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1.本试卷共6页，满分为150分，考试时间为100分钟。考试结束后，请将答题卡交回。
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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
下列各数中无理数是
A.-√2
B.0.311
C.22
D.8
2.在平面直角坐标系中，点P(一5,10)所在的象限是
A.第一象限
B.第二象限
C.第三象限
D.第四象限
多，如图，测量运动员跳远成绩时应该选取的量是
H
P
板
落地点
跳
(第3题)
A.线段PA的长度
B.线段PB的长度
C.线段PM的长度
D.线段PH的长度
÷.如果点P(m+3,m+1)在x轴上，那么点P的坐标是
A.(0,-1)
B.(-1,0)
C.(2,0)
D.(0,-2)
5.对于命题“若a+b<0,则a<0,b<0”,下列能说明该命题是假命题的是
A.a=6,b=8
B.a=-6,b=8
C.a=6,b=-8
D.a=-6,b=-8
5.
我国古代数学著作《九章算术》中有这样一题，原文是：“今有大器五小器一容三斛，
大器一小器五容二斛，问大小器各容几何？”意思是：有大小两种盛酒的桶，已知5
七年级数学第1页（共6页）
小補可以盛酒2斛.问：1个大桶、1个小桶各盛酒多少槲？若设1个大桶可以盛酒x斛，
1个小桶可以盛酒y斛，则列方程组是
A5x+y=2,
B.5x+y=3,
5x+y=3,
C
5x=y+3,
x+5y=3
x+5y=2
x=5y+2
x+5y=2
7.如图，给出下列条件：①∠1=∠2：②∠3=∠4：③AB∥CE且∠ADC=∠B:④AB∥CE
且∠BCD=∠BAD.利用其中一个条件能推出BC∥AD的是
A.①②
B.②④
C.②③
D.②③④
4
3
4-3-2-月
0
1
2
6:
(第7题)
(第10题)
8.将点A(α，b)先向左平移2个单位长度，再向下平移3个单位长度得到点B(一b,a),
则点B的坐标是
A3-2B(-32
c2
D.32
9.若【x】表示实数x的整数部分，
=1,<√2>=√2-1，则<3-V5>+【万】的值是
A.4-V3
B.1-V13
C.6-3
D.V13-1
10.如图，在平面直角坐标系中，动点A从(1,0)出发，向上运动1个单位长度到达点B
(1,1),分裂为两个点，分别向左、右运动到点C(0,2),D(2,2),此时称动
点A完成第一次跳跃；再分别从C,D点出发，每个点重复上面的运动，到达点G
(一1,4)，H(1,4),1(3,4),此时称动点A完成第二次跳跃：依此规律跳跃
下去，动点A完成第2023次跳跃时，最右边一个点的坐标是
A.(2023,4046)B.(2023,22023)C.（2024,4046)D.(2024,22023)
七年级数学第2页（共6页)2022~2023 学年度第二学期期中学业质量监测试卷
七年级数学
参考答案及评分标准
说明：本评分标准每题给出的解法仅供参考，如果考生的解法与本解答不同，参照本评分标
准给分．
一、选择题（本大题共 10 小题，每小题 3 分，共 30 分）
题号 1 2 3 4 5 6 7 8 9 10
选项 A B D C C B D A A C
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）
7 2x
11．5 12．< 13．－3 14．
3
11 25
15．4 16．103 17．－1 18． 或
2 4
三、解答题（本大题共 8 小题，共 90 分）
19．（本小题满分 10 分）
解：（1）原式＝2 2 3 + 3 2 ······················································ 3 分
＝ 2 ． ··········································································· 5 分
（2）原式= 4 ( 3)+ 2 1 ··························································· 8 分
= 6+ 2 ． ····································································· 10 分
20．（本小题满分 12 分）
解：（1）把②代入①，得
2x 3(x 1) =1．
解这个方程，得
x = 2． ················································ 3 分
把 x = 2代入②，得
y =1 ． ··············································· 5 分
所以这个方程组的解是
x = 2，
···················································· 6 分
y =1.
（2）原方程组可化为
七年级数学试卷参考答案和评分标准 第 1 页（共5页）
4m 6n =13，①
···································· 7 分
4m 3n = 7.②
①－②，得
－3n=6．
n=－2． ················································· 9 分
把 n=－2 代入①，得
1
m= ． ··············································· 11 分
4
所以这个方程组的解是
1
m = ，
4 ············································· 12 分
n = 2.
