2018-2019学年广东省东莞市虎门外语学校、宏远外语学校、丰泰外语学校联考七年级(上)期末数学试卷-解析版
2018-2019学年广东省东莞市虎门外语学校、宏远外语学校、丰泰外语学校联考七年级（上）期末数学试卷
一、选择题（本大题共10小题，共30.0分）
1.-2的绝对值是（）
A. B. C. 2 D.
2.港珠澳大桥2018年10月24日上午9时正式通车，这座大桥跨越伶仃洋，东接香港，西接广东珠海和
澳门，总长约55000m，集桥、岛、隧于一体，是世界最长的跨海大桥，数据55000用科学记数法表示为（）
A. B. C. D.
3.单项式-5x2yz2的系数和次数分别是（）
A. 5，4
B. ，5
C. 5，5
D. ，
4.下列运算中，正确的是（）
A. B. C. D.
5.下列方程中，解为x=-2的方程是（）
A. B. C. D.
6.如图，点D是线段AB的中点，C是线段AD的中点，若AB=16cm，
则线段CD=（）cm．
A. 2
B. 4
C. 8
D. 16
7.将“中国梦我的梦”六个字分别写在一个正方体的六个面上，这个正方体的展开图
如图，那么在这个正方体中，和“我”字相对的字是（）
A. 中
B. 国
C. 的
D. 梦
8.若有理数a、b在数轴上的对应点的位置如图所示，则
下列结论中错误的是（）
A. B. C. D.
9.下午2点30分时（如图），时钟的分针与时针所成角的度数为（）
A. B. C. D.
10.现用90立方米木料制作桌子和椅子，已知一张桌子配4张椅子，1立方米木料可做5张椅子或1张桌
子，要使桌子和椅子刚好配套．设用x立方米的木料做桌子，则依题意可列方程为（）
A. B. C. D.
二、填空题（本大题共6小题，共24.0分）
11.比较大小：-3______-4（用“＞”“=”或“＜”表示）．
12.2xy-6xy=______．
13.当代数式2x-2与3+x的值相等时，x=______．
14.已知∠A=56°34′，则∠A的补角的度数是______．
15.已知x-4y=2，那么-5+2x-8y的值为______．
16.已知：A、B、C三个点在同一直线上，若线段AB=8，BC=5，则线段AC=______．
三、计算题（本大题共5小题，共33.0分）
17.解方程：．18.随着中国快递行业整体规模的迅速壮大，分拣机器人系统的应用也呈现智能化、自动化
的发展趋势．每台分拣机器人一小时可以分拣1.8万件包裹，大大提高了分拣效率．某分拣仓库计划平均每天分拣20万件包裹，但实际每天分拣量与计划相比有出入，下表是该仓库10月份第一周分拣包裹的情况（超过计划量记为正、未达计划量记为负）：
（1）该仓库本周内分拣包裹数量最多的一天是星期______，最少的一天是星期______，最多的一天比最少的一天多分拣了______万件包裹；
（2）该仓库本周实际分拣包裹一共多少万件？
19.如图，点B、C把线段MN分成三部分，其比是MB：BC：CN=2：3：4，P是MN的中点，且MN=18cm，
求PC的长．
20.计算：|-12019|+（1-32）×2-（-2）3÷16
21.先化简，再求值3（a2b-ab2）-2（2a2b-1）+3ab2-1，其中a=-2，b=1．
四、解答题（本大题共4小题，共33.0分）
22.2x-5y+3x+y-2
23.某自来水公司按如下规定收取水费：每月用水不超过10吨，按每吨1.5元收费；每月用水超过10吨，
超过部分按每吨2元收费．设每月用水量为x吨．
（1）小红家3月用水10吨，应交水费多少元？
（2）试用x的代数式表示付水费的费用．
（3）小明家4月份的水费是25元，小明家4月份用水多少吨？
24.用火柴棒按下列方式搭建三角形，如图所示：
第一个图形要3根，第二个图形要5根，第三个图形要7根，
第5个图形需要______根火柴棒；
第n个图形需要______根火柴棒；
第几个图形需要用到2019根火柴棒？
25.已知点O是AB上的一点，∠COE=90°，OF平分∠AOE．
（1）如图1，当点C，E，F在直线AB的同一侧时，若∠AOC=40°，求∠BOE和∠COF的度数；
（2）在（1）的条件下，∠BOE和∠COF有什么数量关系？请直接写出结论，不必说明理由；
（3）如图2，当点C，E，F分别在直线AB的两侧时，若∠AOC=β，那么（2）中∠BOE和∠COF的数量关系是否仍然成立？请写出结论，并说明理由．
答案和解析
1.【答案】C
【解析】
解：因为|-2|=2，
故选：C．
根据负数的绝对值等于它的相反数求解．
绝对值规律总结：一个正数的绝对值是它本身；一个负数的绝对值是它的相反数；0的绝对值是0．
2.【答案】C
【解析】
解：55000=5.5×104，
故选：C．
