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第十讲 行程（二）在今天这节课中，我们来研究行程问题中的相遇与追及问题．这一讲就是通过例题加深对行程问题三个基本数量关系的理解，使学生养成画图解决问题的好习惯！ 知识点：1、直线型的相遇与追及问题2、环形上的相遇与追及问题.分析：要求狗走的路程，速度已知，关键是求出狗所走的时间.经过认真审题，不难发现狗行走的时间与甲、乙二人的相遇时间是相等的.这就是一道行程问题应用题.甲、乙二人相遇时间为：50÷（3+2）=10（小时），狗的速度是5千米/时 ，所以，狗所走的路程一共是：5×10=50（千米）.1. 甲、乙二人分别从相距30千米的两地同时出发相向而行，甲每小时走6千米，乙每小时走4千米，问：二人几小时后相遇？分析：出发时甲、乙二人相距30千米，以后两人的距离每小时都缩短6＋4＝10（千米），即两人的速度的和（简称速度和），所以30千米里有几个10千米就是几小时相遇. 30÷（6＋4）＝30÷10＝3（小时）.2. 甲、乙二人都要从北京去天津，甲行驶10千米后乙才开始出发，甲每小时行驶15千米，乙每小时行驶10千米，问：乙经过多长时间能追上甲？你还记得吗教学目标想 挑 战 吗？苏步青教授是我国著名的数学家.有一次在外国，他在电车上碰到一位有名的德国数学家，这位德国数学家出了一道有趣的数学题让他做，这道题是：“两地相距50千米，甲、乙二人同时从两地出发相向而行.甲每小时走3千米，乙每小时走2千米．甲带着一只狗，狗每小时走5千米.这只狗同甲一起出发，碰到乙的时候它就掉转头来往甲这边走，碰到甲时又往乙这边走，直到两人碰头.问这只狗一共走了多少千米路？”苏步青略加思索，未等下电车就把正确答案告诉了这位德国数学家.同学们，你们也来试一试，会解吗？分析：出发时甲、乙二人相距10千米，以后两人的距离每小时都缩短15-10＝5（千米），即两人的速度的差（简称速度差），所以10千米里有几个5千米就是几小时能追上.10÷（15-10）＝10÷5＝2（小时）.在行程问题中涉及到两个或两个以上物体运动的问题，其中最常见的是相遇问题和追及问题.甲从A 地到B 地，乙从B 地到A 地，然后两人在途中相遇，实质上是甲和乙一起走了A,B 之间这段路程，如果两人同时出发，那么A,B 之间的路程=甲走的路程+乙走的路程=甲的速度×相遇时间+乙的速度×相遇时间=（甲的速度+乙的速度）×相遇时间=速度和×相遇时间. 一般地，相遇问题的关系式为：速度和×相遇时间=路程和，即t v S 和和=有两个人同时行走，一个走得快，一个走得慢，当走得慢的在前，走得快的过了一些时间就能追上他.这就产生了“追及问题”.实质上，要算走得快的人在某一段时间内，比走得慢的人多走的路程，也就是要计算两人走的路程之差（追及路程）.如果设甲走得快，乙走得慢，在相同的时间（追及时间）内：追及路程=甲走的路程-乙走的路程=甲的速度×追及时间-乙的速度×追及时间=（甲的速度-乙的速度）×追及时间=速度差×追及时间. 一般地，追击问题有这样的数量关系：追及路程=速度差×追及时间，即t v S 差差=（一） 直线型的相遇问题：【例1】 王老师从甲地到乙地，每小时步行5千米，张老师从乙地到甲地，每小时步行4千米.两人同时出发，然后在离甲、乙两地的中点1千米的地方相遇，求甲、乙两地间的距离.专题精讲分析：画一张示意图（可让学生先判断相遇点在中点哪一侧，为什么？）离中点1千米的地方是A点，从图上可以看出，王老师走了两地距离的一半多1千米，张老师走了两地距离的一半少1千米.从出发到相遇，王老师比张老师多走了2千米王老师比张老师每小时多走（5-4）千米，从出发到相遇所用的时间是2÷（5-4）＝2（小时）.因此，甲、乙两地的距离是（5＋ 4）×2＝18（千米）.[巩固]夏夏和冬冬同时从两地相向而行，两地相距1100米，夏夏每分钟行50米，冬冬每分钟行60米，问两人在距两地中点多远处相遇?分析：根据题意，两人相遇时经过的时间为：1100÷（50+60）=10分钟，10分钟夏夏走了50×10=500（米），两地的中点距离夏夏的出发地距离为：1100÷2=550，所以两人相遇处距离两地中点550-500=50米远.【例2】甲、乙两车分别同时从A、B两地相对开出，第一次在离A地95千米处相遇．相遇后继续前进到达目的地后又立刻返回，第二次在离B地25千米处相遇．求A、B两地间的距离．分析：画线段示意图(实线表示甲车行进的路线，虚线表示乙车行进的路线)：可以发现第一次相遇意味着两车行了一个A、B两地间距离，第二次相遇意味着两车共行了三个A、B两地间的距离．当甲、乙两车共行了一个A、B两地间的距离时，甲车行了95千米，当它们共行三个A、B 两地间的距离时，甲车就行了3个95千米，而这285千米比一个A、B两地间的距离多25千米，可得：95×3-25=285-25=260(千米)．[拓展]甲、乙两列火车同时从东西两镇之间的A地出发向东西两镇反向而行，它们分别到达东西两镇后，再以同样的速度返回，已知甲每小时行60千米，乙每小时行70千米，相遇时甲比乙少行120千米，东西两镇之间的路程是多少千米？分析：教师注意帮助学生画图分析.从出发到甲、乙两列火车相遇，两列火车共同行驶了2个全程.已知甲比乙少行120千米，甲每小时比乙少行（70—60 =）10千米，120÷10 = 12（小时），说明相遇时，两辆车共同行驶了12小时.那么两辆车共同行驶1个全程需要6小时，东西两镇之间的路程是（60 + 70）×6 = 780（千米）【例3】甲、乙、丙三辆车同时从A地出发到B地去，甲、乙两车的速度分别为每小时60千米和48千米，有一辆迎面开来的卡车分别在它们出发后的5小时.6小时，8小时先后与甲、乙、丙三辆车相遇，求丙车的速度.分析：甲车每小时比乙车快60-48＝12（千米）.则5小时后，甲比乙多走的路程为12×5＝60（千米）.也即在卡车与甲相遇时，卡车与乙的距离为60千米，又因为卡车与乙在卡车与甲相遇的6-5＝1小时后相遇，所以，可求出卡车的速度为60÷1-48＝12（千米/小时），卡车在与甲相遇后，再走8-5＝3（小时）才能与丙相遇，而此时丙已走了8个小时，因此，卡车3小时所走的路程与丙8小时所走的路程之和就等于甲5小时所走的路程.由此，丙的速度也可求得，应为：（60×5-12×3）÷8＝33（千米/小时）. 所以卡车的速度：（60-48）×5÷（6－5）－48＝12（千米/小时），丙车的速度：（60×5－12×3）÷8＝33（千米/小时），[拓展] 甲、乙、丙三人，甲每分钟走100米，乙每分钟走80米，丙每分钟走75米.甲从东村，乙、丙从西村同时出发相向而行，途中甲、乙相遇后3分钟又与丙相遇.求东西两村的距离.分析：先画示意图如下：3分钟甲（100米/分）甲、丙相遇甲、乙相遇乙（80米/分）丙（75米/分）东西甲、乙相遇后3分钟，甲、丙相遇.甲、丙在3分钟内共走路程是（100＋75）×3＝525（米）.显然，这就是甲、乙相遇时，乙比丙多走的路程，乙比丙每分钟多走80－75＝5（米）.所以，甲、乙相遇时离出发的时间是525÷（80－75）=105（分钟）.两村间的距离是：（100＋80）×[（100＋75）×3÷（80-75]＝180×（525÷5）＝18×105=18900（米）[趣味数学]皮皮和琪琪乘车从城里到郊区去，琪琪对皮皮说：“我发觉每隔5分钟就有1辆迎面开来的客车和我们擦肩而过，如果两面对来的客车速度一样，在1小时有多少辆客车开到城里？”“那还用说，当然是12辆了，因为60除以5等于12.”皮皮说.但是琪琪不同意他的解答，认为是6辆.你知道他们谁正确吗？分析：当然是琪琪正确.假设皮皮他们所乘的客车从与第一辆对开的客车相遇A点与到第二辆客车相遇B 点相隔5分钟，那么第二辆对开的客车要从B点达到A点好需要5分钟，也就是两辆对开的客车之间的时间间隔为10分钟，60÷10=6（辆）（二）直线型的追及问题【例4】军事演习中，“我”海军英雄舰追击“敌”军舰，追到A岛时，“敌”舰已在10分钟前逃离，“敌”舰每分钟行驶1000米，“我”海军英雄舰每分钟行驶1470米，在距离“敌”舰600米处可开炮射击，问“我”海军英雄舰从A岛出发经过多少分钟可射击敌舰？分析：“我”舰追到A岛时，“敌”舰已逃离10分钟了，因此，在A岛时，“我”舰与“敌”舰的距离为10000米（=1000×10）.又因为“我”舰在距离“敌”舰600米处即可开炮射击，即“我”舰只要追上“敌”舰9400（=10000米-600米）即可开炮射击.所以，在这个问题中，不妨把9400当作路程差，根据公式求得追及时间.即（1000×10-600）÷（1470-1000）=（10000-600）÷470=9400÷470=20（分钟），所以，经过20分钟可开炮射击“敌”舰.[前铺]下午放学时，弟弟以每分钟40米的速度步行回家.5分钟后，哥哥以每分钟60米的速度也从学校步行回家，哥哥出发后，经过几分钟可以追上弟弟？（假定从学校到家有足够远，即哥哥追上弟弟时，仍没有回到家）.分析：若经过5分钟，弟弟已到了A地，此时弟弟已走了40×5=200（米）；哥哥每分钟比弟弟多走20米，几分钟可以追上这200米呢？40×5÷（60-40）=200÷20=10（分钟）【例5】上午8点8分，小明骑自行车从家里出发，8分钟后，爸爸骑摩托车去追他，在离家4千米的地方追上了他.然后爸爸立即回家，到家后又立刻回头去追小明，再追上小明的时候，离家恰好是8千米，这时是几点几分？分析：画一张简单的示意图：图上可以看出，从爸爸第一次追上到第二次追上，小明走了8-4＝4（千米）.而爸爸骑的距离是4＋ 8＝ 12（千米）.这就知道，爸爸骑摩托车的速度是小明骑自行车速度的 12÷4＝3（倍）.按照这个倍数计算，小明骑8千米，爸爸可以骑行8×3＝24（千米）.但事实上，爸爸少用了8分钟，骑行了4＋12＝16（千米）. 少骑行24-16＝8（千米）.摩托车的速度是8÷8=1（千米/分），爸爸骑行16千米需要16分钟.8＋8＋16＝32.所以这时是8点32分.[前铺]小明步行上学，每分钟行70米.离家12分钟后，爸爸发现小明的文具盒忘在家中，爸爸带着文具盒，立即骑自行车以每分钟280米的速度去追小明.问爸爸出发几分钟后追上小明？分析：爸爸要追及的路程：70×12=840（米）,爸爸与小明的速度差：280-70=210（米/分）,爸爸追及的时间：840÷210=4（分钟）.【例6】小红和小蓝练习跑步，若小红让小蓝先跑20米，则小红跑5秒钟就可追上小蓝；若小红让小蓝先跑4秒钟，则小红跑6秒钟就能追上小蓝.小红、小蓝二人的速度各是多少？