江苏省南京市部分学校2021-2022学年度七年级上学期期末调研数学试卷(Word版含答案)
展开2021~2022学年第一学期期末调研
七年级数学学科
一、选择题(本大题共8小题,每小题2分,共16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.下面四个数中比-5小的数是
A.-6 | B.-4 | C.0 | D.1 |
2.下列图形都是由六个相同的正方形组成的,经过折叠不能围成正方体的是
A. | B. | C. | D. |
3.下列合并同类项结果正确的是
A.2a2+3a2=6a2 | B.2a2+3a2=5a2 | C.2xy-xy=1 | D.2x3+3x3=5x6 |
4.目前全球新型冠状病毒肺炎疫情防控形势依旧严峻,我们应该坚持“勤洗手,戴口罩,常通风”.一双没有洗过的手,带有各种细菌约75 000万个,将数据75 000用科学记数法表示是
A.7.5×103 | B.75×103 | C.7.5×104 | D.7.5×105 |
5.如图,把三角形剪去一个角,所得四边形的周长比原三角形的周长小,能正确解释这一现象的数学知识是
A.四边形周长小于三角形周长 | B.两点确定一条直线 |
C.垂线段最短 | D.两点之间,线段最短 |
6.左图中的图形绕虚线旋转一周可得到的几何体是
7.一件夹克衫先按成本价提高50%标价,再将标价打8折出售,结果获利28元.如果设这件夹克衫的成本价是x元,那么根据题意,所列方程正确的是
A.0.8×(1+0.5)x=x-28 | B.0.8×(1+0.5)x=x+28 |
C.0.8×(1+0.5x)=x-28 | D.0.8×(1+0.5x)=x+28 |
8.若x、y、z是三个连续的正整数,若x2=44944,z2=45796,则y2=
A.45 369 | B.45 371 | C.45 465 | D.46 489 |
二、填空题(本大题共10小题,每小题2分,共20分.请把答案填写在答题卡相应位置)
9. 若3x4y2n和-x2my6是同类项,则m+n= ▲ .
10.若x=2是关于x的一元一次方程2x+m-5=0的解,则m= ▲ .
11.若∠α=10°45',则∠α的余角等于 ▲ .
12.若x2-2x=1,则代数式2x2-4x-1的值为 ▲ .
13.从三个不同方向看一个几何体,得到的平面图形如图所示,则这个几何体是 ▲ .
14.如图,点A,B在数轴上,点O为原点,OA=OB.按如图所示方法用圆规在数轴上截取BC=AB,若点C表示的数是15,则点A表示的数是 ▲ .
15.如图,直线 a、b相交于点O,将量角器的中心与点O重合,发现表示60°的点在直线a上,表示135°的点在直线b上,则∠1= ▲ °.
16.已知线段AB,点C在直线AB上,AB=8,BC=4,若点M是线段AC的中点,则线段AM的长为 ▲ .
17.小明在计算1-3+5-7+9-11+13-15+17时,不小心把一个运算符号写错了(“+”错写成“-”或“-”错写成“+”),结果算成了-17,则原式从左往右数,第 ▲ 个运算符号写错了.
18.小淇同学在元旦晚会上表演了一个节目:他准备了♥(红桃)和♠(黑桃)的扑克牌各10张,洗匀后将这些牌的牌面朝下,排成两列:一列m(m>10)张,一列(20-m)张,他立刻报出长的一列中的♠(黑桃)比短的一列中的♥(红桃)多了 ▲ 张.(结果用含有m的代数式表示)
三、解答题(本大题共9小题,满分64分,请在答题卡指定区域内作答,解答应写出文字说明,证明过程或演算步骤)
19.(6分)计算:
(1)-12-2+(-3)×; (2)12×(--).
20.(6分)先化简,再求值:
(3x2-2xy+5y2 )-2(x2-xy-2y2),其中x=-1,y=2.
21.(8分)解方程:
(1)4(x+3)=2x-1; (2)1 - = .
22.(4分)如图,△ABC的三个顶点均在格点处.
(1)过点B画AC的平行线BD;
(2)过点A画BC的垂线AE.
(请用黑水笔描清楚)
23.(6分)
(1)由大小相同的小立方块搭成的几何体如左图,请在右图的方格中画出该几何体的俯视图和左视图.(用阴影部分表示)
(2)若现在你手头还有一些相同的小立方块,如果保持俯视图和左视图不变,则在左图中最多可以再添加 ▲ 个小立方块.
24.(8分)如图,直线 AB 与 CD 相交于点 O,∠AOE=90°.
(1)如果∠AOC=20°,求∠COE 和∠BOD 的度数;
(2)如果∠COE=2∠BOD,求∠BOC 的度数.
25.(7分)列方程解应用题:
为了加强公民的节水意识,某市将要采用价格调控手段达到节水目的,设计了如下的调控方案.
