枫叶国际学校入学测试数学模拟试题
枫叶国际学校秋季入学试题

枫叶国际学校秋季入学试题七年级数学试A 卷姓名:__________成绩:___________/75一、我会填(每空2分,共20分)1、1.5小时=()分2、45和60的最大公因数是()3、5.893保留一位小数是()4、在含盐20%的50千克盐水中,其中盐有()千克5、在 ,3.14,3.1415,722这些数中,最大的数是()6、若12=⋅y x ,则x 和y 成()比例7、以直角三角形的一条直角边为轴,旋转一周,就可以得到一个()8、圆锥的侧面展开图可以得到一个()9、等腰三角形的一个底角是45°,它的顶角是()°10、如图,将右图中的硬纸片沿虚线折起来,便可以成为一个正方体,这个正方体3号面对的是()号面二、对号入座(将正确的答案序号填在括号里)(每题2分,共20分)11、把84900改写成以“万”为单位的数是()A.84.9万B.8.49C.8.49万12大、小两个圆的半径之比为4︰3,它们的面积比是()A.8︰6B.16︰9C.12︰913、下面等式中,()应用了乘法结合律A.6×(18×2)=(6×18)×2B.6×(3×a)=6×(a×3)C.7×(16+18)=7×16+7×1814、a 与b 的和的21用式子表示是()A.a+21bB.21(a+b)C.21a+b 15、在一幅地图上用3厘米长的线段表示150千米的实际距离,这幅图的比例尺是()A.1︰5000000B.1︰50000C.1︰5016、()A.B.C.D.17、把下面各种形状的硬纸按照虚线折叠,能围成一个立方体的是()564321ABC18、下列是轴对称图形的是()A.B.C.19、某小组6名同学的体重如下(单位:千克):47,50,52,53,52,52这组数据的中位数和众数分别是(A.52和53B.52和52C.52和52.520、从一幅扑克牌(54张)中抽取一张牌,抽到牌“Q”的概率是()A.541B.181 C.272三、计算(每题5分,共10分)21、1037594()178178⨯+++22、97141-167-5.70⨯⎥⎦⎤⎢⎣⎡)(四、解方程(每题5分,共10分)23、7x+8x=4524、322121=+x 五、列式计算(5分)25、一个数的54比45的2倍少30,求这个数(用方程解)六、看图完成下面各题(10分)26、(1)学校的位置如果用(0,0)表示,分别说出网吧、图书馆、大众广场、人民医院、青少年活动中心的位置(5分)(2)星期天,李岚的活动路线为(3,4)(3,6)(6,4)(10,8)(7,9),请问她去了哪些地方?兴庆公园青少年活动中心人民医院大众广场图书馆体育馆网吧学校123456789101111109876543210。
海南省海南枫叶国际学校2024年九年级数学第一学期开学综合测试模拟试题【含答案】

海南省海南枫叶国际学校2024年九年级数学第一学期开学综合测试模拟试题题号一二三四五总分得分A 卷(100分)一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、(4分)如图,点A ,B 在反比例函数()20y x x=-<的图象上,连结OA ,AB ,以OA ,AB 为边作OABC ,若点C 恰好落在反比例函数()10y x x=>的图象上,此时OABC 的面积是()A .3B C .D .62、(4分)已知平行四边形,下列条件中,不能判定这个平行四边形为菱形的是()A .B .C .平分D .3、(4分)已知点(﹣2,y 1),(﹣1,y 2),(1,y 3)都在直线y=﹣3x+2上,则y 1,y 2,y 3的值的大小关系是()A .y 3<y 1<y 2B .y 1<y 2<y 3C .y 3>y 1>y 2D .y 1>y 2>y 34、(4分)一次数学测试中,小明所在小组的5个同学的成绩(单位:分)分别是:90、91、88、90、97,则这组数据的中位数是()A .88B .90C .90.5D .915、(4分)如图所示,在直角ABC ∆中,90BAC ∠=︒,8AB =,6AC =,DE 是AB 边的垂直平分线,垂足为D ,交边BC 于点E ,连接AE ,则ACE ∆的周长为()A .16B .15C .14D .136、(4分)一元二次方程x 2-9=0的解为()A .x 1=x 2=3B .x 1=x 2=-3C .x 1=3,x 2=-3D .x 1x 27、(4分)在函数11y x=+中,自变量x 必须满足的条件是()A .1x ≠B .1x ≠-C .0x ≠D .1x >8、(4分)下列二次根式中,属于最简二次根式的是()A .B C D .二、填空题(本大题共5个小题,每小题4分,共20分)9、(4分)对于非零的两个实数a 、b ,规定a ⊕b=,若2⊕(2x ﹣1)=1,则x 的值为.10、(4分)已知四边形ABCD 是平行四边形,且A ,B ,C 三点的坐标分别是()3,3,()8,3,()4,6则这个平行四边形第四个顶点的坐标为______.11、(4分)如果多项式22(2)9x k xy y +-+是一个完全平方式,那么k 的值为______.12、(4分)已知一次函数y =mx +n 与x 轴的交点为(﹣3,0),则方程mx +n =0的解是_____.13、(4分)如图如果以正方形的对角线为边作第二个正方形,再以对角线为边作第三个正方形,如此下去,…,已知正方形的面积为1,按上述方法所作的正方形的面积依次为,…(为正整数),那么第8个正方形的面积__.三、解答题(本大题共5个小题,共48分)14、(12分)已知△ABC ,分别以BC ,AB ,AC 为边作等边三角形BCE ,ACF ,ABD (1)若存在四边形ADEF ,判断它的形状,并说明理由.(2)存在四边形ADEF 的条件下,请你给△ABC 添个条件,使得四边形ADEF 成为矩形,并说明理由.(3)当△ABC 满足什么条件时四边形ADEF 不存在.15、(8分)为了解高中学生每月用掉中性笔笔芯的情况,随机抽查了30名高中学生进行调查,并将调查的数据制成如下的表格:月平均用中性笔笔芯(根)456789被调查的学生数749523请根据以上信息,解答下列问题:(1)被调查的学生月平均用中性笔笔芯数大约________根;(2)被调查的学生月用中性笔笔芯数的中位数为________根,众数为________根;(3)根据样本数据,若被调查的高中共有1000名学生,试估计该校月平均用中性笔笔芯数9根的约多少人?16、(8分)已知反比例函数y =kx的图象经过点(-1,-2).(1)求y 与x 的函数关系式;(2)若点(2,n)在这个图象上,求n 的值.17、(10分)据某市交通运管部门5月份的最新数据,目前该市市面上的共享单车数量已达39万辆,共享单车也逐渐成为高校学生喜爱的“绿色出行”方式之一.某高校为了解本校学生出行使用共享单车的情况,随机调查了某天部分出行学生使用共享单车的情况,并整理成如下统计表.使用次数012 34人数810222614(1)求这天部分出行学生使用共享单车次数的平均数,中位数和众数.(2)若该校这天有720名学生出行,估计使用共享单车次数在2次以上(含2次)的学生数.18、(10分)计算(1-÷.(2)2(1-+-+.B 卷(50分)一、填空题(本大题共5个小题,每小题4分,共20分)19、(4分)如图,将边长为12的正方形ABCD 沿其对角线AC 剪开,再把△ABC 沿着AD 方向平移,得到△A′B′C′,当两个三角形重叠部分的面积为32时,它移动的距离AA′等于________.20、(4分)直线22y x =+与y 轴的交点坐标是________________.21、(4分)公路全长为skm,骑自行车t 小时可到达,为了提前半小时到达,骑自行车每小时应多走_____________.22、(4分)如图,已知菱形ABCD 的一个内角∠BAD=80°,对角线AC ,BD 相交于点O ,点E 在AB 上,且BE=BO ,则∠EOA=___________°.23、(4分)几个同学包租一辆面包车去旅游,面包车的租价为180元,后来又增加了两名同学,租车价不变,结果每个同学比原来少分摊了3元车费.若设原参加旅游的同学有x 人,则根据题意可列方程___________________________.二、解答题(本大题共3个小题,共30分)24、(8分)已知:如图,平面直角坐标系xOy 中,B (0,1),OB =OC =OA ,A 、C 分别在x 轴的正负半轴上.过点C 的直线绕点C 旋转,交y 轴于点D ,交线段AB 于点E .(1)求∠OAB 的度数及直线AB 的解析式;(2)若△OCD 与△BDE 的面积相等,求点D 的坐标.25、(10分)如图,BD ,CE 是△ABC 的高,G ,F 分别是BC ,DE 的中点,求证:FG ⊥DE .26、(12分)在△ABC 中,AH ⊥BC 于H ,D 、E 、F 分别是BC 、CA 、AB 的中点.求证:DE=HF .参考答案与详细解析一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、A【解析】连接AC,BO交于点E,作AG⊥x轴,CF⊥x轴,设点A(a,2a-),点C(m,1m)(a<0,m>0),由平行四边形的性质和中点坐标公式可得点B[(a+m),(2a-+1m)],把点B坐标代入解析式可求a=-2m,由面积和差关系可求解.【详解】解:如图,连接AC,BO交于点E,作AG⊥x轴,CF⊥x轴,设点A(a,2a-),点C(m,1m)(a<0,m>0),∵四边形ABCO是平行四边形,∴AC与BO互相平分,∴点E(21,22a m a m-+ +),∵点O坐标(0,0),∴点B[(a+m),(2a-+1m)].∵点B在反比例函数y=2x-(x<0)的图象上,∴212a m a m -+=-+,∴a=-2m,a=m(不合题意舍去),∴点A(-2m,1 m),∴四边形ACFG是矩形,∴▱OABC的面积=2×S△AOC=3.故选:A.本题考查了反比例函数图象上点的坐标特征,平行四边形的性质,中点坐标公式,解决问题的关键是数形结合思想的运用.2、A【解析】菱形的判定有以下三种:①一组邻边相等的平行四边形是菱形;②四边相等的四边形是菱形;③对角线互相垂直的平行四边形是菱形.据此判断即可.【详解】解:A、由平行四边形的性质可得AB=CD,所以由AB=CD不能判定平行四边形ABCD是菱形,故A选项符合题意;B、一组邻边相等的平行四边形是菱形,故B选项不符合题意.C、由一条对角线平分一角,可得出一组邻边相等,也能判定为菱形,故C选项不符合题意;D、对角线互相垂直的平行四边形是菱形,故D选项不符合题意;故选:A.本题考查菱形的判定方法,熟记相关判定即可正确解答.3、D【解析】k=-3<0,所以函数y随x增大而减小,所以y1>y2>y3,所以选D.4、B【解析】先将题中的数据按照从小到大的顺序排列,然后根据中位数的概念求解即可.【详解】将小明所在小组的5个同学的成绩重新排列为:88、90、90、91、97,所以这组数据的中位数为90分,故选B.本题考查了中位数的概念:将一组数据按照从小到大(或从大到小)的顺序排列,如果数据的个数是奇数,则处于中间位置的数就是这组数据的中位数;如果这组数据的个数是偶数,则中间两个数据的平均数就是这组数据的中位数.5、A【解析】首先连接AE ,由在直角△ABC 中,∠BAC=90°,AB=8,AC=6,利用勾股定理即可求得BC 的长,又由DE 是AB 边的垂直平分线,根据线段垂直平分线的性质,即可得AE=BE ,继而可得△ACE 的周长为:BC+AC .【详解】连接AE ,∵在Rt △ABC 中,∠BAC=90∘,AB=8,AC=6,∴10=∵DE 是AB 边的垂直平分线,∴AE=BE ,∴△ACE 的周长为:AE+EC+AC=BE+CE+AC=BC+AC=10+6=16,故选A.本题考查勾股定理,熟练掌握勾股定理的性质是解题关键.6、C 【解析】先变形得到x 2=9,然后利用直接开平方法解方程.【详解】解:x 2=9,∴x=±1,∴x 1=1,x 2=-1.故选:C .本题考查了直接开平方法:形如x 2=p 或(nx+m )2=p (p ≥0)的一元二次方程可采用直接开平方的方法解一元二次方程.7、B 【解析】由函数表达式是分式,考虑分式的分母不能为0,即可得到答案.【详解】解:∵函数11y x=+,∴10x +≠,∴1x ≠-;故选:B.本题考查了分式有意义的条件,解题的关键是掌握当函数表达式是分式时,考虑分式的分母不能为0.8、C 【解析】满足下列两个条件的二次根式,叫做最简二次根式:(1)被开方数的因数是整数,因式是整式;(2)被开方数中不含能开得尽方的因数或因式.