科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数．确定n的值时，要看把原数变成a时，小数点移动了多少位，n的绝对值与小数点移动的位数相同．当原数绝对值＞10时，n是正数；当原数的绝对值＜1时，n是负数．
此题考查科学记数法的表示方法．科学记数法的表示形式为a×10n的形式，其中1≤|a|＜10，n为整数，表示时关键要正确确定a的值以及n的值．
3.【答案】B
【解析】
解：单项式-5x2yz2的系数和次数分别是-5，5．
故选：B．
根据单项式系数、次数的定义来求解．单项式中数字因数叫做单项式的系数，所有字母的指数和叫做这个单项式的次数．
考查了单项式，解答此题关键是构造单项式的系数和次数，把一个单项式分解成数字因数和字母因式的积，是找准单项式的系数和次数的关键．
4.【答案】C
【解析】解：A、3a与2b不是同类项，不能合并，此选项错误；
B、2a3与3a2不是同类项，不能合并，此选项错误；
C、-4a2b+3ba2=-a2b，此选正确；
D、5a2-4a2=a2，此选项错误；
故选：C．
根据同类项的定义和合并同类项的法则逐一判断即可得．
本题主要考查合并同类项，解题的关键是掌握同类项的定义和合并同类项的法则．
5.【答案】B
【解析】
解：分别将x=-2代入题目中的四个方程：
A、左边=-2-2=-4≠右边，该方程的解不是x=-2，故本选项错误；
B、左边=2-6=-4=右边，该方程的解是x=-2，故本选项正确；
C、左边=-6-1=-7≠右边，该方程的解不是x=-2，故本选项错误；
D、左边=4+6=10≠右边，该方程的解不是x=-2，故本选项错误；
故选：B．
方程的解就是能够使方程左右两边相等的未知数的值，把x=2代入各个方程进行进行检验，看能否使方程的左右两边相等．
本题的关键是正确理解方程的解的定义，就是能够使方程左右两边相等的未知数的值．
6.【答案】B
【解析】
解：由点D是线段AB的中点，得
AD=
AB=×16=8cm，
由C是线段AD的中点，得
CD=
AD=×8=4cm．
故选：B．
根据线段中点的性质，可得答案．
本题考查了两点间的距离，利用了线段的中点分线段相等．
7.【答案】B
【解析】
解：正方体的表面展开图，相对的面之间一定相隔一个正方形，
“中”与“梦”是相对面，
“国”与“我”是相对面，
“梦”与“的”是相对面．
故选：B．
正方体的表面展开图，相对的面之间一定相隔一个正方形，根据这一特点作答．
本题主要考查了正方体相对两个面上的文字，注意正方体的空间图形，从相对面入手，分析及解答问题．
8.【答案】C
【解析】
解：由数轴可知，a＜0，b＞0
选项A，a＜0，b＞0，a＜b，选项正确
选项B，ab＜0，正确
选项C，∵|a|＞|b|，∴a＜-b，选项错误
选项D，∵|a|＞|b|，∴a+b＜0，选项正确
故选：C．
数轴的点左右的数，右边的总比左边的大．
此题考查的是数轴点的规律．数轴上的点右边数比左边的大，位于原点左边的数小于零，右边
的数大于零
9.【答案】B
【解析】
解：∵1个小时在时钟上的角度为180°÷6=30°，
∴3.5个小时的角度为30°×3.5=105°．故选：B．
钟表12个数字，每相邻两个数字之间的夹角为30度．
本题主要考查角度的基本概念．在钟表问题中，常利用时针与分针转动的度数关系：分针每转动1°时针转动（）°，并且利用起点时间时针和分针的位置关系建立角的图形．
10.【答案】A
【解析】
解：设用x立方米的木料做桌子，则用（90-x）立方米的木料做椅子，
依题意，得：4x=5（90-x）．
故选：A．
设用x立方米的木料做桌子，则用（90-x）立方米的木料做椅子，根据制作的椅子数为桌子数的4倍，即可得出关于x的一元一次方程，此题得解．
本题考查了由实际问题抽象出一元一次方程，找准等量关系，正确列出一元一次方程是解题的关键．
11.【答案】＞
【解析】
解：根据有理数大小比较的规律可得两个负数中绝对值大的反而小，-3＞-4．
故答案为：＞．
本题是基础题，考查了实数大小的比较．两负数比大小，绝对值大的反而小；或者直接想象在数轴上比较，右边的数总比左边的数大．
规律总结：（1）在以向右方向为正方向的数轴上两点，右边的点表示的数比左边的点表示的数大．
（2）正数大于0，负数小于0，正数大于负数．
（3）两个正数中绝对值大的数大．
（4）两个负数中绝对值大的反而小．
12.【答案】-4xy
【解析】
解：2xy-6xy=-4xy，
故答案为：-4xy
根据把同类项的系数相加，所得结果作为系数，字母和字母的指数不变解答即可．
此题考查合并同类项，关键是根据合并同类项的法则解答．
13.【答案】5
【解析】
解：根据题意得：2x-2=3+x，
移项合并得：x=5，
故答案为：5．
根据题意列出方程，求出方程的解即可得到x的值．
此题考查了解一元一次方程，熟练掌握运算法则是解本题的关键．
14.【答案】123°26′
【解析】
解：∵∠A=56°34′，
