30 compulsory questions for the primary school Mathematical Olympiad
The following are some of the common types of questions for the primary school Mathematical Olympiad exam: return to one problem, itinerary problem, cow eating grass problem, etc. However, because the reference materials did not completely list the 30 compulsory questions, the contents of all 30 compulsory questions could not be provided. Read more exciting novels for free
An Analysis of the Ten Difficult Questions in Primary School Mathematical Olympiad
The following is an analysis of the ten difficult questions commonly seen in primary school Mathematical Olympiad: ** One, Return to One Problem ** 1. ** Meaning ** - When solving a problem, the first thing to do was to find out how much a portion was (that is, a single amount), and then use the single amount as the standard to find the required amount. 2. ** Number of relationships ** - Total amount/portions = single amount; single amount × portions = the number of portions required; or total amount A/(total amount B/portions B)= portions A. 3. ** Solution train of thought ** - First, find a single quantity. Using a single quantity as the standard, find the required quantity. For example, if you want to buy 5 pens, you need 0.6 yuan. If you want to buy 16 pens, you need to find 0.6 + 5 = 0.12 yuan for one pencil, and 0.12×16 = 1.92 yuan for 16 pens. ** 2. The problem of returning to the main body ** 1. ** Meaning ** - When solving a problem, first find the "total number" and then solve the problem according to the known conditions. The so-called "total quantity" could refer to the total price of the goods, the workload in a few days, the total output of a few acres of land, the total journey in a few hours, and so on. 2. ** Number of relationships ** - 1 serving x number of copies = total amount; total amount/1 serving = number of copies. 3. ** Solution train of thought ** - He would first find the total number before solving the problem. For example, the clothing factory originally used 3.2 meters of cloth to make a set of clothes. Originally, it made 791 sets of clothes, so the total amount of cloth was 3.2×791 = 2531.2 meters. Now, each set of clothes used 2.8 meters of cloth, which could make 2531.2/2.8 = 904 sets. ** 3. Problem of sum and difference ** 1. ** Meaning ** - Given the sum and difference of two quantities, find out what the two quantities are. 2. ** Number of relationships ** - Large numbers =(sum + difference) div2; Decimals =(sum-difference) div2. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, Class A and Class B had a total of 98 students. Class A had 6 more students than Class B. Class A =(98 + 6) div2 = 52 students, Class B =(98 - 6) div2 = 46 students. ** 4. The Problem of Summing Times ** 1. ** Meaning ** - Given the sum of the two numbers and "how many times the large number is a fraction (or how many fraction of the large number is a fraction)", find out what the two numbers are. 2. ** Number of relationships ** - Sums × (Multiple + 1)= Lesser Number; Sums-Lesser Number = Greater Number; or Lesser Number × Multiple = Greater Number. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, there were 248 apricot trees and peach trees in the orchard. The peach trees were three times the number of apricot trees. There were 248/(3 + 1)=62 apricot trees and 62×3 = 186 peach trees. ** 5. The problem of the difference ** 1. ** Meaning ** - Given the difference between the two numbers and "how many times the large number is a decimal (or how many times the small number is a large number)", find out what the two numbers are. 2. ** Number of relationships ** - The difference between the two numbers divided by (multiple- 1)= lesser number; lesser number × multiple = greater number. 3. ** Solution train of thought ** - Simple questions could be solved by applying the formula, while complex questions could be solved by adapting the formula. For example, the number of peach trees in the orchard was three times that of apricot trees, and there were 124 more peach trees than apricot trees. There were 124/(3 - 1)=62 apricot trees and 62×3 = 186 peach trees. ** 6. Multiple ratio problem ** 1. ** Meaning ** - There are two known quantities of the same kind, one of which is several times the other. When solving a problem, first find the multiple, and then use the multiple ratio method to calculate the required number. 