Shinkarev Egor Alexandrovich
Collection of problems
Non-standard movement tasks
Scientific director of the project Natalya Nikolaevna Kudryavtseva
The collection contains detailed solutions to problems conventionally classified into the following groups: circular motion, motion of extended bodies, and offers problems for independent solution. This selection of problems can be used to develop skills in solving problems of this type in preparation for the Unified State Exam and mathematics Olympiads. The collection can be useful for students in grades 8-11, and teachers for organizing the consolidation and repetition of movement tasks both in class and in extracurricular activities.
Abakan 2017
Introduction__________________________________________________________3
Chapter 1. Problems involving movement in a circle
§ 1.1. Problems involving movement in a circle, in one direction, at one time, from one point__________________________________________________________4
§ 1.2. Problems involving movement in a circle, in one direction, at one time from diametrically opposite points___________________________6
§ 1.3. Problems involving movement in a circle, in one direction, at different times from one point.………….7
§ 1.4. Problems involving movement in a circle, in opposite directions, at the same time from one point.………………..8
Chapter 2. Problems on the motion of extended bodies
§ 2.1. Problems involving the motion of two extended bodies in one direction
§ 2.2. Problems on the motion of two extended bodies towards each other
§ 2.3. Problems involving the motion of one extended body relative to another stationary one
§ 2.4. Problems involving the motion of an extended body relative to a fixed point
§ 2.5. Problems on the motion of an extended body and a point towards each other
§ 2.6 problems on the movement of an extended body and a point in one direction______
Introduction
In practice there are a lot of interesting movement problems. Entertaining problems are offered at various Olympiads and final exams. This collection contains only problems conditionally classified into the following groups: problems on motion in a circle, problems on the motion of extended bodies.
Each group has subgroups that differ from each other in their solution methods.
This collection of problems presents a selection of problems of each type with answers. The collection contains detailed solutions to problems of each type and offers tasks for independent solution. This selection of problems can be used to practice skills in solving problems of this type in preparation for the Unified State Exam, Unified State Exam and mathematics Olympiads. The collection can be useful for students in grades 8-11, teachers for organizing the consolidation and repetition of movement tasks, both in class and in extracurricular activities.
Chapter 1
Circular movement problems§1.1 Problems involving movement in a circle, in one direction, at one time, from one point
Task: From one point on a circular track, the length of which is 14 km, two cars started simultaneously in the same direction. The speed of the first car is 80 km/h, and 40 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
Solution:
Speed
Time
Distance
1st car
80 km/h
80*= km
2nd car
X km/h
x ⋅ km
Knowing that in 2/3 of an hour the first car covered a circle, that is, 14 km more than the second, we will create an equation.
X ⋅ +14;
2x=160 −14 ⋅ 3;
x=59 .
Answer: 59 km/h
1. Two runners start simultaneously in the same direction from the same location on a circular track in a multi-lap race. One hour later, when one of them had 1 km left before the end of the first lap, he was informed that the second runner completed the first lap 20 minutes ago. Find the speed of the first runner if it is known that it is 8 km/h less than the speed of the second. (13)( )
2. Two drivers are racing. They will have to drive 60 laps along a 3 km long ring track. Both racers started at the same time, and the first one reached the finish line 10 minutes earlier than the second one. What was the average speed of the second driver, if it is known that the first driver overtook the second driver for the first time after 15 minutes? Give your answer in km/h. (108) ( )
3. Two drivers will have to drive 85 laps around an 8 km long ring track. Both riders started at the same time, and the first one reached the finish line 17 minutes earlier than the second. What was the average speed of the second driver, if it is known that the first driver overtook the second driver for the first time after 48 minutes? Give your answer in km/h.
(150)( )
4. Two drivers will have to drive 68 laps around a 6 km long ring track. Both riders started at the same time, and the first one reached the finish line 15 minutes earlier than the second one. What was the average speed of the second driver, if it is known that the first driver overtook the second driver for the first time after 60 minutes? Give your answer in km/h.
(96 )( )
5.
Two points, moving around a circle in the same direction, meet every 12 minutes, and the first one goes around the circle 10 seconds faster than the second. What part of the circle does each point cover in 1 s? (1/80 and 1/90 of the circle)
)
§1.2. Problems involving movement in a circle, in one direction, at one time from diametrically opposite points
Task: Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 14 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 21 km/h greater than the speed of the other?
Solution:
Speed
Time
Distance
1st motorcyclist
X km/h
t h
xt km
2nd motorcyclist
X + 21 km/h
t h
(x+21)t km
Let motorcyclists be on the road for the same time equal to t
hours. In order for motorcyclists to catch up, the faster one must overcome the distance initially separating them, equal to half the length of the track, that is, 14:2 = 7 km. Therefore, the path traveled by the second motorcyclist is 7 km longer than the path traveled by the first:
(x+21)t−xt=7;
21t=7
t=h
Thus, the motorcyclists will reach each other in t= hours or in 20 minutes.
