资源简介

《2025-2026 学年度初中数学期中检测》参考答案及评分建议
一、选择题（每小题 3 分，满分 30 分）
题号 1 2 3 4 5 6 7 8 9 10
答案 A A C B D C D D D B
二、填空题（每小题 3 分，满分 18 分）
11． 12．3 13．150 度/150
16
14．40 或 80 15．①④ 16． 9
三、解答题（满分 72 分）
17．解：（1） ·················································2 分
如图所示：线段 AB a 2b即为所求；·············································3 分
（2） ····························· 5 分
如图所示： AOB 就是要作的角．······································6 分
18．（1）解：2x x 10 5x 2 x 1 ，············································1 分
去括号得，2x x 10 5x 2x 2，
移项，合并同类项得， 6x 8，·····················································2 分
4
系数化为 1得， x ；······························································3 3 分
0.1x 0.4 1 0.2x 1（2）解： 1.2 0.3 ，
x 4
整理得， 1
2x 10
，····························································12 3 4 分
去分母得， x 4 12 4 2x 10 ，
去括号得， x 4 12 8x 40，
移项，合并同类项得， 7x 56，····················································5 分
系数化为 1 得， x 8．······························································6 分
19．解：由题意，得方程2 2x 4 1 5 x a 的解为 x 4．
把 x 4代入，得 a 1．··································································2 分
2x 4 x 1
将 a 1代入原方程，得 1 ．···········································5 2 3 分
去分母，得2 2x 4 10 5 x 1 ．·················································· 4 分
去括号，得 4x 8 10 5x 5．
移项、合并同类项，得 x 13．
系数化为 1，得 x 13．·································································6 分
20．解：设安排 x人生产 A部件，则安排 (16 x)人生产 B部件，·········· 2 分
依题意得3 1000x 5 600(16 x)，····················································4 分
解得 x 8，················································································ 6 分
∴16 x 8．·············································································· 7 分
答：应安排 8人生产 A部件，8人生产 B部件，才能使每天生产的 A部件和 B部
件配套．··················································································· 8 分
21．（1）解：∵OD 平分 AOC，OE平分 BOC，
DOC 1 AOC, EOC 1 BOC，················································2 2 2 分
DOE DOC EOC 1 AOC BOC 1 180 90 ；····················4 分
2 2
（2） AOD 30 ，OD平分 AOC，
AOC 60 , BOC 120 ，·························································· 6 分
∵OE平分 BOC，
1
BOE BOC 60 ．······························································2 8 分
22．（1）解：根据题意得标价为 1 60% x，售价为0.75 1 60% x，······2 分
则：75% 1 60% x x 8，···························································3 分
解得： x 40，··········································································· 4 分
∴ 1 60% x 1 60% 40 64（元）．··············································5 分
答：这款春联每副的标价是 64 元．················································ 6 分
（3）这次买卖中，这家商店赚了；················································ 7 分
理由如下：
设盈利的灯笼成进价为 a元，亏损的灯笼成进价为 b元，根据题意得：· 8 分
200 a 60%a，b 200 20%b，······················································ 9 分
解得：a 125，b 250，·····························································10 分
200 200 125 250 25（元），······················································ 11 分
∴赚了25元．···········································································12 分
23．（1）解：∵A，B两点间的距离为 10，点 A表示的数为 6，
