工作就用美篇工作版","guide_search_desc":"海量模板范文,一键做同款 工作就用美篇工作版","img_beautify_switch":0,"ad_free":0,"text_direction":2,"template":{"id":9462,"series_id":36,"theme_id":36},"container":{},"redirect":{"redirect_type":1,"img_url":false,"link_url":"/wap/downloadpage/backpage","button_bg_color":false,"button_desc":false,"redirect_desc":false},"edit_date_str":"发布于 2024-04-13","current_time":1744198297,"font_name":"","rcmd_meipian":0,"hide_article_link":0,"from_wechat":false,"topk_keywords":[{"name":"问题","score":0.131549},{"name":"年龄","score":0.111863},{"name":"多少","score":0.103763},{"name":"解题","score":0.099265},{"name":"题目","score":0.086864},{"name":"鸡兔","score":0.085437},{"name":"数量","score":0.084588},{"name":"生长量","score":0.081867},{"name":"草量","score":0.080825},{"name":"头牛","score":0.07444}],"music_name":"","origin_status":0,"password_v2":"","font_id":0,"title_style":"","rich_text_title":"
《学习笔记》分享小学数学部分精品例题归纳&解析
","enable_download":1,"cover_thumb":"https://ss-mpvolc.meipian.me/users/686154/9d4e78c85002d2de81883a91a967e869_crop__jpg.heic","has_video":false,"gift_switch":1,"enable_watermark":2,"content":{"article_id":397813337,"content":[{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"1bc16627b53690943662b511369be431"},"img_del":0,"img_height":818,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin9d4e78c85002d2de81883a91a967e869__jpg.heic","img_width":624,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"从《精品文档 小学数学典型应用题归纳汇总30种题型》中选择了几种常用题型例题,不仅对小学生学习数学有所帮助,还能激发他们对学习数学的兴趣和提高其思维能力。这些题对老年人提高大脑活力和思维能力也有一定好处和帮助,为此对有些不常用的和偏题就没有被纳入,而是有针对性的选择了以下十二种常用题型,通过改编和添加插图保存到该篇中以便于学习和参考。
01归一问题 02归总问题
03和差问题 04和倍问题
05差倍问题 06相遇问题
07追及问题 08年龄问题
09牛吃草问题 10鸡兔同笼题
11幻方问题 12构图布数问题
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"f8bc456f89cf9070c743332d980f4d59"},"img_del":0,"img_height":420,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin689109a8bd29ee2d6a77822bfee73a00.gif","img_width":600,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"1, 归一问题
【含义】 在解题时,先求出一份是多少(即单一量),然后以单一量为标准,求出所要求的数量。这类应用题叫做归一问题。
【数量关系】 总量÷份数=1份数量
1份数量×所占份数=所求几份的数量
另一总量÷(总量÷份数)=所求份数
【解题思路和方法】 先求出单一量,以单一量为标准,求出所要求的数量。
例1 ,买5支铅笔要0.6元钱,买同样的铅笔16支,需要多少钱?
解
(1)买1支铅笔多少钱?
0.6÷5=0.12(元)
(2)买16支铅笔需要多少钱?
0.12×16=1.92(元)
列成综合算式
0.6÷5×16=0.12×16=1.92(元)
答:需要1.92元。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"8b4f7b6d958a593bd1ff09c4cace433b"},"img_del":0,"img_height":818,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin9d4e78c85002d2de81883a91a967e869__jpg.heic","img_width":624,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"2 , 归总问题
【含义】 解题时,常常先找出“总数量”,然后再根据其它条件算出所求的问题,叫归总问题。所谓“总数量”是指货物的总价、几小时(几天)的总工作量、几公亩地上的总产量、几小时行的总路程等。
【数量关系】
1份数量×份数=总量
总量÷1份数量=份数
总量÷另一份数=另一每份数量
【解题思路和方法】
先求出总数量,再根据题意得出所求的数量。
例1 服装厂原来做一套衣服用布3.2米,改进裁剪方法后,每套衣服用布2.8米。原来做791套衣服的布,现在可以做多少套?
解:
(1)这批布总共有多少米?
3.2×791=2531.2(米)
(2)现在可以做多少套?
2531.2÷2.8=904(套)
列成综合算式 :
3.2×791÷2.8=904(套)
答:现在可以做904套。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"d2d46de1079ad43aa5ef7737595062bf"},"img_del":0,"img_height":346,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin3e36bf910601bf214909f36b9388123d__jpg.heic","img_width":242,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"3 , 和差问题
【含义】 已知两个数量的和与差,求这两个数量各是多少,这类应用题叫和差问题。
【数量关系】
大数=(和+差)÷ 2
小数=(和-差)÷ 2
【解题思路和方法】 简单的题目可以直接套用公式;复杂的题目变通后再用公式。
例1 甲乙两班共有学生98人,甲班比乙班多6人,求两班各有多少人?