Let's give another solution
A fast motorcyclist moves relative to a slow one at a speed of 21 km per hour, and must overcome the 7 km separating them. Therefore, it will take him one-third of an hour to do this.
Answer: 20 min
Tasks for independent solution:
6.Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 22 km. In how many minutes will the motorcyclists meet for the first time if the speed of one of them is 20 km/h greater than the speed of the other? (33)(https://www.metod-kopilka.ru/konspekt_uroka_matematiki_po_teme_reshenie_zadach_na_dvizhenie_po_okruzhnosti-59657.htm)
7. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 5 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 5 km/h greater than the speed of the other? (30) (https://www.metodkopilka.ru/konspekt_uroka_matematiki_po_teme_reshenie_zadach_na_dvizhenie_po_okruzhnosti-59657.htm)
8 . Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 14 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 21 km/h greater than the speed of the other? (20)
9 . Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 27 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 27 km/h greater than the speed of the other? (30)
10. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 6 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 9 km/h greater than the speed of the other? (20)
§1.3. Problems involving movement in a circle, in one direction, at different times from one point
Task: A cyclist left point A of the circular track, and 30 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h.
Solution:
Speed
Time
Distance
1 meeting
Cyclist
X km/h
40min=h
Motorcyclist
4X km/h
10 min=h
2 meeting
Cyclist
X km/h
Motorcyclist
4X km/h
By the time of the first overtaking, the motorcyclist has covered the same distance in 10 minutes as the cyclist did in 40 minutes, therefore, his speed is 4 times greater. Therefore, if the speed of the cyclist is taken to be x km/h, then the speed of the motorcyclist will be 4x km/h, and the speed of their approach will be 3x km/h.
On the other hand, the second time the motorcyclist caught up with the cyclist in 30 minutes, during which time he traveled 30 km more. Consequently, their speed of approach is 60 km/h.
So, 3x=60 km/hour, from which the speed of the cyclist is 20 km/hour, and the speed of the motorcyclist is 80 km/hour.
Answer: 80 km/h.
Tasks for independent solution:
11 . From pointA cyclist left the circular track, and 10 minutes later a motorcyclist followed him. 2 minutes after departure, he caught up with the cyclist for the first time, and 3 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 5 km. Give your answer in km/h. (6) ( https://www.metodkopilka.ru/konspekt_uroka_matematiki_po_teme_reshenie_zadach_na_dvizhenie_po_okruzhnosti-59657.htm
12. From pointA cyclist left the circular track, and 40 minutes later a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and another 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h. (60) ( https://www.metodkopilka.ru/konspekt_uroka_matematiki_po_teme_reshenie_zadach_na_dvizhenie_po_okruzhnosti-59657.htm)
13. A cyclist left point “A” of the circular route, and 50 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 18 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 15 km. Give your answer in km/h.(60)
14. A cyclist left point “A” of the circular route, and 30 minutes later a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and another 12 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 15 km. Give your answer in km/h.(95)
15. A cyclist left point “A” of the circular route, and 40 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 36 km. Give your answer in km/h.(75)
§1.4. Problems involving movement in a circle, in opposite directions, at the same time from one point
Z hell acha 1: A certain point A is taken on a circle. Two bodies simultaneously emerge from this point and move uniformly along the given circle in opposite directions. At the moment of their meeting, it turned out that the first body had traveled 10 meters more than the second. In addition, the first body arrived at point A 9 seconds later, and the second one 16 seconds after the meeting. Determine the circumference in meters.
Solution:
Time
Distance
1st point
X km/h
t h
xt km
2nd point
y km/h
t h
Ytkm
Let x be the speed of one point moving clockwise, and y the speed of the second. Then, before meeting, the first point will travel a distance xt, and the second will travel a distance yt.
After meeting the first point to the starting point, you need to go the same distance as the second one went before the meeting, and the first point spends this time equal to 10 s, and the second one, on the contrary, needs to go the distance that the first one walked before the meeting, and it spends this time 16 p. We get the following equalities:
Xt=16y
Yt=9x
Let us express the time it takes for the points to move until they meet t
t= =
Where do we get it from?
x=
According to the condition, the first body traveled 10 m more than the second, that is
16y-9x=10
We replace one of the unknowns in this equation:
16 y-12 y =10
And we find Y=2.5 from where x= .
The total length of the circle is: 70
Answer: the circumference is 70 m.