分析：小红让小蓝先跑20米，则20米就是小红、小蓝二人的路程差，小红跑5秒钟追上小蓝，5秒就是追及时间，据此可求出他们的速度差为20÷5=4（米/秒）；若小红让小蓝先跑4秒，则小红6秒可追上小蓝，在这个过程中，追及时间为6秒，根据上一个条件，由追及差和追及时间可求出在这个过程中的路程差，这个路程差即是小蓝4秒钟所行的路程，路程差就等于4×6=24（米），也即小蓝在4秒内跑了24米，所以可求出小蓝的速度，也可求出小红的速度.综合列式计算如下：小蓝的速度为：20÷5×6÷4=6（米/秒），小红的速度为：6+4=10（米/秒）【例7】张、李、赵三人都从甲地到乙地．上午6时，张、李两人一起从甲地出发，张每小时走5千米，李每小时走4千米；赵上午8时从甲地出发．傍晚6时，赵、张同时到达乙地．那么赵追上李的时间是几时?分析：赵追上李是追及问题，但是赵的速度我们并不清楚，这需要从赵、张同时到乙地来计算．本题的解题过程分三步．第一步：求出甲、乙两地距离．张早上6时出发，晚上6时到，用12小时，每小时5千米，所以甲、乙两地相距5×12=60千米．第二步：求出赵的速度．赵早上8时出发，晚上6时到，用10小时，走了60千米，每小时走60÷10=6千米．第三步：追及问题．赵出发时，李已出发2小时，此时与甲地相距4×2=8千米，赵追上李用8÷(6-4)=8÷2=4小时．所以，赵追上李是上午12时．评注：本题需要逆向思维，根据所需从题目条件中找，分析思考的过程可以说正好与详解的顺序相反，按我们的需要一步步找上去，直到题目满足我们的要求为止．（三）环形上的相遇与追及问题【例8】在300米的环形跑道上，田奇和王强同学同时同地起跑，如果同向而跑2分30秒相遇，如果背向而跑则半分钟相遇，求两人的速度各是多少？分析：同向而跑，2分30秒相遇，这实质是快的追上慢的.起跑后，由于两人速度的差异，造成两人路程上的差异，随着时间的增长，两人间的距离不断拉大，到两人相距环形跑道的半圈时，相距最大.接着，两人的距离又逐渐缩小，直到快的追上慢的，此时快的比慢的多跑了一圈.这就是所谓的追及问题，数量关系为：路程差÷速度差=追及时间，由题意，得知路程差为300米，追及时间为2分30秒，即150秒，因此两人速度差为300÷150＝2(米)背向而跑即所谓的相遇问题，数量关系为：路程和÷速度和=相遇时间，由题意，可以求得两人的速度和为300÷30＝10(米)有了两人的速度和与速度差，即可求得两人的速度：慢者：(10- 2)÷2＝4(米)，快者：10-4＝6(米)[巩固]小张和小王各以一定速度，在周长为500米的环形跑道上跑步.小王的速度是200米/分.（1）小张和小王同时从同一地点出发，反向跑步，1分钟后两人第一次相遇，小张的速度是多少米/分？（2）小张和小王同时从同一点出发，同一方向跑步，小张跑多少圈后才能第一次追上小王？分析：（1）两人相遇，也就是合起来跑了一个周长的行程.小张的速度是500÷1-200=300（米/分）.（2）在环形的跑道上，小张要追上小王，就是小张比小王多跑一圈（一个周长），因此需要的时间是：500÷（300-200）＝5（分）.300×5÷500＝3（圈）.【例9】如图，A、B是圆的直径的两端，小张在A点，小王在B点同时出发反向行走，他们在C点第一次相遇，C离A点80米；在D点第二次相遇，D点离B点6O米.求这个圆的周长.分析：第一次相遇，两人合起来走了半个周长；第二次相遇，两个人合起来又走了一圈.从出发开始算，两个人合起来走了一周半.因此，第二次相遇时两人合起来所走的行程是第一次相遇时合起来所走的行程的3倍，那么从A到D的距离，应该是从A到C距离的3倍，即A到D是80×3＝240（米）.240-60=180（米）.180×2＝360（米）.[拓展]一个圆周长90厘米，3个点把这个圆周分成三等分，3只爬虫A，B，C分别在这3个点上.它们同时出发，按顺时针方向沿着圆周爬行.A的速度是10厘米/秒，B的速度是5厘米/秒，C的速度是3厘米/秒，3只爬虫出发后多少时间第一次到达同一位置？分析：先考虑B与C这两只爬虫，什么时候能到达同一位置.开始时，它们相差30厘米，每秒钟B能追上C（5-3）厘米.30÷（5-3）＝15（秒）.因此15秒后B与C到达同一位置.以后再要到达同一位置，B要追上C一圈，也就是追上90厘米，需要90÷（5-3）＝45（秒）.B与C到达同一位置，出发后的秒数是15，，105，150，195，……再看看A与B什么时候到达同一位置.第一次是出发后30÷（10-5）=6（秒），以后再要到达同一位置是A追上B一圈.需要90÷（10-5）＝18（秒），A与B到达同一位置，出发后的秒数是6，24，42，，78，96，…对照两行列出的秒数，就知道出发后60秒3只爬虫到达同一位置.【例10】实验小学有一条200米长的环形跑道，冬冬和晶晶同时从起跑线起跑，冬冬每秒钟跑6米，晶晶每秒钟跑4米，问冬冬第一次追上晶晶时两人各跑了多少米，第2次追上晶晶时两人各跑了多少圈？分析：这是一道封闭路线上的追及问题，冬冬与晶晶两人同时同地起跑，方向一致.因此，当冬冬第一次追上晶晶时，他比晶晶多跑的路程恰是环形跑道的一个周长（200米），又知道了冬冬和晶晶的速度，于是，根据追及问题的基本关系就可求出追及时间以及他们各自所走的路程.（1）冬冬第一次追上晶晶所需要的时间：200÷（6-4）=100（秒）（2）冬冬第一次追上晶晶时他所跑的路程应为：6×100=600（米）（3）晶晶第一次被追上时所跑的路程：4×100=400（米）（4）冬冬第二次追上晶晶时所跑的圈数：（600×2）÷200=6（圈）（5）晶晶第2次被追上时所跑的圈数：（400×2）÷200=4（圈）专题展望本讲主要讲了行程问题中的相遇与追及问题，在四年级的寒假班我们会继续学习更复杂的行程问题，希望同学们再接再厉，加油！练习十1.（例1）大头儿子的家距离学校3000米，小头爸爸从家去学校接大头儿子放学，大头儿子从学校回家，他们同时出发，小头爸爸每分钟比大头儿子多走24米，50分钟后两人相遇，那么大头儿子的速度是每分钟走多少米？分析：大头儿子和小头爸爸的速度和：3000÷50=60(米／分钟)，小头爸爸的速度：(60+24)÷2=42(米／分钟)，大头儿子的速度：60—42=18(米／分钟).2. （例4）龟兔赛跑同时出发，全程7000米，乌龟以每分30米的速度爬行，兔子每分钟跑330米．兔子跑了10分钟就停下来睡觉，200分钟后醒来，立即以原速往前跑．当兔子追上乌龟时，他们离终点的距离是多少千米?分析：线段图如下：兔子追乌龟的追及路程差为：30×(10+200)-330×10=3000(米)，根据公式t v S 差差 兔子追上乌龟的追及时间为：3000÷(330-30)=10(分)，离终点的距离为：7000-330×(10+10)=400(米).3. （例6）东东、西西二人练习跑步，若东东让西西先跑10米，则东东跑5秒钟可追上西西；若东东让西西先跑2秒钟，则东东跑4秒钟就能追上西西.问：东东、西西二人的速度各是多少？分析 若东东让西西先跑10米，则10米就是东东、西西二人的路程差，5秒就是追及时间，据此可求出他们的速度差为10÷5=2（米/秒）；若东东让西西先跑2秒，则东东跑4秒可追上西西，在这个过程中，追及时间为4秒，因此路程差就等于2×4=8（米），也即西西在2秒内跑了8米，所以可求出西西的速度，也可求出东东的速度.综合列式计算如下：西西的速度为：10÷5×4÷2=4（米/秒），东东的速度为：10÷5+4=6（米/秒）4. （例9）如图，有一个圆，两只小虫分别从直径的两端A 与C 同时出发，绕圆周相向而行.它们第一次相遇在离A 点8厘米处的B 点，第二次相遇在离C 点处6厘米的D 点，问，这个圆周的长是多少?分析：如上图所示，第一次相遇，两只小虫共爬行了半个圆周，其中从A 点出发的小虫爬了8厘米，第二次相遇，两只小虫从出发共爬行了1个半圆周，其中从A 点出发的应爬行8×3=24(厘米)，比半个圆周多6厘米，半个圆周长为8×3—6=18(厘米)，一个圆周长就是：(8×3—6)×2=36(厘米)5. （例10）小新和正南在操场上比赛跑步，小新每分钟跑250米，正南每分钟跑210米，一圈跑道长800米，他们同时从起跑点出发，那么小新第一次超过正南需要多少分钟？第三次超过正南需要多少分钟？6cm 8cm 第二次相遇第一次相遇B DC A分析：小新第一次超过正南是比正南多跑了一圈，根据t v S 差差 ，可知小新第一次超过正南需要：800÷（250-210）=20（分钟），第三次超过正南是比正南多跑了三圈，需要800×3÷（250-210）=60分钟.让偷盗者赛跑“你是抢劫犯!” “你才是抢劫犯呢!”“是怎么回事呀?”警察车良见前面有争论的，就对身边的田大凯说，“走!我们看看去.”当他们俩来到两个争论的人面前，他们仍争论得很厉害，无法区分是抢劫者.“这是怎么回事呀?”车良问.这时，一帮人簇拥着一个老太太过来，一个说：“还是你自己把事情告诉大家吧!” 原来，黄昏时分，老太太提着一个提包从一个胡同出来，突然窜出一个强盗，二话没说，把老太太的提包夺过来就跑.然后又有一个人马上追上去抓住了强盗.老太太也没有看清楚那个人是个什么样子.面对两个人同时出现她也说不清是哪个.“把他们送到警察局去处理吧!”有人提议. “到那里也没有法说清楚呀!，’有人说，“公说公有理，婆说婆有理.这怎么能断得清呀?”车良对吵吵嚷嚷的人大声说：“大家安静下来，这两个人就不用送警察局了.现在，我们就可把抢劫者定下来!到时候，我们会把他带走的!” “你们怎么定啊?”有人不理解地问. “这可不是开玩笑的呀!”有人在提醒. “请大家放心.”车良说，“我们是有办法的.我们让他们来一个百米冲刺，跑一跑就数学故事可以定下来.”车良对大凯说：“大凯，你到前面大约100米的地方计时，我到时候打一枪，你听到枪声就马上计时，把他们的百米跑成绩记下来.”然后，车良大声说：“大家听好，你们两个人也听好.”他用手指指了那两个“抢劫者”，“现在你们这两个人进行百米赛跑，通过赛跑我就可以断定哪一个是抢劫者.谁跑得快，谁就是好人!大家闪一闪.现在赛跑开始!预备!”“叭!”一声枪响，那两个人拼命地向前跑了起来.不一会儿，大凯押着那两个人来到了大家面前. “那个年轻人百米速度为12秒；那个老一点的人百米速度为15秒.”大凯指着他们说.车良对那个年轻人说：“谢谢你!你是一个正直的人.”田大凯对那个年老的人说：“你就是抢劫者，要受到应有的惩罚.走! 跟我们到警察局去!”说完，车良和田大凯押着那个人就走.“哎!奇怪，怎么跑一跑就能断出哪个是抢劫者呢?”人们议论着. “真是不可思议呀!”你明白这个道理了吗?这是因为当老太太的提包被抢了之后，过了一会儿才有一个人追赶，后来被追上的那人却说对方是强盗.从这里不难看出破绽，追抢劫者的人肯定要比抢劫者跑得快，否则后者不会追上前者.所以，让两个人进行百米赛跑测一下速度就行，就可以判断出谁是抢劫者.。

