2026届深圳福田六年级数学小升初分班考试考前模拟试卷第039套强证据校准版（含答案详解与评分标准）

第页2026届深圳福田六年级数学小升初分班考试考前模拟试卷第039套强证据校准版（含答案详解与评分标准）考试名称深圳福田六年级数学小升初分班考试考前模拟试卷考试时间90分钟满分120分适用对象2026届小学六年级小升初备考学校________________班级________________姓名________________考号________________注意事项本卷共18题，满分120分，考试时间90分钟。试题分为选择题、填空题、解答题三部分，参考答案与解析另起新页。选择题请在答题卡中填涂或写出对应选项；填空题只写最终结果，必要时保留分数形式；解答题必须写出主要计算过程、关键公式和结论。涉及单位的结果须写明单位；涉及π的计算，题中未特别说明时可保留π，题中给出π＝3.14时按3.14计算。本卷为Word文本版，可打印可作答，答题区域已预留；请保持卷面整洁，严禁在密封线外写无关内容。卷面结构与分值题型题号分值重点能力选择题1—840分基础概念、运算、函数图象、几何判定填空题9—1220分数形结合、方程、不放回概率、几何面积解答题13—1860分过程计算、综合应用、图形几何、统计方案、动点压轴选择题答题卡12345678（）（）（）（）（）（）（）（）一、选择题（共8题，每题5分，共40分）每题只有一个正确选项。请认真审题，避免因单位、符号或运算顺序失分。1.计算5/8＋3/4÷(1/2)－0.25的结果是（）。A.15/8B.17/8C.9/4D.11/42.三位数4□8既是3的倍数，又是2的倍数。□中可填的最大数字是（）。A.0B.3C.6D.93.若3a＝5b（a、b均不为0），则a:b等于（）。A.3:5B.5:3C.8:15D.15:84.在平面直角坐标系中，一次函数y＝－2x＋3的图象不经过的象限是（）。A.第一象限B.第二象限C.第三象限D.第四象限5.一个三角形三边长分别为5cm、12cm、13cm。下列判断正确的是（）。A.是锐角三角形B.是直角三角形C.是钝角三角形D.不能构成三角形6.某小组5名同学一分钟跳绳个数为132、136、140、144、148。若把每人都提高8个，则这组数据的（）。A.平均数不变B.中位数不变C.极差不变D.众数增加87.一个圆柱底面半径为3cm，高为5cm，它的体积是（）。A.15πcm³B.30πcm³C.45πcm³D.90πcm³8.一件运动外套先提价20%，再按提价后的价格降价20%。与原价相比，现价（）。A.提高4%B.降低4%C.不变D.降低20%二、填空题（共4题，每题5分，共20分）请把答案直接写在横线上，能化简的分数请化为最简分数。9.数轴上点A表示－2.5，点B表示1.75，则线段AB的中点表示的数是__________。答题处：____________________________________________________________10.解方程4(2x－3)＝3x＋18，得x＝__________。答题处：____________________________________________________________11.袋中有3个红球、2个蓝球、1个黄球，除颜色外完全相同。不放回地连续摸2次，第一次摸到红球且第二次摸到蓝球的概率是__________。答题处：____________________________________________________________12.一个梯形的上底为8cm，下底为14cm，高为6cm，它的面积是__________cm²。答题处：____________________________________________________________三、解答题（共6题，每题10分，共60分）解答题要写出必要的文字说明、计算过程、公式依据和最终结论。只写答案且无过程的，按评分标准酌情扣分。13.计算与方程。请写出必要的化简步骤。

（1）(7/8－1/6)÷(5/12)＋0.4×15；

（2）解方程：2/3x＋5＝17。解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________14.行程应用题。深圳福田中心到南山科技园的模拟路线全长25.2km。甲车从福田中心出发，每分钟行720m；乙车从南山科技园出发，每分钟行480m。甲车8:00出发，乙车8:05出发，两车沿同一路线相向而行。

（1）乙车出发后多少分钟两车相遇？

（2）相遇时甲车已经行驶多少千米？

（3）若要求两车在8:20相遇，乙车速度不变，甲车速度应调整为每分钟多少米？解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________15.图形几何应用。某校劳动实践区是一块长24m、宽18m的长方形菜地。靠其中一条长边修建一个半圆形蓄水池，半圆直径为12m，且直径完全在长方形边上。计算时取π＝3.14。

（1）蓄水池面积是多少平方米？

（2）剩余可种植面积是多少平方米？

（3）若沿半圆弧铺设护栏，护栏每米38元，护栏费用约是多少元？解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________16.统计与概率综合。六年级某班进行一次数学分层练习，成绩分布如下表。

请根据表格完成问题。等级A优秀B良好C合格成绩区间85—100分70—84分60—69分人数12人18人6人等级代表分92分78分64分备注全班共40人另有D待提高4人D代表分为52分（1）A等级人数占全班人数的百分之几？（2）若制作扇形统计图，B等级所对应扇形的圆心角是多少度？（3）用“等级代表分”估算全班平均分，并说明这只是估算值的原因。解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________17.方案选择应用。某文具店同一种数学练习本标价12元/本，面向毕业班采购给出两种优惠方案：

