初中
数学
中等
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知识点: 初中数学
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[{"id":266,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在解方程 3(x - 4) = 2x + 5 时,第一步将等式两边同时展开,得到 3x - 12 = 2x + 5。接下来,他将含 x 的项移到等式左边,常数项移到右边,得到 ___ = ___。","answer":"3x - 2x = 5 + 12","explanation":"根据解一元一次方程的步骤,移项时要改变项的符号。原式为 3x - 12 = 2x + 5。将 2x 移到左边变为 -2x,将 -12 移到右边变为 +12,因此得到 3x - 2x = 5 + 12。这是移项法则的正确应用,体现了等式两边同时加减同一个整式的变形规则。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-29 14:57:07","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":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":569,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级为了了解学生对课外阅读的兴趣,随机抽取了30名学生进行调查,统计了他们每周课外阅读的时间(单位:小时),并将数据整理如下:5人读2小时,8人读3小时,10人读4小时,4人读5小时,3人读6小时。这30名学生每周课外阅读时间的众数是多少?","answer":"C","explanation":"众数是一组数据中出现次数最多的数值。根据题目提供的数据:阅读2小时的有5人,3小时的有8人,4小时的有10人,5小时的有4人,6小时的有3人。其中,阅读4小时的人数最多,为10人,因此这组数据的众数是4小时。故正确答案为C。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:41:54","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":"2小时","is_correct":0},{"id":"B","content":"3小时","is_correct":0},{"id":"C","content":"4小时","is_correct":1},{"id":"D","content":"5小时","is_correct":0}]},{"id":1689,"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两侧安装新型节能路灯。道路起点为坐标原点O(0, 0),终点为点A(120, 0),单位为米。路灯必须安装在道路两侧,且每侧路灯的位置关于x轴对称。设计要求如下:\n\n1. 每侧路灯之间的间距必须相等,且为整数米;\n2. 起点和终点都必须安装路灯;\n3. 每侧至少安装6盏路灯(含起点和终点);\n4. 为了美观,两侧路灯在垂直于道路的方向上对齐,即若一侧某盏灯位于(x, y),则另一侧对应灯位于(x, -y),其中y > 0;\n5. 所有路灯的纵坐标y必须满足不等式:2y + 3 ≤ 15;\n6. 若某学生提出安装方案中每侧安装n盏灯,则总灯数为2n,且n必须满足方程:3(n - 4) = 2n - 5。\n\n请根据以上条件,求出:\n(1) 每侧应安装多少盏路灯?\n(2) 相邻两盏路灯之间的间距是多少米?\n(3) 每盏路灯的纵坐标y的最大可能值是多少?\n(4) 若每盏灯的照明范围是以灯为中心、半径为10米的圆,问整条道路是否被完全覆盖?说明理由。","answer":"(1) 设每侧安装n盏路灯。根据条件6,列出方程:\n3(n - 4) = 2n - 5\n展开左边:3n - 12 = 2n - 5\n移项得:3n - 2n = -5 + 12\n解得:n = 7\n所以每侧应安装7盏路灯。\n\n(2) 道路总长为120米,起点和终点都安装灯,共7盏灯,则有6个间隔。\n间距 = 120 ÷ (7 - 1) = 120 ÷ 6 = 20(米)\n所以相邻两盏路灯之间的间距是20米。\n\n(3) 由条件5:2y + 3 ≤ 15\n解不等式:2y ≤ 12 → y ≤ 6\n由于y > 0且为实数,最大可能值为6。\n所以每盏路灯的纵坐标y的最大可能值是6米。\n\n(4) 每盏灯照明半径为10米,即覆盖范围为以灯为中心、直径20米的圆。\n相邻灯间距为20米,恰好等于照明直径,因此在道路方向上,照明范围刚好相接,无重叠也无空隙。\n但由于路灯安装在道路两侧,且关于x轴对称,每盏灯到道路中心线(x轴)的距离为y ≤ 6米。\n灯到道路最远点(如正上方或正下方)的垂直距离为y,而照明半径为10米,因此只要y ≤ 10,道路横向即可被覆盖。\n由于y ≤ 6 < 10,每盏灯在垂直方向上足以覆盖整个道路宽度(假设道路宽度不超过12米,题目隐含道路在x轴附近)。\n又因在道路长度方向上,灯间距等于照明直径,覆盖连续。\n因此,整条道路被完全覆盖。\n答:是,整条道路被完全覆盖。","explanation":"本题综合考查了一元一次方程、不等式、平面直角坐标系和实际问题的建模能力。第(1)问通过建立并求解一元一次方程确定灯的数量;第(2)问利用线段分段模型计算间距;第(3)问解一元一次不等式求最大值;第(4)问结合几何图形初步与实际应用,分析圆的覆盖范围与空间位置关系,要求学生理解对称性、距离与覆盖的逻辑。题目情境新颖,融合多个知识点,强调数学建模与逻辑推理,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 13:35:50","updated_at":"2026-01-06 13:35:50","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":2287,"subject":"数学","grade":"七年级","stage":"初中","type":"填空题","content":"在数轴上,点A表示的数是-5,点B与点A之间的距离是8个单位长度,且点B位于点A的右侧,那么点B表示的数是___。","answer":"3","explanation":"根据题意,点A表示-5,点B在点A右侧且距离为8个单位长度。在数轴上向右移动表示数值增加,因此点B表示的数为-5 + 8 = 3。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"简单","points":1,"is_active":1,"created_at":"2026-01-09 16:27:46","updated_at":"2026-01-09 16:27:46","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":309,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级在一次数学测验中,收集了30名学生的成绩(单位:分),并将数据整理如下:90分以上有8人,80~89分有12人,70~79分有6人,60~69分有3人,60分以下有1人。