第5页做一做答案4是24的因数，24是4的倍数。

13是26的因数，26是13的倍数。

25是75的因数，75是25的倍数。

9是81的因数，81是9的倍数。

练习二答案1、36的因数：1,2,3,4,6,9,12,18,36。

60的因数：1，2，3，4，5，6，10, 12, 15,20,30,60。

2、(1)10的因数：1,2,5,10。

17的因数：1,17。

28的因数：1,2,4,7,14,28。

32的因数：1,2,4,8,16,32。

48的因数：1，2，3，4，6，8，12, 16,24,48。

(2)（答案不唯一）4的倍数：4，8，12，16,20。

7的倍数：7,14,21,28,35。

10的倍数：10,20, 30,40,50。

6的倍数：6,12,18,24, 30。

9的倍数：9,18, 27, 36,45。

3、把5,35,10,55,60,100这6颗星星涂上黄色。

4、15的因数有1，3，5,15。

15是1,3,5,15的倍数。

5、(1)√(2)×(3)√(4)×6、1 2 47、(1)18 (2)1 (3)428、这个数可能是3，6，21,42。

思考题14和21的和是7的倍数；18和27的和是9的倍数。

发现：两个数分别是一个数的倍数，这两个数的和也是这个数的倍数。

第9页做一做答案2的倍数有24，90，106，60，130，280，6018，8100。

5的倍数有35，90，15，60，75，130，280,8100。

既是2的倍数，又是5的倍数：90，60，130,280,8100。

发现：既是2的倍数，又是5的倍数的数的个位一定是0。

第10页做一做答案3的倍数有24，96。

在24后面可放卡片：0，3，6，9。

在58后面可放卡片：2，5，8。

在46后面可放卡片：2，5，8。

在96后面可放卡片：0，3，6，9。

练习三答案1、奇数有33，355，123，881，8089，565,677。

信息的获取-1 BCBBB信息的获取-2 DDCAB信息的获取-3 DBABC信息的获取-4 CBBC实体店购买与网购，实店买可翻阅价格高，网购不能翻阅价格便宜。

信息的获取-5 AD,DAB,登陆百度网，搜索南京，景点，路线，住宿等信息。

信息与信息技术-1 CACDB信息与信息技术-2 DDCDB信息与信息技术-3 BDCAC信息与信息技术-4 CDDBB网络信息检索4-1 ADBDB网络信息检索4-2 BACCB网络信息检索4-3 BBBAD网络信息检索4-4 1.半人马座比邻星2.茉莉花3.mào dié八九十岁的意思。

