hindustan 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>

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>

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. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>

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>

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>

The following are some classic Olympiad math questions and answers for primary school students: ** 1. Profits and losses ** 1. [Question: How many children are there in total? Each child has 10 watermelons, 9 less, 8 more than 7.] - ** Answer **: According to the formula,"First add more or less, then divide by the distribution to find the difference." If there are 9 fewer and 7 more, the first sum is [9 + 7=16]; if each person is allocated 10 and each person is allocated 8, the difference is [10 - 8 = 2]. Finally, divide by the number of people. ** 2. Train itinerary ** 1. [Title: There are two trains, one is 102 meters long and travels 20 meters per second, and the other is 120 meters long and travels 17 meters per second.] The two trains were traveling in the same direction. How many seconds would it take for the first train to catch up with the second train and for the two trains to leave? - ** Answer **: From catching up to leaving, the distance of catching up is the sum of the two train leaders, which is <102 + 120 = 222> m. The speed difference is <20 - 17 = 3> m/s. Time = distance difference/speed difference, so <222> div3 = 74> s. 2. [Title: A person's walking speed is 2 meters per second.] A train came from behind and overtook him by 10 seconds. It was known that the train was 90 meters long. Ask for the speed of the train. - [Answer: The distance that a train can travel is 90 meters, the time is 10 seconds, and the speed of a person is 2 meters per second.] Assuming that the train speed is <v> m/s, according to distance = speed x time,<(v - 2)<times10 = 90>, the solution is <v = 11> m/s. 3. [Title: There are two trains moving in the same direction at the same time. After 12 seconds, the fast train overtakes the slow train.] The express train traveled 18 meters per second, while the slow train traveled 10 meters per second. If the two trains were traveling in the same direction, the express train would overtake the slow train in 9 seconds. Find the length of the two trains. - ** Answer **: When moving head to head, the distance that the express train passes over the slow train is the length of the express train, which is [(18 - 10)<times12 = 96] m; when the rear of the train is moving head to head, the distance that the express train passes over the slow train is the length of the slow train, which is [(18 - 10)<times9 = 72] m. 4. [Title: A train needs 40 seconds to cross a 440-meter bridge, and 30 seconds to cross a 310-meter tunnel at the same speed.] What was the speed and length of this train? - ** Answer **: Let the train's length be x meters and its speed be v meters per second. According to the relationship between the distance and time through the bridge and tunnel, the equation system can be obtained.(\begin{cases} 40v = 440 + x30v = 310 + x\end{cases}}), subtract the second equation from the first equation to get\(10v = 130'), the solution is [v = 13') m/s. Substitute [v = 13')] into [30v = 310 + x], the solution is [390 = 310+x]. 5. ** Title **: In order to measure the speed and length of the train, Xiaoying used a watch to record the time it took for the train to pass in front of her was 15 seconds, and Xiaomin used another watch to record the time it took for the train to pass the first telephone pole from the front to the second telephone pole at the back was 20 seconds. It was known that the distance between two utility poles was 100 meters. Find the length and speed of the train. - ** Answer **: Let the train's length be x meters and its speed be v meters per second. The train passes in front of us, the distance it travels is the length of the train (x = 15v\); The distance from the front of the car to the back of the car is the length of the car plus the distance between the poles.