学年度第一学期期末试卷 第1页（共5页）北京市西城区2022—2023学年度第一学期期末试卷七年级数学答案及评分参考 2023.1一、选择题（共16分，每题2分）二、填空题（共16分，每题2分）9．4.07． 10．125． 11．9−． 12．答案不唯一，如：4−a b ．13．3． 14．5−． 15．(0.950)−a ． 16．3三、解答题（共68分，第17题18分，第18-19题，每题6分，第20题11分，第21题6分，第22-24题，每题7分）17．解：（1）12(6)(28)−+−−−＝18(28)−−− ······································································································· 2分 ＝10． ·················································································································· 4分（2）815()(9)54−⨯÷− ＝8151549⨯⨯ ·········································································································· 3分 ＝23． ···················································································································· 4分 （3）375()(48)16246−−+⨯− ＝91440+− ········································································································· 3分 ＝17−． ················································································································ 5分（4）2273(1)(2)8−+−⨯− ＝19()48−+−⨯ ····································································································· 3分学年度第一学期期末试卷 第2页（共5页）＝192−− ··············································································································· 4分 ＝192−． ··············································································································· 5分 18．解：（1）点C 在直线AB 外； ································ 1分（2）如图所示； ················································ 3分（3）∵DC ＝AD ＋AC ，AD ＝AB ，∴DC ＝AB ＋AC ．∵AB ＋AC ＞ BC ，（ 两点之间，线段最短 ）∴DC ＞ BC ． ····························································································· 6分19．解：2223(2)(37)10−−++x y y x y222363710=−−−+x y y x y ························································································ 2分24=−+x y ． ················································································································· 4分 当14=−x ，5=y 时， 原式214()54=−⨯−+ ···································································································· 5分 125=+26=． ··················································································································· 6分20．（1）7202(33)−=−x x解：去括号，得 72066−=−x x ． ·············································································· 1分移项，得 76620+=+x x ． ·················································································· 3分合并同类项，得 1326=x ． ·················································································· 4分系数化1，得 2=x ． ····························································································· 5分（2）2331152−−=+x x 解：去分母，得 2(23)5(31)10−=−+x x ． ································································ 2分去括号，得 4615510−=−+x x ． ······································································· 3分学年度第一学期期末试卷 第3页（共5页）移项，得 4156510−=−+x x ． ··········································································· 4分合并同类项，得 1111−=x ． ················································································· 5分系数化1，得 1=−x ． ··························································································· 6分21．解：（1）①如图所示； ············································ 1分②∵∠EOD ＝90°，∴∠EOC ＋∠ COD ＝90°． ············ 2分∵∠AOB ＝∠COD ，∴∠EOC ＋∠ AOB ＝90°． ············ 3分∵∠AOC ＝90°，∴∠EOC ＋∠AOE ＝90°．∴∠ AOB ＝∠ AOE ．（ 同角的余角相等 ） ······································· 5分∴OA 平分∠EOB ．（2）OC ，EOD ． ······································································································ 6分22．