美国AMC8数学竞赛试题(含答案)
2019年美国数学竞赛8年级(AMC8)真题(附答案)(电脑版)
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
AMC 系列比赛一共有以下几个比赛：
American Mathematics Competition 8 - C 8
American Mathematics Competition 10/12 - AMC 10/12
American Invitational Mathematics Exam - AIME United States Mathematical Olympiad and Junior Mathematical Olympiad - USA(J)MO
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
2019 年美国数学竞赛 8 年级(AMC8) 真题(附答案 )(电脑版 )
美国数学竞赛 8 年级(AMC8)真题（附答案）
AMC 系列全称 American Mathematics Competitions 的（1950 年开始举办） 的和最负盛名初高中生数学竞赛。 日下午结束。
美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)
美国数学竞赛AMC8 – 2005年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:If x is the number, then 2x=60 and x=30. Dividing the number by 2 yields 15.中文解析：按照Connie的计算，这个数乘以2是60，可知这个数是30. 应该做的计算是30除以2，因而正确答案应该是15. 答案是B。
Problem 2Answer: CSolution:Karl paid 5*2.5=$12.5. 20% of this cost that he saved is 12.5*0.2=$2.5.中文解析：Karl按原价买了5个文件夹，支付的费用是：2.5*5=12.5. 折扣价是：1.25*0.8=10。
如果Karl 等一天，可以省2.5元。
答案是C.Problem 3Answer: DSolution:Rotating square ABCD counterclockwise 45° so that the line of symmetry BD is a vertical line makes it easier to see that 4 squares need to be colored to match its corresponding square.中文解析：如上图所示，以BD为对称轴，标蓝色的方块需要涂黑。
共4块，答案是D。
Problem 4Answer: CSolution:The perimeter of the triangle is 6.1+8.2+9.7=24cm. A square's perimeter is four times its side length, since all its side lengths are equal. If the square's perimeter is 24, the side length is24/4=6, and the area is 6*6=36.中文解析：三角形的周长是：6.1+8.2+9.7=24. 正方形的周长和三角形相等，也是24，则其边长是24/4=6. 其面积是：6*6=36. 答案是C。
amc8试题
amc8试题AMC 8试题第一题1. 在一盘巨大的沙漏中，向上面的玻璃瓶中积分有10多克，此后倒转沙漏，沿相同的管道均匀地下降。
五分钟后，沙漏的底部以每分钟多少克的速度吐出沙子？ (A) 2 (B) 4 (C) 5 (D) 10 (E) 12解答：这道题考察的是单位换算和平均速度的概念。
首先要注意题目中的单位换算，从“分钟”转换为“克”。
沙漏中积分的10克流失掉后，下降的总时间是五分钟，所以平均每分钟下降的克数就是10克除以五分钟，即2克/分钟。
因此，答案是 (A) 2。
第二题2. 将1到999的所以整数从小到大写下来，在所有数字中有多少个数字出现了至少一次？ (A) 2701 (B) 3024 (C) 4000 (D) 5004 (E) 7381解答：这道题考察的是数字的计数和求和。
要找到所有1到999的整数中出现了至少一次的数字个数，我们可以分别计算每个位上的数字。
从个位开始，0出现了100次（0至999共100个数字），1出现了在个位上9次（10次）（1至991共100个数字，再加上100次），以此类推，9出现了100次（9至999共100个数字）。
因此，个位上的数字总共出现了10\*100次，即1000次。
同样的道理，我们可以得出十位上的数字总共出现了10\*100次，即1000次，百位上的数字总共出现了10\*100次，即1000次。
所以，所有1到999的整数中出现了至少一次的数字个数为1000+1000+1000，即3000个。
因此，答案是(B) 3024。
第三题3. 小明有6支不同颜色的铅笔和8张不同的宣纸。
一张宣纸必须用全相同颜色的铅笔画。
小明可以使用的方法数是多少？ (A) 48 (B) 56 (C) 64 (D) 72 (E) 96解答：这道题考察的是排列组合。
小明有6种选择颜色的铅笔，对于每种颜色，有8张宣纸可供选择。
所以，总的方法数为6种颜色的选择乘以8张宣纸的选择，即6乘以8，得到答案为 (C) 48。
AMC 美国数学竞赛试题+详解 英文版
2013 AMC8 Problems1.Danica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?2.A sign at the fish market says, "50% off, today only: half-pound packages for just $3 perpackage." What is the regular price for a full pound of fish, in dollars?What is the value of?3.4.Eight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill. What was the total bill? 5.Hammie is in thegrade and weighs 106 pounds. His quadruplet sisters are tiny babiesand weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these five children or the median weight, and by how many pounds?6.The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example, . What is the missing number in the top row?7.Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass, Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clear the crossing at a constant speed. Which of the following was the most likely number of cars in the train?8.A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?9.The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then on which jump will he first be able to jump more than 1 kilometer?10.What is the ratio of the least common multiple of 180 and 594 to the greatest common factor of 180 and 594?11.Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?12.At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?13.When Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?14.Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?15.If , , and , what is the product of , , and ?16.A number of students from Fibonacci Middle School are taking part in a community serviceproject. The ratio of -graders to -graders is , and the the ratio of -graders to-graders is . What is the smallest number of students that could be participating in the project?17.The sum of six consecutive positive integers is 2013. What is the largest of these six integers?18.Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?19.Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?20.A rectangle is inscribed in a semicircle with longer side on the diameter. What is thearea of the semicircle?21.Samantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?23.Angle of is a right angle. The sides of are the diameters of semicirclesas shown. The area of the semicircle on equals , and the arc of the semicircle onhas length . What is the radius of the semicircle on ?24.Squares , , and are equal in area. Points and are the midpointsof sides and , respectively. What is the ratio of the area of the shaded pentagonto the sum of the areas of the three squares?25.A ball with diameter 4 inches starts at point A to roll along the track shown. The track iscomprised of 3 semicircular arcs whose radii are inches, inches, andinches, respectively. The ball always remains in contact with the track and does notslip. What is the distance the center of the ball travels over the course from A to B?2013 AMC8 Problems/Solutions1. ProblemDanica wants to arrange her model cars in rows with exactly 6 cars in each row. She now has 23 model cars. What is the smallest number of additional cars she must buy in order to be able to arrange all her cars this way?Solution:In order to have her model cars in perfect, complete rows of 6, Danica must have a number ofcars that is a multiple of 6. The smallest multiple of 6 which is larger than 23 is 24, so she'll need to buy more model car.2.A sign at the fish market says, "50% off, today only: half-pound packages for just $3 per package." What is the regular price for a full pound of fish, in dollars?ProblemSolution: The 50% off price of half a pound of fish is $3, so the 100%, or the regular price, of a half pound of fish is $6. Consequently, if half a pound of fish costs $6, then a whole pound of fish is dollars.What is the value of?3. ProblemNotice that we can pair up every two numbers to make a sum of 1:SolutionTherefore, the answer is .4. ProblemEight friends ate at a restaurant and agreed to share the bill equally. Because Judi forgot her money, each of her seven friends paid an extra $2.50 to cover her portion of the total bill.What was the total bill?Each of her seven friends paidto cover Judi's portion. Therefore, Judi's portion mustbe. Since Judi was supposed to payof the total bill, the total bill must be.Solution5.Hammie is in thegrade and weighs 106 pounds. His quadruplet sisters are tiny babiesand weigh 5, 5, 6, and 8 pounds. Which is greater, the average (mean) weight of these fivechildren or the median weight, and by how many pounds?ProblemLining up the numbers (5, 5, 6, 8, 106), we see that the median weight is 6 pounds. SolutionThe average weight of the five kids is .Therefore, the average weight is bigger, bypounds, making the answer.6. The number in each box below is the product of the numbers in the two boxes that touch it in the row above. For example,. What is the missing number in the top row?ProblemSolutionLet the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.Solution 1: Working BackwardsWe see that, making.It follows that, so.Another way to do this problem is to realize what makes up the bottommost number. Thismethod doesn't work quite as well for this problem, but in a larger tree, it might be faster. (In this case, Solution 1 would be faster since there's only two missing numbers.)Solution 2: Jumping Back to the StartAgain, let the value in the empty box in the middle row be , and the value in the empty box in the top row be . is the answer we're looking for.We can write some equations:Now we can substitute into the first equation using the two others:7. Trey and his mom stopped at a railroad crossing to let a train pass. As the train began to pass,Trey counted 6 cars in the first 10 seconds. It took the train 2 minutes and 45 seconds to clearthe crossing at a constant speed. Which of the following was the most likely number of cars inthe train?ProblemIf