2011启智杯真题

2011启智杯真题

1.今天是2011年12月03日，请在20111203八个数字之间添加＋、－、×、÷四种运算符号中的某些符号，使得下面等式成立。

2□0□1□1□1□2□0□3=9

2.汉字“数”、“学”、“好”分别表示不同的数字，根据下列所给算式，则“数”表示数字，“学”表示数字“好”表示数字

学好

学好

学好

学好

＋学好

数字

3．“启”、“智”、“杯”各表示一个数字，同时满足下列等式：

（l ）启+智+杯=22

（2 ）启-智+杯=4

（ 3 ）启+智-杯=6

则“启”表示，“智”表示，“杯”表示。

4．有一个孤岛，那里的人们在商品交易时有如下特殊的要求：

（ l )所有商品价格（单位：元）为整数，目不超过31元

（ 2 )消费者支付款项时每—砷币值的钱币最多只能使用1枚，而且商家不找零钱。

问：为了保证公平交易（照实支付），至少应该生产种不同面值的钱币，请具体列举出。

5．按照下图规律，写出第四个图中x ，y ，z 所表示的三位数x= ,y= ,z= . 说明理由。

请写出你发现的规律：

第一个图 第二个图 第三个圈 第四个图

6.在同一个布袋里有红色、黄色、蓝色袜子各15只，最少要拿 只才能保证其中至少有2双颜色不同的袜子（注意：袜子不分左右）。

7.如图所示是2011

年12月份的日期，现用一矩形在日历中任意框出6个数

，请

用一个等式表示a 、

b 、

c 、

d 、

e 、

f 这6个数之间的关系。

将答案写在下列横线上：

8、如图所示是，长方形的长：宽=4：3，将该长方形划分为四个三角形，其面积分别是S 1、S 2、、S 3、S 4。若S 1=S 2、+ S 3=S 4 ，则S 2 : 、S 3 = .

9、a 、b 、c 、d 为整数，满足等式,4330

1

1

1

1=+++d

c b a 则d= 。

10、观察下面的算式：

0×0=0-0，1×31

21

31

21

,32

232

2,21121-=?-=?-=，……

根据算式反映出的规律，再写出满足这个规律的两个算式。

将答案写在下列横线上： ， 。

D

C

11、半径为1的圆沿着半径为3的圆的内侧与外侧无滑动地滚动（如下图），当他们回到开始滚动的位置时：

（1）哪个圆滚动的圈数多？为什么？

（2）若点A 是内侧滚动圆上的一个定点（如左图），请在甲图中画出点A 的运动路线。

（甲） （乙）

12、有一家公司董事长从他四个得力助手中挑选一名担任策划部经理，他给他们出了一下问题；把一闲置的圆形土地平均分成四块，要求每块都与其他三块相连（即有公共边），请你在下列所给图中至少画出2个符合要求的分割方法。

13、A 、B 、C 、D 四个盒子中分别放有6，5，4，3个球，第一个小朋友找到放球最少的盒子，从其它的盒子中各取1个球放入这个盒子中，然后第二个小朋友又找到一个放球最少的盒子，从其他的盒子中各取1个球放入这个盒子中，……如此进行下去，当第2011个小朋友放完后，A 、B 、C 、D 四个盒子中的球数依次是多少个？

将答案写在下列横线上： 。

A

A

微生物期末考试试题附答案

《微生物学教程》考试试卷 （B 卷） （本试卷共 3 页） 要求：开卷考试；答案写在答题纸上，标清楚题号；答案和试题全交。 一、解释名词（每个2分，本题满分20分） 1、菌落 2、病毒 3、生长因子 4、微生物典型生长曲线 5、灭菌 6、质粒 7、营养缺陷型 8、水体自净作用 9、微生物的种 10、培养基 二、填空（每个空0.5分，40个满分20分） 1、微生物与生物环境间的相互关系分为五种： （1） 、 （2） 、 （3） 、 （4） 、 （5） 。 2、单细胞微生物典型生长曲线由 （1） 、 （2） 、（3） 、（4）四个时期组成。 3、防止霉腐微生物引发食品腐败的常用措施有： （1） 、 （2） 、 （3） 、 （4） 、 （5） 。 4、微生物的五大共性包括： （1） 、 （2） 、 （3） 、 （4） 、 （5） 。 5、微生物根据对氧的需求可分为5种类型： （1） 、（2） 、（3） 、 （4） 、 （5） 。 6、葡萄糖经EMP 代途径10步反应产生的3种产物是： （1） 、 （2） 、 （3） 。 7、真菌（霉菌类）无性孢子有多种类型。比如（1） 、 （2） 、 （3） 、 （4） 。 8、列举对微生物学发展做出过巨大贡献的科学家名字： （1） 、 （2） 、 （3） 。 第1页 9、三域学说将整个生物划分为3个域，即 （1） 、 （2） 、 （3） 。