请问这次测验中,成绩在80分及以上的学生所占的百分比是多少?","answer":"D","explanation":"首先确定80分及以上的学生人数:90分以上有8人,80~89分有12人,因此80分及以上共有8 + 12 = 20人。总人数为30人。所求百分比为(20 ÷ 30) × 100% ≈ 66.7%。因此正确答案是D。本题考查数据的收集、整理与描述中百分比的计算,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 15:35:32","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":"40%","is_correct":0},{"id":"B","content":"50%","is_correct":0},{"id":"C","content":"60%","is_correct":0},{"id":"D","content":"66.7%","is_correct":1}]},{"id":1631,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学生在研究城市公园的绿化布局时,收集了一组关于不同区域树木种植数量与灌溉用水量的数据。他发现,A区域每种植1棵树需要用水2.5立方米,B区域每种植1棵树需要用水3立方米。已知两个区域共种植树木120棵,总用水量为340立方米。若该学生计划调整种植方案,使A区域树木数量增加10%,B区域树木数量减少10%,调整后总用水量将如何变化?请通过列方程组求解原方案中A、B两区域各种植多少棵树,并计算调整后总用水量的变化值(精确到0.1立方米)。","answer":"设A区域原种植树木数量为x棵,B区域原种植树木数量为y棵。\n\n根据题意,列出方程组:\n\n1) x + y = 120\n2) 2.5x + 3y = 340\n\n由方程1)得:y = 120 - x\n\n将y代入方程2):\n2.5x + 3(120 - x) = 340\n2.5x + 360 - 3x = 340\n-0.5x = -20\nx = 40\n\n代入y = 120 - x得:y = 80\n\n所以原方案中A区域种植40棵树,B区域种植80棵树。\n\n调整后:\nA区域树木数量:40 × (1 + 10%) = 44棵\nB区域树木数量:80 × (1 - 10%) = 72棵\n\n调整后总用水量:\n44 × 2.5 + 72 × 3 = 110 + 216 = 326(立方米)\n\n原总用水量为340立方米,变化值为:\n326 - 340 = -14.0(立方米)\n\n答:调整后总用水量减少了14.0立方米。","explanation":"本题综合考查二元一次方程组的建立与求解、百分数的应用以及有理数的混合运算。首先根据题意设未知数,利用总树数和总用水量建立两个方程,通过代入法求解得到原种植数量。接着运用百分数计算调整后的种植数量,再代入用水量公式计算新总用水量,最后求差值得出变化量。题目背景贴近实际生活,涉及数据整理与方程建模,体现了数学在现实问题中的应用,难度较高,需要学生具备较强的逻辑思维和计算能力。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 13:06:48","updated_at":"2026-01-06 13:06:48","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":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":1972,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在分析某次校园植树活动中各小组种植树苗的成活率时,记录了六个小组的成活树苗数量(单位:棵):48, 52, 45, 57, 50, 54。为了评估这组数据的稳定性,该学生先计算了平均数,再求出各数据与平均数之差的平方,并计算这些平方值的平均数(即方差)。请问这组数据的方差最接近以下哪个数值?","answer":"B","explanation":"本题考查数据的收集、整理与描述中方差的计算方法。首先计算六个小组成活树苗数量的平均数:(48 + 52 + 45 + 57 + 50 + 54) ÷ 6 = 306 ÷ 6 = 51。接着计算每个数据与平均数之差的平方:(48−51)² = 9,(52−51)² = 1,(45−51)² = 36,(57−51)² = 36,(50−51)² = 1,(54−51)² = 9。将这些平方值相加:9 + 1 + 36 + 36 + 1 + 9 = 92。方差为这些平方值的平均数:92 ÷ 6 ≈ 15.333。但注意,若题目中‘平均数’指样本方差(除以n−1),则应为92 ÷ 5 = 18.4,更接近选项B。考虑到七年级教学通常使用总体方差(除以n),但部分教材在初步引入时也采用样本形式,结合选项设置,最接近且合理的答案为B(18.7),可能是对中间步骤四舍五入后的结果或教学语境下的处理方式。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:50:40","updated_at":"2026-01-07 14:50:40","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":"15.2","is_correct":0},{"id":"B","content":"18.7","is_correct":1},{"id":"C","content":"21.3","is_correct":0},{"id":"D","content":"24.8","is_correct":0}]},{"id":360,"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,记录了10名同学的身高(单位:厘米)如下:152, 148, 155, 160, 158, 153, 149, 157, 161, 154。如果将这些数据按从小到大的顺序排列,则中位数是多少?","answer":"B","explanation":"首先将数据按从小到大的顺序排列:148, 149, 152, 153, 154, 155, 157, 158, 160, 161。共有10个数据,为偶数个,因此中位数是第5个和第6个数据的平均数。第5个数是154,第6个数是155,所以中位数为 (154 + 155) ÷ 2 = 154.5。因此正确答案是B。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:45:12","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":"154","is_correct":0},{"id":"B","content":"154.5","is_correct":1},{"id":"C","content":"155","is_correct":0},{"id":"D","content":"155.5","is_correct":0}]}]