Tasks for independent solution:
16. Two bodies moving in different directions along a circle 500 m long at constant speeds meet every 125 seconds. When moving in one direction, the first body catches up with the second every 12.5 seconds. Find the speed of each body. (22 and 18)
17. From point A of a circular track, two bodies simultaneously begin to move uniformly in opposite directions. By the time they meet, the first body travels 100 meters more than the second and returns to point A 9 minutes after the meeting. Find the length of the path in meters if the second body returns to point A 16 minutes after the meeting. (700)
18. Two bodies moving around a circle in one direction meet every 112 minutes, and moving in opposite directions - every 16 minutes. In the second case, the distance between the bodies decreased from 40 m to 26 m in 12 s. How many meters per minute does each body travel and what is the circumference? (1120 m; 40 m/min, 30 m/min)
19. V 2.4
20. V 2.4
Chapter 2
Problems on the motion of extended bodies
§2.1. Problems involving the movement of two extended bodies in one direction
Task: Two dry cargo ships travel along the sea in parallel courses in one direction: the first is 130 meters long, the second is 120 meters long. At first, the second cargo ship lags behind the first, and at some point in time the distance from the stern of the first cargo ship to the bow of the second is 600 meters. 11 minutes after this, the first cargo ship lags behind the second so that the distance from the stern of the second cargo ship to the bow of the first is 800 meters. How many kilometers per hour is the speed of the first cargo ship less than the speed of the second? (http://www.ug.ru/method_article/519)
Solution:
Time
Distance
2 - 1
X m/min
11 min
600+130+120+800= 1650 m
The distance traveled by the bow 2 of the cargo ship is equal to: the initial distance from the bow 2 of the cargo ship to the stern 1(600) + length 1(130) + length 2(120) + final distance from the bow 1 to the stern 2(800) = 1650 m
V=S:t
V = 1650: 11= 150 m/min =9 km/h
Answer: 9 km/h
Tasks for independent solution:
21. Passenger and freight trains travel along two parallel railway tracks in the same direction at speeds of 80 km/h and 50 km/h, respectively. The length of a freight train is 800 meters. Find the length of the passenger train if the time it takes to pass the freight train is 2 minutes. Give your answer in meters. (200)
22. Two dry cargo ships follow parallel courses in the same direction across the sea: the first is 110 meters long, the second is 70 meters long. At first, the second cargo ship lags behind the first, and at some point in time the distance from the stern of the first cargo ship to the bow of the second is 200 meters. 8 minutes after this, the first cargo ship lags behind the second so that the distance from the stern of the second cargo ship to the bow of the first is 500 meters. How many kilometers per hour is the speed of the first cargo ship less than the speed of the second? (6.6)
( )
23. Two barges follow parallel courses in the same direction across the sea: the first is 70 meters long, the second is 30 meters long. First, the second barge lags behind the first, and at some point in time the distance from the stern of the first barge to the bow of the second is 250 meters. 14 minutes after this, the first barge lags behind the second so that the distance from the stern of the second barge to the bow of the first is 350 meters. How many kilometers per hour is the speed of the first barge less than the speed of the second? (3)
( )
24. Two barges follow parallel courses in the same direction across the sea: the first is 60 meters long, the second is 40 meters long. First, the second barge lags behind the first, and at some point in time the distance from the stern of the first barge to the bow of the second is 200 meters. 18 minutes after this, the first barge lags behind the second so that the distance from the stern of the second barge to the bow of the first is 300 meters. How many kilometers per hour is the speed of the first barge less than the speed of the second? (2.1)
( )
25 . Two dry cargo ships follow parallel courses in the same direction across the sea: the first is 120 meters long, the second is 80 meters long. At first, the second cargo ship lags behind the first, and at some point in time the distance from the stern of the first cargo ship to the bow of the second is 400 meters. 12 minutes after this, the first cargo ship lags behind the second so that the distance from the stern of the second cargo ship to the bow of the first is 600 meters. How many kilometers per hour is the speed of the first cargo ship less than the speed of the second? (6)
( )
§3
Problems for writing numbers digitally
Task 1: Find the smallest four-digit number that is a multiple of 11 and whose product of its digits is 12.
Solution:
The number must be a multiple of 11, that is, the difference between the digits in even positions and the digits in odd positions is a multiple of 11. Consider the case when their difference is 0. Note that 0 should not occur, since when multiplied by 0 we get 0 Since the number is the smallest, let's take the first digit 1. The number will take the form 1bcd. And so 1 + c = b + d and c×b×d=12. Moreover, if we imagine 12 as a product of 3 numbers, we get 12 = 2×3×2, while 2+2 = 3+1 and we get 1232
Answer: 1232
Tasks for independent solution:
26. Find a four-digit number that is a multiple of 22, the product of whose digits is 40. In your answer, indicate one such number.
27. Find a four-digit number that is a multiple of 22, the product of whose digits is 60. In your answer, indicate one such number.
28. Find a four-digit number that is a multiple of 18, the product of whose digits is 24. In your answer, indicate one such number.
29. Find a four-digit number that is a multiple of 33, the product of whose digits is 40. In your answer, indicate one such number.
30. Find the smallest four-digit number, a multiple of 11, whose product of its digits is 12
Task 2: Find a six-digit natural number that is written only as 1 and 0 and is divisible by 24.
Solution:
For a number to be divisible by 24 it must be divisible by 3 and 8.
