初中
数学
中等
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知识点: 初中数学
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[{"id":263,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生将一个三位数的个位数字与百位数字交换位置,得到的新数比原数大396。已知原数的十位数字是5,且原数的个位数字比百位数字大4,那么原数是____。","answer":"155","explanation":"设原三位数的百位数字为x,则个位数字为x+4(因为个位比百位大4),十位数字已知为5,因此原数可表示为100x + 10×5 + (x+4) = 101x + 54。交换个位与百位后,新数为100(x+4) + 50 + x = 101x + 450。根据题意,新数比原数大396,列方程:(101x + 450) - (101x + 54) = 396,化简得396 = 396,恒成立。说明只要满足个位比百位大4且十位为5即可。由于是三位数,x为1到9的整数,且x+4 ≤ 9,故x ≤ 5。尝试x=1时,原数为155,交换后为551,551 - 155 = 396,符合条件。因此原数是155。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-29 14:55:35","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":566,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"在一次环保知识竞赛中,某班级共收集了120份有效问卷。统计结果显示,有45份问卷支持‘垃圾分类’,有38份支持‘节约用水’,其余支持‘绿色出行’。请问支持‘绿色出行’的问卷数量是多少?","answer":"A","explanation":"题目考查的是数据的收集、整理与描述中的基本运算能力。已知总问卷数为120份,其中支持‘垃圾分类’的有45份,支持‘节约用水’的有38份,其余为支持‘绿色出行’的问卷。因此,支持‘绿色出行’的问卷数量为:120 - 45 - 38 = 37(份)。计算过程为:120 - 45 = 75,75 - 38 = 37。故正确答案为A。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 19:33:53","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":"37","is_correct":1},{"id":"B","content":"42","is_correct":0},{"id":"C","content":"47","is_correct":0},{"id":"D","content":"53","is_correct":0}]},{"id":766,"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欢的运动项目数据时,发现喜欢篮球的人数占总人数的30%,喜欢足球的人数占总人数的25%,喜欢跳绳的人数占总人数的15%,其余同学喜欢其他项目。如果班级共有40名学生,那么喜欢其他项目的学生有___人。","answer":"12","explanation":"首先计算喜欢篮球、足球和跳绳的学生人数:篮球人数为40 × 30% = 12人,足球人数为40 × 25% = 10人,跳绳人数为40 × 15% = 6人。将这三部分人数相加:12 + 10 + 6 = 28人。总人数为40人,因此喜欢其他项目的人数为40 - 28 = 12人。本题考查数据的收集与整理,涉及百分数的基本计算,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 23:43:26","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":1371,"subject":"数学","grade":"七年级","stage":"初中","type":"解答题","content":"某学校组织七年级学生开展‘校园植物多样性调查’活动。调查小组在校园内选取了5个不同区域进行植物种类统计,并将数据整理如下表。已知每个区域的植物种类数均为正整数,且满足以下条件:\n\n1. 区域A的植物种类数比区域B多3种;\n2. 区域C的植物种类数是区域D的2倍;\n3. 区域E的植物种类数比区域A少5种;\n4. 五个区域植物种类总数为67种;\n5. 区域D的植物种类数比区域B少2种;\n6. 所有区域的植物种类数都不超过20种。\n\n请根据以上信息,求出每个区域的植物种类数。","answer":"设区域B的植物种类数为 x 种。\n\n根据条件1:区域A = x + 3\n根据条件5:区域D = x - 2\n根据条件2:区域C = 2 × (x - 2) = 2x - 4\n根据条件3:区域E = (x + 3) - 5 = x - 2\n\n根据条件4,五个区域总数为67:\nA + B + C + D + E = 67\n代入表达式:\n(x + 3) + x + (2x - 4) + (x - 2) + (x - 2) = 67\n合并同类项:\nx + 3 + x + 2x - 4 + x - 2 + x - 2 = 67\n( x + x + 2x + x + x ) + (3 - 4 - 2 - 2) = 67\n6x - 5 = 67\n6x = 72\nx = 12\n\n代回各区域:\n区域B:x = 12 种\n区域A:x + 3 = 15 种\n区域D:x - 2 = 10 种\n区域C:2x - 4 = 2×12 - 4 = 20 种\n区域E:x - 2 = 10 种\n\n验证总数:15 + 12 + 20 + 10 + 10 = 67,正确。\n验证条件6:所有数值均 ≤ 20,满足。\n\n答:区域A有15种,区域B有12种,区域C有20种,区域D有10种,区域E有10种植物。","explanation":"本题综合考查了二元一次方程组的思想(虽未显式列出两个方程,但通过多个等量关系建立一元一次方程)、整式的加减运算、有理数的四则运算以及数据的整理与分析能力。解题关键在于合理设元,将多个文字条件转化为代数表达式,再通过列方程求解。题目设置了多个约束条件,包括总数限制和范围限制(不超过20种),要求学生在解出答案后进行验证,体现了数学建模与逻辑推理的结合。情境贴近生活,考查学生从实际问题中抽象出数学模型的能力,属于综合性较强的困难题。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-06 11:12:47","updated_at":"2026-01-06 11:12:47","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":2214,"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5℃,记作+5℃;第二天又下降了8℃,应记作____℃。","answer":"-8","explanation":"根据正数和负数表示相反意义的量的知识点,气温上升用正数表示,下降则用负数表示。