Trey saw, then he saw.Solution 12 minutes and 45 seconds can also be expressed asseconds.Trey's rate of seeing cars,, can be multiplied byon the top andbottom (and preserve the same rate):. It follows that the most likely number of cars is.2 minutes and 45 seconds is equal to.Solution 2Since Trey probably counts around 6 cars every 10 seconds, there are groups of 6cars that Trey most likely counts. Since, the closest answer choice is.8. A fair coin is tossed 3 times. What is the probability of at least two consecutive heads?ProblemFirst, there areways to flip the coins, in order.Solution The ways to get two consecutive heads are HHT and THH. The way to get three consecutive heads is HHH.Therefore, the probability of flipping at least two consecutive heads is .9. The Incredible Hulk can double the distance he jumps with each succeeding jump. If his first jump is 1 meter, the second jump is 2 meters, the third jump is 4 meters, and so on, then onwhich jump will he first be able to jump more than 1 kilometer?ProblemThis is a geometric sequence in which the common ratio is 2. To find the jump that would be over a 1000 meters, we note that. SolutionHowever, because the first term isand not, the solution to the problem is10. What is the ratio of the least common multiple of 180 and 594 to the greatest common factorof 180 and 594?ProblemTo find either the LCM or the GCF of two numbers, always prime factorize first. Solution 1The prime factorization of . The prime factorization of .Then, find the greatest power of all the numbers there are; if one number is one but not the other, use it (this is ). Multiply all of these to get 5940.For the GCF of 180 and 594, use the least power of all of the numbers that are in bothfactorizations and multiply. = 18. Thus the answer = =.We start off with a similar approach as the original solution. From the prime factorizations, the GCF is 18.Similar SolutionIt is a well known fact that. So we have,.Dividing by 18 yields .Therefore, .11. Ted's grandfather used his treadmill on 3 days this week. He went 2 miles each day. On Monday he jogged at a speed of 5 miles per hour. He walked at the rate of 3 miles per hour on Wednesday and at 4 miles per hour on Friday. If Grandfather had always walked at 4 miles per hour, he would have spent less time on the treadmill. How many minutes less?ProblemWe use that fact that . Let d= distance, r= rate or speed, and t=time. In this case, letrepresent the time.SolutionOn Monday, he was at a rate of . So,.For Wednesday, he walked at a rate of . Therefore,.On Friday, he walked at a rate of. So,. Adding up the hours yields++=.We now find the amount of time Grandfather would have taken if he walked atperday. Set up the equation,.To find the amount of time saved, subtract the two amounts: -=.To convert this to minutes, we multiply by 60.Thus, the solution to this problem is12. At the 2013 Winnebago County Fair a vendor is offering a "fair special" on sandals. If you buy one pair of sandals at the regular price of $50, you get a second pair at a 40% discount, and a third pair at half the regular price. Javier took advantage of the "fair special" to buy three pairs of sandals. What percentage of the $150 regular price did he save?ProblemFirst, find the amount of money one will pay for three sandals without the discount. We have.SolutionThen, find the amount of money using the discount: .Finding the percentage yields .To find the percent saved, we have13. ProblemWhen Clara totaled her scores, she inadvertently reversed the units digit and the tens digit of one score. By which of the following might her incorrect sum have differed from the correct one?Let the two digits be and. SolutionThe correct score was . Clara misinterpreted it as. The difference between thetwo iswhich factors into. Therefore, since the difference is a multiple of 9,the only answer choice that is a multiple of 9 is.14.Abe holds 1 green and 1 red jelly bean in his hand. Bea holds 1 green, 1 yellow, and 2 red jelly beans in her hand. Each randomly picks a jelly bean to show the other. What is the probability that the colors match?ProblemThe