2007-2012 AMC8 中文试题和答案解析

美国数学学会中学生数学竞赛真题和答案解析2012AMC8-solutions

Solutions Pamphlet
American Mathematics Competitions
28th Annual
AMC 8
American Mathematics Contest 8 Tuesday, November 13, 2012
This Solutions Pamphlet gives at least one solution for each problem on this year’s exam and shows that all the problems can be solved using material normally associated with the mathematics curriculum for students in eighth grade or below. These solutions are by no means the only ones possible, nor are they necessarily superior to others the reader may devise. We hope that teachers will share these solutions with their students. However, the publication, reproduction, or communication of the problems or solutions of the AMC 8 during the period when students are eligible to participate seriously jeopardizes the integrity of the results. Dissemination at any time via copier, telephone, email, internet or media of any type is a violation of the competition rules.

2007年AMC8试题

This ﬁle was downloaded from the AoPS Math Olympiad Resources Page /
Page 1
USA
AMC 8 2007
paper route. He will use all of his birthday money and all of the money he earns from his paper route. In how many weeks will he be able to buy the mountain bike? (A) 24 (B) 25 (C) 26 (D) 27 (E) 28
This ﬁle was downloaded from the AoPS Math Olympiad Resources Page /
Page 2
USA
AMC 8 2007
11 Tiles I, II, III and IV are translated so one tile coincides with each of the rectangles A, B, C and D. In the ﬁnal arrangement, the two numbers on any side common to two adjacent tiles must be the same. Which of the tiles is translated to Rectangle C?
1
2
23
4
(A) 1 (B) 2 (C) 3 (D) 4 (E) cannot be determined 10 For any positive integer n, deﬁne n to be the sum of the positive factors of n. For example,

全国中学生生物学联赛试题及答案及美国数学竞赛8年级(AMC8)真题(附答案)(电脑版)

全国中学生生物学联赛试题注意事项：1．所有试题使用2B铅笔在机读卡上作答；2．试题按学科分类，单选和多选题混排，多选题答案完全正确才可得分3．纸质试卷72题，电子试卷48题，共计l20题；4．答题时间120分钟。

一、细胞生物学、生物化学、微生物学16题1．癌细胞与正常细胞的不同之处在于A．癌细胞不能合成DNA B．癌细胞被锁定在细胞周期中的S期C．癌细胞能持续分裂尽管彼此紧密相接D．癌细胞始终处于细胞周期中的分裂期2．人的肌肉组织分为快缩纤维和慢缩纤维两种，快缩纤维负责剧烈运动如举重，短跑，易产生酸痛感觉；慢缩纤维负责慢跑，游泳等有氧运动。

以下关于慢缩纤维和快缩纤维的描述，哪个是正确的A．快缩纤维含有的线粒体多，有氧呼吸能产生大量乳酸和ATP供能B．慢缩纤维含有的线粒体多，有氧呼吸不产生乳酸，产生的ATP也少C．快缩纤维含有的线粒体少，主要依靠糖酵解产生ATP供能，因此产生大量乳酸D．慢缩纤维含有的线粒体多，主要依靠糖酵解产生ATP供能3．在光合作用中参与电子传递的载体是A．叶绿素B．磷酸烯醇式丙酮酸C．NADH D．NADPH4．肽链生物合成时，信号肽A．是线粒体的定位信号B．将新生肽链导入内质网C．控制蛋白质分子的最终构象D．处于肽链的C末端5．原核细胞的特征包括A．无核糖体B．无转录后修饰C．无有氧呼吸D．无染色体6．以下糖类中属于寡糖的有(多选2分)A．甘露糖B．麦芽糖C．半乳糖D．蔗糖E．糖原7．以下关于蛋白质变性后所产生现象的描述不正确的是：A．溶解度降低B．疏水侧链基团被包裹在分子内部C．形成沉淀D．失去结晶能力8．真菌中的半知菌是指。

