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第二届初中学生学科素养竞赛
七年级数学
注意事项
考生在答题前请认真阅读本注意事项：
1.本试卷共6页，满分为150分，考试时间为120分钟。考试结束后，请将本试卷和答
题卡一并交回。
2.答题前，请务必将自己的姓名、考试证号用0.5毫米黑色字迹的签字笔填写在试卷及
答题卡上指定的位置。
3.答案必须按要求填涂、书写在答题卡上，在试卷、草稿纸上答题一律无效。
一、选择题（本大题共10小题，每小题3分，共30分.在每小题给出的四个选项中，恰
有一项是符合题目要求的，请将正确选项的字母代号填涂在答题卡相应位置上)
1.下列各组数中，互为相反数的是
A.-(-3)与-3
B.+(+3)与3
C.-(+3)与-3D.+(-3)与-3
2.以“聚力新南通、奋进新时代”为主题的第五届通商大会暨全市民营经济发展大会上
有40个重大项目集中签约，计划总投资约41800000000元.将41800000000用科
学记数法表示为
A.4.18×101
B.4.18×1010
C.0.418×10"
D.418×108
3.对于代数式5+m的值，下列说法正确的是
A.比5大
B.比5小
C.比m大
D.比m小
4.计算（一7）÷（一》）×7的结果为
A.1
B.-7
C.7
D.343
5.已知有理数a满足a=一a,那么在数轴上表示有理数a的点
A.在原点右侧
B.在原点或原点右侧
C.在原点左侧
D,在原点或原点左侧
6.对于多项式xy一3y一4，下列说法正确的是
A.二次项系数是3B.常数项是4
C.次数是3
D.项数是2
7.如果a十b>0,且b<0,那么a,b,一a,一b的大小关系为
A.a<-b<-aC.a8.某商场促销，把单价2500元的空调以八折出售仍可获利400元，则这款空调进价为
A.1375元
B.1500元
C.1600元
D.2000元
七年级数学第1页（共6页）
0000000
9.若a2一4a一12=0，则2a2-8a-8的值为
A.16
B.18
C.20
D.24
10.某窗户的形状如图所示（图中长度单位：cm),其上部是半圆形，下部是由两个相同
的长方形和一个正方形构成.已知半圆的半径为acm,长方形的长和宽分别为bcm
和ccm,给出下面四个结论：
①窗户外围的周长是（πa+3b+2c)cm:
②窗户的面积是（πa2+2bc+b2)cm2:
③b+2c=2a:
④b=3c.
←C
(第10题)
上述结论中，所有正确结论的序号是
A.①②
B.①③
C.②④
D.③④
二、填空题（本大题共8小题，第11~12题每小题3分，第13~18题每小题4分，共
30分.不需写出解答过程，请把答案直接填写在答题卡相应位置上)
11.数轴上表示一2024和表示2025的两点之间的距离为▲
12.若a、b互为倒数，m、n互为相反数，则(m十m)2+2ab=_▲
13.若有理数a,b满足1a-1川+b2=0,则a十b=▲
14.如果x一y=5,m十n=2,则0y十m)-x一)的值是▲
15.如果代数式x2-(36y十y2+1)十y-8合并同类项后不含y项，则k=▲
16.有理数a,b,c在数轴上的位置如图所示：
a
b
0
化简代数式：3lc-a+2b-c-3la+bl=▲
(第16题)
17.如图所示是一组有规律的图案，它们是由边长相同的小正方形组成，其中部分小正
方形涂有阴影，按照这样的规律，第n个图案中有▲个涂有阴影的小正方形（用
含有n的代数式表示).
第1个图案
第2个图案
第3个图案
七年级数学第2页（共6页）
0000000★保密材料 第二届初中学生学科素养竞赛
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七年级数学试题参考答案与评分标准
说明：本评分标准每题只给出了一种解法供参考，如果考生的解法与本解答不同，参照本
评分标准给分．
一、选择题（本大题共 10 小题，每小题 3 分，共 30 分）
题号 1 2 3 4 5 6 7 8 9 10
选项 A B C D D C B C A B
二、填空题（本大题共 8 小题，11－12 每小题 3 分，13－18 每小题 4 分，共 30 分．）
11．40 49 12．2 13． 1 14．－3
1
15． 16．b＋5c 17．4n＋1 18．a－b＋c＝2
3
三、解答题（本大题共 8 小题，共 90 分）
19．（本小题满分 16 分）
解：（1）原式＝20＋7＋2 ········································································· 3 分
＝29 ··················································································· 4 分
5 5 5 5 5 1
（2）原式＝－ × － × － × ······················································· 6 分
7 12 7 12 3 4
5 5 5
＝－ ×( ＋ ＋1)
12 7 7
5 17
＝－ ×
12 7
85
＝－ ················································································ 8 分
84
（3）原式＝－4+32 ·········································································· 10 分
＝28 ··············································································· 12 分
3 7 11
（4）原式＝－36× +36× －36× ···················································· 14 分
4 6 12
＝－27+42－33 ···································································· 15 分
＝－18 ············································································· 16 分
20．（本小题满分 8 分）
1
（1）图略；－2.5＜－1＜0＜1.5＜2 ＜3； ······················································· 4 分
2
（2）如图，以物流中心为原点，向东为正方向，每个单位长度表示 1 km.
