习题练习

[{"id":522,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的课外阅读时间时，收集了以下数据（单位：小时）：3, 5, 4, 6, 3, 7, 5, 4, 3, 6。他将这些数据按从小到大的顺序排列后，发现中位数是4.5。如果再加入一个数据4，那么新的数据组的中位数是多少？","answer":"A","explanation":"原数据有10个数：3, 3, 3, 4, 4, 5, 5, 6, 6, 7。按从小到大排列后，第5个数是4，第6个数是5，中位数是(4+5)÷2=4.5。加入一个4后，新数据组有11个数：3, 3, 3, 4, 4, 4, 5, 5, 6, 6, 7。此时数据个数为奇数，中位数是第6个数，即4。因此新的中位数是4。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:25:10","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"4","is_correct":1},{"id":"B","content":"4.25","is_correct":0},{"id":"C","content":"4.5","is_correct":0},{"id":"D","content":"5","is_correct":0}]},{"id":533,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生在整理班级同学的课外阅读情况时，随机抽取了20名学生，记录了他们每周课外阅读的时间（单位：小时），数据如下：3, 5, 4, 6, 3, 7, 5, 4, 5, 6, 4, 3, 5, 6, 7, 4, 5, 6, 5, 4。为了分析这些数据，该学生制作了频数分布表。请问阅读时间为5小时的学生人数是多少？","answer":"C","explanation":"题目考查的是数据的收集、整理与描述中的频数统计。我们需要从给出的20个数据中，统计出数值为5的个数。原始数据为：3, 5, 4, 6, 3, 7, 5, 4, 5, 6, 4, 3, 5, 6, 7, 4, 5, 6, 5, 4。逐个数出5出现的次数：第2个是5，第7个是5，第9个是5，第13个是5，第17个是5，第19个是5，共出现6次。因此，阅读时间为5小时的学生有6人，正确答案是C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 18:45:20","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"4人","is_correct":0},{"id":"B","content":"5人","is_correct":0},{"id":"C","content":"6人","is_correct":1},{"id":"D","content":"7人","is_correct":0}]},{"id":813,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在整理班级同学最喜爱的运动项目调查数据时，将收集到的原始数据按类别列出后，下一步应该进行的步骤是____。","answer":"分类整理（或整理成频数分布表）","explanation":"在数据的收集、整理与描述这一知识点中，数据处理的流程通常为：收集数据 → 整理数据 → 描述数据 → 分析数据。当原始数据已经收集完毕后，下一步是将数据进行分类、排序或制成频数分布表，以便更清晰地观察数据的分布情况。因此，空白处应填写“分类整理”或“整理成频数分布表”。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 00:28:26","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2228,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"某学生在记录一周内每天气温变化时，发现某天的气温比前一天上升了3℃，记作+3℃；而另一天的气温比前一天下降了2℃，应记作____℃。","answer":"-2","explanation":"根据正数和负数表示相反意义的量的规则，气温上升用正数表示，气温下降则用负数表示。下降2℃应记作-2℃，符合七年级正负数在实际生活中的应用知识点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:27:19","updated_at":"2026-01-09 14:27:19","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1680,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市计划在一条笔直的主干道旁建设一个矩形公园，公园的一边紧贴道路（无需围栏），其余三边需用围栏围起。已知可用于围栏的总长度为60米。为了便于管理，公园被划分为两个区域：一个正方形活动区和一个矩形绿化区，两者共用一条与道路垂直的隔栏。设正方形活动区的边长为x米，矩形绿化区的长为y米（与道路平行），宽与正方形相同。若要求整个公园的总面积最大，求此时正方形活动区的边长x和绿化区的长y各为多少米？并求出最大面积。","answer":"解：\n根据题意，公园紧贴道路的一边不需要围栏，其余三边加上中间的一条隔栏共需围栏。\n围栏总长度 = 正方形的一边（与道路垂直）+ 绿化区的一边（与道路垂直）+ 底边总长（与道路平行）+ 中间隔栏（与道路垂直）\n即：围栏长度 = x + y方向上的两条垂直边 + 底边总长 + 中间隔栏\n但注意：正方形和绿化区共用一条与道路垂直的隔栏，且它们的宽都是x（因为正方形边长为x，绿化区宽也为x）。