silver \u003d 140 silver + 60 copper
Substituting, and having received the final exchange. We will find how much silver has been changed. So - the amount decreased by 250-140 \u003d 110
Reply to 110 coins
6. GLOBE
On the surface of the globe, the marker was taken x parallels and the meridian. For many parts, the lines spent divided the surface of the globe? (Meridian is an arc of a circle connecting the northern and southern poles, and the parallel is the border of the globe cross section with a plane parallel to the equator plane).
DECISION:
Since the parallel EO border of the globe cross section with a plane, then one breaks the globe into 2 parts, two into three parts, x on x + 1 parts
The meridian is an arc of a circle (more precisely, the semicircle) and the meridian split the surface into parts in parts is therefore turned out (x + 1) * in parts.
EXAMPLE
After conducting similar arguments, we will get:
(30 + 1) * 24 \u003d 744 (parts)
Answer: 744 parts
7. Colt
On the stick marked the transverse lines of red, yellow and green. If you cut the stick along the red lines, it will turn out and pieces, if on yellow - in pieces, and if in green - from pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors?
DECISION
To decide to take into account that the number of pieces of 1 more than the number of saws. Now it is necessary to find how many lines are marked on a stick. We get red (A-1), yellow - (B-1), green - (C-1). Finding the number of lines of each color and summing them by obtaining the total number of lines: (A-1) + (B-1) + (C-1). We add a unit to the resulting number (since the number of pieces per longer than the number of saws) get the number of pieces if you saw all the lines.
EXAMPLE
On the stick marked the transverse lines of red, yellow and green. If you cut the stick over the red lines, it will turn out 7 pieces, if on yellow, 13 pieces, and if the green is 5 pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors?
DECISION
We find the number of lines
Red: 7-1 \u003d 6
Yellow: 13-1 \u003d 12
GREEN: 5-1 \u003d 4
Total number of lines: 6 + 12 + 4 \u003d 22
Then the number of pieces: 22 + 1 \u003d 23
Answer: 23 pieces
8. Columns and strings
AT each table of the table was installed along a natural number so that the sum of all numbers in the first column is C1, in the second - C2, in the third - C3, and the sum of the numbers in each line is greater than U1, but less U2. How many rows in the table?
DECISION
Since the numbers in the cell cells do not change, the sum of all numbers of the table is equal to: C \u003d C1 + C2 + C3.
Now we will pay attention to the fact that the table consists of natural numbers, which means the amount of numbers by lines should be integers and to be in the range from (U1 + 1) to (U2-1) (since the amount of strings is limited strictly). Now we can appreciate the number of rows:
C / (U1 + 1) - Maximum number
C / (U2-1) - minimum number
EXAMPLE
AT table three columns and several lines. AT
DECISION
We find the sum of the table
C \u003d 85 + 77 + 71 \u003d 233
Determine the boundaries of the lines sum
12 + 1 \u003d 13 - minimum
15-1 \u003d 14 - maximum
We estimate the number of rows in the table
233/13 \u003d 17.92 Maximum
233/14 \u003d 16.64 Minimum
Within these limits only one integer is concluded - 17
Answer: 17.
9. Refill on ring
and the distance between A and B. - 35 km, between a and B. - 20 km, between in and g - 20 km, between g and a and V.
DECISION
Carefully reading the task, we note that almost the circumference is divided into three arcs AB, VG and AG. Based on this, we will find the length of the entire circumference (ring). For this problem, it is equal to 20 + 20 + 30 \u003d 70 (km).
Now putting all the points on the circle and signing the length of the corresponding arcs, it is easy to determine the desired distance. In this problem, BV \u003d AB-AB, that is, bv \u003d 35-20 \u003d 15
Answer: 15 km
10. Combinations
DECISION
To solve this type of tasks, you should remember what factorial
Factorial numberN.! called the product of consecutive numbers from 1 toN., that is, 4! \u003d 1 * 2 * 3 * 4.
Now back to the task. We find the total number of cubes: 3 + 1 + 1 \u003d 5. Since one color we have three cubes, then the total number of cubes can be found by formula 5! / 3! We obtain (5 * 4 * 3 * 2 * 1) / (1 * 2 * 3) \u003d 5 * 4 \u003d 20
Answer: 20 ways to place
11 . Wells
The owner agreed with the workers that they dig a well on the following conditions: for the first meter he will pay them x rubles, and for each next meter - more than the previous one. How many rubles the owner will have to pay the worker if they dig a well of depthN. meters?