∴6 10 4，
∵动点 P从点 A出发，以每秒 6个单位长度的速度沿数轴向左匀速运动，
∴点 P表示的数是6 6t，
故答案为： 4；6 6t．································································4 分
（2）解：①∵动点Q从点 B出发，以每秒 4个单位长度的速度沿数轴向右匀速
运动，
∴点Q表示的数是 4+4t，
∵点 P与点Q相遇，
∴ 6 6t 4 4t，·········································································· 6 分
解得 t 1，················································································· 7 分
答：当点 P运动 1秒时，点 P与点Q相遇．······································· 8 分
②根据题意得， 6 6t 4 4t 8 ，················································ 9 分
1
解得 t
9
或 ，·········································································5 11 分5
1 9
答：当点 P运动 5或 5秒时，点
P与点Q间的距离为 8个单位长度．····· 12 分
24．解：（1）∵MN 45cm， AB 3cm， AM 18cm
∴ BN MN AB AM 45 3 18 24cm，
∵点 C和点 D分别是 AM ，BN的中点，
AC 1∴ AM 9cm
1
， BD BN 12cm2 ，2
∴ AC BD 21cm．
∴CD AC AB BD 3 21 24cm．
故答案为：24．··········································································2 分
（2）①∵OC和OD分别平分 AOM 和 BON，
AOC 1∴ AOM ， BOD
1
BON．··············································2 2 3 分
1 1 1
∴ AOC BOD AOM BON ( AOM BON )2 2 2 ．
·····················4 分
又∵ MON 150 ， AOB 30 ，
∴ AOM BON MON AOB 150 30 120 ．
∴ AOC BOD 60 ．································································ 5 分
∴ COD AOC AOB BOD 60 30 90 ．································ 6 分
② COD
1
( MON AOB)．························································ 7 分
2
理由如下：
∵OC和OD分别平分 AOM 和 BON，
1
∴ AOC AOM ， BOD
1
BON．··············································2 2 8 分
AOC BOD 1 AOM 1 BON 1∴ ( AOM BON )．·····················2 2 2 9 分
∴ COD AOC AOB BOD
1
( AOM BON ) AOB
2
1
( MON AOB) AOB
2
1
( MON AOB)．··································································2 11 分
（3）∵ MON 150 ， AOB 30 ，
∴ AOM BON 120 ，····························································· 12 分
∵ MOC k AOC， NOD k BOD，
∴ AOM MOC AOC (1 k) AOC，
BON NOD BOD (1 k ) BOD ，········································ 13 分
∴
∴ COD AOC AOB BOD
120
30 ．···································k 1 14 分 2025-2026学年度下学期初一数学期中检测
5。在解方程2红+3=1时，去分母正确的是()
23
考试时间：120分钟：满分：120分。
注意事项：
A.3(x-)-2(2+3x)=1
B.3(x-)+2(2x+3)=1
1.答题前填写好自己的姓名、班级、考号等信息
2.请将答案正确填写在答题卡上
C.3(-)+2(2+3x)=6
D.3(x-1)-2(2x+3)=6
6.
如图，在6：20这一时刻，时钟上的分针与时针之何的夹角为()
一、单选题（共30分）
1。音桥是我国古代经爽建筑之一，它的修建增加了游人在桥上行走的路程，有利于萄人更好地观赏风充，如
112
10
图。本B两地间修建血桥与修建直的桥相比，增如了桥的长度，其中擅含的数学道理是()
E
-8
iy
6
A,60
B.65
C.70
D.75
7。春节将至，某工艺品店用红纸制作春联和福字满种装饰品，2刷春联和3个褐字配成一套销售，该工艺品
店共有红纸90张。一张红纸能制作2幅春联或制作6个栖字，应该怎样分配红纸才能使制作的春联和福字刚
好配套？若设分配x张红纸制作春联，则依题意可列方程为()，
A.两点之间，线段最知
B.平行于同一条直线的两条直线平行
A.3×2.x=6(90-x)
B.2x=6(90-x)
C,垂线段最知
D.两点确定一条直线
C.3×6x=(90-x×2
D.3×2x=2×6(90-x
2.如图。下列表示角的说法，错误的是(》
8。如图是一块弘，扬“杜会主义核心价值观”的帛面宜传展板，该展板的部分示意图如图2所示，它是以O为
圆心，0A,0B长分别为半径，圆心角∠0=120°形成的扇面，若0A=5m,0B=3m,则阴影部分的面积是
()m2
2=0+7
A.∠AOC也可用∠O表示
B.∠I与∠AOB表示同一个角
C,∠B表示的是∠BOG
D.∠AOB和∠OC都不能用LO表示
3.已知线段AB=10em,点C在直线AB上，BC=8cm,点M、N分别是AB、BC的中点，则MW的长度
为)
C.4
A.18cm
B.2cm
C.cm或lcm
D.18cm或2cm
4.下列运用等式的性质变形不正确的是()
9.如图，一副三角尺按不同的位置摆放，摆放位置中∠a与∠B不相等的图形为《)
A.若a=b,则2a+5=2h+5
B.若-x=6,则x=-2
D.若y-x=0,则x=y
第1页共8项
第2页共8页

展开更多......

收起↑