2025-2026学年七年级数学七年级月度质量检测质量检测卷（福建专用版·压轴题突破卷含答案详解与评分标准）

2025-2026学年七年级数学七年级月度质量检测质量检测卷（福建专用版·压轴题突破卷，含答案详解与评分标准）学校：________________班级：________________姓名：________________考号：________________考试时间：120分钟满分：120分范围标签：福建专用版试卷类型：月度质量检测题型结构：选择题（1—10题，共30分）；填空题（11—16题，共18分）；解答题（17—26题，共72分）。注意事项与答题要求1.答题前，请将学校、班级、姓名、考号填写清楚；全卷共26题，考试时间120分钟，满分120分。2.选择题每题只有一个正确答案，请把答案填写在答题栏中；填空题只填写最终结果，必要单位不能遗漏。3.解答题应写出必要的文字说明、运算过程和推理步骤；只写答案而无过程的，按评分标准酌情扣分。4.本卷按七年级月度质量检测要求命制，题目由基础到综合递进，压轴题注重数形结合、方程思想与分类讨论。选择题答题栏题号12345678910答案一、选择题（本大题共10小题，每小题3分，共30分。每小题只有一个正确答案）1.下列四个数中，最小的数是（）A.-3/2B.-1.4C.0D.|-2|2.数轴上点A表示-2，点B表示5，点M是线段AB的中点，则点M表示的数是（）A.1B.1.5C.2D.3.53.化简2(3x-1)-(x+4)，正确的结果是（）A.5x+2B.5x-6C.6x-6D.5x+64.若单项式-3a²bᵐ与5aⁿb³是同类项，则m+n的值为（）A.3B.4C.5D.65.方程3(x-2)=2x+1的解是（）A.x=5B.x=6C.x=7D.x=86.计算35°18′+24°42′，结果为（）A.59°60′B.60°10′C.59°50′D.60°7.某蓄水池原有水20m³，打开进水阀后每分钟增加8m³。若进水t分钟后水量为Vm³，则当t=6时，V的值为（）A.48B.60C.68D.808.一组数据为6，8，9，9，10，x，若这组数据的平均数是9，则x的值为（）A.9B.10C.12D.149.按规律排列的一列数为1，4，9，16，25，…，则第10个数是（）A.81B.90C.100D.12110.定义新运算a☆b=ab+a-b。例如2☆3=2×3+2-3=5。若2☆x=9，则x的值为（）A.5B.6C.7D.8二、填空题（本大题共6小题，每小题3分，共18分）11.计算：(-1)²⁰²⁵+|-4|-2³=__________。12.当x=-2，y=1/3时，代数式3x²y-2xy的值为__________。13.方程(x-1)/3-(x+2)/4=1的解为x=__________。14.单项式-5/6a²b³的系数是__________，次数是__________。15.观察下列图形小圆点个数的规律：第1个图形5个，第2个图形9个，第3个图形13个，按此规律，第n个图形的小圆点个数为__________。16.某地居民用电收费标准为：每月不超过200度时按0.55元/度收费，超过200度的部分按0.75元/度收费。小明家本月电费为155元，且用电量超过200度，则本月用电量为__________度。三、解答题（本大题共10小题，共72分。解答应写出必要过程）17.（6分）计算下列各题：

（1）-6+(-2)²÷4-(-3)；

（2）(-3/4)×(-8/9)+5/6÷(-5/12)。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（6分）先化简，再求值：A=2(3x²-xy)-[4x²-2(xy-y²)]，其中x=3，y=-1。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（6分）解方程：(2x-1)/3-(x+2)/6=1，并写出检验过程。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（8分）学校文具角购买笔记本和中性笔共8件，已知笔记本每本5元，中性笔每支3元，总费用34元。设购买笔记本x本。

（1）请根据题意列出方程；

（2）求购买笔记本和中性笔各多少件。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（8分）如图，A，O，B三点在同一直线上，射线OC在直线AB的上方。OD平分∠BOC，OE平分∠AOC，已知∠COD=28°。求∠EOD的度数，并说明理由。图中射线位置仅用于表示角的关系。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（8分）七年级某班开展“每日阅读”调查，随机记录30名学生某天课外阅读时间，统计表如下。

（1）求这30名学生阅读时间的众数和中位数；

（2）求这30名学生阅读时间的平均数；

（3）若全年级共有600名学生，请估计当天课外阅读时间不少于40分钟的学生人数。阅读时间/分钟2030405060人数481062解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（8分）从学校到科技馆全程9km。小明骑自行车前一段以12km/h行驶，后一段因道路施工以6km/h行驶，全程共用1小时。求两段路程分别是多少千米。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（8分）用火柴棒按如下规律摆连续相连的正方形：摆1个正方形需要4根，摆2个正方形需要7根，摆3个正方形需要10根。

（1）摆n个正方形需要多少根火柴棒？

（2）若有100根火柴棒，最多能摆多少个这样的正方形？

（3）若用2026根火柴棒，能否刚好摆成若干个这样的正方形？说明理由。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________25.（7分）某社区实行阶梯水价：每月用水不超过10t的部分按2.8元/t收费，超过10t的部分按3.6元/t收费。甲家本月用水xt（0<x≤10），乙家比甲家多用水6t，且乙家用水超过10t。若两家本月水费合计71.2元，求甲家本月用水量。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（7分）数轴上点A表示-8，点B表示4。点P从A出发以每秒2个单位长度的速度向右运动，点Q从B出发以每秒1个单位长度的速度向左运动，P、Q同时出发，运动时间为t秒（0≤t≤4）。