A．没有光合作用B．菌丝没有横隔C．没有发现有性生殖阶段D．不能运动。

9．关于维生素A的生理功能，下面的描述中哪个是错误的?A．抑制胆碱酯酶活性B．构成视觉细胞感光物质的成分C．参与上皮组织细胞膜糖蛋白合成D．严重缺乏时导致夜盲症10．磷酸戊糖途径的发生部位在A．细胞质B．线粒体C．叶绿体D．细胞膜11．在C4植物循环中，CO2进入叶肉细胞被固定的最初产物是A．甘油酸-3-磷酸B．苹果酸C．草酰乙酸D．丙酮酸12．在等电聚焦电泳过程中，随着蛋白质样品的迁移，电流的变化为A．越变越大，当样品到达其等电点位置时，电流达到最大值B．越变越小，当样品到达其等电点位置时，电流达到最小值，接近于零C．基本不变，为一恒定值D．不确定13．利用酵母菌进行乙醇发酵时若通入氧气，会导致菌株对葡萄糖利用速度降低，甚至停止生成乙醇，这种现象称为A．呼吸抑制效应B．巴斯德效应C．分子氧效应D．葡萄糖效应14．巴斯德设计的曲颈瓶实验，可以(单选1分)A．证明酒是酵母菌发酵获得B．否定自然发生学说的论点C．说明培养基灭菌后的作用D．验证某细菌是不能培养的15．营养缺陷型菌株是指(单选1分)A．不需要添加生长因子可在基础培养基上生长的菌株B．不需要添加生长因子可在丰富培养基上生长的菌株C．因突变需提供某种营养才能在基础培养基上生长的菌株D．因自发或诱发突变而导致的可抵抗环境不良因素的菌株16．以下哪类化合物属于微生物的次生代谢产物(多选2分)A．脂肪B．色素C．抗生素D．蛋白质E．毒素二、植物和动物的解剖、生理、组织和器官18题17．草履虫、水螅、乌贼、蟾蜍受到刺激后，均可从体内发出一些物质以攻击或防御敌害，在他们身体上，发出这些物质的结构是A．刺丝泡、刺细胞、墨囊、耳后腺B．刺丝泡、刺丝囊、外套腔、唾液腺C．表膜泡、刺丝囊、墨囊、唾液腺D．表膜泡、刺细胞、外套腔、耳后腺18．在动物卵裂时期，由于不同动物受精卵内卵黄多少及其在卵内分布的不同，卵裂方式也有很大差异，海胆、沙蚕、昆虫、乌贼的卵裂方式依次分别为(单选1分) A．完全均等卵裂(等裂)、表面卵裂、螺旋形卵裂、盘裂B．螺旋形卵裂、完全均等卵裂(等裂)、盘裂、表面卵裂C．螺旋形卵裂、完全均等卵裂(等裂)、表面卵裂、盘裂D．完全均等卵裂(等裂)、螺旋形卵裂、表面卵裂、盘裂19．不同动物类群具有独特的特征，现存棘皮动物、海绵动物、哺乳动物、鸟类所特有的特征依次为A．水管系、水沟系、下颌为单一齿骨、羽毛B．后口、水沟系、胎生、飞翔C．后口、骨针、胎生、羽毛D．水管系、骨针、下颌为单一齿骨、飞翔20．节肢动物类群很多，不同类群的排泄器官亦有差异，节肢动物门甲壳纲动物的排泄器官有(多选l分)A．基节腺B．触角腺C．颚腺D．马氏管21．家鸽的一侧体动脉弓退化，雌家鸽的一侧卵巢和输卵管也退化了，退化的这些器官是(单选1分)A．左体动脉弓和右侧的卵巢、输卵管B．左体动脉弓和左侧的卵巢、输卵管C．右体动脉弓和左侧的卵巢、输卵管D．右体动脉弓和右侧的卵巢、输卵管22．在海滨潮间带经常可以见到石鳖和沙蚕，以下不属于它们共同特征的是A．以裂体腔法形成真体腔B．后肾型排泄系统C．具有担轮幼虫期D．开管式循环系统23．以下哪项不是文昌鱼的特征A．具有脊索，背神经管，鳃裂B．有分节的肌肉，有哈氏窝C．有头，有心脏D．有特化的口器24．一家饭店涉嫌出售野生鸟类，检查人员在检查时发现了一种鸟类的足，三趾向前一趾向后，后趾与前面三趾在同一平面上，趾长，基部有蹼相连，这种鸟类是A．鹈形目B．鹳形目C．雁形目D．鹤形目25．以下哪组元素在植物体内参与氧化还原反应(单选2分)A．钼镍铜铁B．铁铜镁钼C．钙镁镍铜D．锰镍镁铜26．盐胁迫条件下，较耐盐的禾本科植物大麦可以通过将盐分局域于以下部位来缓解盐分对植物生长造成的危害(多选l分)A．根系B．幼叶C．叶鞘D．老叶27．关于植物的种子，下列描述正确的是(多选2分)A．种子由胚珠发育而来B．种子表面均有种孔、种脐和种脊的结构C．种子中的胚乳多来源于受精后的中央细胞，也有来自于雌配子体的细胞D．胚是休眠的幼小孢子体E．无胚乳种子在发育过程中没有胚乳形成28．有关被子植物花的叙述，下列哪一个是错误的(单选2分)A．花是适应于繁殖功能的变态短枝B．花托、花萼和花冠被认为是叶的变态C．雄蕊和雌蕊也被认为是叶的变态D．花托、花被、雄蕊和雌蕊均有茎的顶端分生组织产生29．玉米干旱缺水时叶片的内卷主要是失水造成的A．叶肉细胞B．叶表皮的毛状体C．位于上表皮的泡状(运动)细胞D．位于下表皮的泡状(运动)细胞30．有关C4植物，以下说法中正确的是(多选2分)A．叶解剖结构中可观察到“花环结构”B．光合作用CO2的初步固定和同化在不同细胞中进行C．光合作用CO2的初步固定和同化在同一细胞中进行D．在一定范围的强光、高温条件下光合效率高31．心肌细胞有效不应期的长短主要取决于A．静息电位水平B．0期去极化的速度C．阈电位水平D．平台期的长短32．血液中CO2分压升高使呼吸运动加强的最主要途径是(单选2分)A．直接刺激脑桥的呼吸相关神经元B．直接刺激延髓呼吸中枢的神经元C．刺激中枢化学感受器D．刺激颈动脉体和主动脉体感受器33．当去甲肾上腺素与β受体结合时，下列哪一种肌肉收缩或收缩加强(单选1分) A．心室肌B．子宫平滑肌C．小肠平滑肌D．血管平滑肌E．支气管平滑肌34．下列哪种因素可引起人尿量的明显增加的(多选2分)A．献血200ml后B．饮用清水1000ml后C．静脉注射神经垂体激素D．饮用生理盐水100ml后三、动物行为学、生态学15题35．如果一项研究，专注于了解不同生态因子对生物的影响，及生物对它们的耐受，那么这个研究属于哪一层次上的研究A．个体生态学B．种群生态学C．群落生态学D．生态系统生态学36．关于高等动物种群中性别比例，下面论述中错误的是A．大多数种群倾向于使出生性比趋近于l：1 B．老年组往往雌性多于雄性C．出生的时候，往往雄性多于雌性D．种群性比与世代长度直接相关37．社会性寄生是指A．寄生在动物社会中是普遍现象B．寄生只发生在特定社会等级的动物中C．社会性昆虫中发生的寄生行为D．强迫寄主动物为其提供食物或其他利益38．关于外来物种，以下论述错误的是A．所有的外来物种都是入侵种，都是有害的B．外来物种可以依靠风、鸟、昆虫等自然因素入侵C．有些外来物种是人类有意引入的D．入侵物种可能对生态系统造成长久的破坏39．适合度是指A．动物单一行为的适应性B．动物调整自己的行为以适合于生活在当时的环境C．动物的总体繁殖成功性D．最适合动物生活习性、满足营养需求的食物40．以下哪种情况不属于动物的行为节律A．候鸟随季节的迁徙B．哺乳动物晨昏活动习性C．细菌生长速度随营养物浓度起落而快慢变化D．招潮蟹的活动随潮汐变化而变化41．动物的生长和发育是需要一定温度的，下列哪个说法是正确的(单选2分) A．外界温度的高低直接决定了动物机体的体温，进而影响其生长发育B．当外界温度低于某一温度时，昆虫就停止生长发育，这一温度阈值称为发育起点温度C．动物的发育速度总是随环境温度的增高而加快的D．昆虫发育的有效积温是发育历期乘以发育期的平均温度，然后求和42．下列有关水生群落演替的说法中哪个是错误的A．水生群落的演替一般会依次经历裸底期、浮水植物期、沉水植物期、挺水植物期、湿生草本植物期等阶段B．在这一演替过程中池底逐渐变浅，最终向陆地变化C．挺水植物根系往往较发达，可以使水底迅速增高D．浮水植物的叶子漂浮在水面，影响到水下光照，不利于沉水植物生长43．关于固定行为型，下述论述正确的是(多选2分)A．固定行为型被特定的外部刺激所释放B．每一个物种都有物种特异的固定行为型C．固定行为型一旦释放就会持续到底D．固定行为型是一种先天行为44．在动物行为学研究中，严格定义行为类型是研究工作的基础。

精选美国amc中文版试题及答案

2000到2012年A M C10美国数学竞赛P 0 A 0B 0 C0 D 0 全美中学数学分级能力测验(AMC 10)2000年第01届美国AMC10 (2000年2月日时间75分钟)1. 国际数学奥林匹亚将于2001年在美国举办，假设I 、M 、O 分别表示不同的正整数，且满足I ?M ?O =2001，则试问I ?M ?O 之最大值为。

(A) 23 (B) 55 (C) 99 (D) 111 (E) 6712. 2000(20002000)为。

(A) 40 (B) 50 (C) 55 (D) 60 (E) 754. Candra 每月要付给网络公司固定的月租费及上网的拨接费，已知她12月的账单为12.48元，而她1月的账单为17.54元，若她1月的上网时间是12月的两倍，试问月租费是元。

(A) 2.53 (B) 5.06 (C) 6.24 (D) 7.42 (E) 8.775. 如图M ，N 分别为PA 与PB 之中点，试问当P 在一条平行AB 的直在线移动时，下列各数值有项会变动。

(a) MN 长 (b) △PAB 之周长 (c) △PAB 之面积 (d) ABNM 之面积 (A) 0项 (B) 1项 (C) 2项 (D) 3项 (E) 4项6. 费氏数列是以两个1开始，接下来各项均为前两项之和，试问在费氏数列各项的个位数字中，最后出现的阿拉伯数字为。

(A) 0 (B) 4 (C) 6 (D) 7 (E) 97. 如图，矩形ABCD 中，AD =1，P 在AB 上，且DP 与DB 三等分 ?ADC ，试问△BDP 之周长为。

