习题练习

[{"id":301,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"在一次环保活动中，某班级收集废旧纸张的重量记录如下：第一周收集了12.5千克，第二周比第一周多收集了3.7千克，第三周比第二周少收集了1.8千克。请问这三周平均每周收集多少千克废旧纸张？","answer":"B","explanation":"首先计算第二周收集的纸张重量：12.5 + 3.7 = 16.2（千克）。然后计算第三周的重量：16.2 - 1.8 = 14.4（千克）。三周总重量为：12.5 + 16.2 + 14.4 = 43.1（千克）。平均每周收集量为：43.1 ÷ 3 = 14.1（千克）。因此正确答案是B。本题考查有理数的加减乘除混合运算及平均数的计算，属于数据的收集、整理与描述知识点，符合七年级数学课程要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:34:09","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":"13.2千克","is_correct":0},{"id":"B","content":"14.1千克","is_correct":1},{"id":"C","content":"12.9千克","is_correct":0},{"id":"D","content":"15.0千克","is_correct":0}]},{"id":1064,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在一次环保活动中，某学生记录了连续5天每天回收的废纸重量（单位：千克）分别为：2.5、3、2.8、3.2、2.7。为了估算一个月（按30天计算）大约能回收多少千克废纸，他先计算了这5天的平均每天回收量，再用这个平均数乘以30。请问他计算出的月回收量估计值是___千克。","answer":"86.4","explanation":"首先计算5天回收废纸的总重量：2.5 + 3 + 2.8 + 3.2 + 2.7 = 14.2（千克）。然后求平均每天回收量：14.2 ÷ 5 = 2.84（千克\/天）。最后估算一个月（30天）的回收量：2.84 × 30 = 86.4（千克）。本题考查数据的收集、整理与描述中的平均数计算及其应用，属于简单难度，符合七年级数学课程内容。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-06 08:52:13","updated_at":"2026-01-06 08:52:13","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":1045,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级图书角统计中，某学生整理了上周同学们借阅的图书数量：语文类12本，数学类8本，英语类10本，科学类6本。如果将这些数据用扇形统计图表示，那么表示数学类图书的扇形圆心角的度数是___度。","answer":"80","explanation":"首先计算图书总数：12 + 8 + 10 + 6 = 36（本）。数学类图书占总数的比例为 8 ÷ 36 = 2\/9。扇形统计图中整个圆为360度，因此数学类对应的圆心角为 360 × (2\/9) = 80（度）。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 06:23:36","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":2448,"subject":"数学","grade":"八年级","stage":"初中","type":"填空题","content":"在平面直角坐标系中，点A(2, 3)关于直线y = x的对称点为点B，则点B的坐标为____。","answer":"(3, 2)","explanation":"点关于直线y = x对称时，横纵坐标互换。点A(2, 3)对称后坐标为(3, 2)。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 13:54:13","updated_at":"2026-01-10 13:54:13","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":1473,"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7:00至9:00的车辆通过数量（单位：百辆），数据如下：12, 15, 18, 14, 16, 20, 17。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’，若某天的车流量超过该阈值，则启动延长绿灯时间的应急方案。已知该阈值设定为这组数据的中位数与平均数的较大者。同时，为评估调整效果，工程师在平面直角坐标系中绘制了车流量与绿灯延长时间的函数关系图，其中绿灯延长时间 y（单位：秒）与车流量 x（单位：百辆）满足一次函数关系，且当 x = 15 时 y = 10，当 x = 20 时 y = 20。若某天观测到车流量为 19 百辆，且该天启动了应急方案，求该天绿灯延长时间的理论值，并判断该天车流量是否确实超过了设定的高峰阈值。","answer":"第一步：计算7天车流量的平均数。