2017南京秦淮区数学一模(含答案)

2017南京秦淮区数学一模(含答案)
2017南京秦淮区数学一模(含答案)

2016/2017学年度第二学期第一阶段学业质量监测试卷

九年级数学

注意事项:

1.本试卷共7页.全卷满分120分.考试时间为120分钟.

2.答选择题必须用2B铅笔将答题卷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卷上的指定位置,在其他位置答题一律无效.

一、选择题(本大题共6小题,每小题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

..卷.相应位置

....上)1.下列四个数中,是负数的是

A.||

-3B.(-3)2C.-(-3) D.-32

2.据南京市统计局调查数据显示,截至2016年年底,全市汽车拥有量首次进入全国“200万俱乐部”,达到了2 217 000辆.将2 217 000用科学记数法表示是

A.0.2217×106B.0.2217×107C.2.217×106D.2.217×107

3.如图,数轴上的点A表示的数可能是下列各数中的

A.-8的算术平方根B.10的负的平方根

C.-10的算术平方根D.-65的立方根

4.某公司的拓展部有五个员工,他们每月的工资分别是3000元,5000元,7000元,4000元和10000元,那么他们工资的中位数为

A.4000元B.5000元C.7000元D.10000元5.下列长度的三条线段能组成锐角三角形的是

A.2,3,3 B.2,3,4 C.2,3,5 D.3,4,5

二、填空题(本大题共10小题,每小题2分,共20分. 不需写出解答过程,请把答案直接

填写在答题卷相应位置

.......上)

7.-2的倒数是▲;-2的相反数是▲.

8.若式子x+1在实数范围内有意义,则x的取值范围是▲.

9.计算5×12

3

的结果是▲.

10.方程

1

x-2

3

x的解是▲.

-6 -5 -3

-4 -2 -1 1 2 3 4 5 6

A

(第3题)

14.某种商品因换季准备打折出售,如果按定价的七五折出售将赔25元,而按定价的九折

出售将赚20元,则商品的定价是 ▲ 元.

三、解答题(本大题共11小题,共88分.请在答题卷指定区域内作答,解答时应写出文字

说明、证明过程或演算步骤) 17.(6分)解不等式组????

? 2+3(x -3)≥5, 1+2x 3>x -2.

18.(6分)化简 2x 2

-4 -12x -4

20.(8分)脸谱是中国戏曲男演员脸部的彩色化妆.这种脸部化妆主要用于净(花脸)和

丑(小丑),表现人物的性格和特征.现有四张脸谱,如图所示:有两张相同的表现忠勇侠义的净角姜维,有一张表现直爽刚毅的净角包拯,有一张表现阴险奸诈的丑角夏侯婴.

(1)随机抽取一张,获得一张净角脸谱的概率是 ▲ ; (2)随机抽取两张,求获得一张姜维脸谱和一张包拯脸谱的概率.

22.(8分)“智慧南京、绿色出行”,骑共享单车出行已经成为一种时尚.记者随机调查了

一些骑共享单车的秦淮区市民,并将他们对各种品牌单车的选择情况绘制成图①和图②的统计图(A :摩拜单车;B :ofo 单车;C :HelloBike ).请根据图中提供的信息,解答下列问题:

包拯

姜维

姜维

夏侯婴

(第20题)

秦淮区市民对各种品牌单车选择情况统计图

A

C

B 50%

人数/名

60 90 120

120

秦淮区市民对各种品牌单车选择情况统计图 150

(1)在图①中,C部分所占扇形的圆心角度数为▲°;

(2)将图②补充完整;

(3)根据抽样调查结果,请你估计某天该区48万名骑共享单车的市民中有多少名选择摩拜单车?

