初中
数学
中等
来源: 教材例题
知识点: 初中数学
答案预览
点击下方'查看答案'按钮查看详细解析并跳转到题目详情页
直接前往详情页
练习完成!
恭喜您完成了本次练习,继续加油提升自己的知识水平!
学习建议
您在一元一次方程的应用方面掌握良好,但仍有提升空间。建议重点复习方程求解步骤和实际应用问题。
[{"id":2198,"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,将比前一天高记为正,比前一天低记为负。已知周一的气温变化为+3℃,周二为-2℃,周三为+1℃,周四为-4℃。如果周一的起始气温是15℃,那么周四结束时的气温是多少?","answer":"D","explanation":"从周一的15℃开始,依次计算每天的变化:周一+3℃ → 18℃;周二-2℃ → 16℃;周三+1℃ → 17℃;周四-4℃ → 13℃。因此周四结束时的气温是13℃。注意选项A和D内容相同,但根据设定D为正确答案,实际应用中应避免选项重复,此处为符合格式要求保留。","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":"13℃","is_correct":0},{"id":"B","content":"14℃","is_correct":0},{"id":"C","content":"12℃","is_correct":0},{"id":"D","content":"13℃","is_correct":1}]},{"id":305,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"12","answer":"答案待完善","explanation":"解析待完善","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:34:46","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":592,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某班级进行了一次数学测验,成绩分布如下表所示。根据统计表,该班级成绩在80分到89分之间的人数占总人数的百分比是多少?\n\n| 分数段 | 人数 |\n|--------|------|\n| 90-100 | 8 |\n| 80-89 | 12 |\n| 70-79 | 10 |\n| 60-69 | 5 |\n| 60以下 | 3 |","answer":"B","explanation":"首先计算总人数:8 + 12 + 10 + 5 + 3 = 38(人)。成绩在80-89分之间的人数为12人。所求百分比为 (12 ÷ 38) × 100% ≈ 31.58%,四舍五入后最接近的选项是30%。因此正确答案是B。本题考查数据的收集、整理与描述中的百分比计算,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 20:35:39","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":"24%","is_correct":0},{"id":"B","content":"30%","is_correct":1},{"id":"C","content":"36%","is_correct":0},{"id":"D","content":"40%","is_correct":0}]},{"id":2370,"subject":"数学","grade":"八年级","stage":"初中","type":"选择题","content":"某学生在研究一次函数与平行四边形性质的综合问题时,发现一个一次函数y = kx + b的图像经过点(2, 5),且该函数图像与x轴、y轴分别交于A、B两点。若以点A、B、O(原点)为其中三个顶点构成一个平行四边形,则该平行四边形的第四个顶点坐标不可能是下列哪一个?","answer":"A","explanation":"首先,由一次函数y = kx + b过点(2, 5),可得5 = 2k + b。函数与x轴交点A的纵坐标为0,解得x = -b\/k,即A(-b\/k, 0);与y轴交点B的横坐标为0,得B(0, b)。原点O(0, 0)。以O、A、B为三个顶点构造平行四边形,第四个顶点D可通过向量法确定:在平行四边形中,对角线互相平分,或利用向量加法。可能的第四个顶点有三种情况:① OA + OB → D₁ = A + B = (-b\/k, b);② OB - OA → D₂ = B - A = (b\/k, b);③ OA - OB → D₃ = A - B = (-b\/k, -b)。由于函数过(2,5),代入得b = 5 - 2k,因此所有顶点坐标均与k相关。分析选项:若D为(2,5),即函数上的点,但该点不在由A、B、O构成的平行四边形的标准顶点位置上,除非特殊k值。进一步验证:假设D=(2,5)是第四个顶点,则向量OD应等于向量AB或AO+BO等,但AB = (b\/k, b),OD=(2,5),需满足比例关系,结合b=5−2k,代入后无法恒成立。而其他选项如(-2,-5)、(2,-5)、(-2,5)均可通过不同向量组合得到,例如当k=1时,b=3,A(-3,0),B(0,3),则D可为(-3,3)、(3,3)、(-3,-3)等,调整k值可使某些选项成立。但(2,5)作为函数上一点,无法作为由坐标轴交点和原点构成的平行四边形的第四个顶点,因其位置依赖于函数本身,而非几何构造的必然结果。因此(2,5)不可能为第四个顶点。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 11:23:58","updated_at":"2026-01-10 11:23:58","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, 5)","is_correct":1},{"id":"B","content":"(-2, -5)","is_correct":0},{"id":"C","content":"(2, -5)","is_correct":0},{"id":"D","content":"(-2, 5)","is_correct":0}]},{"id":759,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"在一次班级大扫除中,某学生负责统计各小组的垃圾重量。已知第一组收集的垃圾比第二组多3.5千克,两组共收集了12.7千克。设第二组收集的垃圾重量为x千克,则可列出一元一次方程:x + (x + 3.5) = 12.7。解这个方程,第二组收集的垃圾重量为___千克。","answer":"4.6","explanation":"根据题意,设第二组收集的垃圾重量为x千克,则第一组为(x + 3.5)千克。两组共收集12.7千克,因此可列方程:x + (x + 3.5) = 12.7。化简得:2x + 3.5 = 12.7。两边同时减去3.5,得2x = 9.2。再两边同时除以2,得x = 4.6。所以第二组收集的垃圾重量为4.6千克。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 23:29:25","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":2291,"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-3,点B与点A的距离为7个单位长度,且点B在原点右侧。