DECISION:
Since the owner increases the price for each meter, then for the second it will pay (x + y), for the third - (x + 2y), for the fourth (x + 3th), etc. It is not difficult to see that this payment system recalls the arithmetic progression, where A1 \u003d X,d.= Y., n.= N.. Then
Payment for the work is nothing but the amount of this progression:
S.= ( (2a₁ + d (n - 1)) / 2) · n
EXAMPLE:
DECISION
Based on the above we geta.1=4200
d \u003d 1300.
n \u003d 11.
substituting this data in our formula we get
S \u003d ((2 * 4200 + 1300 (11-1) / 2) * 11 \u003d ((8400 + 13000) / 2) * 11 \u003d 10700 * 11 \u003d 117700
Answer: 117700.
12 . Poles and wires
X poles, are condensed with each other with wires, so that everyone has been departed exactly from the wires. How many wires are stretched between pillars?
DECISION
We find how many gaps between the columns. Between the two one gap, between three - two, between four - 3, between X - (x-1).
At each interval of the wires, then (X-1) * This entire wires between the columns.
EXAMPLE
Ten pillars are condensed with each other with wires, so that each departs exactly 6 wires. How many wires are stretched between pillars?
DECISION
Returning to the previous symbols we get:
X \u003d 9 y \u003d 6
Then we get (9-1) * 6 \u003d 8 * 6 \u003d 48
Answer: 48.
13. Saw boards and logs
It was a few log. Made x cuts and turned out to have chumbachkov. How much could you split?
DECISION
When solving, we will make one note: some tasks do not always have a mathematical solution.
Now to the task. When solving, it is necessary to take into account that the logs are more than one and when they cut each log is obtained \u003d 1 piece.
This type of task is to solve more convenient by the selection method:
Let there be two logs then pieces turn out 13 + 2 \u003d 15
Take three get 13 + 3 \u003d 16
And here you can see the dependence that the number of saws and pieces increases the same, that is, the number of logs that need to be cutting equals uh
EXAMPLE
It was a few log. Made 13 cuts and it turned out 20 pots. How much could you split?
DECISION
Returning to our reasoning, we can pick up, or you can simply 20-13 \u003d 7 means only 7 logs
Answer 7.
14 . Fallen pages
A few pages fell out of the book. The first of the pages dropped has the number X, and the number of the latter is written by the same numbers in some other order. How many pages fell out of the book?
DECISION
The numbering of the pages that fell out, begin with an odd number and should end in an even number. Therefore, we, knowing. What the number of the last dropped is written by the same numbers that the first digit to know its last digit. By permuting the remaining numbers and, given that the numbering of the page should be greater than the first dropping, we receive its number. Knowing page numbers, you can count how much they fell, while we take into account that the page x also fell out. So from the received numbers we must deduct the number (X-1)
EXAMPLE
A few pages fell out of the book. The first of the pages dropped number 387, and the number of the latter is written by the same numbers in some other order. How many pages fell out of the book?
DECISION
Relying, we get to our arguments that the number of the last dropping page should end to the figure 8. It means that we have only two options of numbers it is 378 and 738. 378 It does not fit us as it is less than the number of the first pages that means the latter that dropped it 738.
738-(387-1)=352
Answer: 352.
You should add the following: Sometimes they are asked to specify the number of sheets, then the number of pages are divided by half.
15. FINAL GRADE
At the end of a quarter, Vovochka wrote down in a row to the line of his current markers on singing and put a multiplication sign between some of them. The works of the resulting numbers turned out to be equal to x. What mark goes out of the late one of the fourth of singing?
DECISION
When solving this type of tasks, it is necessary to take into account that its estimates must be 2,3,4 and 5. Therefore, we need to decompose the number x on the multipliers of 2,3,4 and 5. And the balance of decomposition should also consist of these numbers.
Example1
At the end of a quarter, Vovochka wrote down in a row to the line of his current markers on singing and put a multiplication sign between some of them. The works of the resulting numbers turned out to be equal to 2007. What mark goes out of a quarter of a fourth?