probability that both show a green bean is. The probability that both show ared bean is . Therefore the probability isSolution15. If ,, and , what is the product of, , and ?ProblemSolutionTherefore,.Therefore,.To most people, it would not be immediately evident that , so we can multiply 6'suntil we get the desired number:, so.Therefore the answer is16. A number of students from Fibonacci Middle School are taking part in a community serviceproject. The ratio of-graders to-graders is, and the the ratio of-graders to-graders is . What is the smallest number of students that could be participating inthe project?ProblemSolutionWe multiply the first ratio by 8 on both sides, and the second ratio by 5 to get the same number for 8th graders, in order that we can put the two ratios together:Solution 1: AlgebraTherefore, the ratio of 8th graders to 7th graders to 6th graders is. Since the ratiois in lowest terms, the smallest number of students participating in the project is.The number of 8th graders has to be a multiple of 8 and 5, so assume it is 40 (the smallest possibility). Then there are 6th graders and7th graders. The numbers ofstudents isSolution 2: Fakesolving17. The sum of six consecutive positive integers is 2013. What is the largest of these six integers?ProblemThe mean of these numbers is. Therefore the numbers are, so the answer isSolution 1Let thenumber be . Then our desired number is.Solution 2Our integers are , so we have that.Let the first term be. Our integers are. We have,Solution 318.Isabella uses one-foot cubical blocks to build a rectangular fort that is 12 feet long, 10 feet wide, and 5 feet high. The floor and the four walls are all one foot thick. How many blocks does the fort contain?ProblemThere arecubes on the base of the box. Then, for each of the 4 layers abovethe bottom (as since each cube is 1 foot by 1 foot by 1 foot and the box is 5 feet tall, there are4 feet left), there arecubes. Hence, the answer is.Solution 1 We can just calculate the volume of the prism that was cut out of the originalbox. Each interior side of the fort will be 2 feet shorter than each side of the outside. Since thefloor is 1 foot, the height will be 4 feet. So the volume of the interior box is.Solution 2The volume of the original box is . Therefore, the number of blockscontained in the fort is19. Bridget, Cassie, and Hannah are discussing the results of their last math test. Hannah shows Bridget and Cassie her test, but Bridget and Cassie don't show theirs to anyone. Cassie says, 'I didn't get the lowest score in our class,' and Bridget adds, 'I didn't get the highest score.' What is the ranking of the three girls from highest to lowest?ProblemIf Hannah did better than Cassie, there would be no way she could know for sure that she didn't get the lowest score in the class. Therefore, Hannah did worse than Cassie. Similarly, ifHannah did worse than Bridget, there is no way Bridget could have known that she didn't getthe highest in the class. Therefore, Hannah did better than Bridget, so our order isSolution20. Arectangle is inscribed in a semicircle with longer side on the diameter. What is thearea of the semicircle?ProblemSolutionA semicircle has symmetry, so the center is exactly at the midpoint of the 2 side on the rectangle, making the radius, by the Pythagorean Theorem,. The area is21. ProblemSamantha lives 2 blocks west and 1 block south of the southwest corner of City Park. Her school is 2 blocks east and 2 blocks north of the northeast corner of City Park. On school days she bikes on streets to the southwest corner of City Park, then takes a diagonal path through the park to the northeast corner, and then bikes on streets to school. If her route is as short as possible, how many different routes can she take?SolutionThe number of ways to get from Samantha's house to City Park is, and the number ofways to get from City Park to school is. Since there's one way to go through CityPark (just walking straight through), the number of different ways to go from Samantha's house to City Park to school22.Toothpicks are used to make a grid that is 60 toothpicks long and 32 toothpicks