【详解】A =3,故A 不是;B 、10,故B 不是;C 、,是;D =D 不是.故选C考查了最简二次根式的概念,熟练掌握最简二次根式所需要满足的条件是解题的关键.二、填空题(本大题共5个小题,每小题4分,共20分)9、.【解析】先根据规定运算把方程转化为一般形式,然后把分式方程转化为整式方程求解,再进行检验即可得解.【详解】解:2⊕(2x ﹣1)=1可化为﹣=1,方程两边都乘以2(2x ﹣1)得,2﹣(2x ﹣1)=2(2x ﹣1),解得x=,检验:当x=时,2(2x ﹣1)=2(2×﹣1)=≠0,所以,x=是原分式方程的解,即x 的值为.故答案为.本题考查了解分式方程,(1)解分式方程的基本思想是“转化思想”,把分式方程转化为整式方程求解.(2)解分式方程一定注意要验根.10、()9,6或()1,6-或()7,0.【解析】根据平行四边形的性质,分别以BC 、AC 、AB 为对角线,分三种情况进行分析,即可求得答案.【详解】解:由平行四边形的性质可知:当以BC 为对角线时,第四个顶点的坐标为D 1()9,6;当以AC 为对角线时,第四个顶点的坐标为D 2()1,6-;当以AB 为对角线时,第四个顶点的坐标为D 3()7,0;故答案为:()9,6或()1,6-或()7,0.本题考查了平行四边形的性质:平行四边形的对边平行且相等.解此题的关键是分类讨论数学思想的运用.11、8或-4【解析】根据完全平方公式的定义即可求解.【详解】()2229x k xy y +-+=()222(3)x k xy y +-+为完全平方公式,故()2k -=±6,即得k=8或-4.此题主要考查完全平方公式的形式,解题的关键是熟知完全平方公式.12、x =﹣1.【解析】直接根据函数图象与x 轴的交点进行解答即可.【详解】∵一次函数y =mx +n 与x 轴的交点为(﹣1,0),∴当mx +n =0时,x =﹣1.故答案为:x =﹣1.本题考查了一次函数与一元一次方程的关系.任何一元一次方程都可以转化为ax +b =0(a ,b 为常数,a ≠0)的形式,所以解一元一次方程可以转化为:当某个一次函数的值为0时,求相应的自变量的值.从图象上看,相当于已知直线y =ax +b 确定它与x 轴的交点的横坐标的值.13、128【解析】由题意可以知道第一个正方形的边长为1,第二个正方形的边长为,第三个正方形的边长为2,就有第n 个正方形的边长为(n-1),再根据正方形的面积公式就可以求出结论.【详解】第一个正方形的面积为1,故其边长为1=2;第二个正方形的边长为,其面积为2=2;第三个正方形的边长为2,其面积为4=2;第四个正方形的边长为2,其面积为8=2;…第n 个正方形的边长为(),其面积为2.当n=8时,S =2,=2=128.故答案为:128.此题考查正方形的性质,解题关键在于找到规律.三、解答题(本大题共5个小题,共48分)14、(1)详见解析;(2)当∠BAC =150°时,四边形ADEF 是矩形;(3)∠BAC=60°时,这样的平行四边形ADEF 不存在.【解析】(1)根据等边三角形的性质得出AC =AF ,AB =BD ,BC =BE ,∠EBC =∠ABD =60°,求出∠DBE =∠ABC ,根据SAS 推出△DBE ≌△ABC ,根据全等得出DE =AC ,求出DE =AF ,同理AD =EF ,根据平行四边形的判定推出即可;(2)当AB =AC 时,四边形ADEF 是菱形,根据菱形的判定推出即可;当∠BAC =150°时,四边形ADEF 是矩形,求出∠DAF =90°,根据矩形的判定推出即可;(3)这样的平行四边形ADEF 不总是存在,当∠BAC =60°时,此时四边形ADEF 就不存在.【详解】(1)证明:∵△ABD 、△BCE 和△ACF 是等边三角形,∴AC =AF ,AB =BD ,BC =BE ,∠EBC =∠ABD =60°,∴∠DBE =∠ABC =60°﹣∠EBA ,在△DBE 和△ABC 中BD BA DBE ABC BE BC =⎧⎪∠=∠⎨⎪=⎩,∴△DBE ≌△ABC ,∴DE =AC ,∵AC =AF ,∴DE =AF ,同理AD =EF ,∴四边形ADEF 是平行四边形;(2)解:当∠BAC =150°时,四边形ADEF 是矩形,理由是:∵△ABD 和△ACF 是等边三角形,∴∠DAB =∠FAC =60°,∵∠BAC =150°,∴∠DAF =90°,∵四边形ADEF 是平行四边形,∴四边形ADEF 是矩形;(3)解:这样的平行四边形ADEF 不总是存在,理由是:当∠BAC =60°时,∠DAF =180°,此时点D 、A 、F 在同一条直线上,此时四边形ADEF 就不存在.本题考查了菱形的判定,矩形的判定,平行四边形的判定,等边三角形的性质,全等三角形的性质和判定的应用,能综合运用定理进行推理是解此题的关键,题目比较好,难度适中.15、(1)6;(2)6,6;(3)100【解析】(1)根据平均数的概念求解;(2)根据中位数的概念求解;(3)用人数×平均数即可求解.【详解】解:(1)月平均用中性笔笔芯数:74+54+69+75+82+9330⨯⨯⨯⨯⨯⨯=6(根);(2)∵共有30名学生,∴第15和16为同学的月用中性笔笔芯数的平均数为中位数:662+=6;被调查的学生月用中性笔笔芯数的众数为:6;(3)1000×330=100(根).本题考查了平均数、中位数和众数等知识,掌握平均数、中位数、众数的概念是解答本题的关键.16、(1)y=2x.(2)n=1.【解析】(1)直接把点(﹣1,﹣2)代入反比例函数y=kx即可得出结论.(2)把(2,n )代入强大的解析式即可求得.【详解】解:(1)∵反比例函数y=kx的图象经过(﹣1,﹣2),∴﹣2=k1-,解得k=2.∴这个函数的解析式为y=2x.(2)把(2,n )代入y=2x得n=22=1.17、(1)中位数是2.5次,众数是3次;(2)558人.【解析】(1)根据平均数、中位数和众数的定义求解可得;(2)用总人数乘以样本中使用共享单车次数在2次以上(含2次)的学生所占比例即可得.【详解】(1)71122261480++++=∴()180101223144 2.3580x =⨯+⨯+⨯+⨯=(次)次数从小到大排列后,中间两个数是2与3∴中位数是2.5次共享单车的使用次数中,出现最多的是3次∴众数是3次(2)22261472055880++⨯=即该校这天使用共享单车次数在2次以上(含2次)的学生约有558人.本题考查了中位数、众数、平均数的概念以及利用样本平均数估计总体.抓住概念进行解题,难度不大,但是中位数一定要先将所给数据按照大小顺序重新排列后再求,以免出错.18、(1)43-;(2)3-【解析】(1)直接利用算术平方根以及立方根性质分别化简再计算即可得出答案.(2)直接利用二次根式的混合运算法则,先用完全平方公式和平方差公式计算,再化简得出答案.【详解】解:(1-126=--÷113=--43=-;(2)2(1-+-+3213=+--3=-.此题主要考查了实数运算,正确化简各数是解题关键.一、填空题(本大题共5个小题,每小题4分,共20分)19、1或8【解析】由平移的性质可知阴影部分为平行四边形,设A ′D=x ,根据题意阴影部分的面积为(12−x)×x ,即x(12−x),当x(12−x)=32时,解得:x=1或x=8,所以AA ′=8或AA ′=1.【详解】设AA ′=x,AC 与A ′B ′相交于点E ,∵△ACD 是正方形ABCD 剪开得到的,∴△ACD 是等腰直角三角形,∴∠A=15∘,∴△AA ′E 是等腰直角三角形,∴A ′E=AA ′=x ,A ′D=AD−AA ′=12−x ,∵两个三角形重叠部分的面积为32,∴x(12−x)=32,整理得,x 2−12x+32=0,解得x 1=1,x 2=8,即移动的距离AA ′等1或8.本题考查正方形和图形的平移,熟练掌握计算法则是解题关键·.20、(0,2)【解析】根据一次函数的性质,与y 轴的交点即横坐标为0,代入即可得解.【详解】根据题意,得当0x =时,2y =,即与y 轴的交点坐标是()0,2故答案为()0,2.此题主要考查一次函数的性质,熟练掌握,即可解题.21、221s t --s t【解析】公路全长为skm ,骑自行车t 小时可到达,则速度为/;skm h t若提前半小时到达,则速度为/.12s km h t -则现在每小时应多走(21212s s s s t t t t -=---)/.km h 22、1【解析】根据∠BAD和菱形邻角和为180°的性质可以求∠ABC的值,根据菱形对角线即角平分线的性质可以求得∠ABO的值,又由BE=BO可得∠BEO=∠BOE,根据∠BOE和菱形对角线互相垂直的性质可以求得∠EOA的大小.【详解】解:∵∠BAD=80°,菱形邻角和为180°∴∠ABC=100°,∵菱形对角线即角平分线∴∠ABO=50°,∵BE=BO∴∠BEO=∠BOE=180502︒-︒=65°,∵菱形对角线互相垂直∴∠AOB=90°,∴∠AOE=90°-65°=1°,故答案为1.本题考查了菱形对角线互相垂直平分且平分一组对角的性质,考查了等腰三角形底角相等的性质,本题中正确的计算∠BEO=∠BOE=65°是解题的关键.23、18018032x x-=+【解析】分析:等量关系为:原来人均单价-实际人均单价=3,把相关数值代入即可.详解:原来人均单价为180x,实际人均单价为1802x+,那么所列方程为18018032x x-=+,故答案为:180180 3.2x x-=+点睛:考查列分式方程;得到人均单价的关系式是解决本题的关键.二、解答题(本大题共3个小题,共30分)24、(1)45°,y=﹣x+1;(2)(0,1 3).【解析】(1)根据A、B的坐标和三角形的内角和定理求出∠OAB的度数即可;设直线AB的解析式为y =kx +b ,把A 、B 的坐标代入得出方程组,求出方程组的解即可;(2)推出三角形AOB 和三角形ACE 的面积相等,根据面积公式求出E 的纵坐标,代入直线AB 的解析式,求出E 的横坐标,设直线CE 的解析式是:y =mx +n ,利用待定系数法求出直线EC 的解析式,进而即可求得点D 的坐标.【详解】解:(1)∵OB =OC =OA ,∠AOB =90°,∴∠OAB =45°;∵B (0,1),∴A (1,0),设直线AB 的解析式为y =kx+b .∴01,k b b +=⎧⎨=⎩解得,11,k b =-⎧⎨=⎩∴直线AB 的解析式为y =﹣x+1;(2)∵S △COD =S △BDE ,∴S △COD +S 四边形AODE =S △BDE +S 四边形AODE ,即S △ACE =S △AOB ,∵点E 在线段AB 上,∴点E 在第一象限,且y E >0,∴1122E AC y OA OB ⨯⨯⨯⨯=,∴1211122E y =⨯⨯⨯⨯,1,2E y =把y 12=代入直线AB 的解析式得:112x =-+,∴12x =设直线CE 的解析式是:y =mx+n ,∵()1110,22C E ⎛⎫- ⎪⎝⎭,,代入得:011,22m n m n -+=⎧⎪⎨+=⎪⎩解得:11,33m n ==,∴直线CE 的解析式为11,33y x =+令x =0,则13y =,∴D 的坐标为10,3⎛⎫ ⎪⎝⎭.本题考查了等腰三角形的性质,用待定系数法求一次函数的解析式,三角形的面积等知识点,综合运用这些性质进行推理和计算是解此题的关键,此题题型较好,综合性比较强,但难度适中,通过做此题培养了学生分析问题和解决问题的能力.25、如图,连接EG ,DG .∵CE 是AB 边上的高,∴CE ⊥AB .在Rt △CEB 中,G 是BC 的中点,∴12EG BC =.同理,12DG BC =.∴EG =DG .又∵F 是ED 的中点,∴FG ⊥DE .【解析】根据题意连接EG ,DG ,利用直角三角形斜边上的中线的性质可得EG =DG ,然后根据等腰三角形“三线合一”的性质即可解决.26、证明见解析.【解析】分析:根据题意知EH是直角△ABH斜边上的中线,DE是△ABC的中位线,所以由相关的定理进行证明.详解:∵D、E分别是BC、CA的中点,∴DE=12AB.又∵点F是AB的中点,AH⊥BC,∴FH=12AB,∴DE=HF.点睛:本题考查了三角形中位线定理、直角三角形斜边上的中线.三角形中位线的性质:三角形的中位线平行于第三边且等于第三边的一半.第21页,共21页。
九年级入学数学考试:国际学校版

九年级入学数学考试:国际学校版
考试说明
欢迎参加九年级入学数学考试,本考试旨在评估学生的数学基础知识和解决问题的能力。
考试分为选择题和解答题两部分,满分100分,考试时间为90分钟。
请仔细阅读题目,并在规定时间内完成考试。
考试内容
选择题(共20题,每题2分)
1. 下列哪个数是正数?