解:
甲班人数=(98+6)÷2=52(人)
乙班人数=(98-6)÷2=46(人)
答:甲班有52人,乙班有46人。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"923000ba1cb0626de1a4490ac11c5ded"},"img_del":0,"img_height":376,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originaaf74ce071a4a62d285f4fcbb7789de0__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"4,和倍问题
【含义】已知两个数的和及大数是小数的几倍(或小数是大数的几分之几),要求这两个数各是多少,这类应用题叫做和倍问题。
【数量关系】
总和÷(几倍+1)较小的数
总和-较小的数 = 较大的数
较小的数 ×几倍 = 较大的数
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 果园里有杏树和桃树共248棵,桃树的棵数是杏树的3倍,求杏树、桃树各多少棵?
解 :
(1)杏树有多少棵?
248÷(3+1)=62(棵)
(2)桃树有多少棵?
62×3=186(棵)
答:杏树有62棵,桃树有186棵。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"f5cb7dd056ee30943d5503fb971a42fd"},"img_del":0,"img_height":300,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originfcb2315ecbdf0cd2b72a5ad90b90f70d__jpg.heic","img_width":600,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"5 ,差倍问题
【含义】 已知两个数的差及大数是小数的几倍(或小数是大数的几分之几),要求这两个数各是多少,这类应用题叫做差倍问题。
【数量关系】
两个数的差÷(几倍-1)=较小的数
较小的数×几倍=较大的数
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 果园里桃树的棵数是杏树的3倍,而且桃树比杏树多124棵。求杏树、桃树各多少棵? 解 :
(1)杏树有多少棵?
124÷(3-1)=62(棵)
(2)桃树有多少棵?
62×3=186(棵)
答:果园里杏树是62棵,桃树是186棵。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"b20c4efce44d5d0375776b37d06b5f1c"},"img_del":0,"img_height":1070,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origincd1756cdd380435109e9a98afbc8e89d__jpg.heic","img_width":1080,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"6,相遇问题
【含义】两个运动的物体同时由两地出发相向而行,在途中相遇。这类应用题叫做相遇问题。
【数量关系】相遇时间=总路程÷(甲速+乙速)
总路程=(甲速+乙速)×相遇时间
【解题思路和方法】 简单的题目可直接利用公式,复杂的题目变通后再利用公式。
………………
例(由于该题比较典型故选此题): 甲乙二人同时从两地骑自行车相对而行,甲每小时行15千米,乙每小时行13千米,两人在距中点3千米处相遇,求两地的距离。
解 : “两人在距中点3千米处相遇”是正确理解本题题意的关键。从题中可知甲骑得快,乙骑得慢,甲过了中点3千米,乙距中点3千米,就是说甲比乙多走的路程是(3×2)千米,因此,相遇时间=(3×2)÷(15-13)=3(小时)
两地距离=(15+13)×3=84(千米)
答:两地距离是84千米。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"d42287306c55cb8eb13bbbb6ba3e4547"},"img_del":0,"img_height":818,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin9d4e78c85002d2de81883a91a967e869__jpg.heic","img_width":624,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"7, 追及问题
【含义】 两个运动物体在不同地点同时出发(或者在同一地点而不是同时出发,或者在不同地点又不是同时出发)作同向运动,在后面的,行进速度要快些,在前面的,行进速度较慢些,在一定时间之内,后面的追上前面的物体。这类应用题就叫做追及问题。
【数量关系】
追及时间=追及路程÷(快速-慢速)
追及路程=(快速-慢速)×追及时间
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 好马每天走120千米,劣马每天走75千米,劣马先走12天,好马几天能追上劣马?
解 :
(1)劣马先走12天能走多少千米?
75×12=900(千米)
(2)好马几天追上劣马?
900÷(120-75)=20(天)
列成综合算式
75×12÷(120-75)
=900÷45=20(天) 答:好马20天能追上劣马。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"5fc9ea4c93b07440930fa7a29a68496f"},"img_del":0,"img_height":898,"img_size":519,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin28f1be94949187103eda55e393910dc5__jpg.heic","img_width":1078,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"……………………………………
例2 小明和小亮在200米环形跑道上跑步,小明跑一圈用40秒,他们从同一地点同时出发,同向而跑。小明第一次追上小亮时跑了500米,求小亮的速度是每秒多少米。
解 :
小明第一次追上小亮时比小亮多跑一圈,即200米,此时小亮跑了(500-200)米,要知小亮的速度,须知追及时间,即小明跑500米所用的时间。又知小明跑200米用40秒,则跑500米用[40×(500÷200)]秒,
所以小亮的速度是
(500-200)÷[40×(500÷200)]
=300÷100=3米/秒
答:小亮的速度是每秒3米。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"01b1bc736816b300832ddd1888f84a13"},"img_del":0,"img_height":372,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originfc24f7f997ca5abce21841924df56143__jpg.heic","img_width":514,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"8, 年龄问题
【含义】这类问题是根据题目的内容而得名,它的主要特点是两人的年龄差不变,但是,两人年龄之间的倍数关系随着年龄的增长在发生变化。
【数量关系】年龄问题往往与和差、和倍、差倍问题有着密切联系,尤其与差倍问题的解题思路是一致的,要紧紧抓住“年龄差不变”这个特点。
【解题思路和方法】 可以利用“差倍问题”的解题思路和方法。
例1 爸爸今年35岁,亮亮今年5岁,今年爸爸的年龄是亮亮的几倍?明年呢?