wide. How many toothpicks are used altogether?ProblemThere are 61 vertical columns with a length of 32 toothpicks, and there are 33 horizontal rowswith a length of 60 toothpicks. An effective way to verify this is to try a small case, i.e. a grid of toothpicks. Thus, our answer isSolution23.Angleof is a right angle. The sides ofare the diameters of semicircles as shown. The area of the semicircle on equals, and the arc of the semicircle onhas length . What is the radius of the semicircle on?ProblemIf the semicircle on AB were a full circle, the area would be 16pi. Therefore the diameter of the first circle is 8. The arc of the largest semicircle would normally have a complete diameter of 17. The Pythagorean theorem says that the other side has length 15, so the radius is.Solution 1We go as in Solution 1, finding the diameter of the circle on AC and AB. Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is , and the middle one is , so the radius is .Solution 224. Squares, , andare equal in area. Pointsandare the midpointsof sidesand, respectively. What is the ratio of the area of the shaded pentagonto the sum of the areas of the three squares?ProblemSolution 1First let(whereis the side length of the squares) for simplicity. We can extenduntil it hits the extension of. Call this point. The area of trianglethen isThe area of rectangleis. Thus, our desired area is. Now, the ratio of the shaded area to the combined area of the three squares is.Solution 2Let the side length of each square be 1.Let the intersection ofandbe .Since, . Sinceand are vertical angles, theyare congruent. We also haveby definition.So we haveby congruence. Therefore,.Since andare midpoints of sides,. This combined withyields.The area of trapezoidis.The area of triangleis.So the area of the pentagon is .The area of the 3 squares is . Therefore, .Solution 3Let the intersection of andbe .Now we haveand .Because both triangles has a side on congruent squares therefore.Becauseand are vertical angles. Also bothand are right angles so .Therefore by AAS (Angle, Angle, Side) . Then translating/rotating the shadedinto the position ofSo the shaded area now completely covers the squareSet the area of a square asTherefore, .25.A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3 semicircular arcs whose radii are inches, inches, andinches, respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from A to B?ProblemThe radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses inches, and it gains inches on B.So, the departurefrom the length of the track means that the answer is .Solution 1The total length of all of the arcs is . Since we want the path fromthe center, the actual distance will be shorter. Therefore, the only answer choice less thanis . This solution may be invalid because the actual distance can be longer if the path the center travels is on the outside of the curve, as it is in the middle bump. Solution 2。
AMC8(美国数学竞赛)历年真题、答案及中英文解析
AMC8（美国数学竞赛）历年真题、答案及中英文解析艾蕾特教育的AMC8 美国数学竞赛考试历年真题、答案及中英文解析：AMC8-2020年：真题 --- 答案---解析（英文解析+中文解析）AMC8 - 2019年：真题----答案----解析（英文解析+中文解析）AMC8 - 2018年：真题----答案----解析（英文解析+中文解析）AMC8 - 2017年：真题----答案----解析（英文解析+中文解析）AMC8 - 2016年：真题----答案----解析（英文解析+中文解析）AMC8 - 2015年：真题----答案----解析（英文解析+中文解析）AMC8 - 2014年：真题----答案----解析（英文解析+中文解析）AMC8 - 2013年：真题----答案----解析（英文解析+中文解析）AMC8 - 2012年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 2010年：真题----答案----解析（英文解析+中文解析）AMC8 - 2009年：真题----答案----解析（英文解析+中文解析）AMC8 - 2008年：真题----答案----解析（英文解析+中文解析）AMC8 - 2007年：真题----答案----解析（英文解析+中文解析）AMC8 - 2006年：真题----答案----解析（英文解析+中文解析）AMC8 - 2005年：真题----答案----解析（英文解析+中文解析）AMC8 - 2004年：真题----答案----解析（英文解析+中文解析）AMC8 - 2003年：真题----答案----解析（英文解析+中文解析）AMC8 - 2002年：真题----答案----解析（英文解析+中文解析）AMC8 - 2001年：真题----答案----解析（英文解析+中文解析）AMC8 - 2000年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1998年：真题----答案----解析（英文解析+中文解析）AMC8 - 1997年：真题----答案----解析（英文解析+中文解析）AMC8 - 1996年：真题----答案----解析（英文解析+中文解析）AMC8 - 1995年：真题----答案----解析（英文解析+中文解析）AMC8 - 1994年：真题----答案----解析（英文解析+中文解析）AMC8 - 1993年：真题----答案----解析（英文解析+中文解析）AMC8 - 1992年：真题----答案----解析（英文解析+中文解析）AMC8 - 1991年：真题----答案----解析（英文解析+中文解析）AMC8 - 1990年：真题----答案----解析（英文解析+中文解析）AMC8 - 1989年：真题----答案----解析（英文解析+中文解析）AMC8 - 1988年：真题----答案----解析（英文解析+中文解析）析）AMC8 - 1986年：真题----答案----解析（英文解析+中文解析）AMC8 - 1985年：真题----答案----解析（英文解析+中文解析）◆AMC介绍◆AMC（American Mathematics Competitions) 由美国数学协会（MAA）组织的数学竞赛，分为 AMC8 、 AMC10、 AMC12 。