(x + 100 = 20v_), substitute_(x = 15v_) into_(x + 100 = 20v_) to get_(15v +100 = 20v_). The solution is_(v = 20_) m/s, the length of the car is_(x = 15, time20 = 300_) m, and the speed is_(20,time3.6 = 72_) km/h. 6. [Title: A train needs 40 seconds to cross a 530-meter bridge, and 30 seconds to cross a 380-meter cave at the same speed.] How many meters is the speed and length of this train? - ** Answer **: Let the train's length be x meters and its speed be v meters per second. We can get the equation set of\(\begin{cases} 40v = 530 + x'30v = 380 + x 'end{cases}). Subtracting the first equation from the second equation gives\(10v = 150), which is solved as\(v = 15) m/s. Substituting\(v = 15) into\(30v = 380 + x) gives\(450 = 380 + x), which is solved as\(x = 70) m. 7. [Title: The two of them walked along the trail beside the railway line, starting from two places and traveling at the same speed.] A train came, and it took 10 seconds for the entire train to pass by A. Three minutes later, B encountered a train. The entire train passed by B in just nine seconds. How long after the train left B did the two meet? - ** Answer **: Let the speed of the train be <v> 1>, the speed of the person be <v> 2>, and the length of the train be <L>. According to the relationship between the train and A, L=(v1v2)<times10>, and the relationship between the train and B, L=(v1v2)<times9>, we can get <(v1v2)<times10>=(v1v2)<times9>, which is simplified to <(v1v2 = 19v2>). 3 minutes after the train leaves A, it meets B. The distance between A and B and the distance between A and B is (180v1 - 180v2). The time after the train leaves B is set to be <t>, then <(v2 + v2)t= 180v1 - 180v2>. Substituting <v1 = 19v2> into <2v2t = 180Times19v2 - 180v2>, the solution is <t = 1710> seconds = 28.5 minutes. 8. [Question: Two trains, one 120 meters long and traveling at 20 meters per second, the other 160 meters long and traveling at 15 meters per second. The two trains are moving in opposite directions. How many seconds does it take for the two trains to meet and leave?] - ** Answer **: The two trains are moving in opposite directions. The distance from the front of the train to the back of the train is the sum of the two train leaders, which is the sum of the speed of the train. The speed is the sum of the speed of the train. The time is the sum of the speed of the train, so the speed of the train is the sum of the speed of the train. 9. [Title: A person's walking speed is 2 meters per second.] A train came from behind and passed him in ten seconds. Given that the train is 90 meters long, find the speed of the train. - ** Answer **: The distance the train can travel over a person is 90 meters, the time is 10 seconds, and the speed of a person is 2 meters per second. Assuming that the train speed is <v> m/s, according to distance = speed x time,<(v - 2)<times10 = 90>, the solution is <v = 11> m/s. 10. [Title: A and B are walking along the railway at the same speed. A train takes 8 seconds to pass by A and meets B 5 minutes later. It only takes 7 seconds to pass by B. How many minutes will it take for B and B to meet each other after B meets the train?] - ** Answer **: Let the speed of the train be <v> 1>, the speed of the person be <v> 2>, and the length of the train be <L>. According to the relationship between the train and A, L=(v1v2)<times8>, and the relationship between the train and B, L=(v1v2)<times7>, we can get <8(v1v2)=7(v1v2)>, which is simplified to <v1v2>. The train will meet B 5 minutes after leaving A. The distance between A and B and the distance between A and B in these 5 minutes is (300v1 - 300v2), from the time B meets the train, the time when A and B meet is set to be (t), then ((v2.1 + v2.1)t = 300v1 - 300v2.1). Substituting <<v1>= 15v2> into <<2v2.1>= 300.1> time15v2 - 300v2>, the solution is <<t = 2100>> seconds = 35 minutes. ** 3. Average problem ** 1. [Question: Cai Chen's average score in the final exam was 89 in politics, Chinese, mathematics, English, and biology.] The average score for politics and mathematics was 91.5. The average score for Chinese and English was 84. The average score for politics and English was 86 points, and English was 10 points higher than Chinese. He asked Cai Chen what his marks for each subject should be for this exam? - ** Answer **: If the Chinese test score is x, then the English test score is x + 10. According to the average score of Chinese and English, the equation could be written as [(x + x + 10)<div2 = 84], and the solution was [x = 79]. The English score was [89]. The average score for politics and English is 86 points, so the political score is 86/89 = 83/. The average score for politics and mathematics is 91.5, so the math score is 91.5'time2 - 83 = 100 '. The biology score was [89 'times5-(79 + 89+83 + 100)=94]. 