习题练习

[{"id":2545,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某圆形花坛的半径为6米，现计划在花坛中心安装一个旋转喷头，其喷洒范围为一个圆心角为120°的扇形区域。若喷头随机旋转，且每次喷洒的起始角度在0°到360°之间均匀分布，则某学生站在距离花坛中心4米的位置时，被水喷洒到的概率是多少？","answer":"A","explanation":"该问题考查概率初步与圆的结合应用。喷头喷洒范围为120°的扇形，而整个圆周为360°。由于喷头起始角度在0°到360°之间均匀随机分布，因此喷洒区域覆盖某一固定方向（如某学生所在位置）的概率等于扇形圆心角占整个圆周的比例。学生位于花坛内部（距离中心4米 < 半径6米），始终处于喷洒半径范围内，因此是否被喷洒仅取决于角度是否落在120°的扇形区域内。故概率为120° \/ 360° = 1\/3。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 17:00:10","updated_at":"2026-01-10 17:00:10","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\/3","is_correct":1},{"id":"B","content":"1\/4","is_correct":0},{"id":"C","content":"1\/6","is_correct":0},{"id":"D","content":"1\/2","is_correct":0}]},{"id":422,"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时间数据时，发现一周内每天阅读时间（单位：分钟）分别为：25，30，28，35，32，27，33。为了分析阅读时间的分布情况，该学生计算了这组数据的平均数。请问这组数据的平均数是多少？","answer":"C","explanation":"要计算这组数据的平均数，需要将所有数据相加，然后除以数据的个数。具体步骤如下：首先，将每天的阅读时间相加：25 + 30 + 28 + 35 + 32 + 27 + 33 = 210（分钟）。然后，用总和除以天数（7天）：210 ÷ 7 = 30（分钟）。因此，这组数据的平均数是30分钟，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 17:32:46","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":"28分钟","is_correct":0},{"id":"B","content":"29分钟","is_correct":0},{"id":"C","content":"30分钟","is_correct":1},{"id":"D","content":"31分钟","is_correct":0}]},{"id":1719,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市为了优化公交线路，对一条主干道的车流量进行了为期7天的观测，记录每天上午8:00至9:00的车辆通过数量（单位：辆），数据如下：120，135，128，142，130，138，145。交通部门计划根据这些数据调整红绿灯时长，并设定一个‘高峰阈值’——若某时段车流量超过该阈值，则启动延长绿灯时间的方案。已知该阈值为这7天数据的平均数向上取整后的值。同时，为评估调整效果，工作人员在实施新方案后又连续观测了5天，得到新的车流量数据：148，152，146，150，154。现要求：\n\n（1）计算原始7天数据的平均数，并确定‘高峰阈值’；\n（2）将原始7天数据与新观测的5天数据合并，求这12天车流量的中位数；\n（3）若规定‘车流量超过高峰阈值的天数占比超过50%’，则认为交通压力显著增大。请判断实施新方案后是否出现这一情况，并说明理由；\n（4）假设每辆车平均占用道路长度为6米，道路有效通行长度为800米，利用不等式估算在高峰阈值下，道路上的车辆是否会发生拥堵（即车辆总长度是否超过道路有效长度），并给出结论。","answer":"（1）原始7天数据之和为：120 + 135 + 128 + 142 + 130 + 138 + 145 = 938。\n平均数为：938 ÷ 7 = 134。\n向上取整后，高峰阈值为135。\n\n（2）合并12天数据并按从小到大排序：\n120，128，130，135，138，142，145，146，148，150，152，154。\n共有12个数据，中位数为第6和第7个数据的平均数：(142 + 145) ÷ 2 = 143.5。\n\n（3）高峰阈值为135。在原始7天中，超过135的数据有：138，142，145（共3天），占比3\/7 ≈ 42.9%，未超过50%。\n在新观测的5天中，所有数据均大于135（148，152，146，150，154），即5天全部超过阈值，占比5\/5 = 100%。\n但题目要求判断的是‘实施新方案后’是否出现‘车流量超过高峰阈值的天数占比超过50%’，应仅针对新观测的5天数据判断。\n由于5天中有5天超过阈值，占比100% > 50%，因此交通压力显著增大。\n\n（4）高峰阈值为135辆，即每小时最多135辆车通过。\n每辆车平均占用6米，则135辆车总长度为：135 × 6 = 810（米）。\n道路有效通行长度为800米。\n因为810 > 800，所以车辆总长度超过道路有效长度，会发生拥堵。\n结论：在高峰阈值下，道路会发生拥堵。","explanation":"本题综合考查了数据的收集、整理与描述中的平均数、中位数、百分比比较，以及有理数运算、不等式在实际问题中的应用。第（1）问考察平均数计算和取整规则；第（2）问要求对12个数据排序并求中位数，注意偶数个数据时取中间两数平均值；第（3）问强调对‘实施新方案后’这一时间范围的准确理解，避免误将全部12天数据纳入判断，体现数据分析的严谨性；第（4）问将实际问题转化为不等式模型，通过比较总长度与道路容量判断是否拥堵，体现数学建模能力。