2. ** Number of relationships ** - Total amount A/quantity A = multiple; quantity B× multiple = total amount B. 3. ** Solution train of thought ** - First, find the multiple, then use the multiple ratio relationship to solve. ** 7. Age problem ** 1. ** Key Points ** - The precession of the equinoxes (the difference in age) never changed, but the sum of years and the multiple of years changed. The precession of the equinoxes, the sum of years, and the difference of sum had to correspond to the multiple and time. ** 8. The problem of cows eating grass ** 1. ** Meaning and Key Points ** - For example, the grassland would grow grass at a constant rate every day. The key to the problem of cows eating grass was to determine how much grass they would grow every day. 2. ** Solution train of thought ** - Usually, a cow was set to eat one unit of grass a day. The original amount of grass and the amount of grass grown per day were calculated by the number of days that different cows ate grass. Then, the number of days that a given number of cows ate grass was calculated. Like a blade of grass that can feed twenty-seven cows for six days (A total of 27×6 = 162 units of grass were eaten, including the original grass and 6-day-old new grass), which could feed 23 cows for 9 days (a total of 23×9 = 207 units of grass were eaten, including the original grass and 9-day-old new grass). The amount of new grass per day could be calculated as (207 - 162)/(9 - 6)=15 units, and the original grass was 162 - 6×15 = 72 units. Then, the number of days that the 21 cows ate grass was calculated. ** 9. Quick and skillful calculations (for lower grades)** 1. ** Meaning ** - For first-year students, calculation was the first problem they encountered in their studies and was the focus of their studies. For second-year students, calculation was also the focus and difficult part of their Mathematical Olympiad studies. ** 10. Enumeration Questions (For Lower Grades)** 1. ** Difficulty analysis ** - For second-year students, orderly and abstract thinking was more difficult. For questions, second-year students were more willing to make up the numbers to try to answer the questions. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Mathematical Olympiad calculations and ingenious calculations in primary school
The following are some common types and methods of quick and clever calculations in primary school Mathematical Olympiad: ** 1. Clever calculations in addition ** 1. ** Rounding Method ** - When adding several numbers, if there were two numbers that could be added together to get a whole ten, a whole hundred, a whole thousand, etc., they could be added first. For example, 24 + 44+56, because 44 + 56 = 100 is a whole hundred, so first calculate 44+56, then add 24, that is, 24+(44 + 56)=24 + 100 = 124. - For an equation like 53+36 + 47, because 53+47 = 100 was a whole hundred,+47 could be moved with the sign to the front of +36, and then (53+47)+36 = 100+36 = 136. 2. ** Split and Rounder Method ** - In the case of 96+15, 15 was split into 15 = 4+11, because 96+4 = 100, which could be rounded up first, that is, 96+15 = 96+(4+11)=(96 + 4)+11 = 100+11 = 111. - For 52+69, since 69+31 = 100, 52 was split into the sum of 21 and 31, and then 31+69 = 100 was rounded up, which was 52+69=(21+31)+69 = 21+(31+69)=21 + 100 = 121. - When calculating 63+18+19, 63 was split into 63 = 60+2+1. Because 2+18 and 1+19 could be rounded up, 63+18+19 = 60+2+1+18+19 = 60+(2+18)+(1+19)=60+20+20 = 100. 3. ** The clever calculation of adding the same number ** - For 28+28+28, you can think of it this way, because 28+2 = 30 can be rounded up, but in the end, you have to subtract the three additional 2s, that is, 28+28+28=(28 + 2)+(28 + 2)+(28+2)-6 = 30+30+30 - 6 = 90 - 6 = 84. ** 2. Clever calculation in the substitution (position swap method)** - In an algorithm, the position of the numbers could be changed, and the order of the calculation of the numbers could be changed. For example, 632 - 156 - 232 = 632 - 232 - 156 = 400 - 156 = 244. ** 3. Clever calculations in multiplication ** 1. **"Tongbu" and "Butong" Quick Calculation Method ** - If the sum of two numbers was equal to 10, then the two numbers were complementary. In the integral multiplication operation, for example, 72×78, where the ten digits of the multiplication