4.齐白石的蛙声十里出山泉网络信息检索4-5 1.搜索引擎最早是yayoo，后来发展壮大，搜索引擎的工作方式和在图书馆里面进行图书查找的工作差不多2.如果可以直接下载，那么右键菜单选择目标另存为，如果不行那么可以选中链接之后采用下载软件下载。

3.例如：优化查找的速度、扩大资料的范围等。

信息的评价5-1 ABAAAA信息的评价5-2 ACDCAB信息的评价5-3 BACCD信息的评价5-4 ABACD(AC多选）信息的评价5-5 CDBC(ACD多选）信息的编程加工6-1 DADDC信息的编程加工6-2 ADCCA信息的编程加工6-3 CBCCC信息的编程加工6-4 CDB信息的编程加工6-5 C, s=a*a，FBACED信息的智能化加工7-1 CACCB信息的智能化加工7-2 CBCDC信息的智能化加工7-3 ADCBC信息的智能化加工7-4 A,D,D ，BD,ACEF,G信息的智能化加工7-5 C, BEF, ACDFG, ACDGH, 计算机，手机，电话等信息的编码8-1 CACDD信息的编码8-2 BCAD 填空题略信息的编码8-3 BBCC文本信息加工9-1 BAACC文本信息加工9-2 CDABC文本信息加工9-3 DCD文本信息加工9-4 BAD文本信息加工9-5 1，复制、字体、字号、艺术字、页面设置、背景、选择性粘贴、样式、图表、目录、三级标题等2，图形化表达带来直观、生动，纯文本枯燥、不形像等。

把1、2、3、4、5、6、7、8、9九个数填在如图的三角形三边上，使每边上的四个数的和都为17（18、19、20、21） 1， 把1、2、3、4、5、6、7、8、9九个数填在如图的三角形三边上，使每边上的四个数的和都为17. 解：设填为如图。

则有 {45171717654321654321=++++++++=+++=+++=+++a a a a a a z y x x a a z z a a y y a a xx+y+z=51-45=6.在1、2、3、4、5、6、7、8、9九个数中的三数之和为6的只有一组。

1+2+3＝6. a 1+a 2=17-3=14; a 3+a 4=17-5=12; a 5+a 6=17-4=13;在余二的4、5、6、7、8、9六个数中二数和为14的有两组5+9＝6+8＝14. 第一、5+9＝14 4+8＝12 6+7＝13. 可填得，下面八个。

第二、6+8＝14. 5+7＝12 4+9＝13. 可填得下面的八个符合条件的共有十六个。

2， 把1、2、3、4、5、6、7、8、9九个数填在如图的三角形三边上，使每边上的四个数的和都为18. 解：设填为如图。

则有a6a 5a 4a 3a 2a 1zyx 456789876532198765432197543219864321123468912345791234567891235678987654a 6a 5a 4a 3a 2a 1zyx 456789876532198765432197543219864321123468912345791234567891235678987654a 6a 5a 4a 3a 2a 1zyx{45181818654321654321=++++++++=+++=+++=+++a a a a a a z y x x a a z z a a y y a a xx+y+z=54-45=9.在1、2、3、4、5、6、7、8、9九个数中的三数之和为9的有两组。