2012年AMC8真题及英文详解

1. Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighborhood picnic?Solution 1Since Rachelle uses pounds of meat to make hamburgers, she uses pounds ofmeat to make one hamburger. She'll need 24 times that amount of meat for 24hamburgers, or .Solution 2Rachelle uses meat for hamburgers in the ratio. If we multiply this ratio on top and bottom by 3 to get 24 hamburgers on thebottom, we get. Therefore, she uses 9pounds of meat for 24 hamburgers, making the answer.2. In the country of East Westmore, statisticians estimate there is a baby born every hours and a death every day. To the nearest hundred, how many people areadded to the population of East Westmore each year?There are births and one death everyday in East Westmore. Therefore, the population increases by people everyday. Thus, thereare people added to the population every year. Rounding, we findthe answer is .3. On February 13 The Oshkosh Northwester listed the length of daylight as 10hours and 24 minutes, the sunrise was , and the sunset as . The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into in order to add easier.Adding, we find that the time of sunsetis .4. Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?Peter ate slices. The pizza has slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peterate of the pizza.5. In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , in centimeters?The figure is the same height on both sides, so the sum of the lengths contributing to the height on the left side will equal the sum of the lengths contributing to the height on the left side.Thus, the answer is .6. A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches?First, we start with a sketch. It's always a good idea to start with a picture, although not as detailed as this one.In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is square inches. The height of the whole frame (including the photograph) would be , and the width of the whole frame, . Therefore, the area of the whole figure wouldbe square inches. Subtracting the area of the photograph from thearea of both the frame and photograph, we find the answer tobe .7. Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?Isabella wants an average grade of on her 4 tests; this also means that shewants the sum of her test scores to be at least (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sumto , which means she needs more points to achieve her desiredaverage. In order to minimize the score on the third test, we assume that Isabella will receive all points on the fourth test. Therefore, the lowest Isabella couldhave scored on the third test would be .8. A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?Solution 1: With AlgebraLet the original price of an item be .First, everything is half-off, so the price is now .Next, the extra coupon applies 20% off on the sale price, so the price after this discount will be of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)The price of the item after all discounts have been applied is . However, we need to find the percentage off the original price, not the currentpercentage of the original price. We then subtract from (the originalprice of the item), to find the answer, .Solution 2: FakesolvingSince the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a totalof9. The Fort Worth Zoo has a number of two-legged birds and a number offour-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?Solution 1: AlgebraLet the number of two-legged birds be and the number of four-legged mammals be . We can now use systems of equations to solve this problem.Write two equations:Now multiply the latter equation by .By subtracting the second equation from the first equation, we findthat . Since there were heads, meaning that therewere animals, there were two-legged birds.Solution 2: Cheating the SystemFirst, we "assume" there are 200 two-legged birds only, and 0 four-legged mammals. Of course, this poses a problem, as then there would onlybe legs.Now we have to do some swapping--for every two-legged bird we swap for afour-legged mammal, we gain 2 legs. For example, if we swapped one bird for one mammal, giving 199 birds and 1 mammal, there would be legs.If we swapped two birds for two mammals, there would be legs. If we swapped 50 birds for 50 mammals, there would be legs. Notice that we must gain legs. This means we must swapout birds. Therefore, there must be birds. Checking our work, we find that , and we are correct.10. How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of , since all of the valid 4-digit number will always be greater than . The best way to solve this problem is by using casework.There can be only two leading digits, namely or .When the leading digit is , you can make such numbers.When the leading digit is , you can make such numbers.Summing the amounts of numbers, we find that there are such numbers.Notice that the first digit cannot be , as the number is greater than . Therefore,there are three digits that can be in the thousands.The rest three digits of the number have no restrictions, and therefore thereare for each leading digit.Since the two 's are indistinguishable, there are suchnumbers .11. The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and are all equal. What is the value of ?Since there must be an unique mode, and is already repeated twice, cannot beany of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So must be , or a new number.Solution 1: Guess & CheckWe can eliminate answer choices and , because of the above statement. Now we need to test the remaining answer choices.Case 1:Mode:Median:Mean:Since the mean does not equal the median or mode, can also be eliminated. Case 2:Mode:Median:Mean:We are done with this problem, because we have found when , the conditionis satisfied. Therefore, the answer is .Solution 2: AlgebraNotice that the mean of this set of numbers, in terms of , is:Because we know that the mode must be (it can't be any of the numbers alreadylisted, as shown above, and no matter what is, either or a new number, it willnot affect being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and equal:12. What is the units digit of ?The problem wants us to find the units digit of , therefore, we can eliminate the tens digit of , because the tens digit will not affect the final result. So our new expression is . Now we need to look for a pattern in the units digit.We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. Theremainder is the power of three that we are looking for, plus one. dividedby leaves a remainder of , so the answer is the units digit of , or . Thus,we find that the units digit of is .13. Jamar bought some pencils costing more than a penny each at the school bookstore and paid . Sharona bought some of the same pencils and paid . How many more pencils did Sharona buy than Jamar?We assume that the price of the pencils remains constant.Convert and to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest commondivisor of and , which is . Therefore, Jamar bought pencilsand Sharona bought pencils. Thus, Sharonabought more pencils than Jamar.14. In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?This problem is very similar to a handshake problem. We use theformula to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.So we have the equation . Solving, we find that the number of teams in the BIG N conference is15. The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?To find the answer to this problem, we need to find the least common multipleof , , , and add to the result. The least common multiple of the four numbersis , and by adding , we find that that such number is . Now we need to find theonly given range that contains . The only such range is answer , and so ourfinal answer is .16. Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choiceswith this property: and . To determine the answer we will have to use estimation and the first two digits of the numbers.For the number that would maximize the sum would start with . The firsttwo digits of (when rounded) are . Adding and , we find that the firstthree digits of the sum of the two numbers would be .For the number that would maximize the sum would start with . The firsttwo digits of (when rounded) are . Adding and , we find that the firstthree digits of the sum of the two numbers would be .From the estimations, we can say that the answer to this problem is .In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of thisare and . The digits can be interchangeable between numbers because we only care about the actual digits.The first digit must be either or . This immediately knocks out .The second digit must be either or . This doesn't cancel any choices.The third digit must be either or . This knocks out and .The fourth digit must be or . This cancels out .This leaves us with .17. A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?The first answer choice , can be eliminated since there must be squares with integer side lengths. We then test the next largest sidelength which is . The square with area can be partitioned into squares with area and two squareswith area , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is .18. What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?The problem states that the answer cannot be a perfect square or have prime factors less than . Therefore, the answer will be the product of at least twodifferent primes greater than . The two smallest primes greaterthan are and . Multiplying these two primes, we obtain the number ,which is also the smallest number on the list of answer choices. So we are done, andthe answer is .19. In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?Let be the number of red marbles, be the number of green marbles, and be the number of blue marbles.If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to . Likewise, the number of red marbles and bluemarbles amount to , and the number of red marbles and the number of greenmarbles amount to .We have three equations:We add all the equations to obtain a fourth equation:Now divide by on both sides to find the total number of marbles:Since the sum of the number of red marbles, green marbles, and blue marbles is thenumber of marbles in the jar, the total number of marbles in the jar is .Notice how we never knew how many of each color there were (there is 1 green marble, 5 blue marbles, and 3 red marbles).20. What is the correct ordering of the three numbers , , and , in increasing order?The value of is . Now we give all the fractions a common denominator.Ordering the fractions from least to greatest, we find that they are in the order listed.Therefore, our final answer is .Instead of finding the LCD, we can subtract each fraction from to get a commonnumerator. Thus,All three fractions have common numerator . Now it is obvious the order of thefractions. . Therefore, our answeris .21. Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?If Marla evenly distributes her square feet of paint between the 6 faces, eachface will get square feet of paint. The surface area of one of the faces of the cube is square feet. Therefore, there willbe square feet of white on each side.22. Let be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?First, we find that the minimum value of the median of will be .We then experiment with sequences of numbers to determine other possible medians.Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Any number greater than also cannot be a median of set .There are then possible medians of set .23. An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?Let the perimeter of the equilateral triangle be . The side length of the equilateral triangle would then be and the sidelength of the hexagon would be .A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio , since the sidelength of the smallequilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is . The area of the hexagon isthen .Let the side length of the equilateral triangle be and the side length of the hexagonbe . Since the perimeters are equal, we must have which reducesto . Substitute this value in to the area of an equilateral triangle toyield .Setting this equal to gives us . Substitue into the area of a regular hexagon to yield .Therefore, our answer is .The area of an equilateral triangle with side length is .The area of a regular hexagon with side length is .24. A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?Draw a square around the star figure. The sidelength of this square is , because the sidelength is the diameter of the circle. The square forms -quarter circles aroundthe star figure. This is the equivalent of one large circle with radius , meaning that the total area of the quarter circles is . The area of the square is . Thus, thearea of the star figure is . The area of the circle is . Taking the ratio of thetwo areas, we find the answer is .25. A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length , and the other of length . What is the value of ?The total area of the four congruent triangles formed by the squares is .Therefore, the area of one of these triangles is . The height of one of these triangles is and the base is . Using the formula for area of the triangle, wehave . Multiply by on both sides to find that the value of is .To solve this problem you could also use algebraic manipulation.Since the area of the large square is , the sidelength is .We then have the equation .We also know that the sidelength of the smaller square is , since its area is . Then,the segment of length and segment of length form a right triangle whose hypotenuse would have length .So our second equation is .Square both equations.Now, subtract, and obtain the equation . We can deduce that the value of is .。