题目情境真实，逻辑层层递进，难度较高，符合困难等级要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:12:28","updated_at":"2026-01-06 14:12:28","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":295,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某学生调查了班级同学最喜欢的运动项目，收集数据后绘制成条形统计图。图中显示喜欢篮球的有12人，喜欢足球的有8人，喜欢乒乓球的有6人，喜欢跳绳的有4人。请问喜欢球类运动（包括篮球、足球和乒乓球）的学生共有多少人？","answer":"C","explanation":"题目要求计算喜欢球类运动的学生总人数，球类运动包括篮球、足球和乒乓球。根据题意，喜欢篮球的有12人，喜欢足球的有8人，喜欢乒乓球的有6人。将这些人数相加：12 + 8 + 6 = 26（人）。因此，喜欢球类运动的学生共有26人，正确答案是C。本题考查数据的收集与整理，重点在于理解分类并正确进行加法运算，符合七年级‘数据的收集、整理与描述’知识点要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:33:09","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":"20人","is_correct":0},{"id":"B","content":"24人","is_correct":0},{"id":"C","content":"26人","is_correct":1},{"id":"D","content":"30人","is_correct":0}]},{"id":2441,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生测量了一块直角三角形草地的两条直角边，分别为√12米和√27米。他计划在斜边上每隔1米种一棵树，包括两个端点。若每棵树占地忽略不计，则最多可以种多少棵树？","answer":"B","explanation":"首先，利用勾股定理计算斜边长度。已知两条直角边分别为√12米和√27米。将根式化简：√12 = 2√3，√27 = 3√3。根据勾股定理，斜边c满足：c² = (2√3)² + (3√3)² = 4×3 + 9×3 = 12 + 27 = 39，因此c = √39米。接下来，计算在长度为√39米的线段上，每隔1米种一棵树（包括两个端点）最多可种多少棵。由于√36 = 6，√49 = 7，所以6 < √39 < 7，即斜边长度约为6.24米。从起点开始，每隔1米种一棵树，位置为0米、1米、2米、…、6米，共7个点（因为6 ≤ √39 < 7，第7棵树在6米处仍在线段上）。因此最多可种7棵树。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:26:53","updated_at":"2026-01-10 13:26:53","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":"6棵","is_correct":0},{"id":"B","content":"7棵","is_correct":1},{"id":"C","content":"8棵","is_correct":0},{"id":"D","content":"9棵","is_correct":0}]},{"id":2517,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"如图，一个圆锥形帐篷的底面半径为3米，母线长为5米。一名学生站在帐篷正前方2米处，视线恰好与帐篷顶部相切。若该学生眼睛离地面高度为1.6米，则帐篷的高为多少米？","answer":"A","explanation":"本题综合考查圆、相似三角形和勾股定理的应用。圆锥底面半径r=3米，母线l=5米，设圆锥高为h。由勾股定理得：h² + 3² = 5²，解得h = √(25 - 9) = √16 = 4米。题目中给出的观察者位置和视线相切的信息用于验证合理性：从眼睛到帐篷顶的视线与圆锥侧面相切，形成直角三角形，利用相似三角形可验证高为4米时，视线斜率与圆锥母线斜率一致，符合几何关系。因此帐篷高为4米。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:47:48","updated_at":"2026-01-10 15:47:48","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":"√7","is_correct":0},{"id":"C","content":"2√5","is_correct":0},{"id":"D","content":"3.2","is_correct":0}]},{"id":1757,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某校七年级组织学生参加数学实践活动，要求将学生分成若干小组，每组人数相同。已知若每组安排6人，则最后一组只有4人；若每组安排8人，则最后一组只有6人；若每组安排9人，则最后一组只有7人。问：该校七年级参加活动的学生至少有多少人？请通过建立方程或不等式模型，并结合整除性质进行分析求解。","answer":"设参加活动的学生总人数为x人。\n\n根据题意，可列出以下同余关系：\n\nx ≡ 4 (mod 6) ——（1）\n\nx ≡ 6 (mod 8) ——（2）\n\nx ≡ 7 (mod 9) ——（3）\n\n观察发现，每个余数都比除数少2：\n\n即：x + 2 ≡ 0 (mod 6)\n\nx + 2 ≡ 0 (mod 8)\n\nx + 2 ≡ 0 (mod 9)\n\n说明 x + 2 是 6、8、9 的公倍数。\n\n先求6、8、9的最小公倍数：\n\n分解质因数：\n\n6 = 2 × 3\n\n8 = 2³\n\n9 = 3²\n\n取各质因数最高次幂：2³ × 3² = 8 × 9 = 72\n\n所以 x + 2 是72的倍数，即 x + 2 = 72k（k为正整数）\n\n因此 x = 72k - 2\n\n当k = 1时，x = 72 - 2 = 70\n\n验证：\n\n70 ÷ 6 = 11组余4人 → 符合（1）\n\n70 ÷ 8 = 8组余6人 → 符合（2）\n\n70 ÷ 9 = 7组余7人 → 符合（3）\n\n当k = 2时，x = 144 - 2 = 142，也满足，但题目要求“至少”有多少人。