A number is divisible by 8 if its last three digits form a number divisible by 8.
The number you are looking for is written only with zeros and ones, which means it ends in 000. A number is divisible by 3 if its sum of the digits of the number is divisible by 3. Since the last three digits of the number are zeros, the first three must be ones. Thus, the only number that satisfies the conditions of the problem is the number 111,000.
Answer: 111000
Tasks for independent solution:
31. Find a six-digit natural number that is written only as 2 and 0 and is divisible by 120. Indicate one such number in your answer.
32. Find a six-digit natural number that is written only as 1 and 5 and is divisible by 45. Indicate one such number in your answer.
33. Find a six-digit natural number that is written only as 2 and 3 and is divisible by 6.
34. Find a six-digit natural number that can only be written as 7 and 3 and is divisible by 11.
35. Find a six-digit natural number that can only be written as 3, 4, 9 and 5 and is divisible by 9.
36. Find the smallest natural number divisible by 36 whose notation contains all 10 digits.
37. Find a six-digit natural number divisible by 47 that can only be written as 2, 8, and 0.
Task 3: The sum of the digits of a three-digit natural number A is divided by 12. The sum of the digits of the number A+6 is also divided by 12. Find the smallest possible number A.
Solution: For convenience, let's call our number abc. Each letter denotes a separate digit of the number A: a - hundreds, b - tens, c - units. The sum of the digits a + b + c must be divisible by 12. Let's assume that this is the case, and let's try to choose a number A + 6 such that the sum of its digits is also divisible by 12. Note that the sum of the digits of the number A + 6 must be different from the sum of the digits of the number A by 12, 24, ... Otherwise, it will not be divisible by 12. Consider all possible options:
Option 1. If c<4 (разряд единиц не переполнится), то новое число будет равно: A + 6 = ab(c + 6) Сумма его цифр a + b + c + 6 отличается от суммы изначального числа abc на 6. Поэтому такой вариант не подходит.
Option 2. If c ≥ 4 and b<9 (чтобы не было переполнения разряда десятков), то новое число будет равно: A + 6 = a(b + 1)(c - 4) Разряд единиц получен следующим образом: c + 6 - 10 = c - 4 То есть к c мы прибавляем 6 и получаем число, превышающее 10. 10 уходит в разряд десятков, поэтому в разряде единиц остается только c - 4. Сумма цифр этого числа равна a + b + 1 + c - 4 = a + b + c - 3 Она отличается от суммы числа A на 3, поэтому такой вариант также не подойдет.
Option 3. If c ≥ 4, b = 9, a<9 (чтобы разряд сотен не переполнился), тогда новое число будет равно: A + 6 = (a + 1)0(c - 4) Сумма цифр нового числа равна: a + 1 + 0 + c - 4 = a + c - 3 Сумма цифр числа A при b = 9 равна: a + 9 + c получается, что 2 этих числа отличаются на 12 (9 - (-3)). Такой вариант подойдет.
Option 4. If c ≥ 4, b = 9, a = 9, then the new number A + 6 will be equal to: A + 6 = 100(c - 4) The sum of the digits of this number is: 1 + 0 + 0 + c - 4 = c - 3 The sum of the digits of the number A with a = 9 and b = 9 is equal to: 9 + 9 + c = c + 18 It turns out that these 2 numbers differ by 21 (18 - (-3)). This option will not work. So the digits of abc must match c ≥ 4, b = 9, a< 9. Чтобы сумма цифр числа abc делилась на 12, нужно чтобы она была равна 12 или 24 (Сумма цифр трехзначного числа не может быть больше 27 = 9 + 9 + 9). Поскольку b = 9, а c ≥ 4 у нас уже получается число, больше 13. Значит сумма цифр числа abc должна быть равна 24. Поскольку b = 9, на a + c остается 24 - 9 = 15. Рассмотрим возможные варианты: c = 4 и a = 11 - не подходит, так как в одном разряде может быть только цифра c = 5 и a = 10 - тоже c = 6 и a = 9, то есть число равно 996 c = 7 и a = 8, то есть число равно 897 c = 8 и a = 7, то есть число равно 798 c = 9 и a = 6, то есть число равно 699. Минимальным из подобранных чисел является 699. Проверим, что мы все сделали правильно: 6 + 9 + 9 = 24; 24 / 12 = 2; 699 + 6 = 705; 7 + 0 + 5 = 12; 12 / 12 = 1
Answer: 699
Tasks for independent solution:
38. The sum of the digits of a natural three-digit number A is divisible by 13. The sum of the digits of the number A+5 is also divisible by 13. Find such a number A.
39. The sum of the digits of a natural three-digit number A is divisible by 12. The sum of the digits of the number A+6 is also divisible by 12. Find the smallest number A that satisfies the condition A › 700.