and the multiplication were the same, and the single digits were complementary (this kind of formula was called "same head, complementary tail" type), and 26×86, where the ten digits of the multiplication and the multiplication were complementary, and the single digits were the same (this kind of formula was called "complementary head, same tail" type), there were very simple and direct fast calculation methods. They were called the "same-complementing" fast calculation method and the "complementing the same" fast calculation method. ** 4. Other methods ** 1. ** Borrowing Method ** - Borrowing a number from a number, turning it into a whole ten or a whole hundred or a whole thousand, and finally, deducting the borrowed number. For example, 9+99 +999+9999=(10 - 1)+(100 - 1)+(1000 - 1)+(10000 - 1)=10+100 + 1000 + 10000 - 4 = 11106. 2. ** Choose a reference number ** - Among all the numbers, find that everyone is close to this number (this number is the benchmark number), first multiply the number of numbers by this benchmark number, and finally, subtract the excess. For example, 489+487+483+485+484+486+488. First find the base number 490 among these numbers, then use 490×7. Finally, subtract the excess, which is 490×7 - 1 - 3 - 7 - 5 - 6 - 4 - 2 = 3430 - 28 = 3402. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Two Formula of Substitute for Mathematical Olympiad in Primary School
In elementary school Mathematical Olympiad, equivalent substitution meant that two identical quantities could be replaced with each other. For example, if 1 × = 2 ×, 2 × = 4 ×, then 1 × = 4 ×. However, there was no fixed formula named " two quantity substitution formulas ". It was mainly based on the relationship of equality to replace the quantity. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Sixth Grade Mathematical Olympiad Questions
The following are some of the 2022 primary school sixth grade Olympiad math questions: 1. A primary school mathematics competition was held in a certain city. The number of people involved was as follows: the number of people who scored no less than 80 points was four times more than the number of people who scored less than 80 points. The number of people who passed was 22 more than the number of people who scored no less than 80 points, and the number of people who passed was exactly six times the number of people who failed. 2. The original price of each movie ticket was a few yuan. Now, each ticket was sold at a lower price of 3 yuan. The audience increased by half, and the income increased by one-fifth. 3. A and B had a total of 9600 yuan in the bank. If the two of them took out 40% of their own deposits and then withdrew 120 yuan from A's deposit to B, then the two people's money was equal and they asked for B's deposit. 4. The milk candy and chocolate candy are mixed into a pile of candy. If 10 milk candies are added, the chocolate candy accounts for 60% of the total; if 30 chocolate candies are added, the chocolate candy accounts for 75% of the total. Find the number of milk candies and chocolate candies in the original mixed candy. 5. Xiao Ming and Xiao Liang each had some glass balls. Xiao Liang said,"You have a quarter less than me!" "If you can give me 1/6 of yours, I'll have 2 more than you." Ask Xiaoming for the number of glass balls he has. 6. Moving the goods in a warehouse, A needed 10 hours, B needed 12 hours, and C needed 15 hours. A was in warehouse A, B was in warehouse B, and they started to move the goods at the same time. C started to help A move the goods, and then turned to help B move the goods halfway. Finally, the goods in the two warehouses were moved at the same time. How long did C help A and B? 7. If A completed a task alone in 73 days, then B would join in after A completed it for a day. After working together for two days, C would also work together, and the three of them would work together for another four days to complete 1/3 of the work. After another eight days, 5/6 of the work would be completed. If the rest of the work was completed by C alone, how many more days would it take? 8. The quotient of two numbers is four, the remainder is two, and the sum of the two numbers is forty-two. Find these two numbers. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the recommendations for innovative thinking in primary school Mathematical Olympiad?