\n数据：12, 15, 18, 14, 16, 20, 17\n总和 = 12 + 15 + 18 + 14 + 16 + 20 + 17 = 112\n平均数 = 112 ÷ 7 = 16（百辆）\n\n第二步：求中位数。\n将数据从小到大排列：12, 14, 15, 16, 17, 18, 20\n共7个数据，中位数为第4个数，即16（百辆）\n\n第三步：确定高峰阈值。\n阈值为中位数与平均数的较大者：max(16, 16) = 16（百辆）\n\n第四步：建立绿灯延长时间 y 与车流量 x 的一次函数关系。\n设函数为 y = kx + b\n已知当 x = 15 时 y = 10，当 x = 20 时 y = 20\n代入得方程组：\n10 = 15k + b ...(1)\n20 = 20k + b ...(2)\n(2) - (1) 得：10 = 5k ⇒ k = 2\n将 k = 2 代入 (1)：10 = 15×2 + b ⇒ 10 = 30 + b ⇒ b = -20\n所以函数为：y = 2x - 20\n\n第五步：当 x = 19 时，求 y 值。\ny = 2×19 - 20 = 38 - 20 = 18（秒）\n\n第六步：判断是否超过高峰阈值。\n车流量为19百辆，阈值为16百辆，19 > 16，因此确实超过了阈值，启动应急方案合理。\n\n最终答案：该天绿灯延长时间的理论值为18秒，且车流量确实超过了高峰阈值。","explanation":"本题综合考查了数据的收集、整理与描述（平均数、中位数）、实数运算、一次函数（二元一次方程组应用）以及不等式比较。解题关键在于：首先通过统计方法确定‘高峰阈值’，这需要准确计算平均数和中位数并比较大小；其次利用两个已知点建立一次函数模型，通过解二元一次方程组求出函数表达式；最后代入具体数值求解并做出逻辑判断。题目情境真实，融合了统计与函数知识，要求学生具备较强的综合分析与计算能力，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 11:52:51","updated_at":"2026-01-06 11:52:51","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":2192,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","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5℃，记作+5℃。如果第二天的气温又比当天下降了8℃，那么第二天的气温变化应记作多少？","answer":"B","explanation":"气温下降应使用负数表示。题目中明确指出‘下降了8℃’，因此变化量应记为-8℃。选项B正确。其他选项中，A表示上升，C和D是数值计算错误或符号错误，不符合题意。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 14:25:31","updated_at":"2026-01-09 14:25:31","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":"+8℃","is_correct":0},{"id":"B","content":"-8℃","is_correct":1},{"id":"C","content":"+3℃","is_correct":0},{"id":"D","content":"-3℃","is_correct":0}]},{"id":2358,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究轴对称图形时，发现一个等腰三角形ABC，其中AB = AC，且∠BAC = 120°。他将该三角形沿底边BC上的高AD折叠，使点A落在点A'处，且A'恰好落在BC的延长线上。已知BD = 3，则折痕AD的长度为多少？","answer":"C","explanation":"本题综合考查轴对称、等腰三角形性质和勾股定理。由于△ABC是等腰三角形，AB = AC，且∠BAC = 120°，则底角∠ABC = ∠ACB = (180° - 120°) \/ 2 = 30°。AD是底边BC上的高，因此AD ⊥ BC，且D为BC中点（等腰三角形三线合一），故BD = DC = 3，BC = 6。在Rt△ABD中，∠ABD = 30°，BD = 3。根据30°-60°-90°直角三角形的边长比例关系（1 : √3 : 2），对边BD（30°所对）为3，则高AD（60°所对）为3√3，斜边AB为6。折叠后点A落在A'，且A'在BC延长线上，说明折痕AD是AA'的垂直平分线，但这不影响AD本身的长度计算。因此AD = 3√3。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:10:08","updated_at":"2026-01-10 11:10:08","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":"√3","is_correct":0},{"id":"B","content":"2√3","is_correct":0},{"id":"C","content":"3√3","is_correct":1},{"id":"D","content":"6","is_correct":0}]},{"id":1333,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市地铁系统计划在两条平行轨道之间修建一条新的联络线，用于列车调度。