江苏省南通市如东县部分学校2023-2024学年七年级下学期期中考试数学试卷

七年级数学期中试题第2页（共6页）七年级数学期中试题第1页（共6页）2023-2024学年度第二学期期中学情调研七年级数学试题（总分150分，考试时间120分钟）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）1.64的平方根是A.±8 B.±4 C.8 D.42.下列各数中，无理数是A.25 B.8 C.2273.下列调查中，适宜采用全面调查方式的是A.了解全国中学生的视力情况B.调查某批次日光灯的使用寿命C.调查市场上矿泉水的质量情况D.调查机场乘坐飞机的旅客是否携带了违禁物品4.在平面直角坐标系中，点P(m，−4−mA.第一象限 B.第二象限 C.第三象限 D.第四象限已知a，b是两个连续整数，a<3−3<b，则a+b的值是

2 B.3 C.4 D.5若x>y，则下列不等式中一定成立的是

A.x2>y2 B.−3x>−3y C.x7.不等式组x−1＞04−2x≥0的解集在数轴上表示为

A. B.C. D.8.若关于x，y的方程组3x−4y=8mx+(2m−1)y=7的解也是二元一次方程x+2y=1A.152 B.132 9.《孙子算经》是中国古代重要的数学著作，成书大约在一千五百年前，其中一道题，原文是：“今三人共车，两车空；二人共车，九人步．问人与车各几何？”意思是：现有若干人和车，若每辆车乘坐3人，则空余两辆车；若每辆车乘坐2人，则有9人步行．问人与车各多少？设有x人，y辆车，可列方程组为

A.x3=y+2x2+9=y B.x3=y−2x10.已知关于x，y的方程组x−y=1x+y=2a+3中x，y均大于0.若a与正数b的和为4A.−6<a−b<4 B.−6<a−b≤4C.−5<a−b<3二．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）11.11的相反数是▲.12.比较大小：−213.空气是由多种气体混合而成的，为例直观地介绍空气各成分的百分比，最合适使用的统计图是▲(从“条形图，扇形图，折线图和直方图”中选一个）14.已知点A坐标为(1,2)，点B在第四象限，直线AB//y轴.若线段AB=3，则点B的坐标为▲.15.若x+2y+3z=5，4x+3y+2z=10，则x+y+z的值是▲.16.某次知识竞赛共有20道题，每一题答对得10分，答错或不答都扣5分，小明得分要超过90分，他至少要答对▲道题.17.对于实数x，我们规定[x]表示不大于x的最大整数，例如已知[1.5]=1，[−2.7]=−3.若实数a满足[2−a]+1=0，则实数a的取值范围是18.已知关于x，y的方程m−2x+m−1y=3m+a，不论m是怎样的常数，总有一组解为x=2y=b(其中a，b是常数)三、解答题（本大题共8小题，共90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）19.(本小题12分)（1）计算：；（2）解方程组：.七年级数学期中试题第3页（共6页）(本小题8分)解不等式组，并写出它的所有非负整数解.(本小题10分)在等式y=ax2+bx+c中，当x=1时，y=−2；当x=−1时，y=20；当x=32与x=13(本小题10分)某校在暑假期间开展“心怀感恩，孝敬父母”的实践活动，并随机抽取了部分学生就暑假“平均每天帮助父母干家务所用时长”进行了调查，列出了下面的频数分布表：时长(分钟)0≤t<1010≤t<2020≤t<3030≤t<4040≤t<50频数6648m304百分比32n12%(1)在本次随机抽取的样本中，调查的学生人数是______人；(2)如表中，m的值为______，n的值为______；(3)如果该校共有学生3000人，请你估计该校暑假“平均每天帮助父母干家务的时长不少于30分钟”的学生大约有多少人？七年级数学期中试题第4页（共6页）(本小题10分）阅读下面材料：