∴∠A的补角=180°-∠A=180°-56°34′=123°26′．
故答案为：123°26′
根据补角的定义可解．
本题主要考查了补角的定义以及角的度分秒运算，正确理解补角的定义是解题的关键．
15.【答案】-1
【解析】
解：把x-4y=2代入-5+2x-8y=-5+2（x-4y）=-5+4=-1，
故答案为：-1
根据题意x-4y=2，再变形后代入求出即可．
本题考查了求代数式的值的应用，能整体代入是解此题的关键．
16.【答案】13或3
【解析】
解；如图①：AC=AB+BC=5+8=13，
如图②：AC=AB-BC=8-5=3．
故答案为：13或3．
根据题意画出图形，分两种情况：①C在AB的右边；②C在AB之间．
此题主要考查了两点间的距离，关键是考虑到两种情况．17.【答案】解：去分母得：5（y-1）=20-2（y+2），
去括号得：5y-5=20-2y-4，
移项合并得：7y=21．
解得：y=3．
【解析】
方程去分母，去括号，移项合并，将y系数化为1，即可求出解．
此题考查了解一元一次方程，其步骤为：去分母，去括号，移项合并，将未知数系数化为1，求
出解．
18.【答案】六日15
【解析】
解：（1）∵-8＜-4＜-3＜-1＜+5＜+6＜+7，
∴分拣最多的一天是星期六，分拣最少的一天是星期天，
分拣最多的一天比分拣最少的一天多分拣了：+7-（-8）=15（万件），
故答案为：六，日，15；
（2）∵-8-4-3-1+5+6+7=2（万件），
∴20×7+2=142（万件）
答：该仓库本周实际分拣包裹142万件．
（1）比较分拣情况的出入数，用不等号连接起来，得到分拣包裹数量的最多最少，最多一天减去最少一天得多分拣包裹数；
（2）把每天的分拣包裹数相加即可．
主要考查正负数在实际生活中的应用及有理数大小的比较．看懂图表是关键．
19.【答案】解：设MB=2x，则BC=3x，CN=4x，
因为P是MN中点，
所以MP=MN=×（2x+3x+4x）=x=9．
解得x=2，
∴PC=MC-MP=2x+3x-x=0.5x=1．
【解析】
设MB的长为2x，分别表示出BC=3x，CN=4x，进一步利用线段中点的意义和线段的和与差解决问题．
本题考查了两点间的距离，线段中点的定义，本题根据比例用x表示出三条线段求解更简便．
20.【答案】解：|-12019|+（1-32）×2-（-2）3÷16
=1+（1-9）×2-（-8）×
=1+（-8）×2+
=1+（-16）+
=-14．
【解析】
根据有理数的乘除法和加减法可以解答本题．
本题考查有理数的混合运算，解答本题的关键是明确有理数混合运算的计算方法．
21.【答案】解：原式=3a2b-3ab2-4a2b+2+3ab2-1=-a2b+1，
当a=-2，b=1时，原式=-3．
【解析】
原式去括号合并得到最简结果，把a与b的值代入计算即可求出值．
此题考查了整式的加减-化简求值，熟练掌握运算法则是解本题的关键．
22.【答案】解：2x-5y+3x+y-2
=2x+3x-5y+y-2
=5x-4y-2．
【解析】
根据合并同类项解答即可．
此题考查合并同类项，关键是根据合并同类项解答．
23.【答案】解：（1）依题意得：10×1.5=15（元）
答：应交水费15元．
（2）①当0≤x≤10时，需要付水费：1.5x．
②当x＞10时，需要付水费：15+2（x-10）=2x-5．
（3）因为25＞10×1.5，
所以小明家4月份用水超过10吨，
依题意得：2x-5=25
解得x =15
答：小明家4月份用水15吨．
【解析】
（1）根据收费标准“每月用水不超过10吨，按每吨1.5元收费”计算．（2）需要对x的取值范围进行分类讨论：0≤x≤10，x＞10，结合收费标准解答．
（3）收费25元时，按照“每月用水超过10吨，超过部分按每吨2元收费”解答．
本题考查了一元一次方程的应用，解题关键是要读懂题目的意思，根据题目给出的条件，找出
合适的等量关系列出方程，再求解．
24.【答案】11 2n+1
【解析】
解：设第n个图形需要a n（n为正整数）根火柴棒，
观察，发现规律：a1=3，a2=5，a3=7，a4=9，…，
∴a n=2n+1．
当n=5时，a5=2×5+1=11．
当2n+1=2019时，解得：n=1009．
故答案为：11；2n+1．
设第n个图形需要a n（n为正整数）根火柴棒，根据给定图形找出部分a n的值，根据数值的变化找出变化规律“a n=2n+1”，依此规律即可得出结论．
本题考查了规律型中的图形的变化类，解题的关键是找出变化规律“a n=2n+1”．本题属于基础题，难度不大，解决该题型题目时，根据给定图形中的数据找出变化规律是关键．
25.【答案】解：（1）∵∠COE=90°，∠AOC=40°，
∴∠BOE=180°-∠AOC-∠COE=180°-40°-90°=50°，
∠AOE=∠AOC+∠COE=40°+90°=130°，