本田故障码查询10-1 IAT 电路输入过低10-2 IAT （进气温度）传感器电电路电压过高10-3 IAT （进气温度）传感器性能10-4 IAT 传感器1电路范围/性能问题10-5 IAT （进气温度）传感器2电电路电压过低10-6 IAT （进气温度）传感器2电电路电压过低10-8 IAT 传感器2电路范围/性能问题11-1 IMA 电路电压过低11-2 IMA 电路电压过高12-4 EGR 控制电路范围/性能12-2 EGR （废气再循环）升程传感器电压过高12-3 EGR （废气再循环）阀升程不够12-5 EGR （废气再循环）升程传感器电压过高12-6 EGR （废气再循环）系统性能12-7 EGR （废气再循环）系统性能12-8 EGR （废气再循环）系统电路12-10 EGR 传感器电压过低12-11 EGR 传感器电路故障12-12 EGR 传感器电路电压过高12-13 EGR 控制电路范围/性能12-14 EGR 控制电路范围/性能12-15 EGR （废气再循环）阀电路电压过低12-16 EGR （废气再循环）阀电路电压过高12-17 EGR （废气再循环）控制电路故障13-1 BARO 传感器电压过低13-2 BARO 传感器电压过高13-3 BARO （大气压力）传感器性能13-5 BARO （大气压力）传感器电压过低13-6 BARO （大气压力）传感器电路电压过高13-7 BARO （大气压力）传感器电路范围/性能问题13-9 BARO 传感器电路高出最大值14-1 IAC 阀电路故障14-2 怠速空气控制系统14-3 IACV 电路故障14-4 IACV 电路故障14-5 IACV 电路故障14-6 IACV 电路故障14-7 IACV 电路故障14-8 IACV 电路故障14-10 怠速空气控制系统低于期望值14-11 怠速进气控制系统高于期望值14-12 怠速进气控制系统故障15-1 点火输出控制故障15-2 点火脉冲问题15-5 点火输出信号故障15-6 点火输出信号故障15-7 点火输出信号故障15-8 点火输出信号故障15-9 点火电路故障15-10 点火电路故障15-11 进气侧点火线圈电路故障15-12 排气侧点火线圈电路故障16-1 喷油器控制故障17-1 VSS（车速信号）无脉冲17-2 VSS无脉冲或噪声17-6 VSS（车速信号）无脉冲17-7 车速传感器B17-8 车速传感器A无信号18-1 IG调节器电压过低18-2 IG调节器电压过高19-1 A/T（自动变速箱）锁定电磁阀A故障19-2 A/T（自动变速箱）锁定电磁阀B故障19-3 液力变扭器离合器故障20-1 ELD 电压过低20-2 ELD 电压过高21-1 VTEC 电磁阀B2故障21-2 VTEC 电磁阀电路电压过高21-3 VTEC 电磁阀电路电压过低21-4 VTEC 电磁阀电路故障21-5 气缸随需响应VTEC 电磁阀电路电压过低21-6 气缸随需响应VTEC 电磁阀电路电压过高21-7 可变气缸管理电磁阀电路电压过低21-8 可变气缸管理电磁阀电路电压过高22-1 VTEC 压力开关故障22-2 VTEC 机油压力开关电路电压过高22-3 VTEC 机油压力开关电路电压过低22-4 VTEC 系统故障22-5 机油压力传感器电压过低22-6 机油压力传感器电压过高22-7 机油压力传感器性能22-9 VTEC 机油压力开关电路故障22-10 VTEC 油压B2开关电路故障23-1 爆震传感器1无信号23-2 爆震传感器电路故障23-3 爆震传感器1电压过低23-4 爆震传感器1电压过高30-1 AFSA电压过低30-2 AFSA电压过高30-3 A/T（自动变速箱）FI数据线路故障30-5 TAC （节气门自动控制）模块通信30-6 与TCM （变速箱控制模块）失去通信31-1 AFSB电压过低31-2 AFSB电压过高31-3 ATSDLB 无信号31-4 ATSDLB 信号失效34-1 PCM 备用电压过低34-2 ECM/PCM 电源电路意外电压34-3 VBU电压过低34-4 VBU电压过高34-8 充电系统电压过低34-9 充电系统电压过高35-1 TCSTB 线路故障35-2 FPTDR 信号输入过高35-3 TCSDL （牵引控制串行数据链）（FPTDR ）信号失效35-4 TCSDL （牵引控制串行数据链）信号失效36-1 TCSTB 信号输入过低36-2 TCSTB 信号输入过高36-4 IMA （集成电机辅助）MOTSTB 无信号37-1 APP（油门踏板）传感器1故障37-2 APP（油门踏板）传感器2故障37-3 AP传感器1-2 相关性37-10 APP传感器A电压过低37-11 APP传感器A电压过高37-12 APP传感器B电压过低37-13 APP传感器B电压过高37-14 APP传感器A-B 相关性37-15 APP传感器A/B（TP传感器D/E ）电压相关性38-2 VTEC 电磁阀2电路故障38-1 VTEC 电磁阀2故障39-1 串行通信故障40-1 TVC 电机指令1故障40-2 TVC 电机指令2故障40-3 节气门位置偏离40-4 节气门关闭位置不对40-5 节气门作动器控制模块性能40-6 节气门作动器系统故障40-7 油门执行器控制系统未学习到40-8 节气门默认弹簧故障40-9 节气门作动器控制模块性能40-10 节气门作动器控制模块继电器故障40-11 节气门回位弹簧故障40-12 节气门执行器控制范围/性能40-13 节气门执行器控制范围/性能40-15 节气门执行器电流范围/性能40-16 电子节气门控制系统（ETCS ）学习偏差40-17 节气门默认位置故障40-18 节气门默认位置故障40-19 TACM 继电器卡在打开位状态40-20 TACM 继电器卡在关闭位状态41-1 HO2S（B2）（S1）加热器故障41-2 HO2S（S1）加热器故障41-3 HO2S S1 加热器故障41-4 HO2S S1 加热器故障41-5 A/F传感器S1加热器故障41-6 A/F传感器S1加热器故障41-7 HO2S S1 电路不工作41-10 HO2S S1 加热器故障42-1 HO2S B1 加热器故障42-2 HO2S S1 加热器故障43-1 燃油供油过浓43-2 燃油供油过稀45-1 燃油测量过浓45-2 燃油测量过稀46-1 燃油系统过浓（右列（气缸列1））46-2 燃油系统过稀（右列（气缸列1））48-1 A/F传感器（S1）电压过低48-2 A/F传感器（S1）电压过高48-3 HO2S（A/F 传感器）（S1）故障48-5 A/F传感器S1 AFS 线路电压过高48-6 HO2S（S1）故障48-7 HO2S（S1）故障48-8 A/F传感器S1信号保持稀48-9 HO2S（A/F 传感器）S1 VS 线路电压过低48-10 HO2S（A/F 传感器）S1 VCENT 线路电压过低48-11 A/F传感器S1 IP 线路电压过低48-12 A/F传感器S1 VS 线路电压过高48-13 A/F传感器S1 VCENT 线路电压过高48-14 A/F传感器S1 IP 线路电压过高48-15 A/F传感器S1标记电路电压过低48-16 A/F传感器S1标记电路电压过高48-17 A/F传感器S1 AFS 负极线路电压过低48-18 A/F传感器S1 AFS 正极线路电压过低49-1 BBP （制动升压）传感器电压过低49-2 BBP （制动升压）传感器电压过高49-3 BBP 传感器性能49-5 BBP （制动升压）传感器电压过低49-6 BBP （制动升压）传感器电压过高50-1 MAF （质量空气流量）传感器电压过低50-2 MAF 传感器电压过高50-3 MAF 传感器故障50-5 MAF 低于期望值50-6 MAF 高于期望值51-1 VTEC 电磁阀B1故障52-1 VTEC 压力开关B1故障52-4 VTEC 系统B1故障52-10 VTEC 机油压力开关电路故障53-1 爆震传感器2无信号53-3 爆震传感器2电压过低53-4 爆震传感器2电压过高54-1 CKP（CKF ）传感器无信号54-2 CKP（CKF ）传感器间歇中断54-3 CYP （曲轴位置）传感器B无脉冲54-4 CKP 传感器B噪声54-6 CKP 传感器B间断55-1 EGT 传感器电压过低55-2 EGT （排气温度）传感器电压过高56-1 CMP （凸轮轴位置传感器）系统作动器（VTC 作动器）电路故障56-2 CMP （凸轮轴位置传感器）系统（VTC 控制）性能56-3 VTC系统保持在提前的位置57-1 CMP （凸轮轴位置传感器）传感器A无脉冲57-2 CMP （凸轮轴位置传感器）传感器噪声57-3 CMP （凸轮轴位置传感器）传感器（VTC 控制）性能58-1 TDC （上止点）传感器2无脉冲58-2 TDC 传感器2噪声58-5 CMP （凸轮轴位置传感器）传感器B故障58-6 CMP （凸轮轴位置）传感器B范围/性能58-7 CMP （凸轮轴位置传感器）传感器B无脉冲59-1 CMP （凸轮轴位置）传感器B无信号59-2 CMP （凸轮轴位置）传感器B间歇性中断60-1 空气泵系统故障60-2 空气泵电机故障60-3 空气控制系统故障60-4 二次空气喷射管故障60-5 空气泵电机电压过低60-6 空气泵电机电压过高60-7 空气泵电机电路故障60-8 空气泵电机电路故障60-9 空气泵故障61-1 HO2S（A/F 传感器）S1反应迟缓61-2 HO2S（A/F 传感器）S1反应迟缓61-3 HO2S（A/F 传感器）S1怠速范围问题61-4 HO2S（A/F 传感器）S1稀薄范围问题61-5 HO2S（A/F 传感器）S1稀油灵敏度61-7 HO2S（A/F 传感器）S1电路不工作61-8 HO2S S1 故障61-9 A/F传感器B1S1反应迟缓61-10 A/F传感器B1S1性能61-12 A/F传感器电路超出范围过高62-1 HO2S B1 S1 反应迟缓63-1 HO2S S2 电压过低63-2 HO2S S2 电压过高63-3 HO2S S2 反应迟缓63-4 HO2S S2 电路故障63-7 HO2S S2 电路不工作63-8 HO2S S2 电路63-11 HO2S S2 信号保持过稀63-12 HO2S S2 信号保持过浓63-13 HO2S S2 电路电压过低63-14 HO2S S2 电路电压过高63-15 HO2S S2 电路不工作63-16 HO2S S2 电路信号保持稀63-17 HO2S S2 电路信号保持浓63-18 HO2S S2 电路电压过低63-19 HO2S S2 电路电压过高63-20 HO2S S2 电路反应迟缓64-1 HO2S B1 S2 电压过低64-2 HO2S B1 S2 电压过高64-3 HO2S B1 S2 反应迟缓65-1 HO2S S2 加热器故障65-2 HO2S S2 加热器故障65-3 HO2S S2 加热器故障65-10 HO2S S2 加热器故障66-2 HO2S B1 S2 加热器故障67-1 催化转化器系统效率低于极限值68-1 触媒B1性能变坏69-5 IMA （集成电机辅助）系统故障69-1 IMA MOTFSA 信号失效69-2 IMA （集成电机辅助）MOTFSB 信号失效69-3 IMA （集成电机辅助）系统故障69-4 IMA （集成电机辅助）系统故障70-1 ATCHK 线路信号故障70-2 自动横置驱动桥70-3 A/T（自动变速箱）系统故障71-10 1号气缸进气侧点火线圈电路故障71-11 1号气缸排气侧点火线圈电路故障71-13 1号气缸点火线圈电路故障71-1 1号气缸缺火71-2 1号气缸IG缺火71-3 1号气缸缺火燃油71-5 1号喷油器电路断路71-6 1号喷油器电路短路71-7 1号喷油器电路短路72-1 2号气缸缺火72-2 2号气缸IG缺火72-3 2号气缸缺火燃油72-5 2号喷油器电路断路72-6 2号喷油器电路短路72-7 2号喷油器电路短路72-10 2缸进气侧点火线圈电路故障72-11 2号气缸排气侧点火线圈电路故障72-13 2号气缸点火线圈电路故障73-1 3号气缸缺火73-2 3号气缸IG缺火73-3 3号气缸缺火燃油73-5 3号喷油器电路断路73-6 3号喷油器电路短路73-7 3号喷油器电路短路73-10 3缸进气侧点火线圈电路故障73-11 3号缸排气侧点火线圈电路故障73-13 3号气缸点火线圈电路故障74-1 4号气缸缺火74-2 4号气缸IG缺火74-3 4号气缸缺火燃油74-5 4号喷油器电路断路74-6 4号喷油器电路短路74-7 4号喷油器电路短路74-10 4缸前点火电路故障74-11 4缸后点火电路故障74-13 4号气缸点火线圈电路故障75-1 5号气缸缺火75-2 5号气缸IG缺火75-3 5号气缸缺火燃油75-4 无规则缺火75-5 5号喷油器电路断路76-1 6号气缸缺火76-2 6号气缸IG缺火76-3 6号气缸缺火燃油76-4 无规则缺火76-5 6号喷油器电路断路77-4 无规则缺火77-9 后部列侧缺火79-1 火花塞电压检测电路故障（左侧气缸列（气缸列2））79-2 火花塞电压检测电路故障（右侧气缸列（气缸列1））79-3 火花塞电压检测模块复位电路故障（左侧气缸列（气缸列2））79-4 火花塞电压检测模块复位电路故障（右侧气缸列（气缸列1））80-1 EGR （废气再循环）检测气流量不足80-5 废气再循环（EGR ）流量不足83-1 电流传感器电压过低83-2 电流传感器电压过高84-1 空气控制电磁阀电压过高84-2 空气控制电磁阀电压过低86-1 ECT 传感器1范围/性能86-2 ECT 传感器1范围/性能86-4 ECT （发动机冷却液温度）传感器性能问题86-5 ECT 传感器范围/性能故障86-6 闭环燃油控制冷却液温度不足87-1 冷却系统故障87-2 节温器范围/性能88-1 IMA 实际扭矩输入过低88-2 IMA 实际扭矩输入过高88-3 IMA （集成电机辅助）QBATT 输入过低88-4 IMA （集成电机辅助）QBATT 输入过高88-5 燃油喷射/电机信号线故障88-6 与驱动电机控制模块失去通信89-1 A/T齿轮多档89-2 A/T齿轮无档90-1 EV AP （蒸发排放控制系统）泄漏（油箱系统）90-2 EV AP （蒸发排放控制系统）泄漏（碳罐部分）90-3 EV AP （蒸发排放控制系统）净化流量小90-4 EV AP （蒸发排放控制系统）泄漏（系统故障）90-5 EV AP 系统少量泄漏90-6 EV AP 系统很少量泄漏90-7 检测到EV AP 系统净化流量小90-8 EV AP 泄漏（燃油加注盖松脱）90-9 EV AP 系统大量泄漏90-10 EV AP 系统很少量泄漏91-1 EV AP FTP 传感器输入电压过低91-2 EV AP （蒸发排放控制系统）FTP传感器输入电压过高91-3 EV AP （蒸发排放控制系统）FTP传感器性能91-5 EV AP （蒸发排放控制系统）FTP传感器性能92-2 EV AP （蒸发排放控制系统）净化流量错误92-4 EV AP 碳罐净化阀电路故障92-5 EV AP PCS（EV AP 净化控制系统）（CPV ）电压过低92-6 EV AP PCS（EV AP 净化控制系统）（CPV ）电压过高92-7 EV AP PCS（EV AP 净化控制系统）（CPV ）净化流量过高92-8 92-1 EV AP （蒸发排放控制系统）净化流量错误94-1 A/C加热器信号输入过低94-2 A/C加热器信号输入过高95-1 燃油油管压力传感器电路电压过低95-2 燃油油管压力传感器电路电压过高95-3 燃油油管压力传感器电路范围/性能问题95-11 FP传感器电压过高95-10 FP传感器电压过低96-1 FT传感器电压过低96-2 FT传感器电压过高96-3 燃油温度传感器电路电压过低96-4 燃油温度传感器电路电压过高97-1 油箱压力（FTP）传感器电路电压过低97-2 油箱压力（FTP）传感器电路电压过高98-1 油箱温度（FTT）传感器电路电压过低98-2 油箱温度（FTT）传感器电路电压过高。