A. -3
B. 0
C. 5
D. -2
2. 两个整数的和是8,它们的差是5,请问这两个整数分别是多少?
3. 下列哪个比例式是正确的?
A. 3:4 = 12:16
B. 5:6 = 15:18
C. 7:8 = 21:24
D. 9:10 = 27:30
...(此处省略17题)
20. 下列哪个数是立方根为5的数?
A. 125
B. 64
C. 81
D. 27
解答题(共4题,每题10分)
1. 已知一个正方形的边长为6cm,求它的面积和周长。
2. 解方程:2x - 5 = 3x + 1
3. 已知直角三角形的两个直角边分别为8cm和15cm,求斜边
的长度。
4. 某商店举行打折活动,原价为1200元的商品打8折后售出,求售出价格。
考试要求
1. 请使用黑色签字笔或钢笔填写答案,字迹要清晰。
2. 选择题请在答题卡上对应题号处填写答案,解答题请在答题
纸上作答。
3. 考试过程中,禁止使用计算器、手机等电子设备。
4. 考试结束后,请将试卷和答题卡一并交回。
祝您考试顺利!。
初中国际学校入学数学试卷

一、选择题(每题2分,共20分)1. 下列各数中,是质数的是()A. 16B. 15C. 13D. 182. 已知x + y = 7,x - y = 3,则x的值为()A. 5B. 6C. 7D. 83. 一个长方形的长是12厘米,宽是8厘米,它的周长是()A. 32厘米B. 40厘米C. 48厘米D. 56厘米4. 如果一个数的平方根是3,那么这个数是()A. 9B. 6C. 3D. 275. 下列分数中,最小的是()A. 1/2B. 1/3C. 1/4D. 1/56. 下列方程中,解为x = 2的是()A. 2x + 3 = 9B. 3x - 4 = 5C. 4x + 5 = 11D. 5x - 6 = 137. 一个圆的半径是r,它的面积是()A. πr²B. 2πr²C. 4πr²D. 8πr²8. 下列图形中,不是平行四边形的是()A. 矩形B. 菱形C. 正方形D. 三角形9. 一个数的5倍与这个数的3倍相差15,这个数是()A. 5B. 6C. 7D. 810. 已知等差数列的第一项是2,公差是3,那么第10项是()A. 28B. 29C. 30D. 31二、填空题(每题2分,共20分)11. 2的平方根是_______,-3的立方根是_______。
12. 一个数的倒数是1/5,这个数是_______。
13. 下列图形中,轴对称图形是_______(填图形名称)。
14. 一个长方形的长是15厘米,宽是5厘米,它的面积是_______平方厘米。
15. 下列数中,有理数是_______。
16. 下列数中,无理数是_______。
17. 一个等腰三角形的底边长是8厘米,腰长是10厘米,它的周长是_______厘米。
18. 下列方程中,解为x = 0的是_______。
19. 一个圆的直径是10厘米,它的半径是_______厘米。
20. 下列图形中,中心对称图形是_______(填图形名称)。
郑州枫杨外国语中学2025届九年级数学第一学期开学复习检测试题【含答案】

郑州枫杨外国语中学2025届九年级数学第一学期开学复习检测试题题号一二三四五总分得分批阅人A 卷(100分)一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、(4分)如图(1),四边形ABCD 中,AB ∥CD ,∠ADC =90°,P 从A 点出发,以每秒1个单位长度的速度,按A →B →C →D 的顺序在边上匀速运动,设P 点的运动时间为t 秒,△PAD 的面积为S ,S 关于t 的函数图象如图(2)所示,当P 运动到BC 中点时,△APD 的面积为( )A .4B .5C .6D .72、(4分)某校九年级(1)班全体学生2018年初中毕业体育考试的成绩统计如表:成绩(分)35394244454850人数(人)2566876根据如表的信息判断,下列结论中错误的是()A .该班一共有40名同学B .该班学生这次考试成绩的众数是45分C .该班学生这次考试成绩的中位数是44分D .该班学生这次考试最高成绩是50分3、(4分)如图所示,在矩形纸片中,,,折叠纸片使边与对角线重合,点落在点处,折痕为,则的长为( )A .B .C .D .4、(4分)下列计算正确的是( )。
ABCD5、(4分)若化简,则的取值范围是( )A .一切实数B .C .D .6、(4分)如图,菱形ABCD 中,AC 交BD 于点O ,于点E ,连接OE ,若,则( )A .20°B .30°C .40°D .50°7、(4分)已知一元二次方程2﹣5x+1=0的两个根为,,下列结论正确的是( )A .+=﹣B .•=1C .,都是正数D .,都是有理数8、(4分)数学兴趣小组的甲、乙、丙、丁四位同学进行还原魔方练习,下表记录了他们10次还原魔方所用时间的平均值与方差:甲乙丙丁-=3=-==1-25x -x 14x ≤≤1x ≤4x ≥DE BC ⊥140ABC ︒∠=OED ∠=2x 1x 2x 1x 2x 521x 2x 1x 2x 1x 2x x 2S(秒)303028281.21 1.05 1.21 1.05要从中选择一名还原魔方用时少又发挥稳定的同学参加比赛,应该选择( )A .甲B .乙C .丙D .丁二、填空题(本大题共5个小题,每小题4分,共20分)9、(4分)已知四边形ABCD 为菱形,∠BAD=60°,E 为AD 中点,AB=6cm ,P 为AC 上任一点.求PE+PD 的最小值是_______10、(4分)从某玉米种子中抽取6批,在同一条件下进行发芽试验,有关数据如下:种子粒数100400800 1 000 2 000 5 000发芽种子粒数853******** 1 604 4 005发芽频率0.8500.7950.8150.7930.8020.801根据以上数据可以估计,该玉米种子发芽的概率为___________(精确到0.1).11、(4分)在平面直角坐标系中,若点P (2x +6,5x )在第四象限,则x 的取值范围是_________;12、(4分)如图,在矩形ABCD 中,顺次连接矩形四边的中点得到四边形EFGH .若AB =8,AD =6,则四边形EFGH 的周长等于__________.13、(4分)已知一组数据3、x 、4、8、6,若该组数据的平均数是5,则x 的值是______.三、解答题(本大题共5个小题,共48分)x 2S14、(12分)如图,四边形ABCD 是正方形,点E 是BC 边上的点,∠AEF=90°,且EF 交正方形外角的平分线CF 于点F .(1)如图①,当点E 是BC 边上任一点(不与点B 、C 重合)时,求证:AE=EF .(2)如图②当点E 是BC 边的延长线上一点时,(1)中的结论还成立吗? (填成立或者不成立).(3)当点E是BC 边上任一点(不与点B 、C 重合)时,若已知AE=EF ,那么∠AEF 的度数是否发生变化?证明你的结论.15、(8分)如图:矩形ABCD 中,AB=2,BC=5,E 、P 分别在AD 、BC 上,且DE=BP=1.(1)判断△BEC 的形状,并说明理由?(2)判断四边形EFPH 是什么特殊四边形?并证明你的判断;(3)求四边形EFPH 的面积.16、(8分)如图,某住宅小区在施工过程中留下了一块空地,已知AD=4米,CD=3米,∠ADC=90°,AB=13米,BC=12米,小区为美化环境,欲在空地上铺草坪,已知草坪每平方米100元,试问用该草坪铺满这块空地共需花费多少元?17、(10分)如图,中,是边上一点,,,,分别是,边上的动点,且始终保持.ABCD E AD 45A ∠=︒3BE CD ==ED =P Q BC CD 45EPQ ∠=︒(1)求的长;(2)若四边形为平行四边形时,求的周长;(3)将沿它的一条边翻折,当翻折前后两个三角形组成的四边形为菱形时,求线段的长.18、(10分)某养猪场要出售200只生猪,现在市场上生猪的价格为11元/,为了估计这200只生猪能卖多少钱,该养猪场从中随机抽取5只,每只猪的重量(单位:)如下:76,71,72,86,1.(1)计算这5只生猪的平均重量;(2)估计这200只生猪能卖多少钱?B 卷(50分)一、填空题(本大题共5个小题,每小题4分,共20分)19、(4分)已知四边形是平行四边形,且,,三点的坐标分别是,,则这个平行四边形第四个顶点的坐标为______.20、(4分)根据图中的程序,当输入x =2时,输出结果y =________.21、(4分)如图,经过点B (-2,0)的直线与直线相交于点A (-1,-2),则不等式的解集为 .AE ABPE CPQ CPQ BP kg kg ABCD A B C ()3,3()8,3()4,6y kx b =+y 4x 2=+4x 2<kx b<0++22、(4分)如图,Rt △ABC 中,∠ACB=90°,BC =AC =3,点D 是BC 边上一点,∠DAC =30°,点E 是AD 边上一点,CE 绕点C 逆时针旋转90°得到CF ,连接DF ,DF 的最小值是___.23、(4分)如图,一根垂直于地面的木杆在离地面高3m 处折断,若木杆折断前的高度为8m ,则木杆顶端落在地面的位置离木杆底端的距离为________m .二、解答题(本大题共3个小题,共30分)24、(8分)在的方格纸中,四边形的顶点都在格点上.(1)计算图中四边形的面积;(2)利用格点画线段,使点在格点上,且交于点,计算的长53⨯ABCD ABCD DE E DE AC ⊥AC F DF度.25、(10分)如图,,是上的一点,且,.求证:≌26、(12分)如下4个图中,不同的矩形ABCD ,若把D 点沿AE 对折,使D 点与BC 上的F 点重合;(1)图①中,若DE ︰EC=2︰1,求证:△ABF ∽△AFE ∽△FCE ;并计算BF ︰FC ;(2)图②中若DE ︰EC=3︰1,计算BF ︰FC= ;图③中若DE ︰EC=4︰1,计算BF ︰FC= ;(3)图④中若DE ︰EC=︰1,猜想BF ︰FC= ;并证明你的结论090A B ∠=∠=E AB AD BE =12∠=∠Rt ADE V Rt BEC n参考答案与详细解析一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、B 【解析】根据函数图象和三角形面积得出AB+BC=6,CD=4,AD=4,AB=1,当P 运动到BC 中点时,梯形ABCD 的中位线也是△APD 的高,求出梯形ABCD 的中位线长,再代入三角形面积公式即可得出结果.【详解】解:根据题意得:四边形ABCD 是梯形,AB+BC=6,CD=10-6=4,∵AD×CD=8,∴AD=4,又∵AD×AB=2,∴AB=1,当P 运动到BC 中点时,梯形ABCD 的中位线也是△APD 的高,∵梯形ABCD 的中位线长=(AB+CD)=,∴△PAD 的面积 故选B .本题考查了动点问题的函数图象、三角形面积公式、梯形中位线定理等知识;看懂函数图象是解决问题的关键.2、C 【解析】根据总数,众数,中位数的定义即可一一判断;【详解】该班一共有:2+5+6+6+8+7+6=40(人),众数是45分,最高成绩为50分,中位数为45分,故A 、B 、D 正确,C 错误,故选:C .12121252154522;=⨯⨯=此题考查总数,众数,中位数的定义,解题的关键是熟练掌握基本知识,属于中考基础题.3、D 【解析】由题得BD= =5,根据折叠的性质得出△ADG ≌△A′DG ,继而得A′G=AG ,A′D=AD ,A′B=BD-A′G ,再Rt △A′BG 根据勾股定理构建等式求解即可.【详解】解:由题得BD==5,根据折叠的性质得出:△ADG ≌△A′DG ,∴A′G=AG ,A′D=AD=3,A′B=BD-A′G=5-3=2,BG=4-A′G 在Rt △A′BG 中,BG 2=A′G 2+A′B 2可得:,解得A′G=,则AG=,故选:D .本题主要考查折叠的性质,由已知能够注意到△ADG ≌△A′DG 是解决的关键.4、C 【解析】根据二次根式的运算法则即可求出答案.【详解】解:(A )原式A 错误;(B )原式=3,故B 错误;(C )原式,故C 正确;(D )原式 ,故D 错误;故选:C本题考查二次根式,解题的关键是熟练运用二次根式的运算法则,本题属于基础题型.5、B【解析】根据完全平方公式先把多项式化简为|1−x|−|x−4|,然后根据x 的取值范围分别讨论,求出符合题意的x 的值即可.【详解】原式可化简为,当,时,可得无解,不符合题意;当,时,可得时,原式;当,时,可得时,原式;当,时,可得时,原式.据以上分析可得当时,多项式等于.故选B.本题主要考查绝对值及二次根式的化简,要注意正负号的变化,分类讨论6、A 【解析】根据直角三角形的斜边中线性质可得OE=OB=OD ,根据菱形性质可得∠DBE= ∠ABC=70°,从而得到∠OEB 度数,再依据∠OED=90°-∠OEB 即可.【详解】解:∵四边形ABCD 是菱形,∴O 为BD 中点,∠DBE=∠ABC=70°,∵DE ⊥BC ,∴在Rt △BDE 中,OE=OB=OD ,∴∠OEB=∠OBE=70°,∴∠OED=90°-70°=20°,故选A .本题主要考查了菱形的性质、直角三角形斜边中线的性质,解决这类问题的方法是四边形转化为三角形.7、C【解析】|1||4|x x ---10x -≥40x -≥x 10x -≥40x -≤4x ≤143x x =--+=-10x -≤40x -≥4x ≥143x x =--+=10x -≤40x -≤14x ≤≤1425x x x =--+=-14x ≤≤25x -1212先利用根与系数的关系得到x 1+x 21,x 1x 21,然后利用有理数的性质可判定两根的符号.【详解】根据题意得x 1+x 21,x 1x 21,所以x 1>1,x 2>1.∵x ,故C 选项正确.故选C .本题考查了根与系数的关系:若x 1,x 2是一元二次方程ax 2+bx +c =1(a ≠1)的两根,则x 1+x 2,x 1x 2.8、D 【解析】在这四位同学中,丙、丁的平均时间一样,比甲、乙的用时少,但丁的方差小,成绩比较稳定,由此可知,可选择丁,故选D.二、填空题(本大题共5个小题,每小题4分,共20分)9、【解析】根据菱形的性质,可得AC是BD 的垂直平分线,可得AC上的点到D 、B 点的距离相等,连接BE 交AC 与P ,可得答案.【详解】解:∵菱形的性质,∴AC 是BD 的垂直平分线,AC 上的点到B 、D 的距离相等.连接BE 交AC 于P 点,PD=PB ,PE+PD=PE+PB=BE ,在Rt △ABE 中,由勾股定理得故答案为52=>12=>52=>12=>=b a =-c a =BE ===本题考查了轴对称,对称轴上的点到线段两端点的距离相等是解题关键.10、1.2【解析】仔细观察表格,发现大量重复试验发芽的频率逐渐稳定在1.2左右,从而得到结论.【详解】∵观察表格,发现大量重复试验发芽的频率逐渐稳定在1.2左右,∴该玉米种子发芽的概率为1.2,故答案为1.2.考查利用频率估计概率,大量反复试验下频率稳定值即概率.用到的知识点为:频率=所求情况数与总情况数之比.11、﹣3<x <1【解析】根据第四象限内横坐标为正,纵坐标为负可得出答案.【详解】∵点P (2x-6,x-5)在第四象限,∴ 解得-3<x <1.故答案为-3<x <1.本题考查了点的坐标、一元一次不等式组,解题的关键是知道平面直角坐标系中第四象限横、纵坐标的符号.12、20.【解析】分析:连接AC,BD,根据勾股定理求出BD,根据三角形中位线定理,菱形的判定定理得到四边形EHGF 为菱形,根据菱形的性质计算.解答:连接AC,BD 在Rt △ABD 中, ∵四边形ABCD 是矩形,∴AC=BD=10, ∵E 、H 分别是AB 、AD 的中点,∴EH ∥BD,EF=BD=5,同理,FG ∥BD,10,=12FG=BD=5,GH ∥AC,GH=AC=5, ∴四边形EHGF 为菱形,∴四边形EFGH 的周长=5×4=20,故答案为20.点睛:本题考查了中点四边形,掌握三角形的中位线定理、菱形的判定定理是解答本题的关键.13、1【解析】根据算术平均数的计算方法列方程求解即可.【详解】解:由题意得: 解得:.故答案为1.此题考查算术平均数的意义和求法,掌握计算方法是解决问题的关键.三、解答题(本大题共5个小题,共48分)14、(1)见解析;(2)成立,理由见解析;(3)∠AEF=90°不发生变化.理由见解析.【解析】(1)在AB 上取点G ,使得BG=BE ,连接EG ,根据已知条件利用ASA 判定△AGE ≌△ECF ,因为全等三角形的对应边相等,所以AE=EF ;(2)在BA 的延长线上取一点G ,使AG=CE ,连接EG ,根据已知利用ASA 判定△AGE ≌△ECF ,因为全等三角形的对应边相等,所以AE=EF ;(3)在BA 边取一点G ,使BG=BE ,连接EG .作AP ⊥EG ,EQ ⊥FC ,先证AGP ≌△ECQ 得AP=EQ ,再证Rt △AEP ≌Rt △EFQ 得∠AEP=∠EFQ ,∠BAE=∠CEF ,结合∠AEB+∠BAE=90°知∠AEB+∠CEF=90°,从而得出答案.【详解】(1)证明:在BA 边取一点G ,使BG=BE ,连接EG ,∵四边形ABCD 是正方形,∴∠B=90°,BA=BC ,∠DCM═90°,1212348655x ++++=⨯4x =∴BA-BG=BC-BE ,即 AG=CE .∵∠AEF=90°,∠B=90°,∴∠AEB+∠CEF=90°,∠AEB+∠BAE=90°,∴∠CEF=∠BAE .∵BG=BE ,CF 平分∠DCM ,∴∠BGE=∠FCM=45°,∴∠AGE=∠ECF=135°,∴△AGE ≌△ECF (ASA ),∴AE=EF .(2)成立,理由:在BA 的延长线上取点G ,使得AG=CE ,连接EG .∵四边形ABCD 为正方形,AG=CE ,∴∠B=90°,BG=BE ,∴△BEG 为等腰直角三角形,∴∠G=45°,又∵CF 为正方形的外角平分线,∴∠ECF=45°,∴∠G=∠ECF=45°,∵∠AEF=90°,∴∠FEM=90°-∠AEB,又∵∠BAE=90°-∠AEB ,∴∠FEM=∠BAE ,∴∠GAE=∠CEF ,在△AGE 和△ECF 中,∵,∴△AGE ≌△ECF (ASA ),∴AE=EF .