∵OF平分∠AOE，
∴∠EOF=∠AOE=×130°=65°，
∴∠COF=∠COE-∠EOF=90°-65°=25°；
（2）∠BOE=2∠COF．
（3）∠BOE=2∠COF．
理由如下：∵∠COE=90°，∠AOC=β，
∴∠AOE=∠COE-∠AOC=90°-β，
∴∠BOE=180°-∠AOE=180°-（90°-β）=90°+β，
∵OF平分∠AOE，
∴∠AOF=∠AOE=（90°-β）=45°-β，
∴∠COF=β+（45°-β）=45°+β，
∴2∠COF=2（45°+β）=90°+β，
∴∠BOE=2∠COF．
【解析】
（1）求出∠BOE和∠COF的度数即可判断；
（2）由（1）即可求解；
（3）结论：∠BOE=2∠COF．根据角的和差定义即可解决问题．
本题考查角的计算，角平分线的定义等知识，解题的关键是灵活运用所学知识解决问题，属于中考常考题型．
2020年重点初中、外国语学校小升初英语试卷附答案
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2008年广东省广州市中考英语试卷 (满分l 35分，考试时间l 20分钟)第l卷(共105分) 一、听力(共两节，满分35分) 第一节听力理解(共15小题；每小题2分，满分30分) 第二节听取信息(共5小题；每小题l分，满分5分) 二、语言知识及运用(共两节，满分20分) 第一节单项选择(共10小题；每小题l分，满分l0分) 从16-～25各题所给的A、B、C和D项中，选出可以填入空白处的最佳选项。 16. —I hear we'll have a new foreign teacher soon. Do you know when_________? —Sorry, I have no idea. A. he will come B. will he come C. is he coming D. he was coming 17. This is just between you and me. You_________tell anyone about this. A. mustn't B. can C. should D. have not to 18. Last summer I went to Lu Xun's hometown and visited the house_________hewas born. A. that B. there C. which D. where 19. I'm sorry. I started eating before you got here_________J was terribly hungry. A. so B. since then C. because D. so that 20. Miss Green didn't talk much to other people. There was always_________a littlesad about her. A. everything B. anything C. nothing D. something 21. My younger brother couldn't work out the answer and A. so could I B. nor could I C. so can I D. nor I could 22. Though the player is over thirty, he can still run________ some younger players. A. as fast as B. so fast as C. much fast than D. more faster than 23. At the meeting Mr. King didn't say a word from beginning to end, _____ ? A. didn't Mr. King B. did he C. did Mr. King D. didn't he 24. From that time on Mary practised_________the piano every day. A. plays B. playing C. played D. to play 25. Hi Mr. Smith. I didn't know you were in New York. How long_________here? A. have you come B. were you C. have you been D. will you come 第二节语法选择(共10小题；每小题l分，满分l0分)阅读下面短文，按照句子结构的语法性和上下文连贯的要求，从26---35各题所给的A、B、C和D项中选出最佳选项。