40. Find a three-digit number A that has all the following properties:
the sum of the digits of the number A is divided by 6
the sum of the digits of A+3 is also divisible by 6
number A is more than 350 and less than 400
Please indicate one such number in your answer.
§4
Problems involving crossing out and adding numbers
Task 1: Cross out three digits in the number 123456 so that the resulting number is divisible by 27. Indicate the number in your answer.
Solution:
Let's start with numbers that begin with the number 1 so that the order is not broken:
123, 124, 125, 126, 134, 135, 136, 145, 146, 156.
Among these numbers, 135 is divisible by 27 (13–8·5= –27)
Next, we check the numbers that begin with the number 2:
234, 235, 236, 245, 246, 256
We check numbers that start with 3:
345, 346, 356.
No number is divisible by 27.
Let's move on to numbers that start with the number 4.
456: not divisible by 27.
Thus, we get the number 135
Answer: 135
Tasks for independent solution:
41. Cross out three digits in the number 123456 so that the resulting number is divisible by 35. Indicate the number in your answer.
42. Cross out three digits in the number 123456 so that the resulting number is divisible by 5. Indicate the number in your answer.
43. Cross out three digits in the number 85417627 so that the resulting number is divisible by 18. In your answer, indicate exactly one resulting number.
44. Cross out three digits in the number 141565041 so that the resulting number is divisible by 30. The answer is please indicate exactly one resulting number.
45. Cross it out the number 181615121 has three digits so that the resulting number is divisible by 12. In your answer, indicate one such number.
Task 2: To the number 26, add numbers to the left and right so that the resulting number is a multiple of 45.
Solution:
The sum of the digits of this number must be divisible by 9, the number itself must be divisible by 5, which means the last digit is 0 or 5, and then we select the first digit.
1260 and 5265.
Answer: 1260 or 5262
Tasks for independent solution:
46. Add one digit to the left and right of the number 374 so that the resulting number is divisible by 45.
47. Add one digit to the left and right of 1022 so that the resulting six-digit number is divisible by 7, 8, 9.
48. Add one digit to the left and right of the number 15 so that the resulting number is divisible by 15.
49. Add one digit to the left and right of the number 10 to get a number that is a multiple of 72.
50. To the number 2012, add two digits to the right so that the resulting six-digit number is divisible by 36.
Answers to problems:
1. 1125
2. 1044
3. 1245
4. 3225
5. 4312
6. 6
7. 5
8. 3
9. 321 0
10. 3211
11. 11
12. 5
13. 1152
14. 1152
15. 2120
16. 20
17. 20
18. 10
19. 35
20. 10
21. 30
22. 24
23. 25
24. 24
25. 54
26. 1254
27. 2156
28. 3222
29. 2541
30. 1232
31. 222000
32. 111555
33. 333222
34. 377333
35. 333459
36. 1023457896
37. 282000
38. 899
39. 798
40. 369
41. 245
42. 12345
43. 54162
44. 115650
45. 181512
46. 43740
47. 910224
48. 1155
49. 4104
50. 420120
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6) Open bank of Unified State Exam problems in mathematics - portal [Electronic resource]. - Access mode: http://base.mathege.ru/, free. - Title from the screen.
7) Preparation for the Unified State Examination and the Unified State Examination - in-person exams - portal [Electronic resource]. - Access mode: http://worksbase.ru/, free. - Title from the screen.
Solving Option 238 Larin Unified State Exam 2018. Detailed analysis of tasks 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15 from the site alexlarin.net. Alex Larin 238 timings: 7-12)5:34 13)15:15 14)18:05 15)26:51 twitter:https://twitter.com/mrMathlesson
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Examples of tasks: 1) The pool has the shape of a rectangular parallelepiped. Its length, width and depth are 25 m, 12 m and 2 m, respectively. For lining the bottom and walls of the pool, it was decided to purchase tiles at a price of 500 rubles. per square meter. How many rubles will the purchase cost if it is additionally planned to lay a rectangular path 1 m wide from the same tiles around the perimeter of the pool? 2) The graph shows the change in pressure in the steam turbine after startup. The abscissa axis shows time in minutes, and the ordinate axis shows pressure in atmospheres. Determine from the graph how many minutes passed from the start of the turbine to the moment when the pressure first reached its highest value. 3) Find the area of triangle ABC if the side of the cell is 4. 4) There are 8 identical pairs of gloves on the counter, but one pair has a defect inside both gloves that is invisible from the outside. During the fitting, all the gloves were mixed. The seller divided all the gloves randomly into 4 groups of 4 pieces. What is the probability that both defective gloves are in the same group? 5)Solve the equation. If an equation has more than one root, answer with the smaller root. 6) Find the acute angle between the bisectors of the acute angles of the right triangle. Give your answer in degrees. 7) The figure shows the graph y=f"(x) - the derivative of the function f(x) defined on the interval (-4;10). Find the number of points at which the tangent to the graph y=f(x) is parallel to the straight line y= x or coincides with it. 8) The height of a regular triangular pyramid is three times less than the side of the base. Find the angle between the side edge and the plane of the base of the pyramid. Give the answer in degrees. 10) After rain, the water level in the well can increase by measuring the time of fall. t small pebbles into the well and calculating using the formula h = 5t. Before the rain, the time the pebbles fell was 1.4 s. To what minimum height must the water level rise after the rain for the measured time to change by more than 0.2 s? points A of a circular path, two bodies simultaneously begin to move uniformly in opposite directions. By the time they meet, the first body travels 200 m further than the second and returns to point A 25 minutes after the meeting. Find the length of the path in meters if the second body returns. to point A 36 minutes after the meeting. 14) In a triangular pyramid ABCD, the lengths of all edges are equal. Point P is equidistant from vertices A and D, and it is known that PB = PC and straight line PB is perpendicular to the altitude of triangle ACD dropped from vertex D. a) Prove that point P lies at the intersection of the heights of the pyramid ABCD. b) Calculate the volume of the pyramid ABCD, if it is known that Link to the original source of the option:
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The same formulas are true: \[(\large(S=v\cdot t \quad \quad \quad v=\dfrac St \quad \quad \quad t=\dfrac Sv))\]
from one point in one direction with speeds \(v_1>v_2\) .