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 逻辑推理题

amc8 逻辑推理题
AMC8（American Math Competition 8）中的逻辑推理题是数学竞赛中的一种题型，通常涉及到逻辑推理、推理分析和问题解决能力等方面的考察。

以下是一个AMC8逻辑推理题的示例：
题目：有五顶不同的帽子，两顶蓝色的，三顶红色的。

甲、乙、丙、丁、戊五人站成一排，已知甲看到的三个人中戴蓝帽子的人是乙、丙、丁，乙看到的三个人中戴蓝帽子的人是甲、丙、丁，丙看到的三个人中戴蓝帽子的人是甲、乙、丁，丁看到的三个人中戴蓝帽子的人是甲、乙、丙。

戊说：“我看到的三个人都戴红帽子。

”根据以上信息，戊看到的三个人分别是谁？
解答：根据题目描述，甲、乙、丙、丁都看到了三个人戴蓝帽子，这意味着他们四人都看到了彼此。

如果甲或乙看到的另外两个人戴红帽子，那么他们看到的另外两个人必然是甲和乙本身。

同理，丙和丁也是如此。

由于他们看到的另外三个人都是戴蓝帽子的人，所以戊只能看到甲、乙、丙三个人，而甲、乙、丙都能看到戊。

因此，戊看到的另外两个人是甲和乙。

综上所述，戊看到的另外三个人分别是甲、乙和丙。

美国AMC8中文试题和答案

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 。

2007年AMC8试题解析

20071Theresa’s parents have agreed to buy her tickets to see her favorite band if she spends an average of 10hours per week helping around the house for 6weeks.For the ﬁrst 5weeks,she helps around the house for 8,11,7,12and 10hours.How many hours must she work during the ﬁnal week to earn the tickets?(A)9(B)10(C)11(D)12(E)132Six-hundred ﬁfty students were surveyed about their pasta preferences.The choices were lasagna,manicotti,ravioli and spaghetti.The results of the survey are displayed in the bar graph.What is the ratio of the number of students who preferred spaghetti to the number of students who preferredmanicotti?50100150200250L a s a g n a M a n i c o t t i R a v i o l iS p a g h e t t i N u m b e r o f P e o p l e (A)25(B)12(C)54(D)53(E)523What is the sum of the two smallest prime factors of 250?(A)2(B)5(C)7(D)10(E)124A haunted house has six windows.In how many ways can Georgie the Ghost enter the house by one window and leave by a diﬀerent window?(A)12(B)15(C)18(D)30(E)365Chandler wants to buy a $500dollar mountain bike.For his birthday,his grandparents send him $50,his aunt sends him $35and his cousin gives him $15.He earns $16per week for his This ﬁle was downloaded from the AoPS Math Olympiad Resources PagePage 12007paper route.He will use all of his birthday money and all of the money he earns from his paper route.In how many weeks will he be able to buy the mountain bike?(A)24(B)25(C)26(D)27(E)286The average cost of a long-distance call in the USA in 1985was 41cents per minute,and the average cost of a long-distance call in the USA in 2005was 7cents per minute.Find the approximate percent decrease in the cost per minute of a long-distance call.(A)7(B)17(C)34(D)41(E)807The average age of 5people in a room is 30years.An 18-year-old person leaves the room.What is the average age of the four remaining people?(A)25(B)26(C)29(D)33(E)368In trapezoid ABCD ,AD is perpendicular to DC ,AD =AB =3,and DC =6.In addition,E is on DC ,and BE is parallel to AD .Find the area of ∆BEC .ABCD E336(A)3(B)4.5(C)6(D)9(E)189To complete the grid below,each of the digits 1through 4must occur once in each row and once in each column.What number will occupy the lower right-hand square?12234(A)1(B)2(C)3(D)4(E)cannot be determined10For any positive integer n ,deﬁne n to be the sum of the positive factors of n .For example,6=1+2+3+6=12.Find 11.(A)13(B)20(C)24(D)28(E)30200711Tiles I,II,III and IV are translated so one tile coincides with each of the rectangles A,B,Cand D .In the ﬁnal arrangement,the two numbers on any side common to two adjacent tiles must be the same.Which of the tiles is translated to Rectangle C ?8672III IIIIV 341972069351A BC D(A)I (B)II (C)III (D)IV (E)cannot be determined 12A unit hexagon is composed of a regular haxagon of side length 1and its equilateral triangularextensions,as shown in the diagram.What is the ratio of the area of the extensions to the area of the originalhexagon?2007(A)1:1(B)6:5(C)3:2(D)2:1(E)3:113Sets A and B,shown in the venn diagram,have the same number of elements.Thier union has2007elements and their intersection has1001elements.Find the number of elements in A.A B1001(A)503(B)1006(C)1504(D)1507(E)151014The base of isosceles ABC is24and its area is60.What is the length of one of the congruent sides?(A)5(B)8(C)13(D)14(E)1815Let a,b and c be numbers with0<a<b<c.Which of the following is impossible?=a(A)a+c<b(B)a·b<c(C)a+b<c(D)a·c<b(E)bc16Amanda Reckonwith drawsﬁve circles with radii1,2,3,4and5.Then for each circle she plots the point(C;A),where C is its circumference and A is its area.Which of the following could be her graph?(A)AC2007 (B)AC (C)AC (D)AC (E)AC200717A mixture of30liters of paint is25%red tint,30%yellow tint,and45%water.Five liters of yellow tint are added to the original mixture.What is the percent of yellow tint that is the mixture?(A)25(B)35(C)40(D)45(E)5018The product of the two99-digit numbers303,030,303,...,030,303and505,050,505,...,050,505has thousands digit A and units digit B.What is the sum of A and B?(A)3(B)5(C)6(D)8(E)1019Pick two consecutive positive integers whose sum is less than100.Square both of those integers and thenﬁnd the diﬀerence of the squares.Which of the following could be the diﬀerence?(A)2(B)64(C)79(D)96(E)13120Before district play,the Unicorns had won45%of their basketball games.During district play,they won six more games and lost two,toﬁnish the season having won half their games.How many games did the Unicorns play in all?(A)48(B)50(C)52(D)54(E)6021Two cards are dealt from a deck of four red cards labeled A,B,C,D and four green cards labeled A,B,C,D.A winning pair is two of the same color or two of the same letter.What is the probability of drawing a winning pair?(A)27(B)38(C)12(D)47(E)5822A lemming sits at a corner of a square with side length10meters.The lemming runs6.2 meters along a diagonal toward the opposite corner.It stops,makes a90◦right turn and runs 2more meters.A scientist measures the shortest distance between the lemming and each side of the square.What is the average of these four distances in meters?(A)2(B)4.5(C)5(D)6.2(E)723What is the area of the shaded pinwheel shown in the5×5grid?2007(A)4(B)6(C)8(D)10(E)1224A bag contains four pieces of paper,each labeled with one of the digits 1,2,3or 4,with norepeats.Three of these pieces are drawn,one at a time without replacement,to construct a three-digit number.What is the probability that the three-digit number is a multiple of 3?(A)14(B)13(C)12(D)23(E)3425On the dart board shown in the ﬁgure,the outer circle has radius 6and the inner circle hasradius 3.Three radii divide each circle into the three congruent regions,with point values shown.The probability that a dart will hit a given region is proportional to to the area of the region.What two darts hit this board,the score is the sum of the point values in the regions.What is the probability that the score is odd?1112222007(A)1736(B)3572(C)12(D)3772(E)1936The problems on this page are copyrighted by the Mathematical Association of America’s American Mathematics Competitions.。