10、常用的消毒剂有： （1） 、 （2） 、 （3） 。 三、判断题（每小题1分，满分10分） 1、冷冻干燥保藏和液氮保藏是长期有效保藏各种微生物的 基本方法。 2、青霉素可以杀死正在旺盛生长繁殖的细菌细胞，但无法杀死正处于休止状态的营养缺陷型细胞。 3、黑曲霉、红曲霉、黄曲霉、根霉、毛霉，其中有些是单细胞微生物。 4、食品的水活度值越低，越不利于腐败（病原）微生物的生长，因此降低含水量有利于食品的保藏。 5、因为细菌是低等原核生物，所以，它没有有性繁殖，只有无性繁殖形式。 6、大肠杆菌F +细胞通过接合作用向受体细胞转移F 质粒，结果导致原供体细胞失去了F 质粒。 7、国家规定：引用水中的微生物数应低于100个细胞/mL, 大肠菌群数不能超过3个细胞/L 。 8、黄曲霉毒素是由黄曲霉菌产生的剧毒、致癌物质，它要在人或动物体经过代活化后才引起致癌作用。 9、地衣实际上是蓝细菌与真菌或藻类与真菌形成的共生体。 10、青霉菌发酵时，产青霉素的最佳温度与菌体生长最适温度完全一致。 四、简答题（4道题，每小题10分，满分40分 ） 1、简述常用化学消毒剂的种类及其杀菌机制。 2、简述艾姆氏试验的原理和实践意义。 3、比较同型乳酸发酵与异型乳酸发酵在产物、产能和菌种上的差异。 4、常用的高温灭菌方法及其适用对象。 第2页

2011年12月大学六级真题+听力原文+答案详解

2011年12月大学英语六级真题 Part I Writing (30 minutes) Directions:For this part, you are allowed 30 minutes to write a short essay entitled The Way to Success by commenting on Abraham Lincoln's famous remark, "Give me six hours to chop down a tree, and I will spend, the first four sharpening the axe." You should write at least 150 words but no more than 200 words. The Way to Success 注意：此部分试题请在答题卡1上作答。 Part II Reading Comprehension (Skimming and Scanning) (15 minutes) Directions:In this part, you will have 15 minutes to go over the passage quickly and answer thequestions on Answer Sheet 1. For questions 1-7, choose the best answer from the four choices marked A), B), C) and D). For questions 8-10, complete the sentences with the information given in the passage. Google's Plan for World's Biggest Online Library: Philanthropy Or Act of Piracy? In recent years, teams of workers dispatched by Google have been working hard to make digital copies of books. So far, Google has scanned more than 10 million titles from libraries in America and Europe - including half a million volumes held by the Bodleian in Oxford. The exact method it uses is unclear; the company does not allow outsiders to observe the process. Why is Google undertaking such a venture? Why is it even interested in all those out-of-printlibrary books, most of which have been gathering dust on forgotten shelves for decades? Thecompany claims its motives are essentially public-spirited. Its overall mission, after all, is to "organise the world's information", so it would be odd if that information did not include books. The company likes to present itself as having lofty aspirations. "This really isn't about making money. We are doing this for the good of society." As Santiago de la Mora, head of Google Books for Europe, puts it: "By making it possible to search the millions of books that exist today, we hope to expand the frontiers of human knowledge." Dan Clancy, the chief architect of Google Books, does seem genuine in his conviction that thisis primarily a philanthropic (慈善的) exercise. "Google's core business is search and find, soobviously what helps improve Google's search engine is good for Google," he says. "But we havenever built a spreadsheet (电子数据表) outlining the financial benefits of this, and I have neverhad to justify the amount I am spending to the company's founders." It is easy, talking to Clancy and his colleagues, to be swept along by their missionary passion. But Google's book-scanning project is proving controversial. Several opponents have recently emerged, ranging from rival tech giants such as Microsoft and Amazon to small bodies representing authors and publishers across the world. In broad terms, these opponents have levelled two sets of criticisms at Google. First, they have questioned whether the primary responsibility for digitally archiving the world's books should be allowed to fall to a commercial company. In a recent essay in the New