\n因此，围栏包括：\n- 左侧垂直边：x 米\n- 右侧垂直边：x 米\n- 底边总长：x + y 米（正方形底边x，绿化区底边y）\n- 中间隔栏：x 米（将正方形与绿化区分开，垂直于道路）\n所以总围栏长度为：x + x + (x + y) + x = 4x + y\n已知总围栏长度为60米，因此有：\n4x + y = 60 → y = 60 - 4x （1）\n\n整个公园的总面积 S = 正方形面积 + 绿化区面积 = x² + x·y\n将（1）代入：\nS = x² + x(60 - 4x) = x² + 60x - 4x² = -3x² + 60x\n这是一个关于x的二次函数：S(x) = -3x² + 60x\n\n求最大值：二次函数开口向下，最大值在顶点处取得。\n顶点横坐标 x = -b\/(2a) = -60 \/ (2×(-3)) = 10\n代入（1）得：y = 60 - 4×10 = 20\n此时最大面积 S = -3×(10)² + 60×10 = -300 + 600 = 300（平方米）\n\n答：当正方形活动区的边长x为10米，绿化区的长y为20米时，公园总面积最大，最大面积为300平方米。","explanation":"本题综合考查了一元一次方程、整式的加减、二次函数的最值问题（通过配方法或顶点公式）以及实际问题的建模能力。解题关键在于正确分析围栏的组成，建立总长度方程，进而表示出总面积，并将其转化为二次函数求最大值。虽然七年级尚未系统学习二次函数，但可通过列举法或顶点公式初步理解最值问题，此处使用顶点公式是基于拓展思维的要求。题目情境新颖，结合了平面几何与代数建模，符合困难难度要求，且知识点覆盖整式、方程与函数初步思想。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:31:23","updated_at":"2026-01-06 13:31:23","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":2478,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"一个圆形花坛的半径为5米，现要在花坛周围铺设一条宽度相同的环形小路，使得整个区域（花坛加小路）的外圆周长为18π米。求这条小路的宽度。","answer":"D","explanation":"设小路的宽度为x米，则整个区域的外圆半径为(5 + x)米。根据圆的周长公式C = 2πr，可得外圆周长为2π(5 + x)。题目中给出外圆周长为18π米，因此列出方程：2π(5 + x) = 18π。两边同时除以π，得2(5 + x) = 18，即10 + 2x = 18，解得2x = 8，x = 4。因此，小路的宽度为4米，正确答案为D。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:08:22","updated_at":"2026-01-10 15:08:22","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"1米","is_correct":0},{"id":"B","content":"2米","is_correct":0},{"id":"C","content":"3米","is_correct":0},{"id":"D","content":"4米","is_correct":1}]},{"id":1954,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某校七年级组织学生参与校园绿化活动，计划在一块长方形空地上种植花草。已知这块空地的周长是60米，且长比宽的2倍少3米。若设这块空地的宽为x米，则根据题意可列方程为：","answer":"A","explanation":"根据题意，设宽为x米，则长为(2x - 3)米。长方形的周长公式为：周长 = 2 × (长 + 宽)。将长和宽代入公式得：2 × (x + (2x - 3)) = 60，即2(x + 2x - 3) = 60。因此选项A正确。选项B错误，因为长是‘比宽的2倍少3米’，应为减3而非加3；选项C和D未正确应用周长公式，漏乘2或结构错误。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:46:41","updated_at":"2026-01-07 14:46:41","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[{"id":"A","content":"2(x + 2x - 3) = 60","is_correct":1},{"id":"B","content":"2(x + 2x + 3) = 60","is_correct":0},{"id":"C","content":"x + (2x - 3) = 60","is_correct":0},{"id":"D","content":"2x + (2x - 3) = 60","is_correct":0}]},{"id":1309,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级学生在学习平面直角坐标系后，开展了一次校园植物分布调查活动。调查小组在校园内选取了A、B、C三个区域，分别记录其中某种植物的数量，并将每个区域的中心位置用平面直角坐标系中的点表示：A(2, 3)、B(5, 7)、C(8, 4)。已知这三个区域中该植物的总数量为60株，且A区域的植物数量是B区域的2倍少5株，C区域的植物数量比A区域多10株。现计划在校园内新建一个圆形花坛，其圆心位于三角形ABC的重心位置，且花坛半径等于点A到点B的距离的一半（结果保留根号）。求：(1) 每个区域各有多少株植物？(2) 新建花坛的圆心坐标和半径长度。","answer":"(1) 设B区域的植物数量为x株，则A区域的数量为(2x - 5)株，C区域的数量为(2x - 5 + 10) = (2x + 5)株。\n根据题意，总数量为60株，列方程：\nx + (2x - 5) + (2x + 5) = 60\n化简得：x + 2x - 5 + 2x + 5 = 60 → 5x = 60 → x = 12\n因此：\nB区域：12株\nA区域：2×12 - 5 = 19株\nC区域：2×12 + 5 = 29株\n验证：12 + 19 + 29 = 60，符合题意。