AMC8数学竞赛试题及详解
Problem 3What is the value of the expression ?化简的标准和顺序Problem 6If the degree measures of the angles of a triangle are in the ratio , what is the degree measure of the largest angle of the triangle?角度制与弧度制Problem 7Let be a 6-digit positive integer, such as 247247, whose first three digits are the same as its last three digits taken in the same order. Which of the following numbers must also be a factor of ?整除特征Problem 10A box contains five cards, numbered 1, 2, 3, 4, and 5. Three cards are selected randomly without replacement from the box. What is the probability that 4 is the largest value selected?离散概率Problem 11A square-shaped floor is covered with congruent square tiles. If the total number of tiles that lie on the two diagonals is 37, how many tiles cover the floor?tiles英[taɪlz]美[taɪlz]n. 瓦片，瓷砖( tile的名词复数); 扁平的小棋子;Problem 13Peter, Emma, and Kyler played chess with each other. Peter won 4 games and lost 2 games. Emma won 3 games and lost 3 games. If Kyler lost 3 games, how many games did he win?Problem 14Chloe and Zoe are both students in Ms. Demeanor's math class. Last night they each solved half of the problems in their homework assignment alone and then solved the other half together. Chloe had correct answers to only of the problems she solved alone, but overall of her answers were correct. Zoe had correct answersto of the problems she solved alone. What was Zoe's overall percentage of correct answers? C设数Problem 15In the arrangement of letters and numerals below, by how many different paths can one spell AMC8? Beginning at the A in the middle, a path allows only moves from oneletter to an adjacent (above, below, left, or right, but not diagonal) letter. One example of such a path is traced in the picture.乘法原理加法原理Problem 16In the figure below, choose point on so that and have equal perimeters. What is the area of ?D如果在AC上呢Problem 17Starting with some gold coins and some empty treasure chests, I tried to put 9 gold coins in each treasure chest, but that left 2 treasure chests empty. So instead I put 6 gold coins in each treasure chest, but then I had 3 gold coins left over. How many gold coins did I have?盈亏问题图示法Problem 18In the non-convex quadrilateral shown below, is a rightangle, , , , and .What is the area of quadrilateral ?勾股定理与逆定理“凸”定义Problem 19For any positive integer , the notation denotes the product of the integers through . What is the largest integer for which is a factor of the sum ?Problem 20An integer between and , inclusive, is chosen at random. What is the probability that it is an odd integer whose digits are all distinct?5*8*8*7Problem 21Suppose , , and are nonzero real numbers, and . What are the possible value(s) for ?可能性列全或者变个形Problem 22In the right triangle , , , and angle is a right angle. A semicircle is inscribed in the triangle as shown. What is the radius of the semicircle?相切，连接切点和圆心。
美国数学竞赛AMC8 -- 2006年真题解析(英文解析+中文解析)
美国数学竞赛AMC8 – 2006年真题解析(英文解析+中文解析) \Problem 1Answer: DSolution:The three prices round to $2, $5, and $10, which has a sum of 17.中文解析：三件商品价格先近似取整，然后求和：2+5+10=17. 答案是D。
Problem 2Answer: CSolution:As the AMC 8 only rewards 1 point for each correct answer, everything is irrelevant except the number Billy answered correctly，13.中文解析：正确的题目每题1分，错误或没做的题目都是0分，做对13题的得分应该是13. 答案是C。
Problem 3Answer: ASolution:When Elisa started, she finished a lap in 25/10=2.5 minutes. Now, she finishes a lap is 24/12=2 minutes. The difference is 2.5-2=0.5中文解析：开始25分钟游10圈，平均2.5分钟游1圈。
后来24分钟游12圈，平均2分钟游1圈。
速度从2.5分钟提高到2分钟，提高了0.5分钟，即1/2 分钟。
答案是A。
Problem 4Answer: BSolution:If the spinner goes clockwise 2+1/4 revolutions and then counterclockwise 3+3/4 revolutions, it ultimately goes counterclockwise 1+1/2 which brings the spinner pointing east.中文解析：最初方向指向西，转整数圈不改变指针方向。
amc8数学竞赛试题及答案
amc8数学竞赛试题及答案AMC8数学竞赛试题及答案试题一：在一次数学竞赛中，有100名学生参加。
如果前10%的学生获得金牌，前20%的学生获得银牌，前30%的学生获得铜牌，那么获得金牌、银牌和铜牌的学生人数分别是多少？答案一：获得金牌的学生人数为100名的10%，即10名学生。