2012年AMC8真题及英文详解

1. Rachelle uses 3 pounds of meat to make 8 hamburgers for her family. How many pounds of meat does she need to make 24 hamburgers for a neighborhood picnic?Solution 1Since Rachelle uses pounds of meat to make hamburgers, she uses pounds ofmeat to make one hamburger. She'll need 24 times that amount of meat for 24hamburgers, or .Solution 2Rachelle uses meat for hamburgers in the ratio. If we multiply this ratio on top and bottom by 3 to get 24 hamburgers on thebottom, we get. Therefore, she uses 9pounds of meat for 24 hamburgers, making the answer.2. In the country of East Westmore, statisticians estimate there is a baby born every hours and a death every day. To the nearest hundred, how many people areadded to the population of East Westmore each year?There are births and one death everyday in East Westmore. Therefore, the population increases by people everyday. Thus, thereare people added to the population every year. Rounding, we findthe answer is .3. On February 13 The Oshkosh Northwester listed the length of daylight as 10hours and 24 minutes, the sunrise was , and the sunset as . The length of daylight and sunrise were correct, but the sunset was wrong. When did the sun really set?The problem wants us to find the time of sunset and gives us the length of daylight and time of sunrise. So all we have to do is add the length of daylight to the time of sunrise to obtain the answer. Convert 10 hours and 24 minutes into in order to add easier.Adding, we find that the time of sunsetis .4. Peter's family ordered a 12-slice pizza for dinner. Peter ate one slice and shared another slice equally with his brother Paul. What fraction of the pizza did Peter eat?Peter ate slices. The pizza has slices total. Taking the ratio of the amount of slices Peter ate to the amount of slices in the pizza, we find that Peterate of the pizza.5. In the diagram, all angles are right angles and the lengths of the sides are given in centimeters. Note the diagram is not drawn to scale. What is , in centimeters?The figure is the same height on both sides, so the sum of the lengths contributing to the height on the left side will equal the sum of the lengths contributing to the height on the left side.Thus, the answer is .6. A rectangular photograph is placed in a frame that forms a border two inches wide on all sides of the photograph. The photograph measures 8 inches high and 10 inches wide. What is the area of the border, in square inches?First, we start with a sketch. It's always a good idea to start with a picture, although not as detailed as this one.In order to find the area of the frame, we need to subtract the area of the photograph from the area of the photograph and the frame together. The area of the photograph is square inches. The height of the whole frame (including the photograph) would be , and the width of the whole frame, . Therefore, the area of the whole figure wouldbe square inches. Subtracting the area of the photograph from thearea of both the frame and photograph, we find the answer tobe .7. Isabella must take four 100-point tests in her math class. Her goal is to achieve an average grade of 95 on the tests. Her first two test scores were 97 and 91. After seeing her score on the third test, she realized she can still reach her goal. What is the lowest possible score she could have made on the third test?Isabella wants an average grade of on her 4 tests; this also means that shewants the sum of her test scores to be at least (if she goes over this number, she'll be over her goal!). She's already taken two tests, which sumto , which means she needs more points to achieve her desiredaverage. In order to minimize the score on the third test, we assume that Isabella will receive all points on the fourth test. Therefore, the lowest Isabella couldhave scored on the third test would be .8. A shop advertises everything is "half price in today's sale." In addition, a coupon gives a 20% discount on sale prices. Using the coupon, the price today represents what percentage off the original price?Solution 1: With AlgebraLet the original price of an item be .First, everything is half-off, so the price is now .Next, the extra coupon applies 20% off on the sale price, so the price after this discount will be of what it was before. (Notice how this is not applied to the original price; if it were, the solution would be applying 50% + 20 % = 70% off the original price.)The price of the item after all discounts have been applied is . However, we need to find the percentage off the original price, not the currentpercentage of the original price. We then subtract from (the originalprice of the item), to find the answer, .Solution 2: FakesolvingSince the problem implies that the percentage off the original price will be the same for every item in the store, fakesolving is applicable here. Say we are buying an item worth 10 dollars, a convenient number to work with. First, it is clear that we'll get 50% off, which makes the price then 5 dollars. Taking 20% off of 5 dollars gives us 4 dollars. Therefore, we have saved a totalof9. The Fort Worth Zoo has a number of two-legged birds and a number offour-legged mammals. On one visit to the zoo, Margie counted 200 heads and 522 legs. How many of the animals that Margie counted were two-legged birds?Solution 1: AlgebraLet the number of two-legged birds be and the number of four-legged mammals be . We can now use systems of equations to solve this problem.Write two equations:Now multiply the latter equation by .By subtracting the second equation from the first equation, we findthat . Since there were heads, meaning that therewere animals, there were two-legged birds.Solution 2: Cheating the SystemFirst, we "assume" there are 200 two-legged birds only, and 0 four-legged mammals. Of course, this poses a problem, as then there would onlybe legs.Now we have to do some swapping--for every two-legged bird we swap for afour-legged mammal, we gain 2 legs. For example, if we swapped one bird for one mammal, giving 199 birds and 1 mammal, there would be legs.If we swapped two birds for two mammals, there would be legs. If we swapped 50 birds for 50 mammals, there would be legs. Notice that we must gain legs. This means we must swapout birds. Therefore, there must be birds. Checking our work, we find that , and we are correct.10. How many 4-digit numbers greater than 1000 are there that use the four digits of 2012?For this problem, all we need to do is find the amount of valid 4-digit numbers that can be made from the digits of , since all of the valid 4-digit number will always be greater than . The best way to solve this problem is by using casework.There can be only two leading digits, namely or .When the leading digit is , you can make such numbers.When the leading digit is , you can make such numbers.Summing the amounts of numbers, we find that there are such numbers.Notice that the first digit cannot be , as the number is greater than . Therefore,there are three digits that can be in the thousands.The rest three digits of the number have no restrictions, and therefore thereare for each leading digit.Since the two 's are indistinguishable, there are suchnumbers .11. The mean, median, and unique mode of the positive integers 3, 4, 5, 6, 6, 7, and are all equal. What is the value of ?Since there must be an unique mode, and is already repeated twice, cannot beany of the numbers already listed (3, 4, 5, 7). (If it were, the mode would not be unique.) So must be , or a new number.Solution 1: Guess & CheckWe can eliminate answer choices and , because of the above statement. Now we need to test the remaining answer choices.Case 1:Mode:Median:Mean:Since the mean does not equal the median or mode, can also be eliminated. Case 2:Mode:Median:Mean:We are done with this problem, because we have found when , the conditionis satisfied. Therefore, the answer is .Solution 2: AlgebraNotice that the mean of this set of numbers, in terms of , is:Because we know that the mode must be (it can't be any of the numbers alreadylisted, as shown above, and no matter what is, either or a new number, it willnot affect being the mode), and we know that the mode must equal the mean, we can set the expression for the mean and equal:12. What is the units digit of ?The problem wants us to find the units digit of , therefore, we can eliminate the tens digit of , because the tens digit will not affect the final result. So our new expression is . Now we need to look for a pattern in the units digit.We observe that there is a pattern for the units digit which recurs every four powers of three. Using this pattern, we can subtract 1 from 2012 and divide by 4. Theremainder is the power of three that we are looking for, plus one. dividedby leaves a remainder of , so the answer is the units digit of , or . Thus,we find that the units digit of is .13. Jamar bought some pencils costing more than a penny each at the school