（1）当t=2时，分别写出点P、Q表示的数，并求PQ的长；

（2）用含t的式子表示点P、Q表示的数和线段PQ的长；

（3）若M为线段PQ的中点，求当M到原点的距离为1时t的值；

（4）是否存在t，使PQ=2PA？若存在，求t；若不存在，说明理由。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析本答案按题号1-26逐题给出，解答题附采分点。阅卷时可根据学生表达的等价性、步骤完整性和计算准确性在相应分值内赋分。一、选择题答案与解析题号12345678910答案ABBCCDCCCC1.答案：A。解析：在负数中，绝对值越大的数越小，-3/2=-1.5，小于-1.4；0与|-2|=2均大于负数。主要干扰项B容易因只比较数字大小而忽视负号。2.答案：B。解析：数轴上线段AB中点表示的数为(-2+5)/2=3/2=1.5。A项少算了端点距离的一半，C项把中点误看成整数。3.答案：B。解析：2(3x-1)-(x+4)=6x-2-x-4=5x-6。D项常见错误是去括号时没有改变第二个括号内各项符号。4.答案：C。解析：同类项要求所含字母相同且相同字母指数分别相同，所以n=2，m=3，m+n=5。A、B项只取了其中一个指数，D项为指数误加后多算。5.答案：C。解析：3(x-2)=2x+1，去括号得3x-6=2x+1，移项得x=7。6.答案：D。解析：18′+42′=60′=1°，35°+24°+1°=60°。A项没有把60′化成1°。7.答案：C。解析：水量V=20+8t，当t=6时，V=20+8×6=68m³。B项漏加原有水量20m³的一部分。8.答案：C。解析：平均数是9，6个数总和为9×6=54，已知五个数和为6+8+9+9+10=42，所以x=12。9.答案：C。解析：1，4，9，16，25分别为1²，2²，3²，4²，5²，因此第10个数为10²=100。10.答案：C。解析：由定义2☆x=2x+2-x=x+2。令x+2=9，得x=7。干扰项A把2☆x误算成2x-1。二、填空题答案与解析11.答案：-5。解析：(-1)²⁰²⁵=-1，|-4|=4，2³=8，所以原式=-1+4-8=-5。12.答案：16/3。解析：3x²y=3×(-2)²×1/3=4，-2xy=-2×(-2)×1/3=4/3，故代数式值为4+4/3=16/3。13.答案：22。解析：方程两边同时乘12，得4(x-1)-3(x+2)=12，化简得4x-4-3x-6=12，所以x=22。14.答案：-5/6；5。解析：单项式中的数字因数是系数，故系数为-5/6；所有字母指数之和为2+3=5，故次数为5。15.答案：4n+1。解析：小圆点个数依次为5，9，13，每次增加4，可写为5+4(n-1)=4n+1。16.答案：260。解析：前200度费用为200×0.55=110元，超出部分费用为155-110=45元，超出电量为45÷0.75=60度，总用电量为260度。三、解答题答案、解析与评分标准17.答案与解析：（1）-6+(-2)²÷4-(-3)=-6+4÷4+3=-6+1+3=-2。（2）(-3/4)×(-8/9)+5/6÷(-5/12)=2/3+5/6×(-12/5)=2/3-2=-4/3。易错点：乘方应先算(-2)²=4；除以分数要转化为乘它的倒数，并注意符号。评分标准：第（1）小题正确处理乘方、除法、符号并得到-2，给3分；第（2）小题正确转化除法并得到-4/3，给3分。18.答案与解析：A=2(3x²-xy)-[4x²-2(xy-y²)]。先去括号：2(3x²-xy)=6x²-2xy，4x²-2(xy-y²)=4x²-2xy+2y²。因此A=6x²-2xy-(4x²-2xy+2y²)=6x²-2xy-4x²+2xy-2y²=2x²-2y²。当x=3，y=-1时，A=2×3²-2×(-1)²=18-2=16。易错点：中括号前有减号，去括号后括号内各项符号都要改变。评分标准：正确展开两个括号2分，正确去中括号并合并为2x²-2y²得2分，正确代入并求得16得2分。19.答案与解析：方程(2x-1)/3-(x+2)/6=1，两边同乘6，得2(2x-1)-(x+2)=6。化简得4x-2-x-2=6，即3x-4=6，3x=10，所以x=10/3。检验：把x=10/3代入左边，(20/3-1)/3-(10/3+2)/6=(17/3)/3-(16/3)/6=17/9-8/9=1，等于右边，故解为x=10/3。评分标准：正确找到最小公倍数6并去分母2分；正确去括号、移项、合并并求出x=10/3给3分；写出代入检验或合理说明1分。20.答案与解析：设购买笔记本x本，则购买中性笔(8-x)支。根据总费用34元可列方程：5x+3(8-x)=34。解得：5x+24-3x=34，2x=10，x=5。所以笔记本5本，中性笔8-5=3支。检验：5×5+3×3=25+9=34，件数5+3=8，符合题意。评分标准：设未知数并表示中性笔数量8-x给2分；依据总费用列出方程5x+3(8-x)=34给3分；解方程求得x=5给2分；答出两种文具数量并检验或说明合理性给1分。21.答案与解析：因为OD平分∠BOC，且∠COD=28°，所以∠BOC=2×28°=56°。A，O，B三点在同一直线上，所以∠AOC+∠BOC=180°，从而∠AOC=180°-56°=124°。又因为OE平分∠AOC