2 4 6 球测红法2 4 6 球测红法就是将本期中奖红号的第６位分别与第１位及第４位相减，其差数，数差个位的替数、补数、减数（也含它们的邻数及同尾数）视为下期中奖红号的预测号。

替数就是隔５期，如：１与６，２与７、３与８、４与９、５与０。

["jun.he 十位密码对应法"的学术叫法]补数，就是和１０数，如：１与９、２与８３与７、４与６。

减数就是和５数，如：１与４、２与３、６与９、７与８。

邻数，就本身±数，如５的邻数６与４。

同尾数，个位相同数，如：８与１８、２８。

实战如下：例一："双色球"０４００８期中奖红号为：０１、０７、１０、２２、３２、３３。

０４００８期中奖红号第６位是３３，第１位是０１，第４位是２２。

第６位与第１位差数为：３３－０１＝３２，取２。

第６位与第４位差数为：３３－２２＝１１，取１。

２的同尾数２２，２的邻数为１，取０１。

１的替数为６，取同尾数１６。

１的邻数为０，取同尾数１０，１的减数为４，取同尾数２４。

１的补数为９，取０９。

２２、０１、１６、１０、２４、０９为下期中奖红号的预测号。

０４００９期中奖红号是０１、０９、１０、１６、２２、２４（全部测中）例二："双色球"０４０８３期中奖红号为：１４、１６、２７、２８、３０、３３，第６位是３３，第１位是１４，第４位是２８。