分子、分母都是整式，并且分母中含有未知数的不等式叫做分式不等式．

小亮在解分式不等式时，是这样思考的：

根据两数相除，同号得正，异号得负．原分式不等式可转化为下面两个不等式组：

①

或

②

解不等式组①得，

解不等式组②得．

所以原不等式的解集为或．

请你参考小亮思考问题的方法，解分式不等式．(本小题12分)用1块A型钢板可制成1块C型钢板和2块D型钢板；用1块B型钢板可制成2块C型钢板和1块D型钢板.(1)若现需18块C型钢板和21块D型钢板，可恰好用A型钢板、B型钢板各多少块？(2)若A型钢板和B型钢板共14块，且能制成的C型钢板数多于D型钢板数，求A型钢板至多有多少块？七年级数学期中试题第5页（共6页）(本小题14分)如图，平面直角坐标系xOy中，点A，B坐标分别为(a,0)，(0,b)，且a，b满足(1)a=,(2)若点C(m，n)在线段AB上，m，n满足m+n=5，点D在y轴负半轴上，线段CD交x轴于点M，三角形MOD与三角形MAC(3)平移直线AB交x轴于点E，交y轴于点F，点P为直线EF上任意一点，且△PAB面积为15,直接写出OE的长.七年级数学期中试题第6页（共6页）（本小题14分）对于平面直角坐标系xOy中的点A，给出如下定义：若存在点B(不与点A重合，且直线AB不与坐标轴平行或重合)，过点A作直线m//x轴，过点B作直线n//y轴，直线m，n相交于点C.当线段AC，BC的长度相等时，称点B为点A的等距点，称三角形ABC的面积为点A的等距面积.例如：如图，点A(2,1)，点B(5,4)，因为AC=BC=3，所以B为点A的等距点，此时点A的等距面积为92．

(1)点A的坐标是(0,1)，在点B1(1,0)，B2((2)点A的坐标是(3,−①若点A的等距点B的坐标是(-1,a)，求点B的坐标和此时点A的等距面积；②若点A的等距面积不小于258，且点A的等距点B在第一象限，写出此时点B的横坐标t的取值范围.2023-2024学年度第二学期期中学情调研七年级数学试题参考答案选择题题号12345678910答案ABDABDCBCA二、填空题11.；12.＜；13.扇形图；14.（1，-1）；15.3；16.13；17.2＜a≤3；18.-5.三、解答题19．（本小题满分12分）（1）解：原式=·································································3分==··························································································6分（2）解：，得，解得x=6···································································································9分把x=6代入①，得···············································································11分所以原方程组的解是··············································································12分20．（本小题满分8分）解：解不等式①，得x≤2·······················································································2分解不等式②，得x＞-4·····················································································4分所以不等式组的解集为-4＜x≤2··········································································6分非负整数解为：0，1，2.·············································································8分（本小题满分10分）解：由题意得············································································3分①-②，得2b=-22b=-11····························································································5分由③，得77a=-42ba=6································································································7分把a=6,b=-11代入①，得c=3················································································9分所以，······················································································10分22．（本小题满分10分）解：（1）200·······································································································2分（2）52,26％······························································································6分（3）510···································································································10分23．（本小题满分10分）解：原分式不等式可转化为下面两个不等式组：①

或

②·····································································4分解不等式组①得··············································································6分解不等式组②得无解···············································································8分

所以原不等式的解集为·····································································10分24．（本小题满分12分）解：（1）设A型钢板用x块，B型钢板用y块.根据题意，得，·······································································4分解得：···········································································6分答：A型钢板用8块，B型钢板用5块.·······························································7分（2）设A型钢板用a块，依题意有···········································································9分解得．································································································11分答：A型钢板至多6块．···················································································12分25．（本小题满分14分）解：（1）4,6，12····························································································4分（2）由m+n=5,得n=5-m，所以点C(m，5-m)因为S△AOC+S△BOC=S△AOB所以解得m=2所以n=3所以C(2，3）·····························································································7分设D（0，y）由S△MOD=S△MAC，得S△BCD=S△AOB所以解得y=-6所以D（0，-6）·············································································