方法一：3x ，4(2)+x ； ································································································ 2分解：设每台A 型机器一天生产x 件产品．依题意列方程，得 34(2)57+=x x ． ······································································ 4分 解得 40=x ． ·········································································································· 5分所以3245=x ．·········································································································· 6分 答：每台A 型机器一天生产40件产品，每箱装24件产品． ··································· 7分方法二：5x ，7x ； ········································································································ 2分解：设每箱装x 件产品．依题意列方程，得 57234+=x x ．········································································· 4分 解得 24=x ． ·········································································································· 5分所以5403=x ．·········································································································· 6分 答：每台A 型机器一天生产40件产品，每箱装24件产品． ··································· 7分学年度第一学期期末试卷 第4页（共5页）23．解：（1）①∵∠AOB ＝∠AOC ＋∠BOC ，∠AOC ＝4∠BOC ，∴∠AOB ＝4∠BOC ＋∠BOC ＝5∠BOC ．∵∠AOB ＝75°，∴5∠BOC ＝75°．∴∠BOC ＝15°． ································ 2分②∵∠EOC 与∠DOB 互余，∴∠EOC ＋∠DOB ＝90°． ········································································ 3分∵OE 平分∠DOC ，∴∠DOC ＝2∠EOC ． ·················································································· 4分∴∠DOB ＝∠DOC ＋∠BOC ＝2∠EOC ＋15°．∴∠EOC ＋2∠EOC ＋15°＝90°．∴∠EOC ＝25°． ························································································ 5分（2）(902−n )°或 (902+n )°． ············································································· 7分 24．解：（1）1； ······················································································································ 1分（2）∵点D 是点B 关于点A 的“k 倍分点”，∴DB ＝kDA ．∵AD ＝10，点A 表示的数是4−，∴当点D 在线段BA 的延长线上时，点D 表示的数是14−．此时DB ＝2(14)−−＝16，则k ＝DB DA ＝85． ·········································· 3分 当点D 在线段AB 的延长线上时，点D 表示的数是6．此时DB ＝62−＝4，则k ＝DB DA ＝25． ∴k ＝85或25． ·································································································· 4分 （3）12，23,6,8． ································································································· 7分学年度第一学期期末试卷 第5页（共5页）四、选做题（共10分，第25题4分，第26题6分）25．解：（1）正，负， ············································································································ 1分 用较大的绝对值减去较小的绝对值； ···························································· 2分（2）①8−； ·············································································································· 3分 ②答案不唯一．如：[(1)(1)](3)0(3)3−⊗+⊗+=⊗+=+，(1)[(1)(3)](1)(2)1−⊗+⊗+=−⊗+=−，所以[(1)(1)](3)(1)[(1)(3)]−⊗+⊗+≠−⊗+⊗+．此时()()⊗⊗=⊗⊗a b c a b c 不成立． ······················································ 4分26．解：（1）①∵AB ＝1，∴b ＝a ＋1．∵m ＝5，∴BC ＝m ＋3＝8，∴c ＝a ＋1＋8＝a ＋9． ················································································ 1分②∵CD ＝m ＋4＝9，∴d ＝a ＋9＋9＝a ＋18．∴a ＋b ＋c ＋d ＝a ＋(a ＋1)＋(a ＋9)＋(a ＋18)＝4a ＋28＝4(a ＋7)．∵a 为整数，∴a ＋b ＋c ＋d 能被4整除． ······································································· 2分（2）①B ，D ； ·········································································································· 4分 ②a ＝52−−m 或42−−m ． ··············································································· 6分。