2. [Title: A and B cotton fields, with an average yield of 185 catties of seed cotton per mu] The A cotton field had 5 mu, with an average yield of 203 catties of seed cotton per mu. The B cotton field had an average yield of 170 catties of seed cotton per mu. How many mu was the B cotton field? - ** Answer **: Suppose B has a cotton field of x mu. According to the total yield equal-countable equation, the solution is x = 6 mu. 3. [Question: Given that the sum of eight consecutive odd numbers is 144, find these eight consecutive odd numbers.] - ** Answer **: If the first odd number is <x>, then the eight consecutive odd numbers are <x,x + 2,x+4,x + 6,x+8,x + 10,x+12,x + 14>. According to the sum of the 144 equations, x+(x + 2)+(x+4)+(x + 6)+(x+8)+(x + 10)+(x+12)+(x + 14)=144, the solution was x = 7. The eight consecutive odd numbers were 7, 9, 11, 13, 15, 17, 19, and 21. 4. ** Question **: A sugar is 8.8 yuan per kilogram, B sugar is 7.2 yuan per kilogram. How much B sugar should be mixed with 5 kilograms of A sugar to make the price of each kilogram of sugar 8.2 yuan? - ** Answer **: Assuming that you need [B] sugar in a kilogram. According to the constant countable equation of the total price before and after mixing, the solution is [8.8'times5 +7.2x=(5 + x)'times8.2]. 5. [Question: Buy five sheep from the canteen. Take out two sheep each time and weigh them. There are ten different weights (kg):47, 50, 51, 52, 53, 54, 55, 57, 58, 59.] How many kilograms did each of the five sheep weigh? - ** Answer **: Let the weight of the five sheep from light to heavy be a, b, c, d, e. Each sheep was called together with the other four sheep once, so each sheep was called four times. Then, 4(a + b + c + d+e)=(47 + 50+51+52+53+54+55+57+58+59)=536, so 4 (a + b + c + d+e = 134). The lightest combination is a + b = 47, and the heaviest combination is d + e = 59, so c = 134- 47 + 59 =28. Because a + c = 50, then a = 50 - 28 = 22, b = 47 - 22 = 25, c+e = 58, then e = 58 - 28 = 30, d = 59 - 30 = 29. ** 4. Arithmetic Sequence Problem ** 1. [Question: Below is a string of numbers arranged according to a pattern: 2, 5, 8, 11, 14…] What was the 1995th item? - ** Answer **: This is an arithmetic sequence. The first term is <a1 = 2>, and the tolerance is <d = 3>. According to the general term formula of arithmetic sequence,<a_n= a1 +(n - 1)d>, the 1995th term is <a_{1995}=2+(1995 - 1)><times3 = 2+ 1994><times3 = 5984>. 2. [Question: Among the natural numbers starting from 1, what is the 100th number that cannot be completely divided by 3?] - ** Answer **: 1, 2, 3, 4, 5, 6, 7,.... Starting from 1, form a group of three numbers. The first two numbers in each group cannot be divided by 3. If there are two numbers in a group, there will be a group of 100 numbers. Each group has three numbers, and there will be a total of 50. Then the 100th number that cannot be divided by 3 is 150 - 1 = 149. ** 5. The problem of cows eating grass ** 1. [Title: A patch of green grass that grows at a constant rate every day.] This patch of grass could feed twenty-seven cows for six days or twenty-three cows for nine days. How many days could this patch of grass feed twenty-one cows? - ** Answer **: Suppose a cow eats one unit of grass a day. First, find the amount of grass that grows every day. Twenty-seven cows ate for six days, eating a total of 27,000 units of grass; Twenty-three cows ate for nine days, eating a total of 20,000 units of grass. (207- 162 = 45 units. This is the amount of grass that grows in a day, so the grass grows in a day. The original amount of grass is in units of [162 - 6'times15 = 72]. Assuming that 21 cows eat for [x] days, the equation [21x = 72+15x] can be solved for [x = 12] days. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>

