[{"id":2131,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程 2(x - 3) = 4 时，第一步将方程两边同时除以2，得到 x - 3 = 2。接下来他应该进行的正确步骤是：","answer":"B","explanation":"方程 x - 3 = 2 中，为了求出 x，需要将 -3 消去。根据等式性质，应在等式两边同时加上3，得到 x = 5。这是七年级一元一次方程求解中的基本步骤，符合课程标准要求。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 12:56:39","updated_at":"2026-01-09 12:56:39","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":"两边同时减去3，得到 x = -1","is_correct":0},{"id":"B","content":"两边同时加上3，得到 x = 5","is_correct":1},{"id":"C","content":"两边同时乘以3，得到 x = 6","is_correct":0},{"id":"D","content":"两边同时除以3，得到 x = 2\/3","is_correct":0}]},{"id":1718,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某城市计划在一条主干道两侧安装新型节能路灯，道路全长1800米，起点和终点均需安装路灯。设计团队提出两种方案：方案A每隔30米安装一盏路灯；方案B每隔45米安装一盏路灯。为优化成本，最终决定采用混合方案：在道路的前半段（即前900米）采用方案A，后半段（后900米）采用方案B。已知每盏路灯的安装成本为200元，维护费用每年为每盏50元。现需计算：(1) 整条道路共需安装多少盏路灯？(2) 若该路灯系统预计使用10年，总成本（安装费 + 10年维护费）是多少元？(3) 若一名学生提出‘若全程采用方案A，总成本将比混合方案高出多少元？’请验证该说法是否正确，并说明理由。","answer":"(1) 前半段900米采用方案A，每隔30米安装一盏，起点安装，终点也安装。\n路灯数量 = (900 ÷ 30) + 1 = 30 + 1 = 31盏。\n后半段900米采用方案B，每隔45米安装一盏，起点安装，终点也安装。\n路灯数量 = (900 ÷ 45) + 1 = 20 + 1 = 21盏。\n但注意：整条道路的中间点（900米处）是前半段终点和后半段起点，为同一点，不能重复安装。\n因此，总路灯数 = 31 + 21 - 1 = 51盏。\n\n(2) 安装成本 = 51 × 200 = 10200元。\n每年维护费 = 51 × 50 = 2550元。\n10年维护费 = 2550 × 10 = 25500元。\n总成本 = 10200 + 25500 = 35700元。\n\n(3) 若全程采用方案A，每隔30米安装一盏，全长1800米，起点终点均安装。\n路灯数量 = (1800 ÷ 30) + 1 = 60 + 1 = 61盏。\n安装成本 = 61 × 200 = 12200元。\n每年维护费 = 61 × 50 = 3050元。\n10年维护费 = 3050 × 10 = 30500元。\n总成本 = 12200 + 30500 = 42700元。\n混合方案总成本为35700元。\n高出金额 = 42700 - 35700 = 7000元。\n因此，该学生的说法正确：全程采用方案A比混合方案高出7000元。","explanation":"本题综合考查了有理数运算、一元一次方程思想（等距分段）、数据的收集与整理（成本计算）以及实际应用建模能力。第(1)问需注意分段安装时中间点的重复问题，体现几何图形初步中的线段分割思想；第(2)问涉及整式加减与有理数乘法，计算总成本；第(3)问通过对比不同方案，强化不等式与方程的应用意识，同时训练学生逻辑推理与验证能力。题目情境新颖，结合城市规划背景，提升数学建模素养，符合七年级数学课程标准对综合应用能力的要求。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 14:11:59","updated_at":"2026-01-06 14:11:59","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":769,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次班级环保活动中，某学生收集了若干个塑料瓶。若每3个塑料瓶可以兑换1支铅笔，且该学生最终兑换了___支铅笔后，还剩下2个塑料瓶。已知他最初收集的塑料瓶总数不超过20个，且兑换过程没有浪费，则他最初至少收集了___个塑料瓶。","answer":"6；20","explanation":"设该学生兑换了x支铅笔，则他用于兑换的塑料瓶数量为3x个，加上剩下的2个，总瓶数为3x + 2。根据题意，总瓶数不超过20，即3x + 2 ≤ 20，解得x ≤ 6。要使最初收集的瓶数最少，应使x尽可能小，但题目问的是“至少收集了多少个”，结合“兑换了___支铅笔”这一空，需满足兑换后剩2个且总数不超过20。当x = 6时，总瓶数为3×6 + 2 = 20，符合“不超过20”且为最大可能值，但题目要求“至少收集”，需反向思考：若兑换6支铅笔，则必须至少有18个用于兑换，加上剩余2个，共20个，这是满足条件的最小总数（因为若总数少于20，则无法兑换6支）。因此，第一个空填6（兑换铅笔数），第二个空填20（最初至少收集的瓶数）。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:47:55","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":2257,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"在数轴上，点A表示的数是-3，点B表示的数是5。若点C是线段AB的中点，则点C表示的数是___。","answer":"A","explanation":"点C是线段AB的中点，其表示的数为点A和点B所表示数的平均数。计算过程为：(-3 + 5) ÷ 2 = 2 ÷ 2 = 1。因此，点C表示的数是1，对应选项A。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:03:06","updated_at":"2026-01-09 16:03:06","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","is_correct":1},{"id":"B","content":"2","is_correct":0},{"id":"C","content":"-1","is_correct":0},{"id":"D","content":"4","is_correct":0}]},{"id":203,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"一个长方形的长是8厘米，宽是5厘米，它的面积是_空白处_平方厘米。","answer":"40","explanation":"长方形的面积计算公式是：面积 = 长 × 宽。