DECISION
Deciprate the number of 2007 for multipliers
We get 2007 \u003d 3 * 3 * 223
It means its marks: 3 3 2 2 3 Now we will find the average arithmetic estimates for this set this is 2.6 consequently its estimate is three (more than 2.5)
Answer 3.
Example 2.
At the end of a quarter, Vovochka was discharged in a row all his marks according to one of the items, they turned out to be 5, and put the signs of multiplication between some of them. The work of the resulting numbers turned out to be equal to 690. What a mark goes out of a quarter of a quarter about this subject if the teacher puts only the mark 2, 3, 4 and 5 and the final level of the quarter is the average arithmetic of all current marks, rounded according to the rounding rules? (For example: 2.4 rounded up to two; 3.5 - to 4; and 4.8 - to 5.)
DECISION
690 Spread on multipliers so that the balance of decomposition consisted of numbers 2 3 4 5
690=3*5*2*23
Consequently, its ratings: 3 5 2 2 3
We find the arithmetic average of these numbers: (3 + 5 + 2 + 2 + 3) / 5 \u003d 3
It will be its assessment
Answer: 3.
16 . MENU
The restaurant's menu has x species of salads, at the sight of the first dishes, and species of second dishes and in the type of dessert. How many options for lunch from salad, the first, second and dessert can choose visitors of this restaurant?
DECISION
When solving a bit with a slight menu: let it only be a salad and the first of the options to become (x * y). Now add the second dish the number of options increases in and once and become (x * y * a). Well, now add dessert. The number of options will increase in times
Now we get the final answer:
N \u003dX * y * a * in
EXAMPLE
DECISION
Relying on the above we get:
N \u003d 6 * 3 * 5 * 4 \u003d 360
Answer: 360.
17 . We divide without residue
In this section, consider the tasks on a specific example, for greater visibility
Since we have a product of consistently reaching numbers and more than 7, then at least one must share on 7. So we have a work, one of whose multipliers to share for 7, therefore, all the work is also divided into seven, and therefore the balance of division It will be zero, or for the second task, the number of multipliers should be equal to the divider.
18.Turists
This type of task will also consider on a specific example.
To begin with, we define that we need to find: TimeShruta \u003d lift + rest + descent
Rest We know, now it is necessary to find the time of lifting and descent
Reading the task, we see that in both cases (lifting and descent) the time depends as an arithmetic progression, but we still do not know how the height was climbing, although it is easy to find:
H.=(95-50)15+1=4
We found the height of the lift, now we will find the time of lifting as an amount of arithmetic progression: Tpodeema \u003d ((2 * 50 + 15 * (4-1)) * 4) / 2 \u003d 290 minutes
Similarly, we find, considering that now the difference in progression is -10. We obtain TSPS \u003d ((2 * 60-10 (4-1)) * 4) / 2 \u003d 180 minutes.
Knowing all components can be considered the total route time:
Torshotrut \u003d 290 + 180 + 10 \u003d 480 minutes or transferred to hours (divide on 60) we will get 8 hours.
Answer: 8 hours
19.Readugables
There are two types of tasks on rectangles: perimeters and on the square
To solve such a plan of tasks, it is not difficult to prove that when breaking any rectangle with two straightforward cuts, we will get four rectangles for which the following ratios will always be performed:
P1 + P2 \u003d P3 + P4
S1 * S2 \u003d S3 * S4,
where R – perimeter , S. - square
Based on these ratios, we can easily solve the following tasks.
19.1. Perimeters
DECISION
Relying on the above we get
24 + 16 \u003d 28 + x
X \u003d (24 + 16) -28 \u003d 12
Answer: 12.
19.2 Square
The rectangle is divided into four small rectangles with two straightforward cuts. Square of three of them starting from the left upper and then clockwise equal to 18, 12 and 20. Find the fourth rectangle area.
DECISION
For the received rectangles should be performed:
18 * 20 \u003d 12 * x
Then x \u003d (18 * 20) / 12 \u003d 30
Answer: 30.
20. There and here
Snail for a day crashes up the tree on a m, and overnight slides on in m. The height of the tree with m. For how many days, the snail will finish the top of the tree for the first time?