bookstore and paid . Sharona bought some of the same pencils and paid . How many more pencils did Sharona buy than Jamar?We assume that the price of the pencils remains constant.Convert and to cents. Since the price of the pencils is more than one penny, we can find the price of one pencil (in cents) by taking the greatest commondivisor of and , which is . Therefore, Jamar bought pencilsand Sharona bought pencils. Thus, Sharonabought more pencils than Jamar.14. In the BIG N, a middle school football conference, each team plays every other team exactly once. If a total of 21 conference games were played during the 2012 season, how many teams were members of the BIG N conference?This problem is very similar to a handshake problem. We use theformula to usually find the number of games played (or handshakes). Now we have to use the formula in reverse.So we have the equation . Solving, we find that the number of teams in the BIG N conference is15. The smallest number greater than 2 that leaves a remainder of 2 when divided by 3, 4, 5, or 6 lies between what numbers?To find the answer to this problem, we need to find the least common multipleof , , , and add to the result. The least common multiple of the four numbersis , and by adding , we find that that such number is . Now we need to find theonly given range that contains . The only such range is answer , and so ourfinal answer is .16. Each of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 is used only once to make two five-digit numbers so that they have the largest possible sum. Which of the following could be one of the numbers?In order to maximize the sum of the numbers, the numbers must have their digits ordered in decreasing value. There are only two numbers from the answer choiceswith this property: and . To determine the answer we will have to use estimation and the first two digits of the numbers.For the number that would maximize the sum would start with . The firsttwo digits of (when rounded) are . Adding and , we find that the firstthree digits of the sum of the two numbers would be .For the number that would maximize the sum would start with . The firsttwo digits of (when rounded) are . Adding and , we find that the firstthree digits of the sum of the two numbers would be .From the estimations, we can say that the answer to this problem is .In order to determine the largest number possible, we have to evenly distribute the digits when adding. The two numbers that show an example of thisare and . The digits can be interchangeable between numbers because we only care about the actual digits.The first digit must be either or . This immediately knocks out .The second digit must be either or . This doesn't cancel any choices.The third digit must be either or . This knocks out and .The fourth digit must be or . This cancels out .This leaves us with .17. A square with integer side length is cut into 10 squares, all of which have integer side length and at least 8 of which have area 1. What is the smallest possible value of the length of the side of the original square?The first answer choice , can be eliminated since there must be squares with integer side lengths. We then test the next largest sidelength which is . The square with area can be partitioned into squares with area and two squareswith area , which satisfies all the conditions of the problem. Therefore, the smallest possible value of the length of the side of the original square is .18. What is the smallest positive integer that is neither prime nor square and that has no prime factor less than 50?The problem states that the answer cannot be a perfect square or have prime factors less than . Therefore, the answer will be the product of at least twodifferent primes greater than . The two smallest primes greaterthan are and . Multiplying these two primes, we obtain the number ,which is also the smallest number on the list of answer choices. So we are done, andthe answer is .19. In a jar of red, green, and blue marbles, all but 6 are red marbles, all but 8 are green, and all but 4 are blue. How many marbles are in the jar?Let be the number of red marbles, be the number of green marbles, and be the number of blue marbles.If "all but 6 are red marbles", that means that the number of green marbles and the number of blue marbles amount to . Likewise, the number of red marbles and bluemarbles amount to , and the number of red marbles and the number of greenmarbles amount to .We have three equations:We add all the equations to obtain a fourth equation:Now divide by on both sides to find the total number of marbles:Since the sum of the number of red marbles, green marbles, and blue marbles is thenumber of marbles in the jar, the total number of marbles in the jar is .Notice how we never knew how many of each color there were (there is 1 green marble, 5 blue marbles, and 3 red marbles).20. What is the correct ordering of the three numbers , , and , in increasing order?The value of is . Now we give all the fractions a common denominator.Ordering the fractions from least to greatest, we find that they are in the order listed.Therefore, our final answer is .Instead of finding the LCD, we can subtract each fraction from to get a commonnumerator. Thus,All three fractions have common numerator . Now it is obvious the order of thefractions. . Therefore, our answeris .21. Marla has a large white cube that has an edge of 10 feet. She also has enough green paint to cover 300 square feet. Marla uses all the paint to create a white square centered on each face, surrounded by a green border. What is the area of one of the white squares, in square feet?If Marla evenly distributes her square feet of paint between the 6 faces, eachface will get square feet of paint. The surface area of one of the faces of the cube is square feet. Therefore, there willbe square feet of white on each side.22. Let be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?First, we find that the minimum value of the median of will be .We then experiment with sequences of numbers to determine other possible medians.Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Median:Sequence:Any number greater than also cannot be a median of set .There are then possible medians of set .23. An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 4, what is the area of the hexagon?Let the perimeter of the equilateral triangle be . The side length of the equilateral triangle would then be and the sidelength of the hexagon would be .A hexagon contains six equilateral triangles. One of these triangles would be similar to the large equilateral triangle in the ratio , since the sidelength of the smallequilateral triangle is half the sidelength of the large one. Thus, the area of one of the small equilateral triangles is . The area of the hexagon isthen .Let the side length of the equilateral triangle be and the side length of the hexagonbe . Since the perimeters are equal, we must have which reducesto . Substitute this value in to the area of an equilateral triangle toyield .Setting this equal to gives us . Substitue into the area of a regular hexagon to yield .Therefore, our answer is .The area of an equilateral triangle with side length is .The area of a regular hexagon with side length is .24. A circle of radius 2 is cut into four congruent arcs. The four arcs are joined to form the star figure shown. What is the ratio of the area of the star figure to the area of the original circle?Draw a square around the star figure. The sidelength of this square is , because the sidelength is the diameter of the circle. The square forms -quarter circles aroundthe star figure. This is the equivalent of one large circle with radius , meaning that the total area of the quarter circles is . The area of the square is . Thus, thearea of the star figure is . The area of the circle is . Taking the ratio of thetwo areas, we find the answer is .25. A square with area 4 is inscribed in a square with area 5, with one vertex of the smaller square on each side of the larger square. A vertex of the smaller square divides a side of the larger square into two segments, one of length , and the other of length . What is the value of ?The total area of the four congruent triangles formed by the squares is .Therefore, the area of one of these triangles is . The height of one of these triangles is and the base is . Using the formula for area of the triangle, wehave . Multiply by on both sides to find that the value of is .To solve this problem you could also use algebraic manipulation.Since the area of the large square is , the sidelength is .We then have the equation .We also know that the sidelength of the smaller square is , since its area is . Then,the segment of length and segment of length form a right triangle whose hypotenuse would have length .So our second equation is .Square both equations.Now, subtract, and obtain the equation . We can deduce that the value of is .。