\n\n所以最小满足条件的x为70。\n\n答：该校七年级参加活动的学生至少有70人。","explanation":"本题考查学生对同余概念的理解与转化能力，结合整除性质和一元一次方程建模思想。关键在于发现三个条件中余数与除数的关系：余数均为除数减2，从而转化为x + 2是6、8、9的公倍数。通过求最小公倍数得到最小解。题目融合了整数的整除性、最小公倍数、方程建模与逻辑推理，属于典型的困难级别应用题，要求学生具备较强的观察力与抽象思维能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:33:35","updated_at":"2026-01-06 14:33:35","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":2523,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生用一根长为20 cm的铁丝围成一个扇形，扇形的半径为r cm，圆心角为θ（0 < θ ≤ 2π）。若扇形的面积S（cm²）与半径r（cm）满足关系式 S = 10r - r²，则该扇形的最大面积为多少？","answer":"B","explanation":"题目给出扇形面积与半径的关系式：S = 10r - r²。这是一个关于r的一元二次函数，形式为S = -r² + 10r，其图像为开口向下的抛物线，最大值出现在顶点处。顶点横坐标为 r = -b\/(2a) = -10\/(2×(-1)) = 5。将r = 5代入函数得 S = 10×5 - 5² = 50 - 25 = 25。因此，扇形的最大面积为25 cm²。该题综合考查了二次函数的最大值问题和扇形的几何背景，但核心是二次函数求最值，属于九年级学生应掌握的基础内容。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:59:28","updated_at":"2026-01-10 15:59:28","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":"20","is_correct":0},{"id":"B","content":"25","is_correct":1},{"id":"C","content":"30","is_correct":0},{"id":"D","content":"35","is_correct":0}]},{"id":2500,"subject":"数学","grade":"九年级","stage":"初中","type":"选择题","content":"某学生用三根木棒搭建一个直角三角形支架，其中两根木棒的长度分别为3cm和4cm。若他将这个三角形绕长度为4cm的木棒所在直线旋转一周，所形成的几何体的俯视图是以下哪种图形？","answer":"A","explanation":"根据勾股定理，第三边长度为√(4² - 3²) = √7 cm 或 √(3² + 4²) = 5 cm。由于题目说明是直角三角形且已知两边为3cm和4cm，可判断第三边为5cm（斜边）或√7 cm（当4cm为斜边时）。但无论哪种情况，绕长度为4cm的直角边旋转时，另一条直角边（3cm）将作为旋转半径，形成一个圆锥体。圆锥的俯视图是从上往下看，呈现为一个完整的圆。因此正确答案是A。本题考查旋转形成的几何体及其视图，属于投影与视图和旋转知识点的综合应用，难度适中，符合九年级学生认知水平。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-10 15:20:16","updated_at":"2026-01-10 15:20:16","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":"一个圆","is_correct":1},{"id":"B","content":"一个矩形","is_correct":0},{"id":"C","content":"一个三角形","is_correct":0},{"id":"D","content":"一个扇形","is_correct":0}]},{"id":2364,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一个几何问题时，发现一个四边形ABCD满足以下条件：① 对角线AC与BD互相垂直且平分；② ∠ABC = ∠ADC = 90°；③ AB = AD。该学生由此推断四边形ABCD一定是正方形。以下选项中，最能支持这一结论的是：","answer":"C","explanation":"解析：首先，对角线AC与BD互相垂直且平分，根据平行四边形的判定定理，可知四边形ABCD是菱形（对角线互相垂直平分的平行四边形是菱形）。其次，已知∠ABC = 90°，而菱形中若有一个角是直角，则其余角也为直角，因此该菱形实际上是矩形。既是菱形又是矩形的四边形是正方形。选项C准确指出了这一逻辑链条，即从条件推出四边形同时具备菱形和矩形的特征，从而得出正方形结论，是最完整且严谨的支持。选项A忽略了‘平分’这一关键条件对平行四边形判定的作用；选项B的三角形全等虽成立，但不足以直接推出所有角为直角；选项D错误地认为仅凭对角线垂直平分加一组邻边相等就能判定正方形，忽略了角度条件的重要性。因此，正确答案为C。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:14:48","updated_at":"2026-01-10 11:14:48","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":"因为对角线互相垂直平分的四边形是菱形，且有一个角为90°，所以是正方形","is_correct":0},{"id":"B","content":"因为AB = AD且∠ABC = ∠ADC = 90°，所以△ABC ≌ △ADC，从而所有边相等且角为直角","is_correct":0},{"id":"C","content":"由条件可推出四边形ABCD既是菱形又是矩形，因此是正方形","is_correct":1},{"id":"D","content":"对角线互相垂直且平分，说明是平行四边形，再加上一组邻边相等，即可判定为正方形","is_correct":0}]}]