陕西省咸阳市实验中学2021-2022学年八年级上册数学第三次月考试题

这是一份陕西省咸阳市实验中学2021-2022学年八年级上册数学第三次月考试题，共8页。试卷主要包含了本试卷分为第一部分，领到试卷和答题卡后，请用0，如图，直线，比较大小等内容，欢迎下载使用。
试卷类型：A咸阳市实验中学2021~2022学年度第一学期第三次月考八年级数学试题注意事项：1.本试卷分为第一部分（选择题）和第二部分（非选择题）。全卷共4页，总分120分。考试时间120分钟。2.领到试卷和答题卡后，请用0.5毫米黑色墨水签字笔，分别在试卷和答题卡上填写姓名和准考证号，同时用2B铅笔在答题卡上填涂对应的试卷类型信息点（A或B）。3.请在答题卡上各题的指定区域内作答，否则作答无效。4.作图时，先用铅笔作图，再用规定签字笔描黑。5.考试结束，本试卷和答题卡一并交回。第一部分（选择题  共24分）一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.的算术平方根是（    ）A. B. C. D.2.在直角中，，若，，则的值为（    ）A.5 B.7 C.25 D.493.某班七个合作学习小组的人数分别如下：4，5，5，x，6，7，8，已知这组数据的平均数是6，则x的值是（    ）A.7 B.6 C.5.5 D.54.若是关于x，y的二元一次方程，则（    ）A.1 B.2 C.3 D.1或25.已知是关于x，y的二元一次方程的一组解，则k的值是（    ）A.-2 B.2 C.-3 D.36.在平面直角坐标系中，将直线沿x轴向左平移3个单位后恰好经过原点，则k的值为A.2 B.-2 C.-3 D.37.如图，直线（）与（）在第二象限交于点A，交x轴于点B，且，，，则关于x，y的方程组的解为（    ）A. B. C. D.8.如图，直线分别与x、y轴交于点A、B，点C在线段上，将沿翻折，点O落在边上的点D处，则的长为（    ）A.4 B.3 C.2 D.1第二部分（非选择题  共96分）二、填空题（共5小题，每小题3分，计15分）9.比较大小：_________0（填“>”、“=”或“<”）.10.在一次中学生田径运动会上，参加男子跳高的14名运动员成绩如下表所示：成绩/m1.501.611.661.701.751.78人数232151则这些运动员成绩的中位数是_________.11.若关于x，y的方程组的解满足，则_________.12.现用190张铁皮做铁盒，一张可以做8个盒身或22个盒底，1个盒身与2个盒底配一个盒子，问用多少张铁皮制盒身、多少张铁皮制盒底，可制成一批完整的盒子？若设用x张铁皮制盒身，y张铁皮制盒底，列方程组为_________.13.在一个的方格中填写了9个数字，使得每行、每列、每条对角线上的三个数之和相等，得到的的方格称为一个三阶幻方，如图的三阶幻方填写了一些数和字母，则_________.三、解答题（共13小题，计81分.解答应写出过程）14.（5分）计算：.15.（5分）解二元一次方程组：.16.（5分）已知点，，如果点A、B关于y轴对称，求a、b的值.17.（5分）如图，在四边形中，，，，，且.求的度数.18.（5分）如图，在平面直角坐标系中，各顶点的坐标分别为：，，.在图中作出关于x轴对称的，并写出点B的对应点的坐标.19.（5分）下表是一次函数（k，b为常数，）的自变量x与函数y的部分对应值.x-201y3p0（1）根据表格，求一次函数表达式；（2）求出p的值.20.（5分）一个两位数，个位上的数字与十位上的数字之和为9，把这个两位数的十位数字和个位数字对调所得新两位数比原两位数大27，请利用二元一次方程组求原两位数.21.（6分）2021年是中国共产党成立100周年.校团委以此为契机，组织了“讲好党史故事，传承红色基因”系列活动.如表是八年级甲、乙两个班各项目的成绩（单位：分）项目班次党史知识问答比赛讲述先烈故事比赛永远跟党走主题板报创作甲909396乙949191（1）如果根据三项成绩的平均分计算最后成绩，已知甲班的平均分是93分，请计算乙班的平均分，通过计算说明甲、乙两班谁将获胜；（2）如果将党史知识问答比赛、讲述先烈故事比赛、永远跟党走主题板报创作按5∶3∶2的比例确定最后成绩，已知乙班的最后成绩是92.5分，请计算甲班的最后成绩，并说明甲、乙两班谁将获胜.22.（7分）清朝数学家梅文鼎的著作《方程论》中有这样一道题：山田三亩，场地六亩共折实田四亩七分；又山田五亩，场地三亩，共折实田五亩五分，问每亩山田折实田多少，每亩场地折实田多少？译文为：假如有山田3亩，场地6亩，其产粮相当于实田4.7亩；有山田5亩，场地3亩，其产粮相当于实田5.5亩，问每亩山田和每亩场地产粮各相当于实田多少亩？请用方程组的知识解答.23.（7分）已知关于x，y的方程组与的解相同，求m和n的值.24.（8分）某公司销售部有营业员15人，该公司为了调动营业员的积极性，决定实行目标管理，根据目标完成的情况对营业员进行适当的奖励，为了确定一个适当的月销售目标，公司有关部门统计了这15人某月的销售量，如下表所示：每人销售件数1805125211512人数113532（1）求这15位营业员月销售量的平均数、中位数、众数；（2）如果想让一半左右的营业员都能获得奖励，销售部把每位营业员的月销售定额规定为32件，你认为是否合理，为什么？25.（8分）小美打算买一束百合和康乃馨组合的鲜花，在“母亲节”祝福妈妈.已知买1支康乃馨和2支百合共需花费14元，3支康乃馨的价格比2支百合的价格多2元.（1）求买一支康乃馨和一支百合各需多少元？（列二元一次方程组解答）（2）小美准备买康乃馨和百合共12支（2种花都要买），设买这束鲜花所需费用为w元，康乃馨有x支，求w与x之间的函数关系式，若小美共用55元买鲜花，则她可以买康乃馨和百合各多少支？26.（10分）已知M、N两地之间有一条240千米长的公路，甲乙两车同时出发，乙车以40千米/时的速度从Ｍ地匀速开往N地，甲车从N地沿此公路匀速驶往M地，两车分别到达目的地后停止，甲、乙两车相距的路程y（千米）与乙车行驶的时间x（小时）之间的函数关系如图所示.（1）甲车速度为_____________千米/时；（2）求甲、乙两车相遇后的y与x之间的函数关系式；（3）当甲车与乙车相距的路程为140千米时，求出乙车行驶的时间. 
试卷类型：A阳市实验中学2021~2022学年度第一学期第三次月考八年级数学试题参考答案及评分标准一、选择题（共8小题，每小题3分，计24分.每小题只有一个选项是符合题意的）1.D  2.C  3.A  4.C  5.D  6.A  7.B  8.B二、填空题（共5小题，每小题3分，计15分）9.<  10.1.68  11.1  12.  13.5三、解答题（共13小题，计81分.解答应写出过程）14.解：原式········································································（3分）.·················································································（5分）15.解：①②，得，·········································································（2分）解得，把代入①，得，·····································································（4分）所以方程组的解是···································································（5分）16.