启智杯数学思维及应用能力竞赛试卷小学组

参考答案及评分标准 本卷共12题，每题10分，满分120分。答题时间120分钟。 在下面的算式中，不同的汉字代表1—9中不同的数字，那么，“为了一切学生”的各字分别代表什么数字？写出一种答案，说明你的分析过程。 为了一切学生 一切为了学生 +为了学生一切 987654 “为了一切学生”的各字分别代表“372415” “切”必为偶数，最小为2. “切”= 2，则“生”= 1 或6。 1）若“切”= 2，“生”= 1，则结合百位，则“了”=3，此时个位、百位、万位均不进位。再分析十位、千位、十万位，可得“为”=“一”=3，“学”=1，数字重复，不符合题意，舍去。 2）若“切”= 2，“生”= 6，则结合百位，则“了”= 8，此时个位、百位、万位均进位1。再分析十位、千位、十万位，可得“为”=3，“一”=2，“学”=1，数字重复，不符合题意，舍去。 “切”= 4，则“生”= 5，结合百位，则“了”=7，此时个位、百位、万位均进位1。再分析十位、千“为”= 3，“一”=2，“学”=1。符合要求。 “为了一切学生”的各字分别代表“372415”，原式为 372415 243715 +371524 987654 答案及分析正确给满分；只写出正确答案而未加说明，给5分；基本思路正确，而答案5分；其他情况酌情给分。 从1、2、3、4、5、6、7、8、9这9个数字中选出8个不同的数字分别填入下面两个算式的方框内（每个数字只许用一次），使它们都成立，简述理由。 ?+? - ? = ?；???÷? = ? : 1）4+5 - 1 = 8 ，3?6 ÷ 2 = 9 ；（2）5+7 - 9 =3，1?8 ÷ 4 = 2.等等 4个数字必须满足两个之积等于另两个之积；而在加减算式中，所4个数字必须满足两个之和等于另两个之和。 在9个数字中，有多种可能性，比如3?6 =2?9；4?6=3?8；1?8=2?4；1?6=2?3；2?6=3?4。 4、6；3、8，在余下的5个数字1、2、 5、7、9中，任何4个数字都不可以取作加减运算。 3、6；2、9，在余下的5个数字1、 4、 5、7、8中，1、8；4、5以及4，8；7,5都可以取作加 1）4+5 - 1 = 8 ，3?6 ÷ 2 = 9 ；（2）8+1 - 4 = 5，3?6 ÷ 9 = 2. 3）5+7 - 8 = 4 ，2?9 ÷ 3 = 6 ；（4）4+8 - 5 =7，9?2 ÷ 6 = 3。等等。 5分；分析合理得5分；其它情况酌情给

2019年12月大学英语四级考试真题完整版(第一套)

Part I Writing (30 minutes) Directions: For this part, you are allowed 30 minutes to write a letter to a foreign friend who wants to learn Chinese. Please recommend a university to him. You should write at least 120 words but no more than 180 words. Part II Listening Comprehension (25 minutes) Section A Directions: In this section, you will hear three news reports. At the end of each news report, you will hear two or three questions. Both the news report and the questions will be spoken only once. After you hear a question, you must choose the best answer from the four choices marked A) , B) , C) and D) . Then mark the corresponding letter on Answer Sheet 1 with a single line through the centre. Questions I and 2 are based on the news report you have just heard. 1.A)Many facilities were destroyed by a wandering cow. B) A wandering cow knocked down one of its fences. C)Some tourists were injured by a wandering cow. D) A wandering cow was captured by the police. 2.A) It was shot to death by a police officer. B)It found its way back to the park' s zoo. C)It became a great attraction for tourists. D)It was sent to the animal control department. Questions 3 and 4 are based on the news report you have just heard. 3.A) It is the largest of its kind. B)It is going to be expanded. C)It is displaying more fossil specimens. D)It is staring an online exhibition. 4.AJA collection of bird fossils from Australia. B)Photographs of certain rare fossil exhibits. C)Some ancient wall paintings from Australia. D)Pictures by winners of a wildlife photo contest. Questions 5 to 7 are based on the news report you have just heard. 5.A) Pick up trash. B)Amuse visitors. C)Deliver messages. D)Play with children. 6.A) They are especially intelligent. 8)They are children' s favorite. C They are quite easy to tame. D) They are clean and pretty. 7.A) Children may be harmed by the rooks. B)Children may be tempted to drop litter. C)Children may contract bird diseases.