2017秦淮区初三第一阶段学业质量监测学生问卷调查

同学好:

为了帮助贵校做好中考复习备考,控制学业负担,请如实填写以下问卷内容。问卷要求如下:

①问卷不需要署名。

②问卷中,第1题至第5题是单选题,其他各题为多项选择题。

③请根据学校及你本人的实际情况,将最符合自己实际想法的选项与答题卡对应,并规范、正确的

填涂。

一、单选题(每题只选择一个最符合或最接近您实际情况的选项)

1.你觉得你自己的学习负担

A.非常重

B.比较重

C.正常

D.比较轻

2.你每天下午放学时间大约是

A.5:30前

B.6:00前

C. 6:30前

D.7:00以后

3.每天晚上,做完功课后,你上床睡觉的时间是

A. 10点半前

B.11点前

C.11点半前

D.12点以后

4.你每天完成老师布置的书面家庭作业总共需要的时间是

A.3小时以上

B.2.5小时左右

C.2小时左右

D.1小时左右

5.到目前为止,本学期你的语文老师认真批改且有评语的大作文篇数是

A.3篇

B.5篇

C.6篇

D.8篇以上

二、多选题(每题选项可以选填多个选项。其中第7题调查问题,您没有这样的经历,可以不选)

6.中考复习阶段,你认为作业负担最重的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

7.中考复习阶段,你在校外参加社会培训和家庭辅导的学科有

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

8.中考复习阶段,你认为复习作业少,而且复习效果好的学科有

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

9.复习课中,能充分使用复习学案,有效指导和帮助学生进行复习备考的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

10.在中考复习开始前,教师能有计划指导学生制定复习计划的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

11.在中考复习过程中,教师始终重视对学生进行复习方法指导的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

12.通过前一阶段的中考复习,你认为获得明显进步的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

13.在复习课中,非常重视题型归类、解题方法与思路指导的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

14.针对复习课中学生暴露的问题,课余时间能认真辅导学生且效果好的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

15.复习阶段,教师布置家庭作业或课外练习,经常不批改的学科是

A.语文

B. 数学

C.英语

D.物理

E.化学

F.政治

G.历史

2016/2017学年度第二学期第一阶段学业质量监测试卷

九年级数学参考答案及评分标准

说明:本评分标准每题给出了一种或几种解法供参考,如果考生的解法与本解答不同,参照评分标准的精神给分.

一、选择题(本大题共6小题,每小题2分,共12分)

1 2 3 4 5 6 D

C

B

B

A

C

二、填空题(本大题共10小题,每小题2分,共20分)

7.-1

2;2 8.x ≥-1 9.2 5 10.x =3 11.22.5

12.60 13.2 14.300 15.①③⑥ 16.6 三、解答题 (本大题共11小题,共88分) 17.(本题6分)

解:解不等式①,得x ≥4. ……………………………………………………………2分

解不等式②,得x <7. ……………………………………………………………4分 所以,不等式组的解集是4≤x <7. ……………………………………………6分

18.(本题6分)

解:2x 2-4 -12x -4

=2(x +2)(x -2)-1

2(x -2)

……………………………………………………………2分

=4

2(x +2)(x -2)- x +22(x +2) (x -2)

…………………………………………………4分

-(x -2)

2(x +2) (x -2)

…………………………………………………………………5分

=-1

2(x +2)

.………………………………………………………………………6分

19.(本题6分)

证明:∵E 是AC 的中点,∴AE =CE . ………………………………………………1分

∵EF =DE ,………………………………………………………………………2分 ∴四边形ADCF 是平行四边形. …………………3分 ∵D 、E 分别是AB 、AC 的中点,

∴DE ∥BC .…………………………………………4分 ∴∠AED =∠ACB .

∵∠ACB =90°,∴∠AED =90°,即AC ⊥DF . ……………………………………………………… 5分

∴□ADCF 是菱形. ………………………………………………………… 6分

A

B

C

D

E

F

20.(本题8分)

解:(1)3

4

. ……………………………………………………………………………3分

(2)记第一张姜维脸谱为1,第二张姜维脸谱为2,包拯脸谱为3,夏侯婴脸谱为

4.随机抽取两张,所有可能出现的结果有:(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)共6种,它们出现的可能性相同.所有的结果中,满足“随机抽取两张,获得一张姜维脸谱和一张包拯脸谱”(记为事件A )的结果有2种,所以P (A )=26=1

3. (8)

21.(本题8分)