题目中给出的长是8厘米，宽是5厘米，因此面积为 8 × 5 = 40 平方厘米。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 14:39:26","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":920,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次环保知识竞赛中，某班级共收集到有效问卷120份，其中男生填写的问卷数量是女生的2倍。设女生填写的问卷数量为x份，则可列出一元一次方程：_ = 120，解得x = _。","answer":"x + 2x；40","explanation":"根据题意，女生填写的问卷数量为x份，男生填写的是女生的2倍，即为2x份。总问卷数为120份，因此可列出方程：x + 2x = 120，合并同类项得3x = 120，解得x = 40。所以女生填写了40份问卷。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 02:42:11","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":2146,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在解方程时，将方程 2x + 3 = 9 的解题步骤写为：第一步，两边同时减去3，得到 2x = 6；第二步，两边同时除以2，得到 x = 3。这名学生使用的解方程依据是___。","answer":"B","explanation":"该学生在解方程过程中，第一步使用了等式的基本性质：两边同时减去3，保持等式成立；第二步两边同时除以2（不为0），也符合等式的基本性质。因此正确依据是选项B所描述的内容。选项C和D虽然也是方程变形中的方法，但不是本题中直接体现的依据。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 13:00:46","updated_at":"2026-01-09 13:00:46","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":"等式两边同时加上同一个数，等式仍然成立","is_correct":0},{"id":"B","content":"等式两边同时减去同一个数，等式仍然成立，且等式两边同时除以同一个不为0的数，等式仍然成立","is_correct":1},{"id":"C","content":"移项时符号要改变","is_correct":0},{"id":"D","content":"合并同类项法则","is_correct":0}]},{"id":631,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保知识竞赛，共收集了120份有效问卷。在整理数据时，发现有15%的学生选择了‘垃圾分类’作为最关注的环保问题，有40人选择了‘节约用水’，其余学生选择了‘减少塑料使用’。请问选择‘减少塑料使用’的学生人数是多少？","answer":"C","explanation":"首先计算选择‘垃圾分类’的学生人数：120 × 15% = 120 × 0.15 = 18人。已知选择‘节约用水’的有40人。那么选择‘减少塑料使用’的人数为总人数减去前两项：120 - 18 - 40 = 62人。因此正确答案是C。本题考查数据的收集与整理，涉及百分数的基本计算和简单减法运算，符合七年级数学中‘数据的收集、整理与描述’知识点，难度为简单。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 21:55:43","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":"52","is_correct":0},{"id":"B","content":"58","is_correct":0},{"id":"C","content":"62","is_correct":1},{"id":"D","content":"68","is_correct":0}]},{"id":742,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"在一次环保活动中，某学生记录了5个家庭一周内节约用水的量（单位：升），分别为：12，8，15，10，_。已知这5个数据的平均数是11升，则第五个家庭节约的用水量是____升。","answer":"10","explanation":"根据平均数的定义，5个数据的总和等于平均数乘以数据的个数。已知平均数是11，共有5个数据，因此总和为 11 × 5 = 55 升。前四个数据分别为12、8、15、10，它们的和为 12 + 8 + 15 + 10 = 45 升。所以第五个数据为 55 - 45 = 10 升。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:14:58","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":594,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级进行了一次数学测验，老师将成绩整理成频数分布表。已知成绩在80分到89分之间的学生有12人，占总人数的30%。那么，参加这次测验的学生总人数是多少？","answer":"B","explanation":"题目中给出成绩在80分到89分之间的学生有12人，占总人数的30%。设总人数为x，则可列方程：30% × x = 12，即0.3x = 12。解这个一元一次方程，两边同时除以0.3，得到x = 12 ÷ 0.3 = 40。因此，参加测验的学生总人数是40人。本题考查了数据的收集与整理中的百分比计算以及一元一次方程的应用，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 20:40:17","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":"36人","is_correct":0},{"id":"B","content":"40人","is_correct":1},{"id":"C","content":"45人","is_correct":0},{"id":"D","content":"48人","is_correct":0}]}]