amc8历年真题答案解析

amc8历年真题答案解析【Amc8历年真题答案解析】Amc8，全名为American Mathematics Competition 8，是一项面向美国中学生的数学竞赛。

每年11月份举行的Amc8竞赛，旨在通过一系列挑战性的数学问题，培养学生的数学思维能力、创造力和解决问题的能力。

以下是Amc8历年真题的答案解析。

一、2000年真题：1. 答案：C解析：问题要求选择一个在100和999之间的正整数，满足各个数字的平方和等于该数本身。

首先明确数字的范围是100到999，因此只需要考虑三位数。

假设一个三位数为$abc$，那么它满足条件$a^2+b^2+c^2=abc$。

考虑到$a,b,c$的范围在1到9之间，可以通过穷举法逐个尝试。

当$a=2$时，$b=5$和$c=1$，满足条件$2^2+5^2+1^2=251$，因此答案为C。

2. 答案：D解析：问题给出了一个小测试，Tom回答7道题目，其中4道题回答正确。

考虑到每道问题只有两个选择（正确或错误），因此回答7道题目的所有可能性为$2^7=128$。

在这些可能性中，只有一种情况Tom回答了4道题目正确，因此答案为D。

3. 答案：A解析：问题给出了一个图形，包含了一系列长方形。

我们需要计算图形中所有（不重叠）长方形的个数。

首先考虑长方形最小边长为1的情况，可以在图形中找到4个这样的长方形。

然后考虑最小边长为2的情况，可以找到3个长方形。

以此类推，最小边长为3、4、5的情况下，分别可以找到2个、1个、0个长方形。

因此，总共可以找到的长方形个数为$4+3+2+1=10$，答案为A。

4. 答案：C解析：问题给出了一个5乘5的正方形网格，要求从左上角走到右下角，只能向右或向下移动，且不能重复经过同一格。

我们需要计算从左上角到右下角有几条路径。

可以发现，从左上角到右下角，一共需要移动4次向右，4次向下。

因此可以将问题转化为在8个移动中选择4个向右，即$C_8^4=\frac{8!}{4!4!}=70$，答案为C。

2007_2012AMC8中文试题和答案解析

2007年美国AMC8(2007年11月日时间40分钟)1.如果希瑞莎能够持续6周，平均每周花10小时帮忙照顾房子，她的父母就帮她买她喜爱乐团的入场券。

在前五周她分别花了 & 11、7、12及10小时照顾房子。

在最后一周，她必须要花多少小时去照顾房子才能获得入场券？ (A) 9 (B) 10 (C) 11 (D) 12 (E) 13 。

2.调查650位学生对面食种类的偏好。

选项包含：卤味面、起司肉燥面、水饺、意大利面，调查结果如长条图所示。

试问偏好意大利面的学生数与偏好起司肉燥面的学生数之比值为多少？2 15 5 5 (A) 2 (B) 1 (C) 5 (D) 5 (E) 5。

524323. 250的最小两个质因子之和为多少？ (A) 2 (B) 5 (C) 7 (D) 10 (E) 12。

4. 某间鬼屋有六个窗子。

小精灵乔治丛一个窗子进入屋内，而从不同的另一个窗子出来的方法共有多少种？ (A) 12 (B) 15 (C) 18 (D) 30 (E) 36。

5.姜德想买一辆价值美金500元的越野脚踏车。

在他生日时，祖父母给他美金50元，姑姑给他美金35元，表哥给他美金15元。

他送报纸每周可赚美金16元。

若用他生日得到的所有礼金及送报纸所有赚得的钱去买越野脚踏车，他需要送几周的报纸才能有足够的钱？(A) 24 (B) 25 (C) 26 (D) 27 (E) 28。

6. 在1985年美国的长途电话费是每分钟 41分钱，在2005年的长途电话费是每分钟7分钱＜试求每分钟长途电话费下降的百分率最接近下列哪一项？(A) 7 (B) 17 (C) 34 (D) 41(E) 807. 房间内5个人的平均年龄为30岁。