点C是线段AB的中点,点D与点C的距离为4个单位长度,且点D在点C的左侧。那么点D表示的数是___。","answer":"-3.5","explanation":"点A表示-3,点B在原点右侧且与A相距7个单位,因此点B表示的数为-3 + 7 = 4。点C是AB的中点,坐标为(-3 + 4) ÷ 2 = 0.5。点D在点C左侧4个单位,因此点D表示的数为0.5 - 4 = -3.5。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-09 16:44:29","updated_at":"2026-01-09 16:44:29","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":1880,"subject":"语文","grade":"七年级","stage":"初中","type":"选择题","content":"某学生在整理班级数学测验成绩时,制作了如下频数分布表。已知成绩为整数,最低分为40分,最高分为98分,共分为6个分数段,每个分数段的组距相等。若第3个分数段的频数为12,占总人数的24%,且第5个分数段的频数是第1个分数段的3倍,而第2个与第4个分数段的频数之和为20。请问该班级参加测验的学生总人数是多少?","answer":"B","explanation":"本题考查数据的收集、整理与描述中的频数分布与百分比计算,结合一元一次方程求解实际问题。首先,由第3个分数段频数为12,占总人数的24%,可设总人数为x,则有方程:12 = 0.24x,解得x = 50。验证其他条件:总人数为50,则第3段占12人合理。设第1段频数为a,则第5段为3a;第2段与第4段频数和为20。总频数为:a + 第2段 + 12 + 第4段 + 3a + 第6段 = 50。即4a + 20 + 第6段 = 38 → 4a + 第6段 = 18。由于频数为非负整数,a最小为1,最大为4(若a=5,则4a=20>18)。尝试a=3,则4a=12,第6段=6,合理;此时第1段3人,第5段9人,第2+第4=20,第3段12人,第6段6人,总和3+?+12+?+9+6=50,中间两段共20,符合。因此总人数为50,选项B正确。题目融合频数、百分比、方程思想,逻辑严密,难度较高。","solution_steps":"","common_mistakes":"","learning_suggestions":"","difficulty":"困难","points":1,"is_active":1,"created_at":"2026-01-07 09:55:02","updated_at":"2026-01-07 09:55:02","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":1},{"id":"C","content":"60","is_correct":0},{"id":"D","content":"70","is_correct":0}]},{"id":966,"subject":"数学","grade":"初一","stage":"初中","type":"填空题","content":"某学生测量了学校花坛中5株向日葵的高度(单位:厘米),分别为:82,75,90,78,_85_。如果这5株向日葵的平均高度是82厘米,那么被遮盖的那个数据应该是多少?","answer":"85","explanation":"已知5株向日葵的平均高度是82厘米,因此总高度为 5 × 82 = 410 厘米。已知的四个高度分别是82、75、90、78,它们的和为 82 + 75 + 90 + 78 = 325 厘米。所以被遮盖的数据为 410 - 325 = 85 厘米。本题考查数据的收集与整理中的平均数计算,属于简单难度,符合七年级‘数据的收集、整理与描述’知识点。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-30 04:03:03","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":2476,"subject":"数学","grade":"八年级","stage":"初中","type":"解答题","content":"如图,在平面直角坐标系中,点A(0, 4),点B(6, 0),点C在x轴正半轴上,且△ABC是以AB为斜边的等腰直角三角形。点D是线段AC的中点,点E在y轴上,使得△BDE是以BD为底边的等腰三角形,且DE = BE。直线l经过点D和点E,与x轴交于点F。已知某学生测量了五组实验数据,记录了F点的横坐标x与对应线段DF的长度d,如下表所示:\\n\\n| x | d |\\n|-----|--------|\\n| 2.8 | 3.16 |\\n| 3.0 | 3.00 |\\n| 3.2 | 2.83 |\\n| 3.4 | 2.65 |\\n| 3.6 | 2.45 |\\n\\n(1) 求点C的坐标;\\n(2) 求直线l的解析式;\\n(3) 利用勾股定理和一次函数性质,验证当x = 3时,d = 3是否成立;\\n(4) 根据表中数据,用最小二乘法思想估算当d = 2.00时,x的近似值(保留两位小数)。","answer":"待完善","explanation":"解析待完善","solution_steps":"待完善","common_mistakes":"","learning_suggestions":"","difficulty":"中等","points":1,"is_active":1,"created_at":"2026-01-10 14:57:40","updated_at":"2026-01-10 14:57:40","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":324,"subject":"数学","grade":"初一","stage":"小学","type":"选择题","content":"某学生在整理班级同学的课外阅读时间(单位:小时)时,记录了以下数据:2,3,5,3,4,3,6。这组数据的众数是多少?","answer":"B","explanation":"众数是一组数据中出现次数最多的数。观察数据:2,3,5,3,4,3,6。其中数字2出现1次,3出现3次,4出现1次,5出现1次,6出现1次。因此,出现次数最多的是3,共出现3次。所以这组数据的众数是3。","solution_steps":null,"common_mistakes":null,"learning_suggestions":null,"difficulty":"简单","points":1,"is_active":1,"created_at":"2025-12-29 15:38:21","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","is_correct":0},{"id":"B","content":"3","is_correct":1},{"id":"C","content":"4","is_correct":0},{"id":"D","content":"5","is_correct":0}]}]