七年级数学参考答案及评分标准一、选择题（本题共30分，每小题3分）三、解答题（本题共27分，第21、23题每小题6分，其余每小题5分）19． 2311,54.3x x x x +<+⎧⎪+⎨>⎪⎩解：解不等式①，得8x <． …………………………………………………………2分解不等式②，得2x >-． …………………………………………………………4分 把不等式①和②的解集在数轴上表示出来．所以原不等式组的解集为28x -<<． …………………………………………5分20．解：（1） ……………………………2分（2）2(21)(21)(3)x x x +---22(2)1(69)x x x =---+ ……………………………………………………4分224169x x x =--+-23610x x =+-． ……………………………………………………………5分21．解：（1）①右，3，上，5； ……………………………………………………………2分 ②（6，3）； …………………………………………………………………4分 （2）如图，过点B 作BD ⊥x 轴于点D ，则点D 的坐标为（6，0）．∵点A ，B ，C 的坐标分别为（0，4），（6，3），（4，0），∴ABC AOC BCD AODB S S S S ∆∆∆=--梯形① ②111()222AO BD OD AO OC CD BD =+⋅-⋅-⋅111(43)644(64)3222=⨯+⨯-⨯⨯-⨯-⨯10=． ………………………………6分 22．解：∵AD 是△ABC 的角平分线，∴∠1=∠2． ………………………………1分 ∵DE ∥AB ，∴∠1=∠3． ………………………………2分 ∴∠2=∠3． ………………………………3分 ∵∠DEC 是△ADE 的外角，∴∠DEC =∠2+∠3=2∠3． ………………………………………………………4分 ∵∠DEC =100°，∴∠3=50°，即∠ADE =50°． ……………………………………………………5分23．解：（1）12，8； ………………………………………………………………………2分（2）设购进A 型电动汽车x 辆．根据题意，得 (16.816)(29.428)(20)19.3x x -+-->． …………………4分解得 1142x <． (5)分∵x 为正整数，∴x 最大为14． ……………………………………………6分 答：A 型电动汽车最多购进14辆．四、解答题（本题共19分，第26题7分，其余每小题6分） 24．解：（1）6%，5%； ..............................................................................2分 （2）补全折线图如图所示； (4)分（3）答案不唯一，理由支撑相应数据即可． ……………………………………6分如：180，预计2017至2018增长的倍数与之前相比可能会减小．解：（1）如图1所示； …………………………………………………………………4分 （说明：画对1~3块，每对1块得1分；全部画对得4分）（2）答案不唯一．如：图2所示． ………………………………………………6分2014~2017年除夕微信红包收发总量统计图图1 图2解：（1）- ………………………………………………………………………2分（2）17-； …………………………………………………………………………4分（3）∵1(34)4(1) 5m n n -<---<． ∴1 3mn <<．∵m ，n 都为整数， ∴2mn =． ∴3m n +=或3-． …………………………………………………………6分 26．解：（1）①补全图形如图1所示； ………………1分 ②30； ………………………………………2分（2）如图2．∵∠CDA =∠CAB ，∠CDF =∠CAD ，∴∠CDA —∠CDF =∠CAB —∠CAD ，即∠1=∠2． ………………………………3分∴FD ∥AB ．∴∠AFD ＋∠FAB =180°． ………………4分（3）∵在△CAB 中，∠CAB +∠CBA +∠C =180°， 在△CAD 中，∠CAD +∠CDA +∠C =180°，又∵∠CDA =∠CAB ， ∴∠CAD =∠CBA ．①当点P 在线段AB 上时，如图3．∵在△DPB 中，∠DPB +∠DBP +∠BDP =180°，∴∠DPB +∠CAD +∠BDP =180°． ………………………………………6分②当点P 在线段AB 的延长线上时，如图4． ∵在△DPB 中，∠DBA =∠BDP +∠DPB ，∴∠CAD =∠BDP +∠DPB ． ……………………………………………7分图1图2图4《。
2022-2023学年北京市西城区七年级（下）期末数学试卷一、选择题（共16分，每题2分）第1-8题均有四个选项，符合题意的选项只有一个。
1．实数3.1415，√23，−57，√9中，无理数是（ ） A ．3.1415B ．√23C ．−57D ．√92．若m ＜n ，则下列各式中正确的是（ ） A ．m ﹣n ＞0B ．m ﹣9＞n ﹣9C ．m +n ＜2nD ．−m 4＜−n 43．如图，直线AB ，CD 相交于点O ，EO ⊥AB ，垂足为O ，∠DOE ＝37°，∠COB 的大小是（ ）A ．53°B ．143°C ．117°D ．127°4．下列命题中，是假命题的是（ ）A ．如果两个角相等，那么它们是对顶角B ．同旁内角互补，两直线平行C ．如果a ＝b ，b ＝c ，那么a ＝cD ．负数没有平方根5．在平面直角坐标系中，点A （1，5），B （m ﹣2，m +1），若直线AB 与y 轴垂直，则m 的值为（ ） A ．0B ．3C ．4D ．76．以下抽样调查中，选取的样本具有代表性的是（ ） A ．了解某公园的平均日客流量，选择在周末进行调查 B ．了解某校七年级学生的身高，对该校七年级某班男生进行调查C ．了解某小区居民坚持进行垃圾分类的情况，对小区活动中心的老年人进行调查D ．了解某校学生每天体育锻炼的时长，从该校所有班级中各随机选取5人进行调查7．以某公园西门O 为原点建立平面直角坐标系，东门A 和景点B 的坐标分别是（6，0）和（4，4）．如图1，甲的游览路线是：O →B →A ，其折线段的路程总长记为l 1，如图2，景点C 和D 分别在线段OB ，BA 上，乙的游览路线是：O →C →D →A ，其折线段的路程总长记为l 2，如图3，景点E 和G 分别在线段OB ，BA 上，景点F 在线段OA 上，丙的游览路线是：O →E →F →G →A ，其折线段的路程总长记为l 3．下列l 1，l 2，l 3的大小关系正确的是（ ）A ．l 1＝l 2＝l 3B ．l 1＜l 2且l 2＝l 3C ．l 2＜l 1＜l 3D ．l 1＞l 2且l 1＝l 38．有8张形状、大小完全相同的小长方形卡片，将它们按如图所示的方式（不重叠）放置在大长方形ABCD 中，根据图中标出的数据，1张小长方形卡片的面积是（ ）A ．72B ．68C ．64D ．60二、填空题（共16分，每题2分）9．若{x =3y =−2是方程ax +y ＝10的解，则a 的值为 ．10．在平面直角坐标系中，已知点P 在第四象限，且点P 到两坐标轴的距离相等，写出一个符合条件的点P 的坐标： ． 11．若一个数的平方等于964，则这个数是 ．12．如图，在三角形ABC 中，∠C ＝90°，点B 到直线AC 的距离是线段 的长，BC ＜BA 的依据是 ．13．点M ，N ，P ，Q 在数轴上的位置如图所示，这四个点中有一个点表示实数√5−1，这个点是 ．14．解方程组{3x +4y =16①5x −6y =33②，小红的思路是：用①×5﹣②×3消去未知数x ，请你写出一种用加减消元法消去未知数y 的思路：用 消去未知数y ．15．如图，四边形纸片ABCD ，AD ∥BC ，折叠纸片ABCD ，使点D 落在AB 上的点D 1处，点C 落在点C 1处，折痕为EF ．若∠EFC ＝102°，则∠AED 1＝ °．16．小明沿街心公园的环形跑道从起点出发按逆时针方向跑步，他用软件记录了跑步的轨迹，他每跑1km 软件会在运动轨迹上标注相应的路程，前5km 的记录如图所示．已知该环形跑道一圈的周长大于1km ． （1）小明恰好跑3圈时，路程是否超过了5km ？答： （填“是”或“否”）； （2）小明共跑了14km 且恰好回到起点，那么他共跑了 圈．三、解答题（共68分，第17题6分，第18题14分，第19题7分，第20题9分，第21-24题，每题8分）解答应写出文字说明、演算步骤或证明过程。
