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期中模拟卷(福建)2023-2024学年七年级数学上学期期中模拟考试试题及答案(含答题卡)
展开2023-2024学年上学期期中模拟考试
七年级数学
一、选择题:(本题共10小题,每小题4分,共40分)。
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
B | B | C | C | B | A | C | C | D | B |
二、填空题:(本题共6小题,每小题4分,共24分)。
11.5 12.0.060 13.4
14. 15. 16.1﹣
三、解答题:(本题共9小题,共86分。其中:17-21每题8分,22-23每题10分,24题12分,25题14分)。
17.【解析】(1)40+×12
=40+×12﹣×12+×12
=40+2﹣8+9
=43;································································4分
(2)(﹣1)2021+|﹣9|×+(﹣3)÷
=(﹣1)+9×+(﹣3)×5
=(﹣1)+6+(﹣15)
=﹣10.······························································8分
18. 【解析】(1)原式
;···························································4分
(2)原式
.···················································8分
19.【解析】∵,,,
∴将各数在数轴上表示如下:
∴.····························8分
20.【解析】原式=
=
=
=;·······················································4分
∵,
∴,
解得:,···················································6分
∴原式=.············8分
21.【解析】(1)∵,,
∴
;····················································3分
(2)∵,,
∴;···5分
(3)∵的值与y的取值无关,
∴,
∴,
∴.····························8分
22.【解析】(1)数轴上表示4和1的两点之间的距离是,
表示和2的两点之间的距离是;····························1分
(2)数轴上表示a和3的两点之间的距离表示为,
数轴上表示a和的两点之间的距离表示为;···········2分
(3)∵数轴上表示a和的两点之间的距离是5,
∴,
解得:或;···············································3分
(4)∵数轴上表示a的点位于与2之间,
∴,
∴;································5分
(5)当时,,
解得:;
当时,,
无解;
当时,,
解得:;
故a的值为或;················································8分
(6)表示数轴上与,2和7的距离之和,
∴当时,的值最小,
最小值为.···················10分
23.【解析】(1)∵,
∴A站是树兜站;···············································5分
(2)
,
千米.
答:小王志愿服务期间乘坐地铁行进的总路程约是48千米.··············10分
24. 【解析】(1),
,
故答案为:11,22.·················································4分
(2).
故答案为:.·················································7分
(3)∵,
∴,
∴,
又,
∵,
∴.··································12分
25.【解析】(1)∵(a+10)2+|b﹣2|=0.
∴a+10=0且b﹣2=0,
∴a=﹣10,b=2;·················································2分
(2)①∵动点P、Q分别同时从A、C出发,分别以每秒6个单位和3个单位的速度,时间是t,
∴AP=6t,CQ=3t,
∵M为AP的中点,N在CQ上,且CN=CQ,
∴AM=AP=3t,CN=CQ=t,
∵点A表示的数是﹣10,C表示的数是6,
∴M表示的数是﹣10+3t,N表示的数是6+t;··························8分
②∵OM=|﹣10+3t|,BN=BC+CN=6﹣2+t=4+t,OM=2BN,
∴|﹣10+3t|=2(4+t)=8+2t,
由﹣10+3t=8+2t,得t=18,
由﹣10+3t=﹣(8+2t),得t=,
故当t=18秒或t=秒时OM=2BN.································14分
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