\n\n(2) 先求三角形ABC的重心坐标。\n重心坐标公式为：((x₁ + x₂ + x₃)\/3, (y₁ + y₂ + y₃)\/3)\nA(2,3), B(5,7), C(8,4)\n横坐标：(2 + 5 + 8)\/3 = 15\/3 = 5\n纵坐标：(3 + 7 + 4)\/3 = 14\/3\n所以圆心坐标为(5, 14\/3)\n\n再求AB的距离：\nAB = √[(5 - 2)² + (7 - 3)²] = √[3² + 4²] = √[9 + 16] = √25 = 5\n半径为AB的一半：5 ÷ 2 = 5\/2\n\n答：(1) A区域19株，B区域12株，C区域29株；(2) 花坛圆心坐标为(5, 14\/3)，半径为5\/2。","explanation":"本题综合考查了二元一次方程组（通过设未知数列一元一次方程解决）、平面直角坐标系中点的坐标运算、两点间距离公式以及三角形重心的计算方法。第一问通过设B区域数量为x，用代数式表示其他区域数量，建立一元一次方程求解；第二问先利用重心坐标公式计算圆心位置，再利用勾股定理计算AB距离并取其一半作为半径。题目融合了数据统计背景与几何坐标计算，强调数学在实际问题中的应用，难度较高，需要学生具备较强的代数运算能力和空间观念。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:50:43","updated_at":"2026-01-06 10:50:43","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":990,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生测量了教室中5个矩形窗户的长和宽，并将数据整理成如下表格。已知每个窗户的周长计算公式为：周长 = 2 × (长 + 宽)。若其中一个窗户的长为1.2米，宽为0.8米，则这个窗户的周长是___米。","answer":"4","explanation":"根据题目给出的周长公式：周长 = 2 × (长 + 宽)，将长1.2米和宽0.8米代入计算：2 × (1.2 + 0.8) = 2 × 2.0 = 4（米）。因此，该窗户的周长是4米。本题考查的是有理数的加法与乘法运算在实际问题中的应用，属于几何图形初步中的矩形周长计算，符合七年级数学知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:40:10","updated_at":"2025-12-30 11:11:27","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]},{"id":1249,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究平面直角坐标系中的几何问题时，发现一个有趣的规律：若将一个点P(x, y)先向右平移3个单位，再向上平移2个单位，得到点P'；然后将点P'绕原点逆时针旋转90°，得到点P''。已知点P''的坐标为(-5, 4)，求原点P的坐标(x, y)。此外，若该点P满足不等式组：2x - y > 1 且 x + 3y ≤ 10，请验证所求得的点P是否满足该不等式组。","answer":"解：\n\n第一步：设原点P的坐标为(x, y)。\n\n根据题意，点P先向右平移3个单位，再向上平移2个单位，得到点P'。\n平移变换规则：向右平移a个单位，横坐标加a；向上平移b个单位，纵坐标加b。\n因此，P'的坐标为：\n P' = (x + 3, y + 2)\n\n第二步：将点P'绕原点逆时针旋转90°，得到点P''。\n旋转90°逆时针的坐标变换公式为：\n 若点A(a, b)绕原点逆时针旋转90°，则新坐标为(-b, a)\n\n对P'(x + 3, y + 2)应用该公式：\nP'' = (-(y + 2), x + 3) = (-y - 2, x + 3)\n\n题目已知P''的坐标为(-5, 4)，因此列出方程组：\n -y - 2 = -5\n x + 3 = 4\n\n解第一个方程：\n -y - 2 = -5\n → -y = -3\n → y = 3\n\n解第二个方程：\n x + 3 = 4\n → x = 1\n\n所以，原点P的坐标为(1, 3)。\n\n第三步：验证点P(1, 3)是否满足不等式组：\n 2x - y > 1\n x + 3y ≤ 10\n\n代入x = 1，y = 3：\n\n第一式：2(1) - 3 = 2 - 3 = -1\n -1 > 1？ 不成立。\n\n第二式：1 + 3×3 = 1 + 9 = 10\n 10 ≤ 10？ 成立。\n\n由于第一式不满足，因此点P(1, 3)不满足整个不等式组。\n\n最终答案：\n点P的坐标为(1, 3)，但该点不满足给定的不等式组。","explanation":"本题综合考查了平面直角坐标系中的平移变换、旋转变换、二元一次方程组的建立与求解，以及不等式组的验证。解题关键在于掌握坐标变换的代数表示：平移是坐标的加减，旋转90°逆时针的公式为(a, b) → (-b, a)。通过逆向推理，从P''的坐标反推出P'，再反推出P。最后将所得坐标代入不等式组进行验证，体现了数形结合与逻辑推理能力。题目设计新颖，融合了多个知识点，要求学生具备较强的综合运用能力，符合困难难度要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:31:09","updated_at":"2026-01-06 10:31:09","sort_order":0,"source":null,"tags":null,"analysis":null,"knowledge_point":null,"difficulty_coefficient":null,"suggested_time":null,"accuracy_rate":null,"usage_count":0,"last_used":null,"view_count":0,"favorite_count":0,"options":[]}]