因此,气温下降8℃应记作-8℃。","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":479,"subject":"数学","grade":"初一","stage":"初中","type":"选择题","content":"某班级组织了一次环保活动,收集废旧纸张进行回收。活动结束后,统计了5个小组的回收量(单位:千克),分别为:8.5、7.2、9.0、6.8、8.5。请问这5个小组回收量的中位数是多少?","answer":"B","explanation":"要找出中位数,首先需要将数据按从小到大的顺序排列:6.8、7.2、8.5、8.5、9.0。由于共有5个数据(奇数个),中位数就是正中间的那个数,即第3个数。排序后第3个数是8.5,因此中位数是8.5。选项B正确。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 17:58:08","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":"7.2","is_correct":0},{"id":"B","content":"8.5","is_correct":1},{"id":"C","content":"8.0","is_correct":0},{"id":"D","content":"9.0","is_correct":0}]},{"id":9,"subject":"化学","grade":"初三","stage":"初中","type":"填空题","content":"电解水的化学方程式为______,反应类型为______反应。","answer":"2H₂O → 2H₂↑ + O₂↑, 分解","explanation":"电解水生成氢气和氧气,是一种分解反应。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":2,"is_active":1,"created_at":"2025-08-29 16:33:04","updated_at":"2025-08-29 16:33:04","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":260,"subject":"数学","grade":"初一","stage":"小学","type":"填空题","content":"某学生在解方程 3(x - 2) + 5 = 2x + 7 时,第一步将方程展开为 3x - 6 + 5 = 2x + 7,第二步合并同类项得到 3x - 1 = 2x + 7,第三步将 2x 移到左边,-1 移到右边,得到 ___ = 8,最后解得 x = 8。","answer":"x","explanation":"根据题意,第三步是将 2x 从右边移到左边变为 -2x,同时将 -1 从左边移到右边变为 +1,因此左边变为 3x - 2x = x,右边变为 7 + 1 = 8,所以空格处应填 x。此题考查一元一次方程的移项与合并同类项,属于七年级代数基础内容,步骤清晰,难度适中。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"中等","points":1,"is_active":1,"created_at":"2025-12-29 14:55:11","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":2160,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在数轴上标出三个有理数 a、b、c,其中 a 与 b 关于原点对称,c 是 a 与 b 之间距离的一半,且 a > 0。若 a = 6,则 c 的值是多少?","answer":"D","explanation":"因为 a = 6 且 a 与 b 关于原点对称,所以 b = -6。a 与 b 之间的距离为 |6 - (-6)| = 12。c 是该距离的一半,即 12 ÷ 2 = 6 个单位长度。但题目中 c 是位于 a 与 b 之间距离的一半位置,即从 a 向左移动 6 个单位或从 b 向右移动 6 个单位,最终都到达原点 0。因此 c = 0。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-09 13:35:36","updated_at":"2026-01-09 13:35:36","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":"-3","is_correct":0},{"id":"C","content":"6","is_correct":0},{"id":"D","content":"0","is_correct":1}]},{"id":1963,"subject":"数学","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在研究自家阳台盆栽植物的生长情况时,记录了连续6周每周植株的高度增长量(单位:厘米):2.3, 3.1, 1.8, 2.9, 3.5, 2.7。为了评估这6周植株高度增长量的波动程度,该学生计算了这组数据的方差。已知方差是各数据与平均数之差的平方的平均数,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概念与计算。首先计算6周高度增长量的平均数:(2.3 + 3.1 + 1.8 + 2.9 + 3.5 + 2.7) ÷ 6 = 16.3 ÷ 6 ≈ 2.717。然后计算每个数据与平均数之差的平方:(2.3−2.717)²≈0.174,(3.1−2.717)²≈0.147,(1.8−2.717)²≈0.841,(2.9−2.717)²≈0.034,(3.5−2.717)²≈0.613,(2.7−2.717)²≈0.0003。将这些平方值相加:0.174 + 0.147 + 0.841 + 0.034 + 0.613 + 0.0003 ≈ 1.8093。最后求平均得方差:1.8093 ÷ 6 ≈ 0.3015,最接近选项B(0.35)。注意:虽然精确值略小于0.35,但在四舍五入和估算范围内,0.35是最合理的选项。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-07 14:47:44","updated_at":"2026-01-07 14:47:44","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":"0.28","is_correct":0},{"id":"B","content":"0.35","is_correct":1},{"id":"C","content":"0.42","is_correct":0},{"id":"D","content":"0.50","is_correct":0}]}]