解:(1)方法一:由题意得图像的顶点坐标为(2,1), 设函数的表达式为y =a (x -2)2+1. ………………………………2分

由题意得函数的图像经过点(0,5),

所以5=a ·(-2)2+1. ……………………………………………3分

所以a =1. …………………………………………………………4分 所以函数的表达式为y =(x -2)2+1(或y =x 2-4x +5).………5分 方法二:因为函数y =ax 2+bx +c 的图像经过点(1,2)、(2,1)、(0,5),

所以,?????c =5,

a +

b +

c =2,4a +2b +c =1.………………………………………………3分

解得?????a =1,

b =-4,

c =5.

………………………………………………………4分

所以函数的表达式为y =x 2-4x +5.………………………………5分

(2)0<x <4.…………………………………………………………………8分

22.(本题8分)

解:(1)30. ……………………………………………………………………………2分

(2)图略,A 为100名. …………………………………………………………5分

(3)120÷50%=240(名).

48×100

240

=20(万名). ………………………………………………………7分

所以估计某天该区48万名骑共享单车的市民中有20万名选择摩拜单车.………………………………………………………………………………8分

23.(本题8分)

解:设保温杯的定价应为x 元.…………………………………………………………1分

根据题意,得(x -80)[1000-5(x -100)]=60500. ………………………………5分 化简,得x 2-380x +36100=0.

解得x 1=x 2=190.……………………………………………………………………7分

答:保温杯的定价应为190元.……………………………………………………8分

24.(本题8分)

解:过点A 作AF ⊥CE ,交CE 于点F . ………………………………………………1分

设AF 的长度为x m . ∵∠AED =45°,

∴△AEF 是等腰直角三角形. ∴EF =AF =x .

在Rt △ADF 中,∵tan ∠ADF =AF DF ,

∴DF =

AF tan ∠ADF =x tan80.5°=x

6

. …………………………………………………2分

∵DE =18.9, ∴x

6+x =18.9.…………………………………3分 解得x =16.2. …………………………………4分 过点B 作BG ⊥AF ,交AF 于点G .…………5分 易得BC =GF =15,∠CBG =90°.

∴AG =AF -GF =16.2-15=1.2.……………6分 ∵∠ABC =120°,

∴∠ABG =∠ABC -∠CBG =120°-90°=30°. 在Rt △ABG 中,

∵sin ∠ABG =AG

AB ,

∴AB =

AG sin ∠ABG = 1.2

sin30°

=2.4. …………………………………………………7分

答:灯杆AB 的长度为2.4 m .………………………………………………………8分

25. (本题9分)

(1)证明:连接OD . ∵OD =OB ,∴∠ODB =∠OBD . ∵AB =AC ,∴∠ACB =∠OBD . ∴∠ACB =∠ODB .

∴OD ∥AC .…………………………………………………………………………2分 ∴∠DEC =∠ODE .

∵DE ⊥AC ,∴∠DEC =90°.

∴∠ODE =90°,即OD ⊥DE .……………………………………………………3分

∵DE 过半径OD 的外端点D ,……………………………………………………4分

∴DE 是⊙O 的切线.………………………………………………………………5分

D A

B

C F E

G

A

C

B

O

D

E

(2)解:连接AD .

∵AB 为半圆O 的直径, ∴∠ADB =90°.∵DE ⊥AC , ∴∠DEC =∠ADB =90°. ∵AB =AC ,BC =6,

∴CD =BD =1

2BC =3. ………………………………………………………6分

又∵∠ECD =∠DBA ,

∴△CED ∽△BDA .……………………………………………………………7分 ∴

CE BD =CD

BA

. ∵CE =1,∴13=3

BA

∴AB =9.………………………………………………………………………8分 ∴半圆O 的半径的长为4.5.…………………………………………………9分

26.(本题11分)

解:(1)中心对称图形,对角线的交点是它的对称中心.……………………………1分 (2)同一底上的两个角相等.………………………………………………………2分

(3)对角线相等.……………………………………………………………………3分 (4)∠ABC =∠DCB ,∠BAD =∠CDA ,AC =BD . ……………………………4分 方法一:

证明:过点D 作DE ∥AB ,交BC 于点E . …………………………………5分 ∴∠ABE =∠DEC . ∵ AD ∥BC ,

∴四边形ABED 是平行四边形.………………………………………………6分 ∴AB =DE . 又∵AB =DC , ∴DE =DC . ∴∠DCE =∠DEC .