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>

There were many ways to solve the question of two squares in the elementary school Mathematical Olympiad. The following were some common situations: ** 1. Area related ** 1. ** Area difference problem ** - If you know the difference in the perimeter of two squares, you can first find the difference in the length of the sides. For example, the difference between the circumference of two squares is 32cm. Since the circumference of a square = 4 x the length of a side, the difference in the length of a side is 32 × 4 = 8cm. If the area difference was known, it could be solved by making an auxiliary line. For example, the shadow part (the area difference part) was cut into several parts, and the area of the special figure (such as a small square) constructed from it was used to calculate the area of the rectangular shape, and then other unknown quantities such as the length of the side were calculated. 2. ** Overlapping Area Problem ** - When two squares overlapped, the area of the shadow (usually the non-overlapping part) needed to be calculated according to the known conditions (such as the length of the side, the length of the line segment of the overlapping part, etc.) through the combination and decomposition of the figures. He might use the area formula of a triangle, a echelon, and other shapes, as well as the idea of equal-area transformation to convert the area of an irregular shape into the area of a regular shape for calculation. ** 2. Edge length related ** 1. ** Proportional relationship of side length ** - If the proportional relationship between the sides of the two squares and other related conditions (such as the sum of the area, the sum of the perimeter, etc.) were given, one could set the side of one of the squares as an unknown number and solve the equation according to the conditions. For example, let the side length of a small square be x, and the side length of a large square be kx (k is a known proportional coefficient). Then, according to the perimeter sum or area sum conditions, the equation is listed to solve the side length. 2. ** The relationship between length of side and line segment ** - When two squares have a specific position relationship and the length of some line segments is given (such as the line segment connecting the two square vertexes, the distance between the two squares, etc.), the Pythagorean theorem or the similarity of geometric figures can be used to establish the relationship between the length of the line and these line segments, thereby solving the length of the line or other related quantities. ** 3. Combination Diagram ** 1. ** Construct a special pattern ** - For example, when calculating the area of the shadow part in the combination of two squares, one could construct a echelon by connecting the diagonal lines, and use the properties of the echelon (such as the equal area of the wings, that is, the area of the two non-adjacent triangle that is divided into four by the diagonal of the echelon is equal) to convert and calculate the area. 2. ** Relationship between the whole and the parts ** - Take the graph composed of two squares as a whole, analyze the relationship between the various parts (such as squares, triangle, echelon, etc.), and calculate the unknown variables such as the area and side length of the whole or part according to the known conditions. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>

The 'Hindustan Gamer Cartoon' is likely to feature local heroes and mythological elements. It may have multiplayer options for added fun and social interaction. The animation quality could be top-notch, making the gaming experience immersive.

For the five subject competitions (Mathematics, Physics, Chemistry, Biology, and Information Science): - Provinces and cities preliminary/preliminaries (city level), league/semi-finals/national preliminary (provincial level): You can register through the official website of the relevant discipline society, the official website of the Science and Technology Association, relevant public accounts, and the middle school where the examinee is. However, many provinces and cities required candidates to register through their secondary schools. They did not accept individual registration. They had to pay attention to the policies of their provinces and cities. - National finals: The provincial associations will submit the list of participants (provincial team list), or students with provincial team qualifications will fill in the registration information themselves. For the National Middle School Essay Competition, you can register individually or collectively through <anno data-annotation-id ="2fd88445 - 4fd2 - 4f16 - 4f16 - 8f1d8f111124"></anno>. For the National Middle School Students 'innovative essay competition, schools or individuals could register through <anno data-annotation-id ="23333340 - 4440 - 4410 - 4410 - 9998 - 999999999999"></anno></anno>. For the "Foreign Language Research Society Cup" National Middle School Students 'Foreign Language Accomplishment Competition, you can download the designated platform "Foreign Research University" App (home version) on your mobile phone and log in to the competition area to register. For the 2025 IEO International Economics Olympiad, if you want to form a team to participate, you can choose Hanlin (it has a rich team resource pool and will conduct a preliminary examination. After teammates recognize each other, they will confirm the team). The article did not mention whether there were special channels for individual registration. It can be speculated that it is possible to register through official channels according to the convention. The registration methods for different Olympiad competitions were different, and the registration operation had to be carried out according to the requirements of the specific competition. While watching the Olympics, you can also read the wonderful novels related to the Olympics!