21．（本小题满分 10 分）
解：内错角相等，两直线平行 ········································································· 2 分
∠AGF ································································································ 4 分
两直线平行，同位角相等 ········································································· 6 分
∠AGF ································································································ 8 分
内错角相等，两直线平行 ······································································· 10 分
22．（本小题满分 8 分）
解：∵∠BOD＝25°，
∴∠AOC＝∠BOD＝25°． ·································································· 2 分
∵OF⊥CD，
∴∠COF＝90°．
∴∠AOC+∠AOF＝90°． ···································································· 4 分
∴∠AOF＝90°－∠AOC=90°－25°＝65°． ·············································· 6 分
∵OF平分∠AOE，
∴∠EOF＝∠AOF＝65°． ··································································· 8 分
23. （本小题满分 12 分）
解：（1）如图，△A1OC1即为所求． ························································· 2 分
A1（2，4），C1（3，2）． ························································ 4 分
（2）A1． ····················································································· 6 分
（3）P（2，3）或（2，5）． ························································· 12 分
七年级数学试卷参考答案和评分标准 第 2 页（共5页）
（第 23 题）
24. （本小题满分 12 分）
解：（1）设篮球的单价为 x 元，足球的单价为 y 元，得 ································ 1 分
4x +3y = 530，
·········································· 3 分
2x + 7y = 650.
解这个方程组，得
x = 80，
·················································· 5 分
y = 70.
答：篮球的单价为 80 元，足球的单价为 70 元． ··································· 6 分
（2）设该班购买篮球 m 个和足球 n 个，依题意，得
0.8（80m+70n）=960． ······································ 8 分
7n
∴m =15 ．
8
∵m，n 均为正整数，
m = 8， m =1，
∴ 或
n = 8 n =16.
答：该班购买篮球 8 个和足球 8 个或者篮球 1 个和足球 16 个． ·············· 12 分
25. （本小题满分 13 分）
解：（1）7，D． ·················································································· 6 分
（2）①当 k 2时，
∵点 M 与点 N 为相关点，
∴ 3k 5 = 2．
7
∴ k = 或 k =1（不符合题意，舍去）． ········································ 9 分
3
②当 k 2时，
∵点 M 与点 N 为相关点，
七年级数学试卷参考答案和评分标准 第 3 页（共5页）
∴ 3k 5 = k ．
5 5
∴ k = （不符合题意，舍去）或 k = ． ····································· 12 分
2 4
7 5
综上所述 k = 或 k = ． ························································· 13 分
3 4
26. （本小题满分 13 分）
（第 26 题图 1） （第 26 题图 2）
解：（1）AB∥CD． ·················································· 2 分
理由如下：
∵EM平分∠AEF，
∴∠AEM＝∠FEM．
∵∠FEM＝∠FME，
∴∠AEM＝∠FME．
∴AB∥CD． ·················································································· 5 分
（2）①∵EH平分∠FEG，
1
∴∠HEF＝ ∠FEG．
2
∵EM平分∠AEE，
1
∴∠FEM＝ ∠AEF．
2
1
∴∠HEM＝∠HEF+∠FEM= ∠AEG．
2
∵HN∥EM，
∴∠HEM＝∠EHN＝α．
∵AB∥CD，
∴∠GEB＝∠EGF＝β．
1
∴α= (180°－β）．
2
∴β＝180°－2α＝80°． ······································································· 9 分
七年级数学试卷参考答案和评分标准 第 4 页（共5页）
（第 26 题备用图）
②α和 β之间的数量关系为 β＝2α或 β＝180°－2α． ································ 10 分
理由如下：
当点 G在点 F的右侧，由①得 α＝180°－2β． ······································ 11 分
当点 G在点 F的左侧时，如图 2．
∵EM平分∠AEF，
∴∠AEF＝2∠FEM．
∵EH平分∠FEH，
∴∠GEF＝2∠HEF．
∴∠AEG＝∠AEF－∠GEF=2∠FEM－2∠HEF=2∠HEM．
∵AB∥CD，
∴∠AEG＝β．
∵HN∥EM，
∴∠HEM＝α．
∴β＝2α． ······················································································ 13 分
综上所述 α和 β之间的数量关系为 β＝2α或 β＝180°－2α．
七年级数学试卷参考答案和评分标准 第 5 页（共5页）
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