Then if \(l\) is the length of the circle, \(t_1\) is the time after which they will find themselves at one point for the first time, then:
That is, in \(t_1\) the first body will travel a distance \(l\) greater than the second body.
If \(t_n\) is the time after which they will end up at the same point for the \(n\)th time, then the formula is valid: \[(\large(t_n=n\cdot t_1))\]
\(\blacktriangleright\) Let two bodies begin to move from different points in the same direction with speeds \(v_1>v_2\) .
Then the problem easily reduces to the previous case: you first need to find the time \(t_1\) after which they will end up at the same point for the first time.
If at the moment of starting the movement the distance between them \(\buildrel\smile\over(A_1A_2)=s\), That:
Task 1 #2677
Task level: Easier than the Unified State Exam
Two athletes start in the same direction from diametrically opposite points on a circular track. They run at different, inconsistent speeds. It is known that at the moment when the athletes first caught up, they stopped training. How many more laps did the athlete run at a higher average speed than the other athlete?
Let's call the athlete with the higher average speed first. First, the first athlete had to run half a circle to reach the starting point of the second athlete. After that, he had to run as much as the second athlete ran (roughly speaking, after the first athlete ran half a circle, before the meeting he had to run every meter of the track that the second athlete ran, and the same number of times as the second athlete ran this meter ).
Thus, the first athlete ran \(0.5\) more laps.
Answer: 0.5
Task 2 #2115
Task level: Easier than the Unified State Exam
The cat Murzik runs in a circle from the dog Sharik. The speeds of Murzik and Sharik are constant. It is known that Murzik runs \(1.5\) times faster than Sharik and in \(10\) minutes they run two laps in total. How many minutes will it take Sharik to run one lap?
Since Murzik runs \(1.5\) times faster than Sharik, then in \(10\) minutes Murzik and Sharik in total run the same distance that Sharik would run in \(10\cdot (1 + 1.5) ) = 25\) minutes. Consequently, Sharik runs two circles in \(25\) minutes, then Sharik runs one circle in \(12.5\) minutes
Answer: 12.5
Task 3 #823
Task level: Equal to the Unified State Exam
From point A of the circular orbit of a distant planet, two meteorites simultaneously flew out in the same direction. The speed of the first meteorite is 10,000 km/h greater than the speed of the second. It is known that for the first time after departure they met 8 hours later. Find the length of the orbit in kilometers.
At the moment when they first met, the difference in the distances they flew was equal to the length of the orbit.
In 8 hours the difference became \(8 \cdot 10000 = 80000\) km.
Answer: 80000
Task 4 #821
Task level: Equal to the Unified State Exam
A thief who stole a handbag runs away from the owner of the handbag along a circular road. The thief's speed is 0.5 km/h greater than the speed of the owner of the handbag, who is running after him. In how many hours will the thief catch up with the owner of the handbag for the second time, if the length of the road along which they are running is 300 meters (assume that he caught up with her the first time after the theft of the handbag)?
First way:
The thief will catch up with the owner of the handbag for the second time at the moment when the distance he will run becomes 600 meters greater than the distance that the owner of the handbag will run (from the moment of theft).
Since his speed is \(0.5\) km/h higher, then in an hour he runs 500 meters more, then in \(1: 5 = 0.2\) hours he runs \(500: 5 = 100\) meters more. He will run 600 meters more in \(1 + 0.2 = 1.2\) hours.
Second way:
Let \(v\) km/h be the speed of the owner of the handbag, then
\(v + 0.5\) km/h – the speed of the thief.