The following are some recommended books for primary school Mathematical Olympiad students: 1. " Mathematical Olympiad Star's Classic Question Bank of innovative thinking," published by China Forest Press on January 1, 2008. The author was Liu Xianguo. 2. Learning and Thinking, Thinking and Creation, also known as the Big White Book, was a set of books with more difficult questions in the learning and thinking system. 3. The 1998 Guangdong-Hong Kong Mathematical Olympiad Invitational Competition for the fourth grade examination paper can also be used as reference material. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The Mathematical Olympiad questions of which grade are interesting
Mathematical Olympiad questions were different from person to person. Mathematical Olympiad questions of different grades had their own unique charm. For the Mathematical Olympiad questions of the lower grades (such as the first and second grades), they were usually more intuitive, mainly based on simple numerical relationships, graphic cognition, and basic logical reasoning. For example, simple mathematical problems, basic queuing problems, etc. These questions were suitable for cultivating beginners 'interest in mathematics and basic thinking ability. It was like a game to let students feel the wonders of mathematics. The third-year Mathematical Olympiad questions had deepened on the foundation of the first and second grades. They began to involve some more complicated applied questions, simple geometry problems, and so on. This question required the students to think more deeply about the relationship between numbers. It could train the students 'ability to establish a more complicated logical chain. In the fourth and fifth grades, the Mathematical Olympiad questions would cover more knowledge points, such as engineering problems, cows eating grass, and so on. The difficulty and complexity of these problems were further increased. Students needed to use a variety of mathematical knowledge and skills to conduct a comprehensive analysis. It was very interesting for students who liked to challenge high difficulty and explore the depth of mathematics. Sixth grade Mathematical Olympiad questions were a comprehensive reflection of primary school Mathematical Olympiad questions. It would integrate all kinds of knowledge and thinking methods learned before into some extremely comprehensive questions. Solving such questions would bring a great sense of accomplishment to the students and also lay a solid foundation for junior high school mathematics learning. In short, every grade's Mathematical Olympiad questions had their own interesting aspects. It depended on the student's personal mathematical foundation, hobbies, and thinking ability. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Fourth grade primary school mathematics Olympiad questions and answers
The following are some of the fourth grade primary school math Olympiad questions and answers: 1. ** and times problem **: The quotient of two numbers is four, the remainder is two, and the sum of the two numbers is forty-two. Find these two numbers. - If the smaller number is x, and the quotient of the larger number divided by the smaller number is 4 and the remainder is 2, then the larger number is x. The sum of the two numbers is 42, which gives the equations: x+(4x + 2)=42, 5x+2 = 42, 5x=40. The solution is x = 8, and the larger number is 44. 2. ** Divide with known dividends, quotient, and remainder **: Divider is 3320, quotient is 150, remainder is 20. - According to the formula, we can get the result: <<(3320 - 20)> div150 = 22>. 3. ** Successive natural number sum problem **: 3998 is the sum of four consecutive natural numbers, find the smallest number. - Let the smallest number be {x}, then the other three numbers are {x + 1},{x+2}, and {x + 3}, which gives the equation {x+(x + 1)+(x+2)+(x + 3)=3998},{4x+6 = 3998}, and {4x=3992}. The solution is {x = 998}. 4. ** Number and Decimals Adding Problem **: There is a two-digit number. Add a decimals point in front of one of its digits and add it to the two-digit number. The result is 20.9. Find the two-digit number. - Let this two-digit number be <x>, because the result after adding is <20.9>, we can know that this decimal is <0.1x>, then <x+0.1x = 20.9>,<1.1x = 20.9>, the solution is <x = 19>. 