获得银牌的学生人数为100名的20%，即20名学生。
获得铜牌的学生人数为100名的30%，即30名学生。
试题二：一个长方形的长是宽的两倍，如果长方形的周长是24厘米，那么这个长方形的长和宽分别是多少厘米？答案二：设长方形的宽为x厘米，那么长为2x厘米。
周长公式为2（长+宽）=24厘米。
代入长宽关系，得到2(2x+x)=24，解得x=4厘米。
因此，宽为4厘米，长为2*4=8厘米。
试题三：一个班级有40名学生，其中1/4的学生喜欢数学，1/3的学生喜欢英语，1/6的学生喜欢科学。
如果班级中有2名学生喜欢数学和英语，1名学生喜欢英语和科学，没有学生同时喜欢数学、英语和科学，那么喜欢科学的学生有多少人？答案三：喜欢数学的学生有40*1/4=10人，喜欢英语的学生有40*1/3≈13.33人（取整数为13人），喜欢科学的学生有40*1/6≈6.67人（取整数为6人）。
因为2名学生喜欢数学和英语，1名学生喜欢英语和科学，所以喜欢科学的学生人数为6-1=5人。
试题四：一个数字的平方加上这个数字本身等于2017，这个数字是什么？答案四：设这个数字为x，根据题意有方程x^2 + x = 2017。
这是一个一元二次方程，可以分解为(x+43)(x-47)=0。
解得x=-43或x=47。
因为数字不能为负，所以这个数字是47。
试题五：在一个圆内接正六边形的边长为a，求这个圆的半径。
答案五：正六边形可以被分成6个等边三角形，每个三角形的边长为a。
圆的半径等于正六边形的中心到顶点的距离，也就是等边三角形的高。
等边三角形的高可以通过公式h=√3/2*a计算得出。
美国数学竞赛AMC8 -- 2005年真题解析(英文解析+中文解析)
美国数学竞赛AMC8 – 2005年真题解析(英文解析+中文解析)Problem 1Answer: BSolution:If x is the number, then 2x=60 and x=30. Dividing the number by 2 yields 15.中文解析：按照Connie的计算，这个数乘以2是60，可知这个数是30. 应该做的计算是30除以2，因而正确答案应该是15. 答案是B。
Problem 2Answer: CSolution:Karl paid 5*2.5=$12.5. 20% of this cost that he saved is 12.5*0.2=$2.5.中文解析：Karl按原价买了5个文件夹，支付的费用是：2.5*5=12.5. 折扣价是：1.25*0.8=10。
如果Karl 等一天，可以省2.5元。
答案是C.Problem 3Answer: DSolution:Rotating square ABCD counterclockwise 45° so that the line of symmetry BD is a vertical line makes it easier to see that 4 squares need to be colored to match its corresponding square.中文解析：如上图所示，以BD为对称轴，标蓝色的方块需要涂黑。
共4块，答案是D。
Problem 4Answer: CSolution:The perimeter of the triangle is 6.1+8.2+9.7=24cm. A square's perimeter is four times its side length, since all its side lengths are equal. If the square's perimeter is 24, the side length is24/4=6, and the area is 6*6=36.中文解析：三角形的周长是：6.1+8.2+9.7=24. 正方形的周长和三角形相等，也是24，则其边长是24/4=6. 其面积是：6*6=36. 答案是C。
2023年数学竞赛AMC8真题A卷(含答案)
Middle Primary DivisionQuestions 1 to 10 are worth 3 marks each.1-10题，每题3分1.What is the total number of petals on all 5 flowers?图中5朵花总计有多少片花瓣?(A)10 (B)15 (C)20 (D)25 (E)502.2+3+7+8=(A)10 (B)20 (C)30 (D)40 (E)503.Which one of these shapes is a rectangle?哪个选项是长方形?(A) (B)(C)(D) (E)4.Which digital clock time matches the time shown on the clock face?哪个选项中的数字时钟显示的时间与钟面相同?(A)3:09 (B)[ 9:03 (C) 9:15(D)10:03 (E)10:155.Emma has a bag containing 5 red,4 yellow,1 black and 2 blue buttons.When she chooses 1 button at random,what colour is it most likely to be?艾玛有一袋纽扣，里面有5个红色纽扣，4个黄色纽扣，1个黑色纽扣和2个蓝色纽扣。
从中随机选出一个纽扣，这个纽扣最有可能是什么颜色?(A)green 绿色(B)blue 蓝色(C)black 黑色(D)yellow 黄色(E)red 红色6.What fraction of the circle is part A?A 部分占整个圆的几分之几?(A) one-half 二分之一(B) one-third 三分之一(C) two-thirds 三分之二(D) one-quarter 四分之一(E) three-quarters 四分之三7.In a board game,Nik rolls three standard dice,one at a time.He needs his three rolls to add to 12.His first two dice rolls are 5 and 3.What does he need his third roll to be?棋盘游戏中，尼克投掷三枚标准的骰子，每次掷出一枚。
2023年数学竞赛AMC8真题E卷(含答案)
Senior DivisionQuestions 1 to 10 are worth 3 marks each.1-10题，每题3分1.What is the value of 2023-3202?2023-3202的值为哪个选项?(A)-1221 (B)-1179 (C)1179 (D)12212.A parallelogram PQRS has an area of 60 cm²and side PQ of length 10 cm. Which length is 6cm?下图中的平行四边形PQRS 的面积为60cm², 边PQ 的长度为10cm。
哪条线段的长度是6cm?(A)RQ (B)RS (C)Q1 (D)PT3.Which one of these is equal to 57×953?57×953的值等于以下哪项?(A)321 (B)4321 (C)54321 (D)6543214.What is the difference between 25 and 5²?2⁵和5²相差多少?(A)0 (B)] (C)3 (D)55.What is the value of the angle O in the diagram?如图所示，角θ的度数是多少?(A)100° (B)110° (C)120° (D)130°6.The shaded square is inscribed in the larger square as shown.(E)5225(E)QS (E)7654321(E)7(E)140°What is the ratio of shaded to unshaded area in the diagram?下图中阴影正方形内接于较大的正方形中。
阴影部分的面积与非阴影部分的面积之比是多少?(A)5:4 (B)25:24 (C)3:2 (D)7:4 (E)12:77.Jemmy multiplies together all the integers from 1 to 18.What are the last three digits of the result?杰米将1到18之间的所有整数相乘。
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2001 年 美国AMC8 (2001年 月 日 时间40分钟)1. 卡西的商店正在制作一个高尔夫球奖品。
他必须给一颗高尔夫球面上的300个小凹洞着色， 如果他每着色一个小凹洞需要2秒钟，试问共需多 分钟才能完成他的工作。
(A) 4 (B) 6 (C) 8 (D) 10 (E) 12 。
2. 我正在思考两个正整数，它们的乘积是24且它们的和是11，试问这两个数中较大的数是什 么 。
(A) 3 (B) 4 (C) 6 (D) 8 (E) 12 。
3. 史密斯有63元，艾伯特比安加多2元，而安加所有的钱是史密斯的三分之一，试问艾伯特 有 元。
(A) 17 (B) 18 (C) 19 (D) 21 (E) 23 。
4. 在每个数字只能使用一次的情形下，将1，2，3，4及9作成最小的五位数，且此五位数为 偶数，则其十位数字为 。