Let \(t\) h be the time after which the thief will catch up with the owner of the handbag for the second time, then
\(v\cdot t\) – the distance that the owner of the handbag will run in \(t\) hours,
\((v + 0.5)\cdot t\) – the distance that the thief will cover in \(t\) hours.
The thief will catch up with the owner of the handbag for the second time at the moment when he runs exactly 2 laps more than her (that is, \(600\) m = \(0.6\) km), then \[(v + 0.5)\cdot t - v\cdot t = 0.6\qquad\Leftrightarrow\qquad 0.5\cdot t = 0.6,\] whence \(t = 1.2\) h.
Answer: 1.2
Task 5 #822
Task level: Equal to the Unified State Exam
Two motorcyclists start simultaneously from one point on a circular track in different directions. The speed of the first motorcyclist is twice that of the second. An hour after the start, they met for the third time (consider that the first time they met after the start). Find the speed of the first motorcyclist if the length of the road is 40 km. Give your answer in km/h.
At the moment when the motorcyclists met for the third time, the total distance they traveled was \(3 \cdot 40 = 120\) km.
Since the speed of the first is 2 times greater than the speed of the second, then out of 120 km he traveled a part 2 times greater than the second, that is, 80 km.
Since they met for the third time an hour later, the first one drove 80 km in an hour. Its speed is 80 km/h.
Answer: 80
Task 6 #824
Task level: Equal to the Unified State Exam
Two runners start simultaneously in the same direction from two diametrically opposite points on a circular track 400 meters long. How many minutes will it take for the runners to meet each other for the first time if the first runner runs 1 kilometer more in an hour than the second?
In an hour, the first runner runs 1000 meters more than the second, which means he will run 100 meters more in \(60: 10 = 6\) minutes.
The initial distance between runners is 200 meters. They will be equal when the first runner runs 200 meters more than the second.
This will happen in \(2 \cdot 6 = 12\) minutes.
Answer: 12
Task 7 #825
Task level: Equal to the Unified State Exam
A tourist left city M along a circular road 220 kilometers long, and 55 minutes later a motorist followed him from city M. 5 minutes after departure he caught up with the tourist for the first time, and another 4 hours after that he caught up with him for the second time. Find the speed of the tourist. Give your answer in km/h.
First way:
After the first meeting, the motorist caught up with the tourist (for the second time) 4 hours later. By the time of the second meeting, the motorist had driven a circle more than the tourist had covered (that is, \(220\) km).
Since during these 4 hours the motorist overtook the tourist by \(220\) km, the speed of the motorist is \(220: 4 = 55\) km/h greater than the speed of the tourist.
Let now the speed of the tourist be \(v\) km/h, then he managed to walk before the first meeting \ the motorist managed to pass \[(v + 55)\dfrac(5)(60) = \dfrac(v + 55)(12)\ \text(km).\] Then \[\dfrac(v + 55)(12) = v,\] from where we find \(v = 5\) km/h.
Second way:
Let \(v\) km/h be the speed of the tourist. When using quantities in the exercise that are related to distance (speed, circle length), they can be solved by reducing them to movement in a straight line. The greatest difficulty for schoolchildren in Moscow and other cities, as practice shows, is caused by problems on circular motion in the Unified State Examination, the search for an answer in which involves the use of an angle. To solve the exercise, the circumference can be specified as a part of a circle. You can repeat these and other algebraic formulas in the “Theoretical Help” section. In order to learn how to apply them in practice, solve exercises on this topic in the “Catalogue”.
Let \(w\) km/h be the speed of the motorist. Since \(55\) minutes \(+ 5\) minutes \(= 1\) hour, then
\(v\cdot 1\) km is the distance that the tourist traveled before the first meeting. Since \(5\) minutes \(= \dfrac(1)(12)\) hours, then
\(w\cdot \dfrac(1)(12)\) km – the distance that the motorist traveled before the first meeting. The distances they traveled before their first meeting are: \
Over the next 4 hours, the motorist drove more than the tourist covered in a circle (by \(220\)
\
\
to the 8th mathematical class of GBOU Lyceum No. 1535. Stage 1
1) Find the value of the expression:
Solution:
The figure shows a schedule for the movement of a tourist from city A to city B, and he made a stop along the way. Define:
a) At what distance (in km) from city A did the tourist stop?
b) What was the tourist’s speed (in km/h) after the stop?
c) What was the average speed of the tourist (in km/h) when moving from A to B? 
Solution: a) answer: 9; b) 18-9=9, 7-5=2, which means 9:2=4.5 km/h; c) 18:5=3.6 km/h.
3) Bring the polynomial (p+3)(p+4)(p-4)-p((1-p)(-p)-16) to standard form/
Solution: (p+3)(p+4)(p-4)-p((1-p)(-p)-16)=(p+3)(p 2 -16)-p(p 2 - p-16)=p 3 +3p 2 -16p-48- p 3 +p 2 +16p=4p 2 -48
4) Find the root of the equation of the expression: 8 15: x=4 17 2 6
Solution: 
5) Using the data in the figure, find the degree measure of angle α 
Solution: 136°+44°=180°, which means the lines are parallel. Therefore, ∠ CBA=44°, ∠ BCA=56°, which means ∠α=180°-44°-56°=80°.