故答案为:成立.(3)∠AEF=90°不发生变化.理由如下:在BA 边取一点G ,使BG=BE ,连接EG .分别过点A 、E 作AP ⊥EG ,EQ ⊥FC ,垂足分别为点P 、Q ,∴∠APG=∠EQC=90°,由(1)中知,AG=CE ,∠AGE=∠ECF=135°,∴∠AGP=∠ECQ=45°,G CEF AG CE GAE CEF ∠∠⎧⎪⎨⎪∠∠⎩===∴△AGP ≌△ECQ (AAS ),∴AP=EQ ,∴Rt △AEP ≌Rt △EFQ (HL ),∴∠AEP=∠EFQ ,∴∠BAE=∠CEF ,又∵∠AEB+∠BAE=90°,∴∠AEB+∠CEF=90°,∴∠AEF=90°.此题是四边形综合题,主要考查的是正方形的性质、全等三角形的判定和性质,正确作出辅助线、灵活运用全等三角形的判定定理和性质定理是解题的关键,解答时,注意类比思想的正确运用.15、(1)△BEC 是直角三角形,理由见解析(2)四边形EFPH 为矩形,理由见解析(3)【解析】(1)△BEC 是直角三角形,理由略(2)四边形EFPH 为矩形证明:在矩形ABCD 中,∠ABC=∠BCD=900∴PA=, PD=2 ∵AD=BC=5∴AP 2+PD 2=25=AD 2 ∴∠APD=900 (3分)同理∠BEC=900∵DE=BP ∴四边形BPDE 为平行四边形∴BE ∥PD (4分)∴∠EHP=∠APD=900,又∵∠BEC=900∴四边形EFPH 为矩形 (5分)(3)在RT △PCD 中∠Ff PD∴PD·CF=PC·CD ∴CF==∴EF=CE-CF=-= (7分)∵PF==5855⊥5224⨯54555455522CF PC -585∴S 四边形EFPH=EF·PF= (1)根据矩形性质得出CD=2,根据勾股定理求出CE 和BE ,求出CE2+BE2的值,求出BC2,根据勾股定理的逆定理求出即可;(2)根据矩形的性质和平行四边形的判定,推出平行四边形DEBP 和AECP ,推出EH ∥FP ,EF∥HP ,推出平行四边形EFPH ,根据矩形的判定推出即可;(2)根据三角形的面积公式求出CF ,求出EF ,根据勾股定理求出PF ,根据面积公式求出即可.16、2400元【解析】试题分析:连接AC ,根据勾股定理求出AC ,根据勾股定理的逆定理求出∠ACB=90°,求出区域的面积,即可求出答案.试题解析:连结AC ,在Rt △ACD 中,∠ADC=90°,AD=4米,CD=3米,由勾股定理得:(米),∵AC 2+BC 2=52+122=169,AB 2=132=169,∴AC 2+BC 2=AB 2,∴∠ACB=90°,该区域面积S=S △ACB ﹣S △ADC =×5×12﹣×3×4=24(平方米),即铺满这块空地共需花费=24×100=2400元.考点:1.勾股定理;2.勾股定理的逆定理.17、(1)2);(3)或3或.【解析】(1)先根据题意推出△ABE 是等腰直角三角形,再根据勾股定理计算即可.(2)首先要推出△CPQ 是等腰直角三角形,再根据已知推出各边的长度,然后相加即可.(3)首先证明△BPE ∽△CQP ,然后分三种情况讨论,分别求解,即可解决问题.585=1212(1)∵四边形ABCD 是平行四边形,∴AB=CD ,∵BE=CD=3,∴AB=BE=3,又∵∠A=45°,∴∠BEA=∠A=45°,∠ABE=90°,根据勾股定理得=;(2)∵四边形ABCD 是平行四边形,∴AB=CD ,∠A=∠C=45°,又∵四边形ABPE 是平行四边形,∴BP ∥AB ,且AE=BP ,∴BP ∥CD ,∴,∵∠EPQ=45°,∴∠PQC=∠EPQ=45°,∴∠PQC=∠C=45°,∠QPC=90°,∴,QC=2,∴△CPQ 的周长;(3)解:如图,作BH ⊥AE 于H ,连接BE .∵四边形ABCD 是平行四边形,∴AB=CD=3,AD=BC=AE+ED=,∠A=∠C=45°,∴,HE=AD -AH -∴∠EBH=∠HEB=∠EBC=45°,∴∠EBP=∠C=45°,∵∠BPQ=∠EPB+∠EPQ=∠C+∠PQC ,∠EPQ=∠C ,∴∠EPB=∠PQC ,∴△BPE ∽△CQP .①当QP=QC 时,则BP=PE ,∴∠EBP=∠BEP=45°,则∠BPE=90°,∴四边形BPEF 是矩形,,②当CP=CQ 时,则BP=BE=3,③当CP=PQ 时,则BE=PE=3,∠BEP=90°,∴△BPE为等腰三角形,∴BP 2=BE 2+PE2,∴BP=综上:或3或.本题利用平行四边形的性质求解,其中运用了分类讨论的思想,这是解题关键.18、(1)78.4(千克);(2)172480(元).【解析】(1)根据平均数的计算可得这5只生猪的平均重量;(2)根据用样本估计总体的思想可估计这200只生猪每只生猪的平均重量,由(1)中的平均数可得.【详解】解:(1)这5只生猪的平均重量为千克;(2)根据用样本估计总体的思想可估计这200只生猪每只生猪的平均重量约为千克;根据题意,生猪的价格为11元,故这200只生猪能卖元.11200⨯⨯767172868778.4(5++++=)78.4/kg 78.411200172480(⨯⨯=)本题主要考查的是通过样本估计总体.统计的思想就是用样本的信息来估计总体的信息.一、填空题(本大题共5个小题,每小题4分,共20分)19、或或.【解析】根据平行四边形的性质,分别以BC 、AC 、AB 为对角线,分三种情况进行分析,即可求得答案.【详解】解:由平行四边形的性质可知:当以BC 为对角线时,第四个顶点的坐标为D 1;当以AC 为对角线时,第四个顶点的坐标为D 2;当以AB 为对角线时,第四个顶点的坐标为D 3;故答案为:或或.本题考查了平行四边形的性质:平行四边形的对边平行且相等.解此题的关键是分类讨论数学思想的运用.20、2【解析】∵x=2时,符合x>1的条件,∴将x=2代入函数y=−x+4得:y=2.故答案为2.21、【解析】()9,6()1,6-()7,0()9,6()1,6-()7,0()9,6()1,6-()7,02<x<1--分析:不等式的解集就是在x 下方,直线在直线上方时x 的取值范围.由图象可知,此时.22【解析】先依据条件判定△ACE ≌△BCF ,可得∠CBF =∠CAE =30°,即可得到点F 在射线BF 上,由此可得当DF ⊥BF 时,DF 最小,依据∠DBF=30°,即可得到DF =BD 【详解】由旋转可得,FC =EC ,∠ECF =90°,又∵∠ACB =90°,BC =AC =3,∴∠CAE =∠CBF ,∴△ACE ≌△BCF ,∴∠CBF =∠CAE =30°,∴点F 在射线BF 上,如图,当DF ⊥BF 时,DF 最小,又∵Rt △ACD 中,∠CAD =30°,AC =3=BC ,∴CD ,∴BD =3又∵∠DBF =30°,∴DF = BD ,.4x 2<kx b<0++y kx b =+y 4x 2=+2<x<1--1212本题考查了旋转的性质,等腰直角三角形的性质,全等三角形的判定与性质,垂线段最短的性质,作辅助线构造出全等三角形是解题的关键,得到点F 的运动轨迹是本题的难点.23、4【解析】由题意得,在直角三角形中,知道了两直角边,运用勾股定理即可求出斜边,从而得出木杆顶端落在地面的位置离木杆底端的距离.【详解】一颗垂直于地面的木杆在离地面处折断,木杆折断前的高度为,.故答案为:.此题考查了勾股定理的应用,主要考查学生对勾股定理在实际生活中的运用能力.二、解答题(本大题共3个小题,共30分)24、(1);(2)【解析】(1)先证明是直角三角形,然后将四边形分为可得出四边形的面积;(2)根据格点和勾股定理先作出图形,然后由面积法可求出DF 的值。
枫叶国际学校入学测试数学模拟试题

- --MATH PROFICIENCY TESTName _________________________Date _____________________Record your answer on the answer sheet provided.1. Simplify: (2x 4y 5) 9. Evaluate −2a 3b −4a 4 for a= −3 and b= −2a) 6x 12y 15 b) 2x 12y 15 c) 8x 64y 125a) 432b) −432 c) −128 d)144 e) −144d) 8x 12y 15e) 8x 7y 810. Solve: 3x −7 ≥ −222. Rewrite 4 − 3using positive exponents only and evaluate the result.a) x ≤ 5 b) x ≤ −5 c) x ≤ 28 − d) x ≥ −5 e) x ≥ 531 1 a)=b) −3 41 = −12c) 43 =644643d) − 43 = −64e) 43 =1211. Solve: 15 2x −3< 7 a) x > 18b) x < 18c) x < −3 d) x> −3 e) x < 33Evaluate:1 24−4912. Simplify: (4x −8xy+2y ) −(9x −4xy −7y ) a) −5x −4xy −5y b) −5x −4xy + 9ya)7 2b) 7 − c)227 d) 2 − e)7− 2 49 c ) −5x + 12xy + 4y d) −5x + 4xy + 9ye) −5x −12xy + 9y 4 Simplify and rewrite using only positive exponents:13. How long is BC?−24x232y( )a) x 44y6b) 4y 6x4c)4y 6−x2d) 1 y 6e) y 6a) 22x - 19x- 23 b) -16x +8x -5 c) 16x -8x - 5 d) 16x -24x+55. Write an algebraic equation for the following " After 7 e) 16x -8x +5 is added to a number the sum is divided by4. The result is 12." 14. Simplify:( −7a −19c) −( −12a+30c) −( −21a −11c)- --7n n 7a) =12 b) + 4 =12 c) n + =124 7 4n n + 7d) 7 12 12+ = e) =4 4a) 26a −38c b) −26a −22c c) −16a−60cd) 2a + 22c e) 26a + 38c15. When 6x y −10x y is completely factored, oneof the factors will be_____6. Solve for x given 4(3x −7)= −7 a) 3x −5y b) 2xy c) 2xya) 76b) 0 c)35−d)1274e)3512d) 3x −5y e) 3x−516. Simplify: ( 2x4y ) (6x y )7. Solve for x given 4 −2(3x −4)=1 −(5x+4).a) 12x12y6 b) 12x7y5 c) 12x4y3 a) −9 b) −1 c) 15 d) 3 e) 7d) 8x12y6 e) 8x7y58. What is the value of the expression −3x2 +4xwhen x= −5 ? 17. Simplify:24x −30x³ +6x66xa) −55 b) −95 c) 50 d) 95 e) 55 a) 4x5 −5x +x b) 4x6 −5x3+1 c) 4x5 −5x+1d) 4x5 −5x e) 18x5 −36xPage 1 of 518. When (4x −1) (2x +7) is expanded and written in 26. Factor completely: xy −5y −2x + 10 the form Ax +Bx + C, what is the value of A + B + C? a) (x + 5) (y −2) b) (x −5) (y + 2) a) 31 b) 27 c) 12 d) −27 e) −31 c) (x + 2) (y + 5 d) (x −5) (y −2)e) (x −2) (y −5 )x x 19. Expand and simplify: 7 9+ −2 2 27. Find the area of the figure, in square meters, to 1decimal place.a) x2 1−−b)x 6 34 2 x24+ −x6 3a) 935.6mb)1691.1mc) x24 −−d)x 6 3x2 − x − 63 c) 362.2md)557.8me) x22 −−x6 3e)271.1m28. Find the volume, in cubic centimeters, of the solid20. Multiply: (4x −3) (3x −5x +3) shown where h =12cm, s=7cm, and d=8cm.Expressa) 12x4 −9x −20x −27x −9b) 12x4 −9x −20x + 27x −9 y our answer to two decimal places.a) 2077cm b) 435.63cmc) 12x4 + 9x −20x + 27x −9d) 12x4 −9x + 20x + 27x −9e) 12x4 + 9x + 20x + 27x −9c) 1742.54cm d) 217.82cme) 1258.96cm29. In the diagram shown, A' B' C is the image of21. Expand and simplify: (3x+4) −3(x+5) ABC. Which type of transformation is shown in thea) 9x + 18x + 1 b) 9x + 21x + 1 illustration?c) 9x + 9x + 1 d) 9x + 21x + 31 a) dilatione) 9x + 9x + 31 b) translationc) slide22. Factor: x −4x −12 d) rotationa) (x + 6) (x −2) b) (x + 12) (x −1) e) reflectionc) (x + 2) (x −6) d) (x + 7) ( x −3)e) (x −4) (x + 3) 30. Find the length of side x.a) 2023. Factor: 2x −7xy −15y b) 34a) (2x + 3y) (x −5y) b) (2x −3y) (x + 5y) c) 26c) (2x −y ) (x −15y) d) (2x −5y) (x + 3y) d) 25e) (2x+ 5y) (x −3y) e) 3524. When 6x + x −15 is factored completely, one of 31. A triangle has vertices A( −5, 2), B( −3, 5), andthe factors will be _____ C( −2, 1). Which of the following translations willa) 2x + 5 b) 2x + 3 c) 2x −3 d) 2x −5 e) 3x −5 place the image triangle in the fourth quadrant?a) 5 units left, 7 units down25. When (4x −7) −10(4x −7)+24 is factored b) 6 units right, 7 units downcompletely, one of the factors will be______ c) 4 units left, 7 units upa) 4x + 13 b) 4x + 3 c) x + 6 d) x + 4 e) 4x −11 d) 6 units right, 7 units upe) 4 units right, 7 units downPage 2 of 532. In the diagram shown, ABC~ DEF, c = 5, e = 7,and f= 9. Find a.d) r = 15cos25 e) r=15cos 25° 34a)c)2312e)3158b)d)3692538. A 2.5 metre ladder leans against a houseforming a30 angle with the house. Exactly how far is thebase ofthe ladder from the house?a) 1.35m b) 1.50mc) 2.25m d) 1.75m33. Given the DEF is congruent to PQR, find the e) 1.25mangle that is congruent to Ra) P b) F 39. The scale shown on a map of Canada is 1cm= c) D d) Q 120km. On the map the distance between Winnipeg and e) E Saskatoon is 5.7cm. How far apart are Winnipeg andSaskatoon to the nearest kilometre?34. Express tan A as a ratio of the variables given in a) 444km b) 516km c) 684kmABC. d) 276km e) 