(A) 1 (B) 2 (C) 3 (D) 4 (E) 9 。
5. 在一个暴风雨的黑夜里，史努比突然看见一道闪光。
10秒钟后，他听到打雷声音。
声音的速 率是每秒1088呎，但1哩是5280呎。
若以哩为单位的条件下，估计史努比离闪电处的距离 最接近下列何者 。
(A) 1 (B) 121 (C) 2 (D) 221 (E) 3 。
6. 在一笔直道路的一旁有等间隔的6棵树。
第1棵树与第4棵树之间的距离是60呎。
试问第1 棵树到最后一棵树之间的距离是 呎。
(A) 90 (B) 100 (C) 105 (D) 120 (E) 140 。
问题7、8、9请参考下列叙述：主题：竞赛场所上的风筝展览7. 葛妮芙为提升她的学校年度风筝奥林匹亚竞赛的质量，制作了一个小风筝与一个大风筝，并陈列在公告栏展览，这两个风筝都如同图中的形状，葛妮芙将小风筝张贴在单位长为一吋(即每两点距离一吋)的格子板上，并将大风筝张贴在单位长三吋(即每两点距离三吋)的格子板上。
试问小风筝的面积是 平方吋。
(A) 21 (B) 22 (C) 23 (D) 24 (E) 25 。
8. 葛妮芙在大风筝内装设一个连接对角顶点之十字交叉型的支撑架子，她必须使用 吋的 架子材料。
(A) 30 (B) 32 (C) 35 (D) 38 (E) 39 。
9. 大风筝要用金箔覆盖。
金箔是从一张刚好覆盖整个格子板的矩形金箔裁剪下来的。
试问从四 个角隅所裁剪下来废弃不用的金箔是 平方吋。
(A) 63 (B) 72 (C) 180 (D) 189 (E) 264 。
10. 某一收藏家愿按二角五分(即41元)银币面值2000%的比率收购银币。
在该比率下，卜莱登现 有四个二角五分的银币，则他可得到 元。
(A) 20 (B) 50 (C) 200 (D) 500 (E) 2000 。
11. 设四个点A ，B ，C ，D 的坐标依次为A (3，2)，B (3，-2)，C (-3，-2)，D (-3，0)。
则四边形 ABCD 的面积是 。
(A) 12 (B) 15 (C) 18 (D) 21 (E) 24 。
12. 若定义a ⊗b =b a b a -+，则(6⊗4)⊗3= 。
(A) 4 (B) 13 (C) 15 (D) 30 (E) 72 。
13. 在黎琪儿班级36位学生中，有12位学生喜爱巧克力派，有8位学生喜爱苹果派，且有6 位学生喜爱蓝莓派。
其余的学生中有一半喜爱樱桃派，另一半喜爱柠檬派。
黎琪儿想用圆形 图显示此项数据。
试问：她应该用 度的扇形表示喜欢樱桃派的学生。
(A) 10 (B) 20 (C) 30 (D) 50 (E) 72 。
14. 泰勒在自助餐店排队，准备挑选一种肉类，二种不同蔬菜，以及一种点心。
若不计较食物 的挑选次序，则他可以有多少不同选择方法？時間 速 率 乙 甲 (A) 時間 速 率 乙 甲 (B) 時間 速 率 乙 甲 (C) 时间 速 率 乙 甲 (D) 时间 速 率 乙 甲 (E) ‧肉类：牛肉、鸡肉、猪肉。
‧蔬菜：烤豆、玉米、马铃薯、蕃茄。
‧点心：巧克力糖、巧克力蛋糕、巧克力布丁、冰淇淋。
(A) 4 (B) 24 (C) 72 (D) 80 (E) 144 。
15. 一堆马铃薯共有44个，已知荷马每分钟可削好3个马铃薯的皮。
他开始削4分钟后， 克莉斯汀加入一起工作。
若克莉斯汀每分钟可削好5个马铃薯的皮。
则当他们完成削皮工 作，克莉斯汀削好多少个马铃薯的皮？ (A) 20 (B) 24 (C) 32 (D) 33 (E) 40 。
16. 把边长4吋的正方形纸张从中间对折，形成两层的矩形纸张，再沿着平行于折线的一半处把两层纸用剪刀剪开，得三个新的矩形，一大二小。
试问其中一个小矩形周长与大矩形周长的比值为 。
(A) 31 (B) 21 (C) 43 (D) 54 (E) 65。
17. 在“谁想成为百万富翁？”的游戏节目中，下表所示者为每一道问题之奖金(以元为 单位，其中K =1000)：试问在那两道问题之间，奖金增加的百分率为最小 。
(A) 从1到2 (B) 从2到3 (C) 从3到4 (D) 从11到12 (E) 从14到15。
18. 投掷两个骰子，掷得两个数字之乘积为5的倍数之机率为 。
(A) 361 (B) 181 (C) 61 (D) 3611 (E) 31。
19. 甲车在一已知时段内以固定速率行进，如下图虚线所示。
在同一距离内，乙车则以两倍速 率行进。
若乙车的速率与时间以实线表示，则下列那一图可描述这种情形 。
20. 甲透露他的考试分数给乙、丙、丁三人知道，但其余的人都隐匿他们的分数。
乙想：“至 少我们四个人之中有两个人分数一样”。
丙想：“我的分数不是最低的”。
丁想：“我的分 数不是最高的”，将乙、丙、丁三人的分数从最低至最高由左而右排列，则下列何者正 确 。
(A) 丁乙丙 (B) 乙丙丁 (C) 乙丁丙 (D) 丙丁乙 (E) 丁丙乙 。
21. 设五个相异正整数的平均数是15，中位数是18，则此五个正整数中的最大者可能之最大值 为 。
(A) 19 (B) 24 (C) 32 (D) 35 (E) 40 。
22. 在一份20道题目的考试中，若答对每题可得5分，未作答者每题得1分，答错每题得0分。
试问下面那一个成绩是不可能的 。
(A) 90 (B) 91 (C) 92 (D) 95 (E) 97 。
S Y Z RX23. 设R ，S ，T 三点为一等边三角形的三个顶点，而X ，Y ，Z 为△RTS 三边的中点。
若用此六个点中的任意三个点作顶点，可画出 类不全等的三角形。
(A) 1 (B) 2 (C) 3 (D) 4 (E) 20 。
24. 右图中心在线半部与下半部都是由3个红色小三角形，5个蓝色小三角形与 8个白色小三角形所组成。
当把上半图沿着中心线往下折迭时，有2对红色小三角形重合，3对蓝色小三角形重合，以及有两对红色与白色小三角形重合，试问有 对白色小三角形重合。
(A) 4 (B) 5 (C) 6 (D) 7 (E) 9。
25. 兹有24个四位数，每一个四位数都是用2，4，5，7四个数字各使用一次所作成。
这些四 位数中只有一个四位数是另一个四位数的倍数。
试问此四位数是下面那一个 。
(A) 5724 (B) 7245 (C) 7254 (D) 7425 (E) 7542 。
简答1. D ,2. D ,3. E ,4. E ,5. C ,6. B ,7. A ,8. E ,9. D , 10. A ,11. C , 12. A , 13. D , 14. C , 15. A , 16. E , 17. B , 18. D , 19. D , 20. A ,21. D , 22. E , 23. D , 24. B , 25. D ,水盆水量 时间 A水盆水量 时间 B 水盆水量 时间 E 水盆水量时间D 水盆水量 时间 C 糖果种类 学 生 人 数 8765432102002年 美国AMC8 (2002年11月 日 时间40分钟)1. 在一张纸上画一个圆与二条相异直线，问这样的图形最多可能会有几个交点？(A) 2 (B) 3 (C) 4 (D) 5 (E) 6 。
2. 利用面额2元的美金纸钞与面额5元的美金纸钞来组成美金17元，不考虑纸钞的排列顺序， 共有几种不同的方法？ (A) 2 (B) 3 (C) 4 (D) 5 (E) 6 。
3. 四个相异的正偶数，其平均值最小可为下列哪一个数？ (A) 3 (B) 4 (C) 5 (D) 6 (E) 7 。
4. 公元2002年是一个回文年(回文是指由左念到右与由右念到左是相同的)，则公元2002年之 后，下一个回文年其各位数字的乘积为下列哪一个数？ (A) 0 (B) 4 (C) 9 (D) 16 (E) 25 。
5. Carlos Montado 出生于公元2002年11月9日，当天是星期六，则他出生后满706天为星期 几？ (A) 星期一 (B) 星期三 (C) 星期五 (D) 星期六 (E) 星期日 。
6. 一个供鸟戏水的水盆会一直以每分钟20毫升的速度注入水，而以每分钟18毫升的速度排出 水，若从开始注水到水盆满溢后继续记录，则下列水盆水量与时间的关系图，哪一个是正确 的？(A) A (B) B (C) C (D) D (E) E 。
7. Sawyer 女士对她班上的学生做了一个调查，发现学生对五种 糖果喜好的人数分布情形如图所示，已知每位学生都只选一 种糖果，则喜爱糖果E 的人数占全班人数的百分之几？ (A) 5 (B) 12 (C) 15 (D) 16 (E) 20 。
第8、9、10题的资料都在下列短文及右表中主题：Juan 搜集的邮票： Juan 将他搜集的邮票依据发 行的国家及年代分类。
他买的价格为：巴西和法国的 邮票每张6分，秘鲁的邮票每张 4分，而西班牙的邮票每张5分。
(巴西和秘鲁位于南美洲，而法 国和西班牙位于欧洲。
) 8. 他共有几张发行于80年代的欧洲邮票？(A) 9 (B) 15 (C) 18 (D) 24 (E) 42 。
9. 他共花了多少钱去买发行于70年代以前(不包含70年代)的南美洲邮票？(1元=100分)(A) 0.40元 (B) 1.06元 (C) 1.80元 (D) 2.38元 (E) 2.64元 。
10. 他的70年代邮票的平均价格最接近下列哪一个答案？(A) 3.5分 (B) 4分 (C) 4.5分 (D) 5分 (E) 5.5分 。
11. 利用一些相同的小正方形地砖可排成一系列的正方形图形，每个正方形图形的边长都比前一个正方形图形的边长多一个地砖的边长，如图所示为此一系列图形的前三个，则第7个正方形图形比第6个正方形图形多几块地砖？ (A) 11 (B) 12 (C) 13 (D) 14 (E) 15 。