价目表 | |
每月用水量 | 单价 |
不超出10吨的部分 | 2.5元/吨 |
超出10吨的部分 | 3元/吨 |
(1)甲户居民五月份用水12吨,则水费为 ▲ 元.
(2)乙户居民八月份缴纳水费40元,则该户居民八月份用水多少吨?(列方程解答)
26.(9分)如图,将一副直角三角板的直角顶点C叠放在一起.
(1)如图(1),若∠DCE=33°,则∠BCD= ▲ ,∠ACB= ▲ .
(2)如图(1),猜想∠ACB与∠DCE的大小有何特殊关系?并说明理由.
(3)如图(2),若是两个同样的直角三角板60°锐角的顶点A重合在一起,则∠DAB与∠CAE的数量关系为 ▲ .
27.(10分)如图,已知直线 AB 和 CD 相交于点O,∠COE=90°,OF 平分∠AOE,∠COF=37°.
(1)求∠EOB的度数.
(2)若射线OF、OD分别绕着点O按顺时针方向转动,两射线同时出发,射线OF每分钟转动6°,射线OD每分钟转动0.5°,多少分钟后,射线OF与射线OD第一次重合.
(3)在(2)的条件下,假设转动时间不超过60分钟,若∠FOD=33°,则两射线同时出发 ▲ 分钟.
2021~2022学年第一学期期末调研
七年级数学学科参考答案及评分标准
说明:本评分标准每题给出了的解法供参考.如果考生的解法与本解答不同,参照本评分标准的精神给分.
一、选择题(本大题共8小题,每小题2分,共16分)
题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 |
答案 | A | D | B | C | D | C | B | A |
二、填空题(本大题共10小题,每小题2分,共20分)
9.5. | 10.1. | 11.79°15' . | 12.1. |
13.圆柱. | 14.-5. | 15.75°. | 16.2或6. |
17.6. | 18.(m-10). |
|
|
三、解答题(本大题共9小题,共64分)
19.(本题6分)
解:(1)原式=-1-2+(-12),·······································2分
=-15.················································3分
(2)原式=1-2-3,·············································5分
=-4.·················································6分
20.(本题6分)
解:原式=3x2-2xy+5y2-2x2+2xy+4y2,·······························1分
=x2+9y2.·················································3分
当x=-1,y=2时,
原式=(-1)2+9×22,············································4分
=37.·························································6分
21.(本题8分)
(1)解:4x+12=2x-1,·············································1分
4x-2x=-1-12,···········································2分
2x=-13,·············································3分
x=-.···············································4分
(2)解:6-2(2x-1)=1+2x,·········································5分
6-4x+2=1+2x,········································6分
-6x=-7,···········································7分
x=.·············································8分
22.(本题4分)
(1)如图,直线BD即为所求;·········································2分
(2)如图,直线AE即为所求.··········································4分
23.(本题6分)
(1)
·········································4分
(2) 2.··························································6分
24.(本题8分)
(1)∵ ∠AOE=90°,∠AOC=20°,······································1分
∴ ∠COE=∠AOE-∠AOC=90°-20°=70°,···························3分
∠BOD=∠AOC=20°.·········································4分
(2)∵ ∠BOD=∠AOC,∠COE=2∠BOD,
∴ ∠COE=2∠AOC.·············································5分
又∵ ∠AOE=90°,
∴ ∠AOE=∠AOC+∠COE=3∠AOC=90°,····························6分
∴ ∠AOC=30°,················································7分
∴ ∠BOC=∠AOB-∠AOC =180°-30°=150°.··························8分
25.(本题7分)
(1)31.··························································2分
(2)该户居民八月份用水x吨,根据题意得:
2.5×10+3(x-10) =40,··········································5分
解得 x=15.
答:该户居民八月份用水15吨. ····································7分
26.(本题9分)
(1)57°,157°.···················································2分
(2)∠ACB=180°-∠DCE,············································4分
理由如下:
∵ ∠ACE=90°-∠DCE,∠BCD=90°-∠DCE,
∴ ∠ACB=∠ACE+∠DCE+∠BCD
=90°-∠DCE+∠DCE+90°-∠DCE
=180°-∠DCE.··············································7分
(3)∠DAB=120°-∠CAE.············································9分
27.(本题10分)
解:(1)∵ ∠COE=90°,∠COF=37°,
∴ ∠EOF=90°-37°=53°.·······································2分
∵ OF 平分∠AOE,
∴ ∠AOE=53°×2=106°.
∴ ∠EOB=180°-106°=74°.·····································4分
(2)∵ ∠COD=180°,∠COE=90°,
∴ ∠EOD=90°.
∴ ∠FOD=90°+53°=143°.······································6分
设x分钟后射线OF与射线OD第一次重合.依题意,得:
6x-0.5x=143,
解得:x=26.
答:26分钟后,射线OF与射线OD第一次重合.························8分
(3)20或32.···················································10分
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