660kma)d)caacb)e)cbabc)ba1a) −34 b) −c)3422534d)34e)34225−140. Evaluate: (3−+ 5−)2 2121335. In ABC, which of the following is equaltoa) sinCb) tan Cc) cosA?41.Evaluate:43 38 −3( × )534( )d) cosCe) tan Aa) 1 b) 6c)127d)13e) 936. Which of the following equations can be used to 42. Find the value of x in this diagram.find the length of x? a) 8.8 b) 12.8a) cos35 =d) cos35 =e) sin35 =20x x20 x1414 b) sin35 =xc) tan55 =x30c) 9.6d) 10.2e) 7.243. Find the area of the shaded region where S = 12.a) 12.65 units b) 30.90 units c) 43.56 units37. Which equation can be used to correctly determine d) 3.90 units the length of side r? e) 115.73 units a) r = 15tan25 b) r = 15sin25 44. A right triangle has one leg of length a and a c) r = 15 sin 25°h ypotenuse of length c. Express the area of this triangle in terms of a and c.Page 3 of 5a)d)+aca22 =b)2AacA=e)2AAa c a( −)= c)2=a c a2 2−ac a2A =c a c2 2−21 1a) (2x −3)= 28 b) x(x −3)=282 21c) x(x + 3)= 28 d) x(x + 3)= 2821e) (2x + 3)= 28245. If Ax + 7x + Bx +30 is evaluated for x = −1 theresult is 54. When the expression is evaluated for x =2the result is 6. What is the value of A?a) 3 b) -3 c) 2 d) -2 e) 451.Simplify:1a) b)23x + 5 6x − 6−4 8−+12x 5c) 16 d) 2 e)10−3x246. The perimeter of a rectangle is 18x + 24 and the 52. Solve ax+ b = cx + d for x.width is 2x - 9. What is the length?a) 7x −3 b) 7x + 3 c) 16x −15d) 7x +21 e) 16x + 33a) x=d) x=b d+a c+d b−c a−b) x=e) x=d b−a c−d b−a c+c) x=d b+a c−47. A guy wire attached to the top of an 84m radioantenna is bolted to the ground 52m from the base of 53. The radius of the circle is 12cm. What is the area ofthe tower. If the wire is tight, how long is the wire? the shaded region?Express your answer to the nearest hundredth of a a) 46.8cm b) 41.1cmmetre. c) 76.8cm d) 68.2cma) 98.79cm b) 93.30cm e) 86.4cmc) 95.56cm d) 89.35cme) 136.00cm54. There are 3 dots in the first figure and 7dots in the 48. A rectangle has both of its sides increased by 20%. second figure. If this pattern is continued, how manyBy what percent dose its area increase? dots will be in the nth Figure?a) 40% b) 44% c) 140%d) 84% e) 144%49. A cone with radius r = 3cm and height h = 9cm justfits inside a cylinder with the same height and radius. 55. If the pattern shown is continued, then what is theH ow many cubic centimetres are in the space betweennumber at the end of the 30th row?the two figures. Round your answer to 2 decimal 1places. 2 3a) 85cm b) 113cm 4 5 6c) 236cm d) 170cm 7 8 9 10e) 57cm50. The base of a triangle is 3units more than the height,the area of the triangle is 28 square units. If the heightis represented by x, which equation could be used tofind the measure of the height of the triangle?Page 4 of 5Fill in the circle that matches your answer.1. ○a ○b ○c ○d ○e 31. ○a ○b ○c ○d ○e2. ○a ○b ○c ○d ○e 32. ○a ○b ○c ○d ○e3. ○a ○b ○c ○d ○e 33. ○a ○b ○c ○d ○e4. ○a ○b ○c ○d ○e 34. ○a ○b ○c ○d ○e5. ○a ○b ○c ○d ○e 35. ○a ○b ○c ○d ○e6. ○a ○b ○c ○d ○e 36. ○a ○b ○c ○d ○e7. ○a ○b ○c ○d ○e 37. ○a ○b ○c ○d ○e8. ○a ○b ○c ○d ○e 38. ○a ○b ○c ○d ○e9. ○a ○b ○c ○d ○e 39. ○a ○b ○c ○d ○e10. ○a ○b ○c ○d ○e 40. ○a ○b ○c ○d ○e11. ○a ○b ○c ○d ○e 41. ○a ○b ○c ○d ○e12. ○a ○b ○c ○d ○e 42. ○a ○b ○c ○d ○e13. ○a ○b ○c ○d ○e 43. ○a ○b ○c ○d ○e14. ○a ○b ○c ○d ○e 44. ○a ○b ○c ○d ○e15. ○a ○b ○c ○d ○e 45. ○a ○b ○c ○d ○e16. ○a ○b ○c ○d ○e 46. ○a ○b ○c ○d ○e17. ○a ○b ○c ○d ○e 47. ○a ○b ○c ○d ○e18. ○a ○b ○c ○d ○e 48. ○a ○b ○c ○d ○e19. ○a ○b ○c ○d ○e 49. ○a ○b ○c ○d ○e20. ○a ○b ○c ○d ○e 50. ○a ○b ○c ○d ○e21. ○a ○b ○c ○d ○e 51. ○a ○b ○c ○d ○e22. ○a ○b ○c ○d ○e 52. ○a ○b ○c ○d ○e23. ○a ○b ○c ○d ○e 53. ○a ○b ○c ○d ○e24. ○a ○b ○c ○d ○e25. ○a ○b ○c ○d ○e 54. ______________________________26. ○a ○b ○c ○d ○e27. ○a ○b ○c ○d ○e 55. ______________________________28. ○a ○b ○c ○d ○e29. ○a ○b ○c ○d ○e30. ○a ○b ○c ○d ○ePage 5 of 5。
天津市滨海新区枫叶国际学校2024年九上数学开学教学质量检测模拟试题【含答案】

天津市滨海新区枫叶国际学校2024年九上数学开学教学质量检测模拟试题题号一二三四五总分得分A 卷(100分)一、选择题(本大题共8个小题,每小题4分,共32分,每小题均有四个选项,其中只有一项符合题目要求)1、(4分)在直角坐标系中,点P (-3,3)到原点的距离是()A .B .C .D .62、(4分)在实数范围内有意义,则x 的取值范围是()A .3x ≥B .3x ≤C .3x ≠D .x <33、(4分)小李家距学校3千米,中午12点他从家出发到学校,途中路过文具店买了些学习用品,12点50分到校.下列图象中能大致表示他离家的距离S (千米)与离家的时间t (分钟)之间的函数关系的是()A .B .C .D .4、(4分)若点()()()112233,,,,,x y x y x y 都是反比例函数21a y x --=的图象上的点,并且1230x x x <<<,则下列各式中正确的是(()A .132y y y <<B .231y y y <<C .321y y y <<D .123y y y <<5、(4分)在“美丽乡村”评选活动中,某乡镇5个村的得分如下:90,88,96,92,96,这组数据的中位数和众数分别是()A .90,96B .92,96C .92,98D .91,926、(4分)下列曲线中,不能表示是的函数的是()A .B .C .D .7、(4分)的结果是A .-2B .2C .-4D .48、(4分)如图,在平面直角坐标系中,菱形ABCO 的顶点O 为坐标原点,边CO在x 轴正半轴上,∠AOC =60°,反比例函数y =x (x >0)的图象经过点A ,交菱形对角线BO 于点D ,DE ⊥x 轴于点E ,则CE 长为()A .1B C .D .﹣1二、填空题(本大题共5个小题,每小题4分,共20分)9、(4分)在学校的卫生检查中,规定各班的教室卫生成绩占30%,环境卫生成绩占40%,个人卫生成绩占30%.八年级一班这三项成绩分别为85分,90分和95分,求该班卫生检查的总成绩_____.10、(4分)已知x +y =0.2,2x +3y =2.2,则x 2+4xy +4y 2=_____.11、(4分)如图,在△ABC 中,BC 的垂直平分线MN 交AB 于点D ,CD 平分∠ACB .若AD =2,BD =3,则AC 的长为_____.12、(4分)若n 边形的内角和是它的外角和的2倍,则n =.13、(4分)如图,在直角坐标系中,已知点A (-3,-1),点B (-2,1),平移线段AB ,使点A 落在A 1(0,1),点B 落在点B 1,则点B 1的坐标为_______.三、解答题(本大题共5个小题,共48分)14、(12分)如图,在平行四边形ABCD 中,E 、F 分别为边AB 、CD 的中点,BD 是平行四边形ABCD 的对角线,//AG BD 交CB 的延长线于点G .(1)求证:四边形BEDF 是平行四边形.(2)若AE DE =,求G ∠的度数.15、(8分)计算:16、(8分)甲车从A 地驶往B 地,同时乙车从B 地驶往A 地,两车相向而行,匀速行驶,甲车距B 地的距离y (km )与行驶时间x (h )之间的函数关系如图所示,乙车的速度是60km/h .(1)求甲车的速度;(2)当甲乙两车相遇后,乙车速度变为a (km/h ),并保持匀速行驶,甲车速度保持不变,结果乙车比甲车晚38分钟到达终点,求a 的值.17、(10分)某景区的水上乐园有一批4人座的自划船,每艘可供1至4位游客乘坐游湖,因景区加大宣传,预计今年游客将会增加.水上乐园的工作人员在去年6月27日一天出租的150艘次4人自划船中随机抽取了100艘,对其中抽取的每艘船的乘坐人数进行统计,并制成如下统计图.(1)求扇形统计图中,“乘坐1人”所对应的圆心角度数;(2)估计去年6月27日这天出租的150艘次4人自划船平均每艘船的乘坐人数;(3)据旅游局预报今年6月27日这天该景区可能将增加游客300人,请你为景区预计这天需安排多少艘4人座的自划船才能满足需求.18、(10分)某中学开展“一起阅读,共同成长”课外读书周活动,活动后期随机调查了八年级部分学生一周的课外阅读时间,并将结果绘制成两幅不完整的统计图,请你根据统计图的信息回答下列问题:(1)本次调查的学生总数为______人,在扇形统计图中,课外阅读时间为5小时的扇形圆心角度数是______;(2)请你补全条形统计图;(3)若全校八年级共有学生900人,估计八年级一周课外阅读时间至少为5小时的学生有多少人?B 卷(50分)一、填空题(本大题共5个小题,每小题4分,共20分)19、(4分)若4,9n n x y ==,则()n xy =_______________.20、(4分)如图,直线y kx b =+经过点()1,2--A 和点()2,0B -,直线2y x =经过点A ,则不等式组20x kx b <+<的解集是______.21、(4分)已知一次函数y =kx +b 的图像过点(-1,0)和点(0,2),则该一次函数的解析式是______。
大连枫叶国际学校入学测试完整版

大连枫叶国际学校入学测试Document serial number【NL89WT-NY98YT-NC8CB-NNUUT-NUT108】Name ______________ Score_________/40Part one:Passage one: Words: 201 Readability: 4.0Early to bed, early to rise, makes a man healthy, wealthy and wise.This is an old English saying. It means that we must go to bed early in the evening and get up early in the morning.Is this true Perhaps it is. The body must have enough sleep. For example, children need ten hours’ sleep every n ight. If we do not go to bed early, we cannot have enough sleep. Then we cannot think or do our work well. We will not be wise or live a comfortable life.Some people go to bed late at night and get up late in the morning. This is not good for them. We must sleep at night when itis dark. When daylight comes, we must get up. This is the time for exercise. Walking, running, jumping, swimming, playing games are all good forms of exercise. Exercise keeps the body strong.Exercise helps blood to move around inside the body. This is very important. Blood takes food to all parts of our bodies. The brains in our heads also need blood. We think with our brains. If we keep our bodies healthy, and do exercises, we can think better!A. Write T for true F for false beside the following statements.(2.5’)____ 1.The passage tells the readers why it is important to be healthy.____ 2.Early to bed and early to rise helps us think and work well.____ 3.To keep healthy, everybody must sleep ten hours a day.____ 4.Daylight is the time for exercise.____ 5.Blood moves to all parts of our bodies.B. Answer the following questions incomplete sentences and in your own words.