若其中一位18岁的人离开了房间，则剩下四个人的平均年龄是几岁？ (A) 25 (B) 26 (C) 29 (D) 33 (E) 36。

8.在梯形 ABCD 中，AD 垂直 DC ，AD = AB =3，DC =6。

2007年AMC8中文版答案

1. (E) 一个立方体有12条边、8个角和6个面。

和是26。

2. (C) 最小的素数是2，它是每个偶数的因数。

因为58是唯一的偶数，它的质因数最小。

3.(D) 因为120克中有30克是填充物，所以30／120 = 25%的汉堡是填充物。

所以100% - 25% = 75%的汉堡不是填充物。

或者有120 - 30 = 90克不是填料。

所以90／120 = 75%不是4. (C) 下表显示答案一定是5辆三轮车。

或者，设b等于自行车的数量，t等于三轮车的数量。

则车辆数量为b+t = 7，车轮数量为2b+3t = 19。

因为b = 7 - t，因此或者，如果每个孩子都有一辆自行车，就会有14个轮子。

由于有19个轮子，其中5辆必须是三轮车。

5. (B) 如果这个数字的20%是12，这个数字必须是60。

那么60的30%是0.30×60 = 18。

或因为这个数的20%是12，所以这个数的10%是6。

所以这个数的30%是18。

square units平方单位7. (A) 布莱克在四项测试中总共得了4×78 = 312分。

珍妮比布莱克多得10 - 10 + 20 + 20 = 40分，所以她的平均分是352／4 = 88，比布莱克高10分。

或Jenny和Blake的测试总分差为10 - 10 + 20 + 20 = 40分。

平均差值是40 ／4= 10分。

8.(A) 因为所有的饼干都有相同的厚度，只有表面的面积，需要考虑它们的形状。

阿特饼干的表面积梯形饼干是1／2·3·8 = 12平方英寸。

因为他做了12块饼干面团的面积是12×12 = 144平方英寸。

罗杰的饼干每个的表面积为2·4 = 8平方英寸;因此,他就能得到144÷8 = 18块饼干。

保罗的平行四边形饼干每个都有2·3 = 6 平方英寸的表面积。

他有144÷6 = 24块饼干特丽莎的三角形饼干每个都有1／2·4·3 = 6 平方英寸的表面积。

2007年AMC8试题解析

20071Theresa’s parents have agreed to buy her tickets to see her favorite band if she spends an average of 10hours per week helping around the house for 6weeks.For the ﬁrst 5weeks,she helps around the house for 8,11,7,12and 10hours.How many hours must she work during the ﬁnal week to earn the tickets?(A)9(B)10(C)11(D)12(E)132Six-hundred ﬁfty students were surveyed about their pasta preferences.The choices were lasagna,manicotti,ravioli and spaghetti.The results of the survey are displayed in the bar graph.What is the ratio of the number of students who preferred spaghetti to the number of students who preferredmanicotti?50100150200250L a s a g n a M a n i c o t t i R a v i o l iS p a g h e t t i N u m b e r o f P e o p l e (A)25(B)12(C)54(D)53(E)523What is the sum of the two smallest prime factors of 250?(A)2(B)5(C)7(D)10(E)124A haunted house has six windows.In how many ways can Georgie the Ghost enter the house by one window and leave by a diﬀerent window?(A)12(B)15(C)18(D)30(E)365Chandler wants to buy a $500dollar mountain bike.For his birthday,his grandparents send him $50,his aunt sends him $35and his cousin gives him $15.He earns $16per week for his This ﬁle was downloaded from the AoPS Math Olympiad Resources PagePage 12007paper route.He will use all of his birthday money and all of the money he earns from his paper route.In how many weeks will he be able to buy the mountain bike?(A)24(B)25(C)26(D)27(E)286The average cost of a long-distance call in the USA in 1985was 41cents per minute,and the average cost of a long-distance call in the USA in 2005was 7cents per minute.Find the approximate percent decrease in the cost per minute of a long-distance call.(A)7(B)17(C)34(D)41(E)807The average age of 5people in a room is 30years.An 18-year-old person leaves the room.What is the average age of the four remaining people?(A)25(B)26(C)29(D)33(E)368In trapezoid ABCD ,AD is perpendicular to DC ,AD =AB =3,and DC =6.In addition,E is on DC ,and BE is parallel to AD .Find the area of ∆BEC .ABCD E336(A)3(B)4.5(C)6(D)9(E)189To complete the grid below,each of the digits 1through 4must occur once in each row and once in each column.What number will occupy the lower right-hand square?12234(A)1(B)2(C)3(D)4(E)cannot be determined10For any positive integer n ,deﬁne n to be the sum of the positive factors of n .For example,6=1+2+3+6=12.Find 11.(A)13(B)20(C)24(D)28(E)30200711Tiles I,II,III and IV are translated so one tile coincides with each of the rectangles A,B,Cand D .In the ﬁnal arrangement,the two numbers on any side common to two adjacent tiles must be the same.Which of the tiles is translated to Rectangle C ?8672III IIIIV 341972069351A BC D(A)I (B)II (C)III (D)IV (E)cannot be determined 12A unit hexagon is composed of a regular haxagon of side length 1and its equilateral triangularextensions,as shown in the diagram.What is the ratio of the area of the extensions to the area of the originalhexagon?2007(A)1:1(B)6:5(C)3:2(D)2:1(E)3:113Sets A and B,shown in the venn diagram,have the same number of elements.Thier union has2007elements and their intersection has1001elements.Find the number of elements in A.A B1001(A)503(B)1006(C)1504(D)1507(E)151014The base of isosceles ABC is24and its area is60.What is the length of one of the congruent sides?(A)5(B)8(C)13(D)14(E)1815Let a,b and c be numbers with0<a<b<c.Which of the following is impossible?=a(A)a+c<b(B)a·b<c(C)a+b<c(D)a·c<b(E)bc16Amanda Reckonwith drawsﬁve circles with radii1,2,3,4and5.Then for each circle she plots the point(C;A),where C is its circumference and A is its area.Which of the following could be her graph?(A)AC2007 (B)AC (C)AC (D)AC (E)AC200717A mixture of30liters of paint is25%red tint,30%yellow tint,and45%water.Five liters of yellow tint are added to the original mixture.What is the percent of yellow tint that is the mixture?(A)25(B)35(C)40(D)45(E)5018The product of the two99-digit numbers303,030,303,...,030,303and505,050,505,...,050,505has thousands digit A and units digit B.What is the sum of A and B?(A)3(B)5(C)6(D)8(E)1019Pick two consecutive positive integers whose sum is less than100.Square both of those integers and thenﬁnd the diﬀerence of the squares.Which of the following could be the diﬀerence?(A)2(B)64(C)79(D)96(E)13120Before district play,the Unicorns had won45%of their basketball games.During district play,they won six more games and lost two,toﬁnish the season having won half their games.How many games did the Unicorns play in all?(A)48(B)50(C)52(D)54(E)6021Two cards are dealt from a deck of four red cards labeled A,B,C,D and four green cards labeled A,B,C,D.A winning pair is two of the same color or two of the same letter.What is the probability of drawing a winning pair?(A)27(B)38(C)12(D)47(E)5822A lemming sits at a corner of a square with side length10meters.The lemming runs6.2 meters along a diagonal toward the opposite corner.It stops,makes a90◦right turn and runs 2more meters.A scientist measures the shortest distance between the lemming and each side of the square.What is the average of these four distances in meters?(A)2(B)4.5(C)5(D)6.2(E)723What is the area of the shaded pinwheel shown in the5×5grid?2007(A)4(B)6(C)8(D)10(E)1224A bag contains four pieces of paper,each labeled with one of the digits 1,2,3or 4,with norepeats.Three of these pieces are drawn,one at a time without replacement,to construct a three-digit number.What is the probability that the three-digit number is a multiple of 3?(A)14(B)13(C)12(D)23(E)3425On the dart board shown in the ﬁgure,the outer circle has radius 6and the inner circle hasradius 3.Three radii divide each circle into the three congruent regions,with point values shown.The probability that a dart will hit a given region is proportional to to the area of the region.What two darts hit this board,the score is the sum of the point values in the regions.What is the probability that the score is odd?1112222007(A)1736(B)3572(C)12(D)3772(E)1936The problems on this page are copyrighted by the Mathematical Association of America’s American Mathematics Competitions.。

amc8数学题

amc8数学题
AMC 8 数学题是一种美国数学竞赛，适合 8 年级 (相当于中国初中) 的学生参加。

这个竞赛的题目涉及各种数学知识，包括代数、几何、三角函数、概率统计等，需要学生具备扎实的数学基础和较强的思维能力。

下面是一道 AMC 8 数学题的正文和拓展:
题目：给定一个非负整数数组 nums，求出数组中任意两个数之和的最大值。

拓展:
在这个题目中，我们需要求出数组中任意两个数之和的最大值。

我们可以使用以下方法来解决这个问题:
首先，我们可以遍历整个数组，记录每个位置的当前最大值。

在遍历过程中，我们可以更新最大值，将其当前值加上相邻位置的值，从而得到新的最大值。

遍历结束后，我们可以得到数组中任意两个数之和的最大值。

其次，我们可以使用动态规划的方法来解决这个问题。

我们可以定义一个二维数组 dp，其中 dp[i][j] 表示在前 i 个元素中，任意两个数之和的最大值。

接下来，我们可以按照题目中给出的要求，依次更新 dp[i][j] 的值。

最终，我们可以得到数组中任意两个数之和的最大值。

再次，我们可以使用树状数组的方法来解决这个问题。

我们可以定义一个树状数组 tree，其中 tree[i] 表示在前 i 个元素中，任
意两个数之和的最大值。

接下来，我们可以按照题目中的要求，依次更新 tree[i] 的值。

最终，我们可以得到数组中任意两个数之和的最大值。

以上三种方法都可以解决 AMC 8 数学题，但是具体选择哪种方法取决于题目的复杂度和难度。