5. ** Family age problem **: A family of three, the total age of the three is 72 years old, mother and father are the same age, mother's age is four times that of the child, please age the three. - If the child's age is considered as a multiple of 1, and the parents are four times the child's age, then the child's age is <72'> div4>(1 + 4+4)=8>, and the mother and father's age is <8'> time4>= 32>. 6. ** Sports allocation problem **: A, B, C and D will participate in basketball, volleyball, football and chess respectively. It was known that A was taller than a volleyball player, that D had lost his legs in an accident a few years ago, and that a soccer player was shorter than C and a basketball player. Ask A, B, C and D to participate in what event. - From the fact that Ding lost his legs, he could tell that Ding was a chess player; A was not a volleyball or football player, so A was a basketball player; C was not a football player, so C was a volleyball player; B was a football player. 7. [Fruit packing problem: 10 fruits must be packed in 6 bags. The number of fruits in each bag must be even, and there must be no fruit or bags left.] - He put two bags in each bag and five bags in the last bag. 8. ** Comparing the amount of money left after spending money **: Naughty had 300 yuan, 56 yuan for books, and 128 yuan for stationery. How much less was Naughty left than before? - The money that was less than the original amount was the money spent. The total amount spent was [56+128 = 184] yuan. 9. [Question of the number of fishes: The two brothers went fishing and caught a total of 23 fishes. The elder brother caught three times more fish than the younger brother. How many fishes did the elder brother and younger brother catch?] - First, calculate the number of fish the younger brother has caught. The number of fish the older brother has caught is [(23 - 3)][Div(3 + 1)][5][1][2][3][3] 10. ** Alien payment problem **: Aliens have 1 cent, 2 cent, 4 cent, and 8 cent coins each. Please pay for 7 cent, 9 cent, 10 cent, 13 cent, 14 cent, and 15 cent items. - \(7 = 1+2+4\)，\(9 = 1+8\)，\(10 = 2+8\)，\(13 = 1+4+8\)，\(14 = 2+4+8\)，\(15 = 1+2+4+8\)。 11. [Fruit distribution problem: There are bananas, apples, and oranges on the plate.] Xiao Gang, Xiao Lin, and Xiao Hong each took a different fruit. "Everyone can only eat one kind of fruit. I don't eat oranges," said Xiao Gang. "I don't eat apples or oranges," said Xiao Lin. Beg who to take what fruit. - Xiao Lin took the bananas, Xiao Hong took the oranges, and Xiao Gang took the apples. 12. ** Four-digit numbers and questions **: Which four-digit numbers have the sum of each number equal to 34? - There were 8899, 8989, 8998, 9889, 9898, 9988, 7999, 9799, 9979, and 9997. 13. ** Price of tables and chairs **: Given that the price of a table is 10 times that of a chair, and knowing that a table is 288 yuan more than a chair, please find the price of a table and a chair. - The price of a chair was $288,000, and the price of a table was $32,000. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
It's best to use that kind of Mathematical Olympiad book in the fifth grade of primary school.
The choice of Mathematical Olympiad books depended on one's interests and level of mathematics. For the fifth grade students, it is recommended to choose a book suitable for the basic level and pay attention to interest and inspiration. Some examples of Mathematical Olympiad books suitable for fifth grade elementary school students include: - Elementary Mathematical Olympiad ( ) - "Primary School Mathematical Olympiad Real Reality Simulation Test Questions"( ) - Math Paradise ( ) These books cover basic mathematics knowledge and provide a variety of topics and challenges suitable for stimulating students 'interest in learning and improving their mathematical ability. Of course, he could also choose some more advanced course books to challenge his own mathematics level. It is recommended that you carefully evaluate your mathematics level and interest before buying.