(2’)1. Give some forms of exercise from the passage._____________________________________________________________________ ___2. You must have some habits which are good for your health. What are they_____________________________________________________________________ _____________________________________________________________________ ______3. Give a proper title to the passage. (0.5’)_____________________________________________________________________ ___Passage Two:Words: 231 Readability: 3.8SealA man moved his boat across the lake. Over his head, a bird flew in the wind and the other rested in a tree. Both watched the man.They knew the man was afraid of the sea. He wanted to swim but he was afraid of drowning. The two birds watched the man return to his home. They saw he didn’t have any confidence in himself. His house was small because he was not sure he could build a larger one. He didn’t cook his food because he was afraid he would burn it. The man did not run fast because he was afraid of falling. One bird said, “How will he succeed” “He needs to find the talents that he had inside himself,” the other answered. “We must choose the correct time and chance carefully,” said the first one. One day, his daughter fell into the water. The man jumped into the sea without thinking. At that time, the two birds changed him into a seal. He saved his daughter. From that day on, the man could change himself into a seal whenever he wanted. He could swim and do things that no other person could do.The two birds were happy. Just like the man, all of us have special ability. It’s just waiting to be changed from dreams into reality, if only we have the courage to try.Answer the following questions in complete sentences. (5’)1. What was the man afraid of when he wanted to swim (1’)_____________________________________________________________________ ___2. Give 2 examples to show the man has no confidence in himself. (1’)_________________________________________________________________3. Why did the man jump into the sea one day (1’)_____________________________________________________________________ ___4. How did the two birds help him Which kind of animal is a seal (1’)_____________________________________________________________________ ___5. What does the story want to tell us (1’)_____________________________________________________________________ ___%Correct ____________=Mastery of Level_________(Yes/No)Part two:Passage one: Words: 242 Readability: 4.9Fiery Visitors from Outer Space1. On the next clear summer night, look up into the sky. If you are lucky, you may see a “shooting star”. They look just like stars falling down from the sky.2. But shooting stars are not stars at all. The real name for a shooting star is “meteor”. There are millions of meteors out in space. They are pieces of rock and metal. When they are in space, meteors do not glow. In fact, we can’t see them at all. But when a meteor comes close to the earth, it starts to fall toward the earth. As it passes through the air, it gets very hot. It starts to glow, and finally it begins to burn. We see the burning meteor as a shooting star.3. Sometimes meteors travel by themselves in space. But many meteors travel in groups called swarms. The swarms have orbits like planets have. They pass close to the earth every year. A famous swarm is the Perseid swarm. It passes close to the earth in summer, usually in August. Each year a few meteors from the Perseid swarm fall to the earth, so August is a good month to watch for shooting stars.4. Most meteors are small, so they burn up before they crash into the earth and are buried in the ground. Once a meteor reaches the earth, we call it a meteorite. The biggest meteorite, which weighs seventy tons, landed in Africa.A.Circle the main idea of paragraph 3.(0.5’)1. Sometimes meteors travel by themselves in space.2. Sometimes meteors travel in swarms.3. A famous swarm is the Perseid swarm.4. August is a good month to watch for shooting stars.B. Match the word on the left with the phrase on the right that fits best by writing letters. (1’)1. ______meteor a. a famous swarm of meteors2. ______swarm b. a meteor that has reached the earth3. ______Perseid c. The real name for a shooting star4. ______Meteorite d. a group of meteorsC. Answer questions in complete sentences and in your own words.1.Why is Augusta good month to watch shooting stars1’_____________________________________________________________________ _____________________________________________________________________ ______2. Where is the biggest meteorite 0.5’_____________________________________________________________________ ___3. There is an old saying that if you make a wish when you see a shooting star, your wish will come true. So when you see a shooting star, what wish will you make Why2’_____________________________________________________________________ ______Passage Two: Words: 270 Readability: 4.2ZanOn the island of Crete, the Great Earth Mother had a child called Zan. The bees of the forest brought honey for little Zan, and thewild goats gave him milk. But children must have toys. So the five servants of the Great Earth Mother made toys for him. Zan played happily on the mountains with his toys and with the animals in the forest.When he grew up into a strong, kind man, Zan wanted to reward his animal friends. So he made a law that the goats could go wherever they wished. He made hives for the bees to protect them from the cold winds and snow. Then, Zan put the toys he had saved in a secret cave where they would be safe. He asked the bees to guard the cave.One day three bad men climbed up the mountain to find treasure. They put on armor to protect themselves from the bees, and they crept through the bushes until they reached the mouth of the cave. The bees buzzed around to attack the men, but their armor protected them and they went inside.Suddenly Zan appeared in the cave. “Stop!”He shouted. “You have come here to rob and kill. I must let you live, for this is the cave of life. But from now on, you will never be able to tell other people where my wonderful cave is.”As Zan finished speaking, the men turned into birds. One turned into a woodpecker, one became a thrush, and one became an owl.That’s why even to day, when the birds speak, no one can understand them.A. Number the events below in the order that they happened. (0.5’*4) __________ Three bad men tried to steal Zan’s toys.__________ The Earth Mother had a child named Zan.__________ Zan turned the men into birds.__________ Zan rewarded his animal friends.B. Answer the following questions in complete sentences. (1’*3)1. Why did the three men come to the mountain_____________________________________________________________________ ________________________________________________________________________ ___2. What happened to the three men at last_____________________________________________________________________ ______3. What kind of person do you think Zen is ( At least 3 adjectives ) _____________________________________________________________________ ___%Correct ____________=Mastery ofLevel_________(Yes/No)Part three:Passage One: Words 312 Readability 6.8Marie CurieMarie Curie was a great scientist. She was born in Warsaw, Poland in 1867. Both of her parents were teachers. When Marie was only 10 years old, her mother died.Marie was a very good student. She loved science, math, and languages. She and her sister Bronya wanted to go to college, but in those days, only men could go to college in Poland. The girls had to go to France to study. There was not enough money for both sisters to go, so Marie worked as a teacher in Poland. She sent money to Bronya to pay for medical school in Paris. After Bronya became a doctor, she helped Marie.When she was 24, Marie became a science student at the Sorbonne, a university in Paris. Even with her sister’s help, she did not have much money. She lived in a small room near college. It had no lights, no water, and no heat. Sometimes Marie only had bread and tea to eat.Marie studied hard and graduated (finished school) in 1894. A year later, she married Pierre Curie. He was also a scientist. They worked hard together for many years. Their most important discovery was radium. Today, doctors use the rays from radium to treat cancer. The Curies won a Nobel Prize for their discovery. This is the highest award for a scientist. Marie Curie was the first woman to receivethis award.When Marie Curie was 39, Pierre died in a road accident. But she continued their work. Curie became the first woman professor in France. In 1911, she won a second Nobel Prize. Unfortunately,years of working with radium harmed her health. She died of cancer in 1934. Her daughter Irene continued Curie’s work. She also received a Nobel Prize. Sadly, Irene also got cancer and died young. Both women gave their lives for their work.Answer the following questions in complete sentences in your own words. 