微生物学考试试题及答案

《微生物学》课程期末考试试题解答及评分标准99b 一．判断改错题（判断下列每小题的正错，认为正确的在题后括号内打“√”； 错误的打“×”，并予以改正，每小题1.5分，共15分） 01.真菌典型营养体呈现丝状或管状，叫做菌丝(√) 02.专性寄生菌并不局限利用有生命力的有机物作碳源。(×) 改正：专性寄生菌只利用有生命力的有机物作碳源 03.根据微生物生长温度范围和最适温度，通常把微生物分成高温性、中温性、低温性三 大类。(√) 04.放线菌、细菌生长适宜的pH范围：最宜以中性偏酸；(×) 改正：放线菌，细菌生长适宜中性或中性偏碱。 05.厌气性微生物只能在较高的氧化还原电位(≥0.1伏)生长,常在0.3-0.4V生长。(×) 改正：厌气性微生物只能在较低的氧化还原电位(≤0.1伏)才能生长,常在 0.1V生长； 06.波长200-300nm紫外光都有杀菌效能，一般以250-280nm杀菌力最强。(√) 07.碱性染料有显著的抑菌作用。(√) 08.设计培养能分解纤维素菌的培养基,可以采用合成培养基。(×) 改正:能分解纤维素菌的培养基,培养基中需加有机营养物：纤维素。

09.液体培养基稀释培养测数法，取定量稀释菌液，经培养找出临界级数，可以间接测定 样品活菌数。(√) 10.共生固氮微生物，二种微生物必须紧密地生长在一起才能固定氨态氮，由固氮的共生 菌进行分子态氮的还原作用。(√) 一．多项选择题（在每小题的备选答案中选出二至五个答案，并将正确的答案填在题干的括号内，正确的答案未选全或有选错的，该小题无分，每小题2分，共20分） 11.放线菌是能进行光合作用的原核微生物，其细胞形态（A；B；C；） A.有细胞壁; B.由分支菌丝组成; C.无核仁; D.菌体无鞭毛; E.菌体中有芽孢。 12.支原体[Mycoplasma],介乎于细菌与立克次体之间的原核微生物,其特点是：（A；B；）A.有细胞壁;B.能人工培养； C.有核仁; D.有鞭毛； E.非细胞型微生物。 13.无机化合物的微生物转化中，其硝化作用包括：（C；D；E；） A.硝酸还原成亚硝酸; B.硝酸还原成NH 3；C.NH 3转化成亚硝酸;D.铵盐转化成亚硝酸； E.亚硝酸盐转化成硝酸盐。 14.单细胞微生物一次培养生长曲线中，其对数生长期的特点：（A；D; E；）

2011年12月大学英语六级真题答案

2011年12月大学英语六级真题答案 快速阅读 1. Google claims its plan for the world’s biggest online library is _____ 【答案】B. to serve the interest of the general public【解析】该题问的是Google 所声称的自己图书馆计划的目的。根据顺序原则该题靠前，同时注意到第三段开头连续并列why，提示第三段很有可能提到原因或目的。用claim定位至第三段第三句，该句motive 引出后面Google声称的原因是“本质上被公众精神驱动”，三段末更提到是为了所有人的知识需求。核心名词public及句意都对应B选项。 2. According to Santiago de la Mora, Google’s book-scanning project will 【答案】B. broaden humanity’s intellectual horizons【解析】该题问的是Santiago de la Mora对Google图书扫面项目的看法。用人名可定位至四段中，之后该人提到Goo gle该项目能expand the frontiers of human knowledge，即拓广人类知识的范围。对应B选项。核心名词knowledge被改为同义词intellectual，frontier被同义替换为horizon，动词expand被同义替换为broaden。 3. Opponents of Google Books believe that digitally archiving the world's books should be controlledby_______. 【答案】C. non-profit organizations【解析】该题问的是反对Google的人对数字知识控制者的看法。用opponents可定位至第7段前后，control可进一步定位至第八段最后R obert Darnton的观点。该人认为只有public, not-for-profit bodies 可以有控制数字知识的权利，对应C选项。Bodies被同义替换为organization。 4.【答案】D. the copyright of the books it scanned【解析】该题问的是Google卷入官司的原因。由legal battle可定位至第9段前后，用之后第10段短首出现的however