DECISION
For one day, the snail can rise to the height (A-C) meters. Since she can rise to height A in one day, then until the last lift, she needs to overcome the height (s - a). Based on this, we get that it will rise (C-A) \\ (A-B) +1 (I add a unit as it rises in one day to height a).
EXAMPLE
DECISION
Returning to our reasoning we get
(10-4)/(4-3)+1=7
Answer for 7 days
It should be noted that this method can solve problems on the filling of anything, when something comes and something follows.
21. Link jumping
The grasshopper jumps along the coordinate direct in any direction on a single segment for the jump. How many different points are on the coordinate direct, in which the grasshopper may be by making x jumps, starting to jump from the start of coordinates?
DECISION
Suppose that the grasshopper makes all jumps in one direction, then he will fall into the point with the coordinate x. Now he jumps forward on (x - 1) jumps and one back: enters the point with the coordinate (x-2). Considering in this way all his jumps can be noted that it will be located at points with coordinates x, (x-2), (x-4), etc. This dependence is nothing more than an arithmetic progression with a difference.d.\u003d -2 and a1 \u003d x, andaN.=- X.. Then the number of members of this progression is the number of points in which it may turn out to be. We find them
aN \u003d A1 + D (N-1)
X \u003d x + d (n-1)
2x \u003d -2 (N-1)
n \u003d x + 1
EXAMPLE
DECISION
Based on the above conclusions we get
10+1=11
Answer 11 points
Tasks for self solutions:
1. Every second, bacteria is divided into two new bacteria. It is known that the entire volume of one glass of bacteria is filled in 1 hour. For how many seconds, the glass will be filled with bacteria half?
2. On the stick marked the transverse lines of red, yellow and green. If you cut off the stick over the red lines, it will turn out 15 pieces if it is yellow - 5 pieces, and if the green is 7 pieces. How many pieces do it turn out if you cut a stick along the lines of all three colors?
3. The grasshopper jumps along the coordinate direct in any direction on a single segment for one jump. The grasshopper begins to jump from the beginning of the coordinates. How many different points are on the coordinate direct, in which the grasshopper may be, making exactly 11 jumps?
4. The basket lies 40 mushrooms: Ryzhiki and freight. It is known that among any 17 mushrooms there are at least one rhyger, and among any 25 mushrooms at least one germ. How many redheads in the basket?
5. Sasha invited Petya to visit, saying that he lives in the seventh entrance in Apartment No. 462, and the floor to say forgot. Going to the house, Petya discovered that the house is seven-story. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
6. Sasha invited Petya to visit, saying that he had lived in the eighth entrance in Apartment No. 468, and forgot the floor. Going to the house, Petya found that the house is twelve-story. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
7. Sasha invited Petya to visit, saying that he lives in the twelfth entrance in Apartment No. 465, and the floor forgot to say. Going to the house, Petya discovered that the house is five-storey. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
8. Sasha invited Petya to visit, saying that he lives in the tenth entrance in Apartment No. 333, and the floor forgot to say. Going to the house, Petya discovered that the house was nine-story. What floor does Sasha live on? (On all floors, the number of apartments is the same, the rooms of the apartments in the house begin with one.)
9. The coach advised Andrey on the first day of occupations to spend 15 minutes on the treadmill, and at each next lesson increase the time spent on the treadmill, for 7 minutes. For how many lessons, Andrei will spend on a treadmill a total of 2 hours of 25 minutes, if you follow the advice of the coach?
10. The doctor prescribed a patient to take a medicine according to such a scheme: on the first day he must take 3 drops, and on each next day - 3 drops more than in the previous one. Having received 30 drops, it drinks 30 drops of drugs for another 3 days, and then daily reduces the reception by 3 drops. How many bubbles medication need to buy a patient for the entire course of reception, if each contain 20 ml of medication (which is 250 drops)?
11. The doctor prescribed a patient to take a medicine according to such a scheme: on the first day he must take 20 drops, and on each next day - 3 drops more than in the previous one. After 15 days of reception, the patient takes a break in 3 days and continues to take a medicine on the conference scheme: in the 19th day he takes the same drops as in the 15th day, and then reduces the dose daily by 3 drops, until the dosage becomes Less than 3 drops per day. How many bubbles medication need to buy a patient for the entire course of reception if 200 drops contain 200 drops?