5’1.What kind of family was Marie born in_____________________________________________________________________ _____________________________________________________________________ 2.What kind of life did Marie Curie have when she studied at university in Paris_____________________________________________________________________ 3.How many times did Marie Curie get the Nobel Prize_____________________________________________________________________ _____________________________________________________________________ 4.Marie was born in Poland. Why did she become the first woman professor in France_____________________________________________________________________ _____________________________________________________________________ 5.What kind of person do you think Marie Currie is Give examples to support your ideas._____________________________________________________________________ _____________________________________________________________________ Passage Two: Words: 412 Readability: 5.6The Choking DogJoanne beat impatiently on the steering wheel of her Mercedes sports car. Now she was stuck in a traffic jam in central Birminghamat 5:30 p.m., and at 6:30 she was expected to hold a meeting of the tennis club.At last, the traffic was moving, and she switched quickly to the fast lane to get home.After she parked her car, she leapt out of the car and ran for the house. As she opened the door, she almost knocked down Sheba, her dog, who was standing behind it.She bent down to stroke the large Alsatian dog's head, "I have to go now” Sheba was coughing or choking as if she was trying to vomit something. She was obviously in real discomfort, and her sad eyes gazed at Joanne helplessly. Joanne examined the dog closely, and Sheba did look very sick. Joanne realized that it is necessary totake her to the vet immediately. Fortunately, the vet's surgery isnot far. Joanne quickly loaded the coughing and choking dog and took her to the vet.As soon as she arrived at Dr. Sterne’s place, he brought her immediately into his office."Something is stuck in her throat," said Dr. Sterne."Doctor, I have to go back to get changed for an emergent meeting. I'll be back in ten minutes to pick her up, OK""Sure,” said the doctor.Joanne hurriedly went back to her home. As she entered the door, the phone rang.“This is Dr. Sterne. I'm coming with the police. Wait outsidefor us.” said t he doctor with fear in his voice. Joanne was confused and also a little frightened.When she ran into the street, two policemen were getting out of their car. After briefly checking that she was the owner of the house, they ran into the house, without explaining anything. Joanne was completely confused and very frightened. Then the doctor arrived."Where's Sheba Is she OK" shouted Joanne."She's fine, Joanne. I extracted the thing which was choking her."Why are the police in my house"Just then, the two policemen reappeared from the house, half-carrying a white-faced man who could hardly walk and with blood all over him."How did he get in there”"I think he must be a burglar," said the doctor. "I knew he was there because when I finally removed what was stuck in Sheba's throat, it turned out to be three human fingers. I don't think he's a very happy burglar."I.Circle the correct letter of the answer. (2’)1.Why was Joanne impatient at the beginning of the storya.She was lost.b. She had lost a client at work.c.She was stuck in a traffic jam.d. Her dog was sick.2.Why did Joanne take the dog to Dr. Sterne's surgerya.It was time for Sheba's checkup.b.The dog couldn't breathe properly.c.She wanted to get her out of the house.d.The doctor had asked to see her.3.What happened as she arrived home for the second timea.The police arrived.b.The phone rang.c.The dog died.d.A burglar was just escaping.4.The story says that the dog "gazed at Joanne helplessly". "Gazed" meansa. staredb. criedc. barkedd. laughedII. Answer the following questions in complete sentences and in your own words. (3’)1. Why did?the doctor ask Joanne to wait outside her house_____________________________________________________________________ _____________________________________________________________________ ____________________2. Use twoor three sentences to describe what happened between thedog and theburglar._____________________________________________________________________ _____________________________________________________________________ ____________________3. What are?the consequences of burgling_____________________________________________________________________ _____________________________________________________________________ ____________________%Correct ____________=Mastery of Level_________(Yes/No)Part Four:Passage One: Words: 358? Readability: 7.5DrumsWhen we think of a drum we usually think of a musical instrument. The drum has also been used for years as an important tool for communication. The human voice is very good for communication, but it cannot be heard very far. As early humans began to live in groups, runners were sent from village to village with important messages, but that took a lot of time. It was discovered that beating on hollow logs could be heard for greater distances, and so a new form of communication called “talking drums” was born.In many early communities, a large drum was set in the middle of the village to help send messages to nearby tribes. The drums could be hit with hands and sticks. Some of the larger drums could be heard up to sixteen miles away!A group in West Africa made an amazing talking drum system. Their drums were made of round pieces of wood that were four to five feet long. A skin (often an elephant ear) was stretched over one end of the wood and was held in place by a rope. The skin was tightened or loosened to make higher and lower sounds. They had learned to use the drums so well that they could actually copy the language that they spoke! The drums sounded like a person speaking in a voice loud enough to be heard far away.Most systems of drum communication used a series of beats that meant something. The messages usually told of danger, death, war, or other news of great importance. The message could be sent very far by a relay system. That is, one village would send a message and a nearby village would hear it. They would then repeat the message on their drums, sending it to their neighbors. Some relay systems worked so well that they could send a message about 200 miles very quickly. Although “talking drums” became a good system of communicating over distances, it had one very big problem. Their enemies could understand the drum messages, so they would know what was being said or planned. Communication by drums certainly wasn’t perfect!A. True