2014年启智杯真题及试题分析(小学组)

2014年第五届启智杯（小学组）真题与详细解析（含评分标准） 1. 观察如下几个等式： （1） 331=；（2）57313+=+；（3）79113135 ++=++；…… 你发现了什么规律？请据此写出第100个式子。 【参考答案】201203...399313 (199) +++=+++ 解：第n 个式子的分母是前n 个连续奇数之和，分子是第n+1至第2n 个连续奇数之和，其等号右端都是3。——————————————————————————6分 第100个式子为 201203...399313...199+++=+++ —————————————————— +4 = 10分 2. 有一个2014位数，其从左到右第2、3位数字分别为2、3，第11、30、2014位数字分别为4、5、6. 如果其任何相邻的五位数字之和全相等，请问该数的第一位数字是几？全部2014位数字之和是多少？写出结果，并说明分析过程。 【参考答案】第一位数字是4；全部2014位数字之和是8055. 解：由于其任何相邻的五位数字之和全相等，所以其中任何连续六位数 abcdef ，总有 a b c d e b c d e f ++++=++++，因此a f =，这说明，这2014位数每五位一循环。 ——————————————————— +2 = 6分 由于11、30、2014被5除余数分别为1、0、4，所以该数的第一位数字是4 （前五位分别是4、2、3、6、5） ———————————————————— + 2 = 8分 由于20145402 4÷=，所以这整个数为 (42365)402(4236)20402158040158055++++?++++=?+=+= —————————————————— + 2 = 10分

2011年12月英语四级真题及答案详解

作文： Nothing succeeds without a strong will There is a prevalent joke around young people saying that ‘quitting smoking is the easiest thing in the world, and I’ve done it for hundreds of times.’ This seemingly funny statement ironically reflects the fact that the determination of most youngsters is oftentimes started with enthusiasm, but the passion becomes increasingly weaker each day, and then diminishes as if there has been no such thing at all. It is obvious that their failure in ‘quitting smoking’ and decline of determination are all ascribed to their lack of will. Initially, every success involves several stages of setbacks and risks, and we need to summon up our will to conquer them. Moreover, there are enormous temptation in our path of pursuit of success. For instance, when we plan to quit smoking, our roommates may smoke freely in front of us; when we are eager to keep fit, our close friends may invite us to have late night snacks. Under these occasions, only strong will can assist us to resist the temptation, and persist in chasing our goals until we triumphantly realize them. In short, no dream will successfully come true if we do not have strong will. An old famous proverb says that ‘ where there is a will, there is a way’. Let us bear this motton in mind no matter how many thunders and thorns are on our roads ahead, then we will be successful with such precious and powerful spirit. [快速阅读] 快速阅读 1. A person of integrity not only sets high moral and ethical standards but also _______. A) sticks to them in their daily life B) makes them known to others C) understands their true values D)sees that others also follow them 选择sticks to them in their daily life 定位在原文第一段，文中说正直(integrity)的关键是一致性(consistency)，不仅要有很多道德和伦理的准则，还要每天都坚守。 2. What role does integrity play in personal and professional relationships? A) It helps to create team spirit B) It facilitates communication C) It is the basis of mutual trust D) It inspires mutual respect 选择It is the basis of mutual trust.

微生物考试试题(附答案)

