2026届七年级数学七年级月度质量检测质量检测卷（陕西专用版·考前适应卷含答案详解与评分标准）

2026届七年级数学七年级月度质量检测质量检测卷（陕西专用版·考前适应卷，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分适用范围：陕西专用版题量：26题题型结构：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.本卷为七年级数学月度质量检测考前适应卷，答题前请先填写学校、班级、姓名和考号。2.选择题每题只有一个正确选项，请把答案填入选择题答题栏；填空题直接把答案写在题后横线上。3.解答题要写出必要的计算过程、推理过程或文字说明；书写应清楚，单位应完整，结果应化到最简。4.全卷满分120分，作答时请注意题号、分值与答题区域，不得在试题主体中标记答案。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。每小题只有一个选项符合题意）1.下列各数中，-5/3的相反数是（）。A.-5/3B.5/3C.-3/5D.3/52.计算(-2)^3+|-6|的结果是（）。A.-14B.-2C.2D.143.比较5/6与0.83的大小，正确的是（）。A.5/6<0.83B.5/6=0.83C.5/6>0.83D.无法比较4.单项式-3/4x^2y的系数和次数分别是（）。A.-3，2B.-3/4，2C.-3/4，3D.3/4，35.化简3a-2b+5a+4b的结果是（）。A.8a+2bB.2a+2bC.8a-6bD.-2a+6b6.方程2x-3=5的解是（）。A.x=1B.x=3C.x=4D.x=57.某班七年级学生共有x人，其中参加数学社团的有18人，未参加数学社团的人数比参加人数的2倍少6人。根据题意可列方程为（）。A.x-18=2×18-6B.x+18=2x-6C.18=2(x-18)-6D.x=2×18-68.已知a+b=3，ab=-2，则2a+2b-3ab的值为（）。A.0B.6C.10D.129.线段AB=12cm，点C是AB的中点，点D在线段CB上，且CD=2cm，则AD的长为（）。A.4cmB.6cmC.8cmD.10cm10.小秦从1月开始每月固定存入零花钱。1月存20元，2月存35元，3月存50元，4月存65元，按此规律，第8月应存（）。A.110元B.120元C.125元D.140元二、填空题（本大题共6小题，每小题3分，共18分）11.-0.25的倒数是__________。12.陕西某景区一季度接待游客约4350000人次，用科学记数法表示为__________人次。13.当x=-2，y=1时，代数式3x^2-2xy+1的值为__________。14.方程5x-1=3x+7的解是x=__________。15.若一个角的度数为37°18′，则它的余角为__________。16.某小组7名同学的数学小测成绩为9，7，8，9，10，8，8，则这组数据的众数是__________。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）-18+(-7)-(-15)+23；

（2）(-3)^2-12÷(-4)+2×(-5)。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.（6分）先化简，再求值：5(2x-3y)-3(4x-y)+2x，其中x=3，y=-1/2。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________19.（7分）解下列方程。

（1）3(x-2)+4=2x+9；

（2）(x-2)/3+(x+1)/2=4。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________20.（7分）为迎接校园读书月，某班计划购买甲、乙两种笔记本共30本。甲种笔记本每本8元，乙种笔记本每本5元，共花费195元。设购买甲种笔记本x本，请列方程并求甲、乙两种笔记本各购买多少本。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________21.（8分）七年级数学兴趣小组记录了某周西安清晨5天的气温变化，以0℃为基准，记录如下表。

请根据表格回答：

（1）这5天清晨气温的最高值与最低值分别是多少？

（2）这5天清晨气温的极差是多少？

（3）这5天清晨气温的平均值是多少？星期周一周二周三周四周五气温/℃-2130-1解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________22.（8分）如图示意：在同一平面内，射线OC在∠AOB内部，∠AOB=150°，∠BOC=2∠AOC。射线OD平分∠AOC，射线OE平分∠BOC。

（1）求∠AOC与∠BOC的度数；

（2）求∠COD与∠DOE的度数。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________23.（8分）某校七年级开展“每天阅读多长时间”调查，随机抽取40名学生，整理结果如下表。

请完成：

（1）求本次调查中每天阅读时间的平均数；

（2）求众数和中位数；

（3）若七年级共有520名学生，请估计每天阅读时间不少于40分钟的学生人数。每天阅读时间/分钟2030405060人数6101284解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________24.（8分）学校计划在操场一角围成一个长方形劳动实践区，长为(3a+2)m，宽为(a-1)m。

（1）用含a的式子表示该长方形的周长；

（2）若a=12，求周长；

（3）若在长方形一侧留出2m宽的入口，其余部分围栏，求当a=12时需要围栏的长度。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________25.（7分）七年级开展秦岭研学活动，共有师生300人参加。计划租用甲、乙两种客车共8辆，甲种客车每辆有45个座位，乙种客车每辆有30个座位，若恰好坐满。

（1）求甲、乙两种客车各租多少辆；

（2）若甲种客车租金为900元/辆，乙种客车租金为650元/辆，租车预算6200元是否够用？说明理由。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________26.（7分）在数轴上，点A表示-6，点B表示12。点P从A出发向右运动，速度为每秒2个单位；点Q从B出发向左运动，速度为每秒1个单位。两点同时出发，运动时间为t秒。