∴∠ABE =∠DCE ,即∠ABC =∠DCB .……………………………………7分 ∵ AD ∥BC ,

∴∠BAD +∠ABC =180°,∠CDA +∠DCB =180°. ∵∠ABC =∠DCB ,

∴∠BAD =∠CDA .……………………………………………………………8分

A

C

B

O

D

E A

B

C D

E

在△ABC 和△DCB 中,

?????AB =DC ,

∠ABC =∠DCB ,BC =CB ,

∴△ABC ≌△DCB .

∴AC =BD .……………………………………………………………………9分 方法二:

证明:分别过点A 、D 作AE ⊥BC 于点E 、DF ⊥BC 于点F . ……………5分 ∴∠AEF =∠DFC =90°. ∴AE ∥DF . ∵AD ∥BC ,

∴四边形AEFD 是平行四边形.……………6分 ∴AE =DF .

在Rt △ABE 和Rt △DCF 中,

???AB =DC ,AE =DF ,

∴Rt △ABE ≌Rt △DCF .

∴∠ABE =∠DCF ,即∠ABC =∠DCB . …………………………………7分 ∵ AD ∥BC ,

∴∠BAD +∠ABC =180°,∠CDA +∠DCB =180°. ∵∠ABC =∠DCB ,

∴∠BAD =∠CDA .……………………………………………………………8分 在△ABC 和△DCB 中,

?????AB =DC ,

∠ABC =∠DCB ,BC =CB ,

∴△ABC ≌△DCB .

∴AC =BD .……………………………………………………………………9分 (5)

………………………………………………………………………………………11分

等腰梯形

四边形

对角线相等的四边形

平行四边形

A

B

C D

E

F

27.(本题10分)

解:(1)900. ……………………………………………………………………………1分 (2)根据图像,得慢车的速度为900

15

=60(km/h ),

快车的速度为900×2-10×60

8=150(km/h ). ………………………………3分

方法一:

所以线段AB 所表示的y 与x 之间的函数表达式为y 1=900-60x . ……5分 所以线段CD 所表示的y 与x 之间的函数表达式为

y 2=(60+150) (x -10)=210x -2100. ………………………………………7分 方法二:

A 点表示快车到达乙地,所以此时快车行驶的时间为

900

150

=6(h ), 两车距离为900-60×6=540(km ),所以A (6,540).

所以设线段AB 所表示的y 与x 之间的函数表达式为y 1=-60x +b . …4分 当x =6时,y 1=540,即-60×6+b =540. 解得b =900.

所以线段AB 所表示的y 与x 之间的函数表达式为y 1=-60x +900.……5分 因为慢车的速度为60 km/h ,快车的速度为150 km/h , 所以两车的速度之和为60+150=210(km/h ).

所以设线段CD 所表示的y 与x 之间的函数表达式为y 2=210x +n .……6分 因为函数图像经过点C (10,0). 得210×10+n =0. 解得n =-2100.

所以线段CD 所表示的y 与x 之间的函数表达式为y 2=210x -2100. ……………………………………………………………………………………7分 (3)①线段OA 所表示的y 与x 之间的函数表达式为y 3=90x (0≤x <6),

令y 3=480,得x =16

3

. ……………………………………………………8分

②线段AB 所表示的y 与x 之间的函数表达式为y 1=-60x +900(6≤x <8), 令y 1=480,得x =7.………………………………………………………9分 ③线段CD 所表示的y 与x 之间的函数表达式为y 2=210x -2100(10≤x <14),

令y 2=480,得x =86

7.

答:慢车出发163h 、7h 、86

7

h 后,两车相距480 km .………………………10分

江苏省南京市鼓楼区2017年中考一模英语试题及参考答案

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