南京市联合体2015年中考二模数学试题及答案(共12页)

1、精选优质文档-倾情为你奉上2015年中考模拟试卷（二）数 学化工园雨花栖霞浦口四区联合体注意事项：1本试卷共6页全卷满分120分考试时间为120分钟考生答题全部答在答题卡上，答在本试卷上无效2请认真核对监考教师在答题卡上所粘贴条形码的姓名、考试证号是否与本人相符合，再将自己的姓名、准考证号用0.5毫米黑色墨水签字笔填写在答题卡及本试卷上3答选择题必须用2B铅笔将答题卡上对应的答案标号涂黑如需改动，请用橡皮擦干净后，再选涂其他答案答非选择题必须用0.5毫米黑色墨水签字笔写在答题卡上的指定位置，在其他位置答题一律无效4作图必须用2B铅笔作答，并请加黑加粗，描写清楚一、选择题（本大题共6小题，每小题

2、2分，共12分在每小题所给出的四个选项中，恰有一项是符合题目要求的，请将正确选项前的字母代号填涂在答题卡相应位置上）1的相反数是 （ ） A. -2B.2 C. -D. 2.下列计算正确的是（ ）A. a3+a4=a7B2a3a4=2a7C（2a4）3=8a7 Da8÷a2=a43.为调查某班学生每天使用零花钱的情况，张华随机调查了20名同学，结果如下表：每天使用零花钱（单位：元）12345人数13655则这20名同学每天使用的零花钱的众数和中位数分别是（ ）A. 3，3B. 3，3.5 C. 3.5，3.5D. 3.5，34小张同学的座右铭是“态度决定一切”，他将这几个字写在一个正

3、方体纸盒的每个面上，其平面展开图如图所示，那么在该正方体中， 和“一”相对的字是（ ）A. 态 B. 度C. 决D. 切（第6题）BADCEF（第5题）ABCO（第4题）5. 如图，O是 ABC的外接圆，OBC=42°，则A的度数是（ ）A. 42° B. 48° C. 52° D. 58°6如图，在矩形ABCD中，AB3，BC5，以B为圆心BC为半径画弧交AD于点E，连接CE，作BFCE，垂足为F，则tanFBC的值为（ ）A. B. C. D. 二、填空题（本大题共10小题，每小题2分，共20分，请在答题卡指定区域内作答）7代数式有意义，则

4、 x的取值范围是 8. 分解因式：a34a 9. 计算2cos30°|1| .10. 反比例函数y的图象经过点（1，6）和（m，-3），则m .11. 如图，在菱形ABCD中，AC2，ABC60°，则BD .BOA1CD（第12题）12. 如图，在O中， AOCD, 130°，劣弧AB的长为3300千米，则O的周长用科学计数法表示为 千米ABCD（第11题）13.某商品原价100元，连续两次涨价后，售价为144元，若平均增长率为x，则x .14直角坐标系中点A坐标为（5，3），B坐标为（1，0），将点A绕点B逆时针旋转90°得到点C，则点C的坐标为 .(

5、第15题)15二次函数yax2+bx+c（a0）的图象如图所示，根据图象可知：方程ax2+bx+ck有两个不相等的实数根，则k的取值范围为 . O(第16题)16.如图，在半径为2的O中，两个顶点重合的内接正四边形与正六边形，则阴影部分的面积为 .三、解答题（本大题共11小题，共88分请在答题卡指定区域内作答，解答时应写出文字说明、证明过程或演算步骤）17（6分）解方程组 18.（6分）化简：（x）÷19.（8分）为了备战初三物理、化学实验操作考试，某校对初三学生进行了模拟训练物理、化学各有3个不同的操作实验题目，物理用番号、代表，化学用字母a、b、c表示测试时每名学生每科只操作一个

6、实验，实验的题目由学生抽签确定（1）小张同学对物理的、和化学的b、c实验准备得较好.请用树形图或列表法求他两科都抽到准备得较好的实验题目的概率；（2）小明同学对物理的、和化学的a实验准备得较好.他两科都抽到准备得较好的实验题目的概率为 .20. （8分）据报道，历经一百天的调查研究，南京PM 2.5源解析已经通过专家论证各种调查显示，机动车成为PM 2.5的最大来源，一辆车每行驶20千米平均向大气里排放0.035千克污染物校环保志愿小分队从环保局了解到南京100天的空气质量等级情况，并制成统计图和表：2014年南京市100天空气质量等级天数统计图 空气质量等级优良轻度污染中度污染重度污染严重污

7、染天数（天）10a12825b2014年南京市100天空气质量等级天数统计表优良轻度重度严重10%25%12%8%n°25%中度（1） 表中a ，b ，图中严重污染部分对应的圆心角n °.（2）请你根据“2014年南京市100天空气质量等级天数统计表”计算100天内重度污染和严重污染出现的频率共是多少？（3）小明是社区环保志愿者，他和同学们调查了机动车每天的行驶路程，了解到每辆车每天平均出行25千米已知南京市2014年机动车保有量已突破200万辆，请你通过计算，估计2014年南京市一天中出行的机动车至少要向大气里排放多少千克污染物？21.（8分）如图， 在ABCD中，E、F