（1）用含t的式子表示t秒后点P、点Q所表示的数；

（2）求点P与点Q相遇时的t值；

（3）当PQ=3时，求t的值。解：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析客观题答案速查题号12345678910答案BBCCACADCC题号111213141516答案-44.35×10^617452°42′8一、选择题解析与评分标准1.答案：B。-5/3的相反数是与它符号相反、绝对值相等的数，即5/3。A是原数，C、D是倒数相关的干扰项，因此选B。评分：选B得3分。2.答案：B。先算乘方和绝对值：(-2)^3=-8，|-6|=6，所以(-2)^3+|-6|=-8+6=-2。易错点是把(-2)^3误算成8。评分：选B得3分。3.答案：C。把5/6化成小数约为0.8333…，而0.83=0.8300…，所以5/6>0.83。也可交叉比较：5×100=500，83×6=498，500>498。评分：选C得3分。4.答案：C。单项式-3/4x^2y的数字因数是-3/4，因此系数为-3/4；字母指数和为2+1=3，因此次数为3。评分：选C得3分。5.答案：A。合并同类项时，a项与a项相加，b项与b项相加：3a+5a=8a，-2b+4b=2b，所以结果为8a+2b。评分：选A得3分。6.答案：C。由2x-3=5，移项得2x=8，两边同时除以2，得x=4。评分：选C得3分。7.答案：A。参加数学社团18人，未参加人数为x-18；题中“未参加人数比参加人数的2倍少6人”，即x-18=2×18-6。评分：选A得3分。8.答案：D。2a+2b-3ab=2(a+b)-3ab。把a+b=3，ab=-2代入，得2×3-3×(-2)=6+6=12。评分：选D得3分。9.答案：C。C是AB的中点，所以AC=CB=6cm。D在线段CB上且CD=2cm，故AD=AC+CD=6+2=8cm。评分：选C得3分。10.答案：C。每月存入的钱数依次增加15元，可写成第n月存20+15(n-1)=15n+5元。第8月为15×8+5=125元。评分：选C得3分。二、填空题解析与评分标准11.答案：-4。-0.25=-1/4，倒数是-4。评分：填-4得3分。12.答案：4.35×10^6。4350000=4.35×1000000=4.35×10^6。科学记数法要求1≤a<10，本题a=4.35。评分：写成4.35×10^6得3分。13.答案：17。把x=-2，y=1代入3x^2-2xy+1，得3×(-2)^2-2×(-2)×1+1=12+4+1=17。评分：填17得3分。14.答案：4。5x-1=3x+7，移项得5x-3x=7+1，即2x=8，所以x=4。评分：填4得3分。15.答案：52°42′。两个角互为余角时和为90°。90°00′-37°18′=52°42′。评分：填52°42′得3分。16.答案：8。数据中8出现3次，9出现2次，7和10各出现1次，出现次数最多的是8，所以众数为8。评分：填8得3分。三、解答题参考解答、关键步骤与评分标准17.（6分）（1）-18+(-7)-(-15)+23=-25+15+23=-10+23=13。（2）(-3)^2-12÷(-4)+2×(-5)=9-(-3)-10=9+3-10=2。知识点：有理数混合运算，先算乘方、绝对值、乘除，再算加减；同级运算按从左到右。易错点：12÷(-4)=-3，前面是减去该结果，因此应变为加3。评分标准：每小题3分。第（1）小题列式处理符号1分，正确计算中间结果1分，结果13得1分；第（2）小题乘方和除法各1分，符号处理与结果2各1分。18.（6分）5(2x-3y)-3(4x-y)+2x=10x-15y-12x+3y+2x。合并同类项，x项为10x-12x+2x=0，y项为-15y+3y=-12y，所以原式=-12y。当y=-1/2时，原式=-12×(-1/2)=6。知识点：去括号、合并同类项、代入求值。易错点：-3(4x-y)=-12x+3y，括号前是负数时括号内每一项都要变号。评分标准：正确去括号2分，合并同类项得-12y2分，代入计算得62分。19.（7分）（1）3(x-2)+4=2x+9，去括号得3x-6+4=2x+9，整理得3x-2=2x+9，移项得x=11。（2）(x-2)/3+(x+1)/2=4，两边同乘6，得2(x-2)+3(x+1)=24。去括号得2x-4+3x+3=24，合并得5x-1=24，移项得5x=25，所以x=5。知识点：一元一次方程的去括号、去分母、移项与合并。易错点：去分母时不要漏乘等号两边的每一项。评分标准：第（1）小题3分，去括号1分，移项合并1分，解得x=111分；第（2）小题4分，正确去分母2分，去括号合并1分，解得x=51分。20.（7分）设购买甲种笔记本x本，则购买乙种笔记本(30-x)本。根据总费用195元列方程：8x+5(30-x)=195。解得8x+150-5x=195，3x=45，x=15。所以甲种笔记本购买15本，乙种笔记本购买30-15=15本。检验：15×8+15×5=120+75=195，数量15+15=30，符合题意。知识点：用一元一次方程解决总量与单价问题。关键是把未知数量