or False. Decide the following statements are true (T) or false (F). (2’)_______1. The main idea of the passage is that drums can be heard over great distances._______2. Drums are only used as a musical instrument._______3. People in West Africa tightened the skin over the drum to make a lower sound._______4. Drums made by people in West Africa could speak.B. Answer the following questions in complete sentences and in your own words. (3’)1. What is a drum “relay system” (1’)_____________________________________________________________________ _____________________________________________________________________ ___.2. Name 4 other forms of communication and which one do you like best and why (2’)_____________________________________________________________________ _____________________________________________________________________ _____________________________________________________________________ _____.Passage Two: Words 360 Readability: 6.0We Like to ShareGeorge and Ellen were successful owners of a restaurant which they kept clean and spotless while serving good inexpensive food. They were very friendly with all their faithful customers who dined at their café because they enjoyed being friendly and helpful.One day an unfamiliar elderly couple came into the thriving restaurant.Although the old man ordered a hamburger and some French fries, his grey-haired wife didn’t order anything to eat.George prepared the food and served it to the man.The old fellow and his wife sat at a dining table and the man cut the hamburger in half.Then he divided the French fries into two equal parts.He put half of the hamburger and half of the French fries infront of his wife.Then he began to eat his half of the hamburger and fries.His wife relaxed and watched while he ate.Ellen observed the old couple.She thought that perhaps they couldn’t afford another hamburger. She felt sorry for them, so she went over to their table and offered them another meal free of charge.The old man was too busy eating to reply, but his wife said, “Oh, No!Don’t worry about us.We like to share.We share everything.”The old fellow nodded and said, “Everything,” before he stuffed more French fries into his mouth.Ellen smiled at them and returned to the counter.She took some orders from other waiting customers and then watched the old couple again.She noticed that the woman was still not eating.Her half of the cooling burger and fries remained sitting on the plate on the table in front of her.She reasoned that perhaps the man’s wife intended to give her share to her husband if he was still hungry after eating his share.Ellen frowned in confusion.She hastily returned to their table.The old woman saw her and commenced telling her not to worry because they had always shared everything.“I know,” responded Ellen.“You’ve already told me that.I was just wondering why you are not eating your half.”“Oh,” said the woman.“I’m just waiting for my turn with the false teeth.” (6.0) Answer the following questions in complete sentences in your own words. 5’1. What kind of food did George and Ellen’s restaurant serve_____________________________________________________________________ ________________________________________________________________________ ___2. Why did Ellen think the old couple couldn’t afford another hamburger_____________________________________________________________________ ________________________________________________________________________ ___3. What is the meaning of intendeda.plannedb.wonderedc. noticedd. wanted4. Write an adjective in each blank._____________________customers ______________________hamburger5. Why was Ellen so confused by the behavior of the oldpeople?_____________________________________________________________________ ________________________________________________________________________ ___%Correct ____________=Mastery of Level_________(Yes/No)。
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6. Solve for x given 4(3x − 7)= − 7 a)
7 6
b) 0
c) −
35 12
d)
7 4
e)
35 12
7. Solve for x given 4 − 2(3x − 4)=1 − (5x+4). a) − 9 b) − 1 c) 15 d) 3 e) 7 8. What is the value of the expression − 3x2 +4x when x= − 5 ? a) − 55 b) − 95 c) 50 d) 95 e) 55
5. Write an algebraic equation for the following " After 7 is added to a number the sum is divided by4. The result is 12."
7n = 12 4 n d) + 7 = 12 4
a)
n + 4 = 12 7 n+7 e) = 12 4
4 Simplify and rewrite using only positive exponents:
4 x2 ( 2 y )3
−2
x4 a) 4 y6
4 y6 b) x4
−4 y 6 c) x2
d)
1 y6
e) y 6 a) 22x²- 19x- 23 c) 16x²-8x - 5 e) 16x²-8x +5 b) -16x²+8x -5 d) 16x²-24x+5
26. Factor completely: xy − 5y − 2x + 10 a) (x + 5) (y − 2) b) (x − 5) (y + 2) c) (x + 2) (y + 5) d) (x − 5) (y − 2) e) (x − 2) (y − 5 ) 27. Find the area of the figure, in square meters, to 1 decimal place. a) 935.6m² b) 1691.1m² c) 362.2m² d) 557.8m² e) 271.1m² 28. Find the volume, in cubic centimeters, of the solid shown where h =12cm, s=7cm, and d=8cm.Express your answer to two decimal places. a) 2077cm³ b) 435.63cm³ c) 1742.54cm³ d) 217.82cm³ e) 1258.96cm³ 29. In the diagram shown, △ A' B' C is the image of △ABC. Which type of transformation is shown in the illustration? a) dilation b) translation c) slide d) rotation e) reflection 30. Find the length of side x. a) 20 b) 34 c) 26 d) 25 e) 35 31. A triangle has vertices A( − 5, 2), B( − 3, 5), and C( − 2, 1). Which of the following translations will place the image triangle in the fourth quadrant? a) 5 units left, 7 units down b) 6 units right, 7 units down c) 4 units left, 7 units up d) 6 units right, 7 units up e) 4 units right, 7 units down
c a a d) c
c b a e) b
b)
c)
b a
40. Evaluate: a) − 34
(3
−2
+ 5−2 )
−1
b) −
1 34
8
c)
225 34
d) 34
e)
34 225
35. In △ABC, which of the following is equal to a) sinC b) tan C c) cosA d) cosC e) tan A
64 125
Date_____________________
9. Evaluate − 2a3b − 4a4 for a= − 3 and b= − 2 c) − 128 d)144 e) − 144 a) 432 b) − 432 10. Solve: 3x − 7 ≥ − 22
2. Rewrite 4 − 3 using positive exponents only and evaluate the result.
38. A 2.5 metre ladder leans against a house forming a 30ºangle with the house. Exactly how far is the base of the ladder from the house? a) 1.35m b) 1.50m c) 2.25m d) 1.75m e) 1.25m 39. The scale shown on a map of Canada is 1cm= 120km. On the map the distance between Winnipeg and Saskatoon is 5.7cm. How far apart are Winnipeg and Saskatoon to the nearest kilometre? a) 444km b) 516km c) 684km d) 276km e) 660km
12 ? 13
41. Evaluate:
(3 × 3 ) (3 )
4 5
−3 4
a) 1
b) 6
c)
a) x ≤ 5
b) x ≤ − 5
c) x ≤ −
28 3
d) x ≥ − 5
e) x ≥ 5
1 1 = 3 4 64 3 d) − 4 = − 64
a)
b) − 3×41 = − 12 e) 4 =12
− 1 2
3
c) 43 =64 11. Solve: a) x > 18
15 − 2 x <7 3
19. Expand and simplify:
2 a) x − 1 x − 6 3
x x + 7 −9 2 2
2 b) x + x − 6 3
4
2
4
Байду номын сангаас
2 c) x − x − 6 3
d) x − x − 63
2
4
2 e) x − x − 6 3
2
20. Multiply: (4x − 3) (3x³ − 5x +3) a) 12x4 − 9x³ − 20x² − 27x − 9 b) 12x4 − 9x³ − 20x² + 27x − 9 c) 12x4 + 9x³ − 20x² + 27x − 9 d) 12x4 − 9x³ + 20x² + 27x − 9 e) 12x4 + 9x³ + 20x² + 27x − 9 21. Expand and simplify: (3x+4)² − 3(x+5) b) 9x² + 21x + 1 a) 9x² + 18x + 1 c) 9x² + 9x + 1 d) 9x² + 21x + 31 e) 9x² + 9x + 31 22. Factor: x² − 4x − 12 a) (x + 6) (x − 2) c) (x + 2) (x − 6) e) (x − 4) (x + 3)
b) x < 18 c) x < − 3 d) x> − 3 e) x < 3
4 3 Evaluate: 49 7 7 a) b) − 2 2
c)
2 7
d) −
2 7
e) −
2 49
12. Simplify: (4x² − 8xy+2y²) − (9x² − 4xy − 7y²) a) − 5x² − 4xy − 5y² b) − 5x² − 4xy + 9y² c) − 5x²+ 12xy + 4y² d) − 5x²+ 4xy + 9y² e) − 5x² − 12xy + 9y² 13. How long is BC?
MATH PROFICIENCY TEST Name_________________________
Record your answer on the answer sheet provided.
1. Simplify: (2x4y5)³ b) 2x12y15 a) 6x12y15 d) 8x12y15 e) 8x7y8 c) 8x y
Page 2 of 5
32. In the diagram shown, △ABC~△DEF, c = 5, e = 7, and f= 9. Find a. a)
d) r = 15cos25º
e) r =
15 cos 25°
3 4
c) 12