12. The work of ten walking in a row of numbers was divided into 7. What can be equal to the residue?
13. How many ways can be put in a row two identical red cubes, three identical green cubes and one blue cube?
14. In the tank, the volume of 38 liters every hour, starting at 12 o'clock, pour a full water bucket with a volume of 8 liters. But in the bottom of the tank there is a small gap, and from it in an hour it flows 3 liters. At what point in time (in the clock) the tank will be filled completely.
15. What the smallest number of people going in a row should be taken to make their work to 7?
16. As a result, the flood of the boiler was filled with water to a level of 2 meters. Construction pump continuously pumps water, lowering its level of 20 cm per hour. Breakdown waters, on the contrary, increase the water level in the pit 5 cm per hour. How many hours of pump work the water level in the pit drops to 80 cm?
17. The restaurant menu has 6 types of salads, 3 types of first dishes, 5 types of second dishes and 4 types of dessert. How many options for lunch from salad, the first, second and dessert can choose visitors of this restaurant?
18. Petroleum company buryur well for oil production, which lies, according to geological exploration, at a depth of 3 km. During the working day, the drillers pass 300 meters deep into the depth, but during the night of the well again "stuck", that is, filled with a soil of 30 meters. For how many working days, oil workers will try the well to the depth of oil occurrence?
19. What the smallest number of people running in a row should be taken to make their work to 9?
20.
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30. {!LANG-6a4ce876c8b396f838923f58192cf808!}
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{!LANG-30efdc59587a4e3019ac1803c5c5f92f!}
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37. {!LANG-04bff8c75f0519ea7b97fb08f42e100e!}
38. {!LANG-f5e6504151dbd1c4e2df7e584bd2a231!}
39. {!LANG-b71d8df0fba9e792ce311b1bd6e16f3a!}
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41. {!LANG-6c42390ef50acab9b6194c21efb3745d!}
42. {!LANG-fd99f5cd4da6b54d7dffd081794cfdcd!}
43. {!LANG-b3fb35b8fb7d3cff4c419a87f338c4f2!}
44. {!LANG-a56bb0f5568a70bacc37bfee1891bc81!}
45. {!LANG-6779bb4bfca13b714dc31ea554ad60a5!}
46. {!LANG-341d13542c3bd36d68566154ab38b982!}
47. {!LANG-54c95bacfa5bbfb88f3b87e49a4e5ebf!}
48. {!LANG-9fdff3780f5ccf1b62e7811bdf1387b8!}
49. {!LANG-21556f66b001c206870cc49dd7b78bfe!}
50. {!LANG-9117bdf501609a789912f852a9c48dd2!}
{!LANG-3dd0c88dd7ddba0aa78bbedf8e3755c8!}
{!LANG-7f3a24a72dba56beaf33e768007aa07f!}
{!LANG-cb1618544653e3607dfd0fede2214590!}
51. {!LANG-21556f66b001c206870cc49dd7b78bfe!}
52. {!LANG-9117bdf501609a789912f852a9c48dd2!}
{!LANG-5128635ab0674752289920d6b47c1af9!}
{!LANG-30997bc3c55bc4d9b21997667c3fc345!}
{!LANG-8d67dacc6fc1df00a29dfb1cee593960!}
53. {!LANG-85a2688e58584c2744fae01d7c94ba09!}
54. {!LANG-d8ad5985031a23f23242ed3936785046!} {!LANG-0f55310bb3751bc16fe1a691d1683ccc!}
55. {!LANG-d8ad5985031a23f23242ed3936785046!} {!LANG-041c36174d0fe017486e31a0e90a5863!}
56. {!LANG-2a67bcfa3a89a72ccea94412593c62ed!}
57.