第６位红号与第１位红号差数为３３－１４＝１９，取个位９。

第６位红号与第４位红号差数为３３－２８＝０５，取个位５，９的邻数为８，取０８。

９的替数为４，取０４，９的补数为１，取０１与同尾数１１、２１。

５的同尾数取２５。

０８、０４、０１、１１、２１、２５为下期中奖红号的预测号。

０４０８４期中奖红号是０１、０４、０８、１１、２１、２５（全部测中）。

例三："双色球"０４０９１期中奖红号为：０９、１３、１４、２１、３０、３３。

０４０９１期中奖红号是第６位是３３。

4、4、5规律公式
4、4、5规律是一个数列规律，其公式可以表示为An = 3n + 1，其中n为正整数，An表示数列中的第n个数。

这个规律的数列依次
为4，8，12，16，20...。

这个公式是如何得出的呢？我们可以通
过观察数列中相邻两项的差来推导出这个公式。

可以发现，相邻两
项之间的差始终为4，即8-4=4，12-8=4，16-12=4，以此类推。

因此，我们可以得出这个数列的通项公式An = 3n + 1。

另外，我们也可以从代数的角度来推导这个公式。

假设数列的
第一个数是a，公差为d，那么数列的通项公式可以表示为An = a
+ (n-1)d。

根据4、4、5规律数列的特点，我们可以得出a=4，d=4，代入通项公式中得到An = 4 + (n-1)4 = 4n。

但是，这个公式并不
完全符合我们观察到的数列规律，因为4、4、5规律数列中的第一
个数并不是4，而是3。

所以我们需要对通项公式做一些调整，即
An = 3n + 1。

除了数学推导外，我们还可以从实际问题中找到4、4、5规律
的应用。

例如，在计算机编程中，这种规律可能被用于生成特定的
数值序列，或者用于数据加工和处理中。

在日常生活中，我们也可
以将这种规律应用于时间、距离等方面的计算中，以解决实际问题。

总之，4、4、5规律的公式An = 3n + 1是通过数学推导和实际应用得出的，它能够描述数列中的每一项，帮助我们理解和应用这种数列规律。

剑桥雅思阅读5翻译及精讲(test4)雅思阅读是块难啃的硬骨头，需要我们做更多的题目才能得心应手。

下面小编给大家分享一下剑桥雅思阅读5test4原文翻译及答案解析，希望可以帮助到大家。

剑桥雅思阅读5原文(test4)READING PASSAGE 1You should spend about 20 minutes on Questions 1-13, which are based on Reading Passage 1 on the following pages.Questions 1-3Reading Passage 1 has three sections, A-C.Choose the correct heading for each section from the list of headings below.Write the correct number i-vi in boxes 1-3 on your answer sheet.List of HeadingsI The expansion of international tourism in recent yearsIi How local communities can balance their own needs with the demands of wilderness tourismIii Fragile regions and the reasons for the expansion of tourism thereIv Traditional methods of food-supply in fragile regionsV Some of the disruptive effects of wilderness tourismVi The economic benefits of mass tourism1 Section A2 Section B3 Section CThe Impact of Wilderness TourismAThe market for tourism in remote areas is booming as neverbefore. Countries all across the world are actively promoting their ‘wilderness’ regions —such as mountains, Arctic lands, deserts, small islands and wetland — to high-spending tourists. The attraction of these areas is obvious: by definition, wilderness tourism requires little or no initial investment. But that does not mean that there is no cost. As the 1992 United Nations Conference on Environment and Development recognized, these regions are fragile (i.e. highly vulnerable to abnormal pressures) not just in terms of their ecology, but also in terms of the culture of their inhabitants. The three most significant types of fragile environment in these respects, and also in terms of the proportion o f the Earth’s surface they cover, are deserts, mountains and Arctic areas. An important characteristic is their marked seasonality, with harsh conditions prevailing for many months each year. Consequently, most human activities, including tourism, are limited to quite clearly defined parts of the year.Tourists are drawn to these regions by their natural landscape beauty and the unique cultures of their indigenous people. And poor governments in these isolated areas have welcomed the new breed of ‘adventure tourist’, grateful for the hard currency they bring. For several years now, tourism has been the prime source of foreign exchange in Nepal and Bhutan. Tourism is also a key element in the economies of Arctic zones such as Lapland and Alaska and in desert areas such as Ayers Rock in Australia and Arizona’s Monument Valley.BOnce a location is established as a main tourist destination, the effects on the local community are profound. When hill-farmers, for example, can make more money in a few weeksworking as porters for foreign trekkers than they can in a year working in their fields, it is not surprising that many of them give up their farm-work, which is thus left to other members of the family. In some hill-regions, this has led to a serious decline in farm output and a change in the local diet, because there is insufficient labour to maintain terraces and irrigation systems and tend to crops. The result has been that many people in these regions have turned to outside supplies of rice and other foods.In Arctic and desert societies, year-round survival has traditionally depended on hunting animals and fish and collecting fruit over a relatively short season. However, as some inhabitants become involved in tourism, they no longer have time to collect wild food; this has led to increasing dependence on bought food and stores. Tourism is not always the culprit behind such changes. All kinds of wage labour, or government handouts, tend to undermine traditional survival systems. Whatever the cause, the dilemma is always the same: what happens if these new, external sources of income dry up?The physical impact of visitors is another serious problem associated with the growth in adventure tourism. Much attention has focused on erosion along major trails, but perhaps more important are the deforestation and impacts on water supplies arising from the need to provide tourists with cooked food and hot showers. In both mountains and deserts, slow-growing trees are often the main sources of fuel and water supplies may be limited or vulnerable to degradation through heavy use.CStories about the problems of tourism have become legion in the last few years. Yet it does not have to be a problem. Although tourism inevitably affects the region in which it takesplace, the costs to these fragile environments and their local cultures can be minimized. Indeed, it can even be a vehicle for reinvigorating local cultures, as has happened with the Sherpas of Nepal’s Khumbu Valley and in some Alpine villages. And a growing number of adventure tourism operators are trying to ensure that their activities benefit the local population and environment over the long term.In the Swiss Alps, communities have decided that their future depends on integrating tourism more effectively with the local economy. Local concern about the rising number of second home developments in the Swiss Pays d’Enhaut resulted in limits being imposed on their growth. There has also been a renaissance in communal cheese production in the area, providing the locals with a reliable source of income that does not depend on outside visitors.Many of the Arctic tourist destinations have been exploited by outside companies, who employ transient workers and repatriate most of the profits to their home base. But some Arctic communities are now operating tour businesses themselves, thereby ensuring that the benefits accrue locally. For instance, a native corporation in Alaska, employing local people, is running an air tour from Anchorage to Kotzebue, where tourists eat Arctic food, walk on the tundra and watch local musicians and dancers.Native people in the desert regions of the American Southwest have followed similar strategies, encouraging tourists to visit their pueblos and reservations to purchase high-quality handicrafts and artwork. The Acoma and San lldefonso pueblos have established highly profitable pottery businesses, while the Navajo and Hopi groups have been similarly successful with jewellery.Too many people living in fragile environments have lost control over their economies, their culture and their environment when tourism has penetrated their homelands. Merely restricting tourism cannot be the solution to the imbalance, because people’s desire to see new places will not just disappear. Instead, communities in fragile environments must achieve greater control over tourism ventures in their regions, in order to balance their needs and aspirations with the demands of tourism. A growing number of communities are demonstrating that, with firm communal decision-making, this is possible. The critical question now is whether this can become the norm, rather than the exception.Questions 4-9Do the following statements reflect the opinion of the writer of Reading Passage 1?In boxes 4-9 on your answer sheet, writeYES if the statement reflects the opinion of the writerNO if the statement contradicts the opinion of the writerNOT GIVEN if it is impossible to say what the writer thinks about this4 The low financial cost of setting up wilderness tourism makes it attractive to many countries.5 Deserts, mountains and Arctic regions are examples of environments that are both ecologically and culturally fragile.6 Wilderness tourism operates throughout the year in fragile areas.7 The spread of tourism in certain hill-regions has resulted ina fall in the amount of food produced locally.8 Traditional food-gathering in desert societies was distributed evenly over the year.9 Government handouts do more damage than tourism does to traditional patterns of food-gathering.Questions 10-13Complete the table below.Choose ONE WORD from Reading Passage 1 for each answer.Write your answers in boxes 10-13 on your answer sheet.The positive ways in which some local communities haveresponded to tourismPeople/Location ActivityS wiss Pays d’EnhautArctic communitiesAcoma and San lldefonsoNavajo and Hopi Revived production of 10……………Operate 11……………businessesProduce and sell 12……………Produce and sell 13……………READING PASSAGE 2You should spend about 20 minutes on Questions 14-26, which are based on Reading Passage 2 below.Flawed Beauty: the problem with toughened glassOn 2nd August 1999, a particularly hot day in the town of Cirencester in the UK, a large pane of toughened glass in the roof of a shopping centre at Bishops Walk shattered without warning and fell from its frame. When fragments were analysed by experts at the giant glass manufacturer Pilkington, which had made the pane, they found that minute crystals of nickel sulphide trapped inside the glass had almost certainly caused the failure.‘The glass industry is aware of the issue,’ says Brian Waldron, chairman of the standards committee at the Glass and Glazing Federation, a British trade association, and standardsdevelopment officer at Pilkington. But he insists that cases are few and far between. ‘It’s a very rare phenomenon,’ he says.Others disagree. ‘On average I see about one or two buildings a month suffering from nickel sulphide related failures,’ says Barrie Josie, a consultant engineer involved in the Bishops Walk investigation. Other experts tell of similar experiences. Tony Wilmott of London-based consulting engineers Sandberg, and Simon Armstrong at CladTech Associates in Hampshire both say they know of hundreds of cases. ‘What you hear is only the tip of the iceberg,’ says Trevor Ford, a glass expert at Resolve Engineering in Brisbane, Queensland. He believes the reason is simple: ‘No-one wants bad press.’Toughened glass is found everywhere, from cars and bus shelters to the windows, walls and roofs of thousands of buildings around the world. It’s easy to see why. This glass has five times the strength of standard glass, and when it does break it shatters into tiny cubes rather than large, razor-sharp shards. Architects love it because large panels can be bolted together to make transparent walls, and turning it into ceilings and floors is almost as easy.It is made by heating a sheet of ordinary glass to about 620°C to soften it slightly, allowing its structure to expand, and then cooling it rapidly with jets of cold air. This causes the outer layer of the pane to contract and solidify before the interior. When the interior finally solidifies and shrinks, it exerts a pull on the outer layer that leaves it in permanent compression and produces a tensile force inside the glass. As cracks propagate best in materials under tension, the compressive force on the surface must be overcome before the pane will break, making it more resistant to cracking.The problem starts when glass contains nickel sulphide impurities. Trace amounts of nickel and sulphur are usually present in the raw materials used to make glass, and nickel can also be introduced by fragments of nickel alloys falling into the molten glass. As the glass is heated, these atoms react to form tiny crystals of nickel sulphide. Just a tenth of a gram of nickel in the furnace can create up to 50,000 crystals.These crystals can exist in two forms: a dense form called the alpha phase, which is stable at high temperatures, and a less dense form called the beta phase, which is stable at room temperatures. The high temperatures used in the toughening process convert all the crystals to the dense, compact alpha form. But the subsequent cooling is so rapid that the crystals don’t have time to change back to the beta phase. This leaves unstable alpha crystals in the glass, primed like a coiled spring, ready to revert to the beta phase without warning.When this happens, the crystals expand by up to 4%. And if they are within the central, tensile region of the pane, the stresses this unleashes can shatter the whole sheet. The time that elapses before failure occurs is unpredictable. It could happen just months after manufacture, or decades later, although if the glass is heated — by sunlight, for example — the process is speeded up. Ironically, says Graham Dodd, of consulting engineers Arup in London, the oldest pane of toughened glass known to have failed due to nickel sulphide inclusions was in Pilkington’s glass research building in Lathom, Lancashire. The pane was 27 years old.Data showing the scale of the nickel sulphide problem is almost impossible to find. The picture is made more complicated by the fact that these crystals occur in batches. So even if, onaverage, there is only one inclusion in 7 tonnes of glass, if you experience one nickel sulphide failure in your building, that probably means you’ve got a problem in more than one pane. Josie says that in the last decade he has worked on over 15 buildings with the number of failures into double figures.One of the worst examples of this is Waterfront Place, which was completed in 1990. Over the following decade the 40-storey Brisbane block suffered a rash of failures. Eighty panes of its toughened glass shattered due to inclusions before experts were finally called in. John Barry, an expert in nickel sulphide contamination at the University of Queensland, analysed every glass pane in the building. Using a studio camera, a photographer went up in a cradle to take photos of every pane. These were scanned under a modified microfiche reader for signs of nickel sulphide crystals. ‘We discovered at least another 120 panes with potentially dangerous inclusions which were then replaced,’ says Barry. ‘It was a very expensive and time-consuming process that took arou nd six months to complete.’ Though the project cost A$1.6 million (nearly ￡700,000), the alternative — re-cladding the entire building — would have cost ten times as much.Questions 14-17Look at the following people and the list of statements below.Match each person with the correct statement.Write the correct letter A-H in boxes 14-17 on your answer sheet.14 Brain Waldron15 Trevor Ford16 Graham Dodd17 John BarryList of StatementsA suggests that publicity about nickel sulphide failure has been suppressedB regularly sees cases of nickel sulphide failureC closely examined all the glass in one buildingD was involved with the construction of Bishops WalkE recommended the rebuilding of Waterfront PlaceF thinks the benefits of toughened glass are exaggeratedG claims that nickel sulphide failure is very unusualH refers to the most extreme case of delayed failureQuestions 18-23Complete the summary with the list of words A-P below.Write your answers in boxes 18-23 on your answer sheet.Toughened GlassToughened glass in favoured by architects because it is much stronger than ordinary glass, and the fragments are not as 18…………… when it breaks. However, it has one disadvantage: it can shatter 19…………… . This fault is a result of the manufacturing process. Ordinary glass is first heated, then cooled very 20…………… . The outer layer 21…………… before the inner layer, and the tension between the two layers which is created because of this makes the glass stronger However, if the glass contains nickel sulphide impurities, crystals of nickel sulphide are formed. These are unstable, and can expand suddenly, particularly if the weather is 22…………… . If this happens, the pane of glass may break. The frequency with which such problems occur is 23…………… by glass experts. Furthermore, the crystals cannot be detected without sophisticated equipment.A numerousB detectedC quicklyD agreedE warmF sharpG expands H slowly I unexpectedlyJ removed K contracts L disputedM cold N moved O smallP calculatedQuestions 24-26Do the following statements agree with the information given in Reading Passage 2?In boxes 24-26 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this24 Little doubt was expressed about the reason for the Bishops Walk accident.25 Toughened glass has the same appearance as ordinary glass.26 There is plenty of documented evidence available about the incidence of nickel sulphide failure.READING PASSAGE 3You should spend about 20 minutes on Questions 27-40, which are based on Reading Passage 3 below.The effects of light on plant and animal speciesLight is important to organisms for two different reasons. Firstly it is used as a cue for the timing, of daily and seasonal rhythms in both plants and animals, and secondly it is used to assist growth in plants.Breeding in most organisms occurs during a part of the year only, and so a reliable cue is needed to trigger breeding behaviour. Day length is an excellent cue, because it provides a perfectly predictable pattern of change within the year. In the temperate zone in spring, temperatures fluctuate greatly fromday to day, but day length increases steadily by a predictable amount. The seasonal impact of day length on physiological responses is called photoperiodism, and the amount of experimental evidence for this phenomenon is considerable. For example, some species of birds’ breeding can be induced even in midwinter simply by increasing day length artificially (Wolfson 1964). Other examples of photoperiodism occur in plants. A short-day plant flowers when the day is less than a certain critical length. A long-day plant flowers after a certain critical day length is exceeded. In both cases the critical day length differs from species to species. Plants which flower after a period of vegetative growth, regardless of photoperiod, are known as day-neutral plants.Breeding seasons in animals such as birds have evolved to occupy the part of the year in which offspring have the greatest chances of survival. Before the breeding season begins, food reserves must be built up to support the energy cost of reproduction, and to provide for young birds both when they are in the nest and after fledging. Thus many temperate-zone birds use the increasing day lengths in spring as a cue to begin the nesting cycle, because this is a point when adequate food resources will be assured.The adaptive significance at photoperiodism in plants is also clear. Short-day plants that flower in spring in the temperate zone are adapted to maximizing seedling growth during the growing season. Long-day plants are adapted for situations that require fertilization by insects, or a long period of seed ripening. Short-day plants that flower in the autumn in the temperate zone are able to build up food reserves over the growing season and over winter as seeds. Day-neutral plants have an evolutionaryadvantage when the connection between the favourable period for reproduction and day length is much less certain. For example, desert annuals germinate, flower and seed whenever suitable rainfall occurs, regardless of the day length.The breeding season of some plants can be delayed to extraordinary lengths. Bamboos are perennial grasses that remain in a vegetative state for many years and then suddenly flower, fruit and die (Evans 1976). Every bamboo of the species Chusquea abietifolio on the island of Jamaica flowered, set seed and died during 1884. The next generation of bamboo flowered and died between 1916 and 1918, which suggests a vegetative cycle of about 31 years. The climatic trigger for this flowering cycle is not yet known, but the adaptive significance is clear. The simultaneous production of masses of bamboo seeds (in some cases lying 12 to 15 centimetres deep on the ground) is more than all the seed-eating animals can cope with at the time, so that some seeds escape being eaten and grow up to form the next generation (Evans 1976).The second reason light is important to organisms is that it is essential for photosynthesis. This is the process by which plants use energy from the sun to convert carbon from soil or water into organic material for growth. The rate of photosynthesis in a plant can be measured by calculating the rate of its uptake of carbon. There is a wide range of photosynthetic responses of plants to variations in light intensity. Some plants reach maximal photosynthesis at one-quarter full sunlight, and others, like sugarcane, never reach a maximum, but continue to increase photosynthesis rate as light intensity rises.Plants in general can be divided into two groups: shade-tolerant species and shade-intolerant species. This classificationis commonly used in forestry and horticulture. Shade-tolerant plants have lower photosynthetic rates and hence have lower growth rates than those of shade-intolerant species. Plant species become adapted to living in a certain kind of habitat, and in the process evolve a series of characteristics that prevent them from occupying other habitats. Grime (1966) suggests that light may be one of the major components directing these adaptations. For example, eastern hemlock seedlings are shade-tolerant. They can survive in the forest understory under very low light levels because they have a low photosynthetic rate.Questions 27-33Do the following statements agree with the information given in Reading Passage 3?In boxes 27-33 on your answer sheet, writeTRUE if the statement agrees with the informationFALSE if the statement contradicts the informationNOT GIVEN if there is no information on this27 There is plenty of scientific evidence to support photoperiodism.28 Some types of bird can be encouraged to breed out of season.29 Photoperiodism is restricted to certain geographic areas.30 Desert annuals are examples of long-day plants.31 Bamboos flower several times during their life cycle.32 Scientists have yet to determine the cue for Chusquea abitifolia’s seasonal rhythm.33 Eastern hemlock is a fast-growing plant.Questions 34-40Complete the sentences.Choose NO MORE THAN THREE WORDS from the passagefor each answer.Write your answers in boxes 34-40 on your answer sheet.34 Day length is a useful cue for breeding in areas where …………… are unpredictable.35 Plants which do not respond to light levels are referred to as…………… .36 Birds in temperate climates associate longer days with nesting and the availability of …………….37 Plants that flower when days are long often depend on …………… to help them reproduce.38 Desert annuals respond to …………… as a signal for reproduction.39 There is no limit to the photosynthetic rate in plants such as …………… .40 Tolerance to shade is one criterion for the …………… of plants in forestry and horticulture.剑桥雅思阅读5原文参考译文(test4)TEST 4 PASSAGE 1 参考译文：The Impact of Wilderness Tourism荒野旅游的影响AThe market for tourism in remote areas is booming as never before. Countries all across the world are actively promoting their ‘wilderness’ regions —such as mountains, Arctic lands, deserts, small islands and wetland — to high-spending tourists. The attraction of these areas is obvious: by definition, wilderness tourism requires little or no initial investment. But that does not mean that there is no cost. As the 1992 United Nations Conference on Environment and Development recognized, these regions are fragile (i.e. highly vulnerable to abnormal pressures)not just in terms of their ecology, but also in terms of the culture of their inhabitants. The three most significant types of fragile environment in these respects, and also in terms of the proportion of the Earth’s surface they cover, are deserts, mountains and Arctic areas. An important characteristic is their marked seasonality, with harsh conditions prevailing for many months each year. Consequently, most human activities, including tourism, are limited to quite clearly defined parts of the year.A偏远地区的旅游市场从未曾像现在这么火爆。