方案A：全部按标价的八折付款；

方案B：每买10本赠送2本，不足10本的部分按原价付款。

（1）六（1）班需要72本，分别按两种方案应付多少元？选择哪种方案更省钱？

（2）若预算为700元，按方案B最多可得到多少本？

（3）请说明在购买数量较多时，为什么不能只看“折扣字样”，还要比较实际得到的本数和付款总额。解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________18.压轴题：动点与面积。直角三角形OAB中，OA＝12cm，OB＝9cm，∠AOB＝90°。点P从O出发沿OA向A运动，速度为1cm/s；点Q从B出发沿BO向O运动，速度为0.5cm/s，两点同时出发。设运动时间为t秒，且0≤t≤12。

（1）用含t的式子表示OP、OQ的长度，并写出△OPQ的面积S与t的关系式；

（2）当S＝20cm²时，求t的值；

（3）求S的最大值，并说明面积随时间变化的特点。解答区：________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

参考答案与解析一、选择题答案与依据（共40分）题号答案解析依据评分说明1A3/4÷1/2＝3/2，0.25＝1/4，5/8＋3/2－1/4＝5/8＋12/8－2/8＝15/8。选对得5分；错选、多选、不选得0分。2D4□8的个位是8，必是2的倍数；要是3的倍数，4＋□＋8＝12＋□应是3的倍数，□可为0、3、6、9，最大为9。选对得5分；错选、多选、不选得0分。3B3a＝5b，两边同除以3b，得a/b＝5/3，所以a:b＝5:3。选对得5分；错选、多选、不选得0分。4Cy＝－2x＋3的截距为3，斜率为负。x＞0且较小时y＞0，过第一象限；x＜0时y＞0，过第二象限；x＞3/2时y＜0，过第四象限；不经过第三象限。选对得5分；错选、多选、不选得0分。5B5²＋12²＝25＋144＝169＝13²，满足勾股定理，故为直角三角形。选对得5分；错选、多选、不选得0分。6C每个数据都增加8，平均数和中位数都增加8，极差仍为148－132＝16；原数据无众数，不能说众数增加8。选对得5分；错选、多选、不选得0分。7C圆柱体积V＝πr²h＝π×3²×5＝45πcm³。选对得5分；错选、多选、不选得0分。8B设原价为1，提价20%后为1.2，再降价20%为1.2×0.8＝0.96，现价比原价降低4%。选对得5分；错选、多选、不选得0分。二、填空题答案与依据（共20分）题号答案解析依据评分说明9－3/8中点数＝(－2.5＋1.75)÷2＝(－0.75)÷2＝－0.375＝－3/8。答案正确得5分；形式等价且最简可得满分；计算小错酌扣1—2分。1064(2x－3)＝3x＋18，8x－12＝3x＋18，5x＝30，x＝6。答案正确得5分；形式等价且最简可得满分；计算小错酌扣1—2分。111/5第一次红球概率为3/6；第一次取走红球后余5个球，其中蓝球2个，第二次蓝球概率为2/5。所求概率为3/6×2/5＝1/5。答案正确得5分；形式等价且最简可得满分；计算小错酌扣1—2分。1266梯形面积＝(上底＋下底)×高÷2＝(8＋14)×6÷2＝66cm²。答案正确得5分；形式等价且最简可得满分；计算小错酌扣1—2分。三、解答题参考答案、采分点与易错提示（共60分）13.计算与方程（10分）（1）(7/8－1/6)÷(5/12)＋0.4×15。先通分：7/8－1/6＝21/24－4/24＝17/24；再除以5/12：17/24÷5/12＝17/24×12/5＝17/10；0.4×15＝6；所以原式＝17/10＋6＝77/10＝7.7。（2）2/3x＋5＝17，2/3x＝12，x＝12×3/2＝18。采分点分值说明分数通分、除法转乘法正确3分含17/24、17/10两个关键结果小数乘法与最终合并正确2分最终为77/10或7.7方程移项正确2分由2/3x＋5＝17得到2/3x＝12方程解与检验意识3分x＝18，代入可成立；书写规范易错点：把“÷5/12”误看成“×5/12”，或把2/3x理解成2/(3x)，都会导致结果错误。14.行程应用题（10分）统一单位：25.2km＝25200m。乙车8:05出发时，甲车已行5分钟，路程为720×5＝3600m，剩余相向距离为25200－3600＝21600m。（1）乙车出发后，两车合速度为720＋480＝1200m/min，相遇时间为21600÷1200＝18分钟。（2）相遇时为8:23，甲车共行23分钟，路程为720×23＝16560m＝16.56km。（3）若要求8:20相遇，则甲车行20分钟，乙车行15分钟。乙车路程为480×15＝7200m，甲车需行25200－7200＝18000m，速度为18000÷20＝900m/min。采分点分值说明单位换算与先行路程2分能把25.2km化为25200m，并求出3600m相遇时间3分列式(25200－3600)÷(720＋480)＝18相遇时甲车路程2分720×23＝16560m＝16.56km调整速度3分列式(25200－480×15)÷20＝900易错点：乙车晚出发5分钟，不能把两车都按同一时间计算；第三问的“8:20”是从甲车8:00开始算20分钟，从乙车8:05开始算15分