试卷一 一．选择题（每个2分，共20分） 1、组成病毒粒子衣壳的化学物质是（） A、糖类 B、蛋白质 C、核酸 D、脂肪 2、属于细菌细胞特殊结构的为（） A、荚膜 B、细胞壁 C、细胞膜 D、核质体 3、噬菌体属于病毒类别中的（） A、微生物病毒 B、昆虫病毒 C、植物病毒 D、动物病毒 4、半固体培养基中，琼脂使用浓度为（）。 A、0 B、0.3—0.5% C、1.5—2.0% D、5% 6．下述那个时期细菌群体倍增时间最快( ) A 稳定期 B 衰亡期 C 对数期 D 延滞期 7.下面关于连续培养的说法中，叙述不正确的是（） A.恒化连续培养的控制对象是菌体密度 B.恒浊连续培养中微生物菌体以最高生长速率生长 C.恒化连续培养的产物是不同生长速率的菌体 D.恒浊连续培养无生长限制因子 8. 下列叙述中正确的说法是（） A.产生抗逆性强的芽孢是产芽孢细菌在不良环境条件下的一种生殖方式。 B.厌氧的微生物需要较低的氧化还原电位。 C.一种细菌的学名是Staphylococcus aureus,其中aureus属名，Staphyloccus是种名 D.类病毒是有侵染性的蛋白颗粒。 9.下述诱变剂中常引起移码突变的是：（） A 亚硝酸 B 烷化剂 C 碱基类似物 D 丫啶类染料 10.下述那个最可能产生一个重组细胞：（） A F+ x F- B Hfr x F+ C Hfr x F- D F+ x F- 二、写出下列名词解释的中文翻译及作出解释（每个2分，共12分） 1.Gram positive bacteria 2.parasporal crystal 3 ,colony 4, life cycle 5,capsule 6,endospore 三、判断改错题（每个2分，共10分） 1.大肠杆菌和枯草芽孢杆菌属于单细胞生物，唾液链球菌和金黄色葡萄球菌属于多细胞生物（） 2.遗传型相同的个体在不同环境条件下会有不同的表现型。（） 3.低剂量照射紫外线，对微生物几乎没有影响，但以超过某一阈值剂量的紫外线照射，则会导致微生物的基因突变。（） 4. 一切好氧微生物都含有超氧化物歧化酶（SOD ）。（） 5.一般认为各种抗性突变是通过适应而发生的，即由其所处的环境诱发出来的。（） 四.名词解释（每题3 分，共计18 分）

2011年12月英语六级全真预测试题及答案解析

之 2011年12月大学英语六级全真预测试题及答案解析

2011年12月英语六级全真预测试题及答案解析 一、阅读理解 第1题： Occasional self-medication has always been part of normal living. The making and selling of drugs have a long history and are closely linked, like medical practice itself, with the belief in magic. Only during the last hundred years or so has the development of scientific techniques made it possible for some of the causes of symptoms to be understood, so that more accurate diagnosis has become possible. The doctor is now able to follow up the correct diagnosis of many illnesses with specific treatment of their causes. In many other illnesses, of which the causes remain unknown, it is still limited, like the unqualified prescriber, to the treatment of symptoms. The doctor is trained to decide when to treat symptoms only and when to attack the cause: this is the essential difference between medical prescribing and self-medication. The advance of technology has brought about much progress in some fields of medicine, including the development of scientific drug therapy. In many countries public health organization is improving and people's nutritional standards have risen. Parallel with such beneficial trends have two adverse effects. One is the use of high-pressure advertising by the pharmaceutical industry, which has tended to influence both patients and doctors and has led to the overuse of drugs generally. The other is the emergence of the sedentary society with its faulty ways of life: lack of exercise, over-eating, unsuitable eating, insufficient sleep, excessive smoking and drinking. People with disorders arising from faulty habits such as these, as well as from unhappy human relationships, often resort to self-medication and so add the taking of pharmaceuticals to the list. Advertisers go to great lengths to catch this market. Clever advertising, aimed at chronic sufferers who will try anything because doctors have not been able to cure them, can induce such faith in a preparation, particularly if steeply priced, that it will produce—by suggestion—a very real effect in some people. Advertisements are also aimed at people suffering from mild complaints such as simple colds and coughs, which clear up by themselves within a short time. These are the main reasons why laxatives, indigestion remedies, painkillers, tonics, vitamin and iron tablets and many other preparations are found in quantity in many households. It is doubtful whether taking these things ever improves a person's health; it may even make it worse. Worse because the preparation may contain unsuitable ingredients; worse because the taker may become dependent on them; worse because they might be taken in excess; worse because they may cause poisoning, and worse of all because symptoms of some serious underlying cause may be masked and therefore medical help may not be sought. 1. The first paragraph is intended to ________. [A] suggest that self-medication has a long history [B] define what diagnosis means exactly [C] praise doctors for their expertise [D] tell the symptoms from the causes