丙 乙 甲
住 住 物流 住
户 户 中心 户
－ 1 . 5 － 1 0 1 2 ····················································· 8 分
（第 20 题）
数学试题参考答案与评分标准 第 1 页（共 4 页）
21.（本小题满分 12 分）
解：（1）原式＝－5a＋3a－2－3a＋7 ······························································· 3 分
＝－5a＋5················································································ 5 分
1 2 3 1
（2）原式＝ x－2x＋ y2－ x＋ y2 ······························································· 7 分
2 3 2 3
＝－3x＋y2． ············································································· 9 分
2
当 x＝－2，y＝ 时，
3
2
原式＝－3×(－2)＋( )2 ······································································ 10 分
3
4
＝6 ． ························································································· 12 分
9
22．（本小题满分 10 分）
1 1
解：（1）s＝a2－ a2－ ×4×b ················································ 4 分
2 2
1
＝ a2－2b． ·················································································· 6 分
2
（2）当 a＝6，b＝2 时，
1
s＝ ×62－2×2 ············································································· 7 分
2
＝14． ························································································ 10 分
23．（本小题满分 8 分）
解：100n＋80(n＋5) ·················································································· 4 分
＝100n＋80n＋400 ··················································································· 6 分
＝180n＋400． ························································································· 7 分
答：李明共需付款(180n＋400)元． ······························································ 8 分
24．（本小题满分 12 分）
解：（1）500； ··························································································· 3 分
（2）200； ··························································································· 6 分
（3）180×5＝900，(260－180)×7＝560，9(a－260)＝9a－2340，
900＋560＋(9a－2340) ···································································· 9 分
＝1460＋9a－2340
＝9a－880． ···················································································· 11 分
答：C 用户该年应缴纳水费(9a－880)元． ······················································· 12 分
数学试题参考答案与评分标准 第 2 页（共 4 页）
25．（本小题满分 12 分）
解：（1）m＝|－1|－2×2＝－3； ····································································· 2 分
（2）①当 b＜3 时，
m＝3＋2b.
∴3＋2b＝b，
∴b＝－3. ····················································································· 4 分
②当 b≥3 时，
m＝3－2b.
∴3－2b＝b，
∴b＝1（舍去）.
综上，b＝－3. ················································································ 6 分
（3）①当 a＞0 时，
∵a＋b＝0，
∴b＜0.
∴m＝|a|＋2b
＝a＋2b
＝a－2a
＝－a.
∴2a－3b＋4m
＝2a＋3a－4a
＝a＞0. ························································································ 9 分
②当 a＜0 时，
∵a＋b＝0，
∴b＞0.
∴m＝|a|－2b
＝－a＋2a
＝a.
∴2a－3b＋4m
＝2a＋3a＋4a
＝9a＜0.
综上，当 a＞0 时，2a－3b＋4m＞0；
当 a＜0 时，2a－3b＋4m＜0. ··················································· 12 分
26．（本小题满分 12 分）
（1）②④ ······························································································· 3 分
（2） ＝1000a＋100b＋10c＋d
＝999a＋99b＋9c＋(a＋b＋c＋d) . ··················································· 5 分
显然，999a，99b，9c 都能被 3 整除，若 a＋b＋c＋d 能被 3 整除，
那么 999a＋99b＋9c＋(a＋b＋c＋d)能被 3 整除，
于是 1000a＋100b＋10c＋d 能被 3 整除，即 能被 3 整除.······················· 7 分
数学试题参考答案与评分标准 第 3 页（共 4 页）
（3）
n n－＝an×10 ＋an－1×10 1＋…＋a1×10＋a0
＝(10n
－
－1) a ＋(10n 1n －1) an－1＋…＋(10－1)a1＋(an＋an－1＋…＋a1＋a0) . ········ 9 分
n －因为(10 －1) an ，(10n 1－1) an－1 ，…，(10－1)a1都能被 3 整除，
若 an＋an－1＋…＋a1＋a0 能被 3 整除，
－
那么(10n－1) an＋(10n 1－1) an－1＋…＋(10－1)a1＋(an＋an－1＋…＋a1＋a0)能被 3
整除，即 能被 3 整除． ···················································· 12 分
数学试题参考答案与评分标准 第 4 页（共 4 页）

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