Primary school Olympiad math questions vertical style complete book
The following are some examples of the types and solutions of the primary school Mathematical Olympiad vertical questions: ** 1. The vertical addition puzzle ** 1. ** When the same number is added up, the same number will still be obtained ** - For example, if a three-digit number plus a three-digit number is equal to a three-digit number, if a triangle plus a triangle is equal to a triangle, it is only satisfied when the triangle is 0, because 0 + 0=0. Then, according to the numerical relationship between the 10th and 100th digits, the circle plus the square on the 10th digit was equal to 0 (in fact, 10 to 100th digit was 1), and the square plus the square plus the carry on the 100th digit was equal to the circle. By transforming the vertical form into the horizontal form, the values of the square and the circle were solved by equivalent substitution. 2. ** Normal addition, find the relationship between numbers vertically ** - For example, when calculating the addition of two numbers, one had to consider the rules of digit alignment and addition, starting from the single digit. If the addition of the single digit had a carry, it had to be carried to the tenth digit, and when the tenth digit was added, the number of the single digit carry had to be added. For example, if two two-digit numbers were added together, the addition of the one-digit numbers was equal to 11, and the addition of the ten-digit numbers plus the one-digit carry was equal to 13. According to this relationship, the one-digit and ten-digit numbers of the two numbers could be obtained respectively. ** 2. Multiplication vertical riddle (Take a five-digit number multiplied by a one-digit number as an example)** 1. ** Confirm the range of key numbers ** - First, determine the possible value of one of the multiplying factors (one digit). For example, in a five-digit number multiplied by a one-digit number equals a five-digit number, the value of the one-digit multiplication factor is determined according to the highest digit without carry, the single-digit calculation condition, the highest digit calculation condition, and so on. For example, this one-digit number could not be 0 or 1. If this one-digit number was 9, there might be conflicts when deducing other digits according to the calculation rules (such as obtaining the same number does not meet the requirements of each Chinese character representing different numbers, etc.). The value of this one-digit number was determined by gradual elimination. 2. ** Derives other numbers based on the determined number ** - After determining this one-digit number, the numbers on the other digits were derived according to the multiplication calculation rules. For example, according to the one digit of the product of the one digit numbers and the carry situation to determine the number on the other five-digit number, then according to the calculation on the ten digit (including the one digit carry) to determine the number on the ten digit, and so on to calculate the number of other digits. These vertical questions were mainly solved by analyzing the operational relationship of the numbers, the carry situation, and the range of the numbers. They were solved by mathematical methods such as elimination and equivalent substitution. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The compulsory questions and answers of the sixth grade mathematics competition in primary school
The following are some examples of the types of questions and answers that may be involved in the sixth grade mathematics competition: ##1. Calculation Class 1. [Question: Calculating 1.25×17.6 + 36.1 × 0.8+2.63×12.5] - Answer: - First of all, the equation can be transformed into: 1.25×17.6+36.1 × 1 = 1.25×17.6 + 36.1× 1 = 26.3 × 1.25. - It was further converted to: 1.25×17.6+45.125 + 26.3×1.25. - Using the distribution law of multiplication: 1.25×(17.6 + 26.3)+45.125. - First, calculate the value in the parenthesis: 1.25×43.9+45.125. - Multiplication: 54.875+45.125 = 100. 2. [Question: Calculating 7.5×2.3+1.9×2.5] - Answer: - Splitting 7.5 into 2.5×3, the equation becomes 2.5×3×2.3+1.9×2.5. - That is, 2.5×(3×2.3)+1.9 × 2.5 = 2.5×6.9+1.9×2.5. - Then, using the distribution law of multiplication: 2.5×(6.9 + 1.9)=2.5×8.8 = 22. ##2. Integer-Based Operations 1. [Question: Calculating 1999+999×999] - Answer: - Divide 1999 into 1000+999, and the formula becomes 1000+999+999×999. - Extracting the common factor 999, you get: 1000+999×(1 + 999). - First, calculate in the parenthesis: 1000+999×1000. - After extracting the common factor 1000, it was 1000×(1+999)=1000×1000 = 1000000. 2. [Question **: Calculating 8+98+998+ 9998 + 9998+99998] - Answer: - Rounding up each number, the original formula =(10 - 2)+(100 - 2)+(1000 - 2)+(10000 - 2)+(100000 - 2). - Removing the parenthesis, it was 10+100+ 1000 + 10000+100000-2×5. - The result was: 111110 - 10 = 111100. ##Three and Four Mixed Operations 1. [Question: Calculating (78.6 - 0.786×25+75%×21.4)/15×1997] - Answer: - First, calculate the formula in the parenthesis. 0.75 = 75%, then the formula becomes: (78.6-0.786×25 + 0.75×21.4)/15×1997. - Distortion of the formula in the parenthesis: 78.6×(1 - 0.25)+21.4 × 0.75 × 15×1997. - That is,(78.6×0.75+21.4×0.75) × 15×1997. - Using the law of multiplication: (78.6 + 21.4)×0.75 div15 ×1997. - First, calculate in the parenthesis: 100×0.75 × 15×1997. - According to the order of calculation: 75 div15 ×1997 = 5×1997 = 9985. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>