{!LANG-d8ad5985031a23f23242ed3936785046!}
{!LANG-5ebb01c801d3df8bfb318ec29af6e78c!}
{!LANG-3857c7056b5cd433b265ce4a14ba9d4c!}
{!LANG-d75a6382f3b8b6ea6c1dd9fd1555d369!}
{!LANG-b87f889b1b921e6e863f74511f8febee!}
58. {!LANG-d8ad5985031a23f23242ed3936785046!} {!LANG-2c7a30dd28a549403466894a341b5e46!}
59. {!LANG-c23fba0eaadbf9e51b7efd6cded40ab1!}
60. {!LANG-b903e71aa03885035278f747319c189a!}
61. {!LANG-ae7a98bb40224fd237cbf33c746a0acb!}
62. {!LANG-c106abeaa0da7d9929ee5124b1a0a807!}
63. {!LANG-df749a499c32aa04c765e8221624ced6!}
64. {!LANG-12a23919dca7f0e5afbdd31f54e2ce73!}
65. {!LANG-09548c5e9cfb6de447f16ea92ebeccca!}
66.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-2bd1882ebd155277b0c7253f1f7a41a1!}
{!LANG-6d896a9ebe22c076302ec7176290c010!}
{!LANG-3e5a1f4092433c8e14037f0f262cf2f2!}
67.
{!LANG-40294b9771a006d7731abb99d270b6ba!}
{!LANG-20893f1735dd946af5ac8b4673c52da8!} 
68.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-f88555260082c5d702c496511192ab7c!}
{!LANG-20eb0ab22105b0688f1a901a3e347f7c!}
{!LANG-f4335adaf77198323f1e569ba1a4a2da!}
69. {!LANG-297cfce4418e1c2482d78c3841d99881!}
70.
{!LANG-c511eba2694dfc3b0e5a716143a36fc3!}
{!LANG-b0aea3f499b10e20763f20e1f4aa8882!}
71. {!LANG-7f613a0d062a2d2dbf6052cd2c4801a9!}
72. {!LANG-932796a02805e8c2adf382190c0a89f0!}
73.
{!LANG-c90fa827b996e0fe461e295760c60aa1!} 
74.
{!LANG-8f736fd9c753c1261f377638a4673be0!} 
75. {!LANG-d8ad5985031a23f23242ed3936785046!} table three columns and several lines. AT {!LANG-49c4189286faea4b640ec946e0e67c38!}
76. {!LANG-f7483214907b6cc530b418604398f400!}
77. {!LANG-b78443d47d2a46a757bd7c9f5ec88c47!}
78.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-c1238b312b09959a5bc898b8df01bfe6!}
{!LANG-43f0965b8d9928372bda0878c6c5db17!}
{!LANG-5d4f5e782402bdecef78ac88921b16fa!}
79.
{!LANG-03b8fb7e5cbc7f446fecbbe479e6f254!}
80. {!LANG-254204636e3dbfa1813a4771a39e5961!} and the distance between A {!LANG-d0941cc4ab94f01e6afa96373b6f6ccd!} {!LANG-9b5c69a371be8bcdd438c33a4c2cfeeb!} {!LANG-6736fefb9dba7922e68223e1fcd83552!} {!LANG-1ddac4d6559f3f75b67edd1c3b55a076!} and g - 20 km, between g and a {!LANG-97787868fe54402f31d866b70bf9b195!} {!LANG-d97d465e0ef9a51ed76d3df5b25f5048!}
81.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-fd831144b46023ddd8ab1cf057661a22!}
{!LANG-b7796bfb47f8f3e7ed58aa4bded846db!}
{!LANG-41e51a1fb324e3b908d24d12229a9535!}
{!LANG-7309e67e7f9296a41fd9ca709a644b54!}
83.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-3de89c22bdcd8d129440c9b4e7ffaab0!}
{!LANG-1a68728025d7eecc12e65ccd5af045e2!}
{!LANG-1ab0d1cfc85e8b4fab63b49e52c6b0d9!}
84.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
{!LANG-db1c1526bba0d2aabd9e45478d26bea1!}
{!LANG-6f2e8d4495391565f83ee7ab6e05376f!}
{!LANG-79b017d7dec9d0512f347046f8de9b5f!}
85.
{!LANG-5ab1f756065348458443e24ccee7fe94!}
86.
{!LANG-d8ad5985031a23f23242ed3936785046!}
exchange can be made one of two operations:
1) for 3 gold coins get 4 silver and one copper;
{!LANG-0e49bee270449ff72e9bba003143ff07!}
{!LANG-ade51f713b05227d6f2818d6df7e76d7!}
{!LANG-9c3d7921e7fbcd498a9f507ac0d6fdf9!}