数字神断基础知识教材书基础知识（1）太极生两仪，两仪生12象（12个数）阴数：5、6、7、8、9、10阳数：1、2、、4、11、12阳中阳：1、3、11、坎阳中阴：2、4、12阴中阳：5、7、9阴中阴：6、8、10、巽6、坤8、兑10（2）数字五行1、2、12为水，1是静水，地下水，2是动不，江湖河流，12是天水、雪、雨、霜、雾。

4、5是木，4是阳林木、木制家具、大树，5是阴木、花草、小树。

6、7为火，6是外火、太阳火，7是内火、7克力大，6克力小，如7克11力大，很危险。

3、8、9为土，3是高山、高山土；平原为路为坟，8是坑。

9是丘陵、沙堆。

3、8、9是相互克的，相互制约。

3和9是对冲数，什么时候都克。

特例：当9与木在一起时，9既论土也论金。

10、11为金，10为阴金，11为阳金（3）五行生克万事三角定律只论生克，而不论刑合。

生克规律与传统的一样，即：1、2、12生4、5,4、5生6、7,6、7生3、8、9，3、8、9生10、11。

1、2、12克6、7,6、7克10、11，10、11克4、5,4、5克3、8、9,3、8、9克1、2、12。

（4）数字颜色1是青色，2是兰色，12是棕色，3是白色，4是紫色，5是绿色，6是浅红，7是老红，8是浅黄，9是老黄，10是浅黑，11是老黑。

简化：水是兰色，木是绿色，火是红色，土是黄色，金是黑色。

（注意：金和水所代表的颜色与传统的不同）如广告牌是白底黑字，是土生金，外生内好，如外克内就不好，底色为外，字为内，用颜色也调病治，以后讲。

（5）数字的力量阳中阳数1、3、11属斥异性，占此数多的人善于表达自己。

只发不收，好动。

阳中阴数，属正异性，先发后收。

阴中阳数，属付异性，先收后发。

阴中阴数，属共性，只收不发。

如两口子打伏，如女人的命局是11－1－13－3，此人全是斥异性数，说明，每次打仗都是女的先动手的。

根据数的属性，可以用来配婚，如：阳中阳的人与阴中阴的人配婚较好。