2013第四届启智杯初中组真题与详细解析

2013年第四届启智杯（初中组）真题与详细解析及评分标准说明：本卷共12题，每题10分，满分120分。答题时间120分钟。 1.完成以下算式：将适当的数字填入下述方框内，使除法算式成立。（不要求理由） 【参考答案】2013÷33 = 61 理由：先确定除数的十位数：因为20 被该数除上6余一位数，说明该除数的十位必为3；再定除数的个位数：根据余数是一位数，个位数只能是2、3、或4,32 6 =192，33 6 =198，34 6 =204；如果除数是32，则余数是8或9,结合后一位3，83或93均不是32的倍数，结论不成立；同样可以否定34；最终确定除数是33. 后面就比较清楚了。 【评分标准】确定除数的十位数得3分；再确定除数的个位数再得3分；确定尚的个位数得2分；全部确定再得2分。 2. 请完成以下两个问题： （1）将999表示成9个连续的奇数之和的形式； （2）将1 7 n表示成7 个连续的奇数之和的形式时，则其中最小的奇数是多少？（用含有字母n 的代数式表示，其中字母n 正整数）

【参考答案】 （1）由于999÷9=111，所取9个奇数的中间数应该是111，这个表示为 999=103+105+107+109+111+113+115+117+119. （2）类似地： )67()47()27(7)27()47()67(7771+++++++-+-+-=?=+n n n n n n n n n 故最小的奇数是67-n 。 【评分标准】两个小题各5分。 【注】这类问题关键找出其平均数，以平均数为中心，向两端延伸。 3.观察下列等式： （1）2 22543=+ （2）2 22221413121110+=++ （3）2 22222227262524232221++=+++ ...... 请按此规律写出第四个等式。 若按照此规律写的第n 个等式的等号左边最小数为2 210，则该等式的等号右边最大数是多少？ 【参考答案】 第四个等式：（4）2 22222222444342414039383736+++=++++ 等号左边最小数为2 210时，等号右边最大数是2 230。 解答方法1： 每个等式的左右两边是连续若干个完全平方数，其个数依次为3、5、7、9个等等，其特点

2011年12月英语四级真题及答案

2011年12月英语四级真题 Part Ⅰ Writing (30 minutes) 注意：此部分试题在答题卡1上． For this part, you are allowed 30 minutes to write a short essay entitled Nothing Succeeds Without a Strong Will by commenting on the humorous saying, "Quitting smoking is the easiest thing in the world. I've done it hundreds of times." You should write at least 120 words but no more than 180 words. Part II Reading Comprehension (Skimming and Scanning) Why Integrity Matters What is Integrity? The key to integrity is consistency--not only setting high personal standards for oneself (honesty, responsibility, respect for others, fairness) but also living up to those standards each day. One who has integrity is bound by and follows moral and ethical standards even when making life's hard choices, choices which may be clouded by stress, pressure to succeed, or temptation. What happens if we lie, cheat, steal, or violate other ethical standards? We feel disappointed in ourselves and ashamed. But a lapse of integrity also affects our relationships with others. Trust is essential in any important relationship, whether personal or professional. Who can trust someone who is dishonest or unfair? Thus, integrity must be one of our most important goals. Risky Business We are each responsible for our own decisions, even if the decision-making process has been undermined by stress or peer pressure. The real test of character is whether we can learn from our mistake, by understanding why we acted as we did, and then exploring ways to avoid similar problems in the future. Making ethical decisions is a critical part of avoiding future problems. We must learn to recognize risks, because if we can't see the risks we're taking, we can't make responsible choices. To identify risks, we need to know the rules and be aware of the facts. For example, one who doesn't know the rules about plagiarism may accidentally use words or ideas without giving proper credit, or one who fails to keep careful research notes may unintentionally fail to quote and cite sources as required. But the fact that such a violation is "unintentional" does not excuse the misconduct. Ignorance is not a defense. "But Everybody Does It" Most people who get in trouble do know the rules and facts, but manage to fool themselves about the risks they're taking by using excuses: "Everyone else does it," "I'm not hurting anyone," or "I really need this grade." Excuses can get very elaborate: "I know I'm looking at another's exam, even though I'm supposed to keep my eyes on my own paper, but that's not cheating because I'm just checking my answers, not copying." We must be honest about our actions, and avoid excuses. If we fool ourselves into believing we're not doing anything wrong, we can't see the real choice we're making--and that leads to bad decisions. To avoid fooling yourself, watch out for excuses and try this test: Ask how you would feel if your actions were public, and anyone could be watching over your shoulder. Would you feel proud or ashamed of your actions? If you'd rather hide your actions, that's a good indication that you're taking a risk and rationalizing it to yourself.