8、、G、H分别为AB、BC、CD、AD的中点，AF与EH交于点M，FG与CH交于点N.（1）求证：四边形MFNH为平行四边形；（2）求证：AMHCNF.ABCDFGEHMN22. （8分）端午节期间，某食堂根据职工食用习惯，购进甲、乙两种粽子260个，其中甲种粽子花费300圆，乙种粽子花费400元，已知甲种粽子单价比乙种粽子单价高20%，乙种粽子的单价是多少元？甲、乙两种粽子各购买了多少个？23.（8分）如图，为了测出某塔CD的高度，在塔前的平地上选择一点A，用测角仪测得塔顶D的仰角为30º，在A、C之间选择一点B （A、B、C三点在同一直线上），用测角仪测得塔顶D的仰角为75

9、6;，且AB间距离为40m（1）求点B到AD的距离； （2）求塔高CD（结果用根号表示）ABCD（第23题）30°75°24（8分）小林家、小华家、图书馆依次在一条直线上小林、小华两人同时各自从家沿直线匀速步行到图书馆借阅图书，已知小林到达图书馆花了20分钟.设两人出发x（分钟）后，小林离小华家的距离为y（米），y与x的函数关系如图所示.（1）小林的速度为 米/分钟 ，a ，小林家离图书馆的距离为 米；（2）已知小华的步行速度是40米/分钟，设小华步行时与自己家的距离为y1（米），请在图中画出y1（米）与x（分钟 ）的函数图象；（3）小华出发几分钟后两人在途中相遇？第题（第

10、24题）x（分钟）y（米）420240Oa25(8分)施工队要修建一个横断面为抛物线的公路隧道，其高度为6米，宽度OM为12米现以O点为原点，OM所在直线为x轴建立直角坐标系（如图所示）（1）求出这条抛物线的函数表达式，并写出自变量x的取值范围；（2）隧道下的公路是双向行车道(正中间有一条宽1米的隔离带)，其中的一条行车道能否行驶宽2.5米、高5米的特种车辆？请通过计算说明；（第25题）26. （10分）如图，已知ABC，AB=6、AC8，点D是BC边上一动点，以AD为直径的O分别交AB、AC于点E、F.（1）如图若AEFC，求证：BC与O相切；（2）如图，若BAC90°，BD长为多

11、少时，AEF与ABC相似.图ADBCEFODBOCFAEABC备用 图27. (10分)已知直角ABC，ACB90°，AC3，BC4，D为AB边上一动点，沿EF折叠，点C与点D重合，设BD的长度为m.（1）如图，若折痕EF的两个端点E、F在直角边上，则m的范围为 ;（2）如图，若m等于2.5，求折痕EF的长度；（3）如图，若m等于，求折痕EF的长度.DFACDBEFACBACB图图图DEFE2015中考数学模拟试卷（二）答案一、选择题（本大题共6小题，每小题2分，共12分）题号123456答案DBBABD二、填空题（本大题共10小题，每小题2分，共20分）7x1 8. a（a2）（a

12、+2） 9. +1 10. 2 11. 2 12.3.96×104 13. （2，4） 14.0.2 15. k2 16. 62三、解答题（本大题共11小题，共88分）17解： ×2得：4x+6y10×3得：9x6y36 +得：13x26解得： x2··························

13、3;·················································

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16、·············5分所以方程组的解为···································&#

17、183;·············································6分18.解原式÷··

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27、83;··············5分x2+x ··································&#

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29、183;···························6分19. （1）画图或列表正确···················

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31、3;·······························4分共有9种等可能结果，期中两科都满意的结果有4种··············

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33、83;·5分P（两科都满意）.··············································

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37、;································8分20. （1）25；20；72°··············

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42、83;5分（3）200×0.035×10000×87500（千克）··········································

43、·································8分21. （1）证明：连接BD，E、F、G、H分别为AB、BC、CD、AD的中点，EH为ABD的中位线，EHBD.同理FGBD.EHFG·····

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46、83;·············2分在ABCD中ADBC，H为AD的中点AHAD，F为BC的中点FCBC，AHFC四边形AFCH为平行四边形，AFCH··························

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48、3;··········································4分又EHFG四边形MFNH为平行四边形····&#

49、183;·················································&#

50、183;····································5分（2）四边形AFCH为平行四边形FADHCB ·········&

51、#183;·················································&

52、#183;···············································6分EHFG,AMHAFN

53、AFCHAFNCNFAMHCNF···············································&#

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55、183;··········7分又AHCFAMHCNF·····································&

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57、#183;·····················8分22.解：设乙种粽子的单价是x元，则甲种粽子的单价为（1+20%）x元，由题意得， +260，····················

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59、·············4分解得：x=2.5，···································&

60、#183;·············································5分经检验：x=2.5是原分式方程的解，·

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62、;·····································6分（1+20%）x=3，则买甲粽子为： 100个，乙粽子为：160个·······

63、·········································7分答：乙种粽子的单价是2.5元，甲、乙两种粽子各购买100个、160个···

64、·····································8分23. （1）作BEAD，垂足为E，ABCD（第23题）30°75°E在RtAEB中，sinA， ，BE20··

65、··············3分（2）DBC是ABD的外角ADBDBCA45°，···············4分在RtDEB中，tanEDB ，1，ED20············

66、;·································5分在RtAEB中，cosEAB ， EA20·············

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69、83;·······7分在RtACD中，sinDAC ， EA10+10············································