已知两条平行轨道分别位于平面直角坐标系中的直线 y = 2 和 y = 6 上。联络线需从点 A(1, 2) 出发，与第一条轨道垂直相交，然后以 45° 角斜向延伸至第二条轨道上的某点 B。同时，为满足安全规范，联络线在斜向延伸段的长度不得超过 4√2 千米。现需确定点 B 的坐标，并验证该设计是否符合长度限制。若不符合，请重新设计一条从 A 点出发、与第一条轨道垂直、且斜向段长度恰好为 4√2 千米的联络线路径，求出此时点 B 的准确坐标。","answer":"第一步：分析题意\n联络线从点 A(1, 2) 出发，首先与第一条轨道 y = 2 垂直。由于 y = 2 是水平线，其垂线为竖直线，因此联络线的第一段为从 A(1, 2) 垂直向上延伸的线段。\n\n第二步：确定斜向延伸方向\n题目要求斜向延伸段与水平方向成 45° 角。由于联络线从 y = 2 向上延伸，斜向段应向右上方或左上方 45° 延伸。考虑到实际调度需求，通常向右延伸更合理，因此假设斜向段沿 45° 方向（即斜率为 1）延伸。\n\n第三步：设点 B 的坐标为 (x, 6)，因为 B 在第二条轨道 y = 6 上。\n斜向段起点为 A 正上方的某点，但由于第一段是垂直的，且 A 已在 y = 2 上，因此斜向段直接从 A(1, 2) 开始斜向延伸。\n\n斜向段从 A(1, 2) 沿 45° 方向延伸，其方向向量为 (1, 1)，因此参数方程为：\nx = 1 + t\ny = 2 + t\n当 y = 6 时，2 + t = 6 ⇒ t = 4\n代入得 x = 1 + 4 = 5\n所以点 B 坐标为 (5, 6)\n\n第四步：计算斜向段长度\n距离 AB = √[(5 - 1)² + (6 - 2)²] = √[16 + 16] = √32 = 4√2（千米）\n\n第五步：验证长度限制\n题目要求斜向段长度不得超过 4√2 千米，而实际长度恰好为 4√2 千米，符合要求。\n\n第六步：结论\n因此，点 B 的坐标为 (5, 6)，设计符合安全规范。\n\n答案：点 B 的坐标为 (5, 6)，联络线斜向段长度为 4√2 千米，符合长度限制。","explanation":"本题综合考查平面直角坐标系、几何图形初步、实数运算及不等式思想。解题关键在于理解‘与轨道垂直’意味着竖直方向，45° 角对应斜率为 1 的直线。利用参数法或坐标差计算点 B 的位置，再通过距离公式验证长度。题目设置了‘不得超过’的条件，引导学生进行验证，体现了不等式在实际问题中的应用。整个过程融合了坐标几何、勾股定理和实际情境建模，难度较高，适合学有余力的七年级学生挑战。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 10:58:21","updated_at":"2026-01-06 10:58:21","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":331,"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身高数据时，制作了如下频数分布表：\n身高区间（cm） | 频数\n150～155 | 4\n155～160 | 7\n160～165 | 10\n165～170 | 6\n170～175 | 3\n请问这组数据的中位数最可能落在哪个身高区间？","answer":"C","explanation":"首先计算总人数：4 + 7 + 10 + 6 + 3 = 30人。中位数是第15和第16个数据的平均值。累计频数：150～155有4人，155～160累计11人，160～165累计21人。第15和第16个数据都落在160～165区间内，因此中位数最可能位于该区间。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:39:24","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":"150～155","is_correct":0},{"id":"B","content":"155～160","is_correct":0},{"id":"C","content":"160～165","is_correct":1},{"id":"D","content":"165～170","is_correct":0}]},{"id":727,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在某次班级大扫除中，学生们被分成若干小组清理教室。如果每组安排5人，则多出3人；如果每组安排6人，则最后一组只有4人。这个班级共有___名学生。","answer":"28","explanation":"设班级共有x名学生。根据题意，当每组5人时，多出3人，说明x除以5余3，即x = 5a + 3（a为组数）。当每组6人时，最后一组只有4人，说明x除以6余4，即x = 6b + 4（b为组数）。寻找同时满足这两个条件的最小正整数。尝试代入：当x=28时，28 ÷ 5 = 5组余3，符合第一种情况；28 ÷ 6 = 4组余4，也符合第二种情况。因此，班级共有28名学生。本题考查一元一次方程的实际应用与整数解问题，属于简单难度。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:02:09","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":[]}]