解：因为点，关于y轴对称，所以··············································································（3分）解得：············································································（5分）17.解：在中，根据勾股定理：，·························································（2分）因为在中，，，所以，············································································（3分）所以为直角三角形，所以.·············································································（5分）18.解：如图所示.····································································（3分）点的坐标为.········································································（5分）19.解：（1）由题意得：解得··············································································（2分）所以一次函数的表达式为.······························································（3分）（2）因为点在一次函数的图象上，所以.·············································································（5分）20.解：设原两位数的十位数字为x，个位数字为y，依题意得：·········································································（2分）解得：············································································（4分）所以.答：原两位数为36.···································································（5分）21.解：（1）乙班的平均成绩是：（分）··················································（2分）因为，所以甲班将获胜.·····································································（3分）（2）甲班的最后成绩是：（分）························································（5分）因为，所以乙班将获胜.·····································································（6分）22.解：设每亩山田产粮相当于实田x亩，每亩场地产粮相当于实田y亩，根据题意得：·······································································（4分）解得：答：每亩山田产粮相当于实田0.9亩，每亩场地产粮相当于实田亩.································（7分）23.解：由已知可得···································································（2分）解得··············································································（4分）把代入剩下的两个方程组成的方程组得解得··············································································（7分）24.解：（1）平均数是：，·····························································（3分）按从小到大顺序排列这组数据，第8个数为21，则中位数为21，·································（5分）21出现的次数最多，则众数为21.·························································（6分）（2）不合理.因为15人中有13人销售额达不到32件，人数达不到一半左右，所以不合理.·························（8分）25.解：（1）设买一支康乃馨需m元，买一支百合需n元，则根据题意得：·····································································（2分）解得：答：买一支康乃馨需4元，买一支百合需5元.················································（4分）（2）根据题意得：，（）·····························································（6分）当时，，解得，所以（支）答：若小美共用55元买鲜花，则她可以买康乃馨5支，百合7支.··································（8分）26.解：（1）80.·····································································（2分）（2）由图可知2小时相遇，因为乙车以40千米/时的速度匀速行驶，所以2小时走的路程：（千米），即甲离目的地80千米，所以甲再用1小时走完全部路程，此时甲、乙相距120千米，乙走完整个路程所用时间：（小时），所以，设线段函数关系式：（），此图象经过，，·····································································（3分）则有解得所以线段函数关系式：，······························································（5分）设线段函数关系式：（），此图象经过，，则有解得所以线段函数关系式：.································································（7分）（3）①相遇前相距140千米，设乙车行驶的时间为x小时，.解得，············································································（8分）②相遇后相距140千米，因为时，甲到目的地，此时两人相距120千米，所以令，即，解得.综上所述：乙车行驶的时间为3.5小时或小时.···············································（10分）