6) What is the root of the equation? 
Solution: multiply all terms by 30, the denominators will cancel: 
7) Find the value of a numerical expression: 
Solution:
8) If one of the adjacent sides of the square is reduced by 2 cm, and the second is increased by 6 cm, then you will get a rectangle, the area of which is equal to the area of the rectangle, which will be obtained from the same original square, if one of its adjacent sides is not changed, and the other increase by 3 cm. What (in square centimeters) is the area of the original square?
Solution. Let x- side of the square. Let's make an equation:
(x-2)(x+6)=x(x+3);
x 2 +4x-12=x 2 +3x;
x=12
The area of the original square is 12 · 12 = 144 cm 2 .
9) Use the formula to define a linear function whose graph in the 0xy coordinate system passes through the point T(209,908) and does not intersect with the graph of the equation 9x+3y=14
Solution. Let's rewrite the equation in the form 
The formula for a linear function in general form is y=kx+b. If the graph of the required equation does not intersect with the graph of this equation, then k=-3. Therefore, 908=-3 209 + b, whence b=1535.
Formula of the desired linear function: y=-3x+1535
10) There is a piece of copper-tin alloy with a total mass of 24 kg, containing 45% copper. How many kilograms of pure tin must be added to this piece of alloy so that the resulting new alloy contains 40% copper?
Solution. If an alloy of copper and tin contains 45% copper, then it contains 55% tin. If a new alloy contains 40% copper, then it contains 60% tin. Let x be the number of kg of pure tin that must be added to the alloy. Let's make an equation:
0.55 24 + x = 0.6 (x+24)
x-0.6x=0.6 24- 0.55 24
0.4x=0.05 24
x=3
Answer: 3 kg.
Math tutor's note: You can read more about methods for solving problems on alloys and mixtures in the article Advantages and disadvantages of various methods for solving problems on alloys and mixtures
11) According to the figure, which shows the graphs of two linear functions and a parabola, find the abscissa of point T. 
Solution. The straight line y=5x and the parabola y=x 2 intersect at two points. Let's find the abscissa of these points using the equation 5x=x 2. Hence x 1 =0; x 2 =5. This means that the ordinate of the intersection point is 25
The straight line on which point T lies passes through the points with coordinates (5;25) and (0;27). The equation of a straight line in general form: y=kx+b. Substituting the coordinates of the points of the line instead of x and y, we obtain a system of equations: 
Point T has an ordinate equal to zero. Hence 
Answer. 67.5.
12) From point A of a circular track, two objects simultaneously begin to move uniformly in opposite directions. By the time they meet, the first object travels 100 meters further than the second and returns to point A 9 minutes after the meeting. Find the length of the path in meters if the second object returns to point A 16 minutes after the meeting.
Note. On the Internet you can find sites where problems of this kind are solved using a quadratic equation. Meanwhile, this work is intended for those entering the 8th grade. That is, solving this problem knowing the quadratic equation, which is taught in the 8th grade, is incorrect. There is no point in changing the 8th grade program to solve a problem addressed to 7th graders. Below is a solution that does not require a quadratic equation
Solution. Let t be the time before the objects meet, v 1 the speed of the first object, v 2 the speed of the second object.
Then v 1 · t - v 2 · t = 100, since at the moment of meeting the first object passed 100 m more. Since v 2 t is the path that the 1st object traveled after the meeting, v 1 is its speed and it returned to point A after 9 minutes, we can create the equation
Likewise
. Three equations form a system of three equations with three unknowns: 
Let's divide the 1st equation by the 2nd. It will turn out: 
where
Thus,
Substituting this expression into the first equation, we get t=12 min
Substituting the last expression and t=12 into the third equation of the system, we get: 
from here 
According to the condition, the length of the route in meters can be determined by adding the path of the first object to the meeting and the path of the second object to the meeting. That is 
Answer. 700 meters
13) An equilateral triangle MPL is constructed on the ML side of the square MNKL, with point P located inside the square. Find the degree measure of angle LPK.
Solution 
According to the condition ML=PL=KL; triangle PLM is equilateral, which means that all angles are equal to 60°, which means ∠ PLK=30°. Thus, ∠LPK=(180°-30°) : 2=75°.
14) Factorize: (solutions are written immediately)
Alexander Anatolyevich, mathematics tutor. 8-968-423-9589. I have successful experience in preparing students for this lyceum, including in the 8th grade of mathematical specialization and in classes of other specializations. For those preparing to enter Lyceum No. 1535, as well as other lyceums, it is important to understand that the real options for the entrance exams are somewhat different from the demonstration ones. Therefore, it is necessary to be able to solve other similar tasks.