解 35÷5=7(倍)
(35+1)÷(5+1)=6(倍) 答:今年爸爸的年龄是亮亮的7倍,明年爸爸的年龄是亮亮的6倍。
例2 母亲今年37岁,女儿今年7岁,几年后母亲的年龄是女儿的4倍?
解 (1)母亲比女儿的年龄大多少岁?
37-7=30(岁)
(2)几年后母亲的年龄是女儿的4倍?
30÷(4-1)-7=3(年)
列成综合算式
(37-7)÷(4-1)-7=3(年)
答:3年后母亲的年龄是女儿的4倍
例3 3年前父子的年龄和是49岁,今年父亲的年龄是儿子年龄的4倍,父子今年各多少岁?
解 今年父子的年龄和应该比3年前增加(3×2)岁,今年二人的年龄和为 :
49+3×2=55(岁) 把今年儿子年龄作为1倍量,则今年父子年龄和相当于(4+1)倍,因此,今年儿子年龄为
55÷(4+1)=11(岁)
今年父亲年龄为
11×4=44(岁)
答:今年父亲年龄是44岁,儿子年龄是11岁。
例4 甲对乙说:“当我的岁数曾经是你现在的岁数时,你才4岁”。
乙对甲说:“当我的岁数将来是你现在的岁数时,你将61岁”。求甲乙现在的岁数各是多少? 解 :这里涉及到三个年份:过去某一年、今年、将来某一年。列表分析:
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"7dddae1bee9196fc7d1a5e18ce46be18"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"这里涉及到三个年份:过去某一年、今年、将来某一年。列表分析:
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"dda846f53437a54bd3f088f730a0f268"},"img_del":0,"img_height":470,"img_size":58,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin81cea68393f06eca20807ecb99cbdd10__jpg.heic","img_width":1080,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"表中二个"□"表示同一个数,表中二个"△"表示同一个数。
因为两个人的年龄差总相等:
□-4=△-□=61-△,也就是4,□,△,61成等差数列,所以,61应该比4大3个年龄差,因此二人年龄差为
(61-4)÷3=19(岁)
甲今年的岁数为
△=61-19=42(岁)
乙今年的岁数为
□=42-19=23(岁)
答:甲今年的岁数是42岁,乙今年的岁数是23岁
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cf029721afd011c72013f47454768240"},"img_del":0,"img_height":346,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin2c467601e0e422fea3164d6f9eb9950c__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"9,“牛吃草”问题
【含义】 “牛吃草”问题是大科学家牛顿提出的问题,也叫“牛顿问题”。这类问题的特点在于要考虑草边吃边长这个因素。
【数量关系】 草总量=原有草量+草每天生长量×天数 【解题思路和方法】 解这类题的关键是求出草每天的生长量。
例1 一块草地,10头牛20天可以把草吃完,15头牛10天可以把草吃完。问多少头牛5天可以把草吃完?
解 草是均匀生长的,所以,草总量=原有草量+草每天生长量×天数。求“多少头牛5天可以把草吃完”,就是说5 天内的草总量要5 天吃完的话,得有多少头牛?
设每头牛每天吃草量为1,按以下步骤解答:
(1)求草每天的生长量
因为,一方面20天内的草总量就是10头牛20天所吃的草,即(1×10×20);另一方面,20天内的草总量又等于原有草量加上20天内的生长量,所以
1x10x20=原有草量÷20天内生长量
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"c4f7a65e18455d056d7bf7b6d8bd12c9"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"同理
1×15×10=原有草量+10天内生长量
由此可知 (20-10)天内草的生长量为 1×10×20-1×15×10=50
因此,草每天的生长量为
50÷(20-10)=5
(2)求原有草量
原有草量=10天内总草量-10内生长量=1×15×10-5×10=100
(3)求5 天内草总量
5 天内草总量=原有草量+5天内生长量=100+5×5=125
(4)求多少头牛5 天吃完草 因为每头牛每天吃草量为1,所以每头牛5天吃草量为5。因此5天吃完草需要牛的头数
125÷5=25(头)
答:需要5头牛5天可以把草吃完。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"ce0ed9cbef67a7cf3c3d08ea7a708f20"},"img_del":0,"img_height":500,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originb356fce34d77c711c3da5dd2d0001d8e__jpg.heic","img_width":750,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2 一只船有一个漏洞,水以均匀速度进入船内,发现漏洞时已经进了一些水。如果有12个人淘水,3小时可以淘完;如果只有5人淘水,要10小时才能淘完。求17人几小时可以淘完?
解 这是一道变相的“牛吃草”问题。与上题不同的是,最后一问给出了人数(相当于“牛数”),求时间。设每人每小时淘水量为1,按以下步骤计算:
(1)求每小时进水量
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cb9ffa1e0c00e501be5e64222518b865"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"因为,3小时内的总水量=1×12×3=原有水量+3小时进水量
10小时内的总水量=1×5×10=原有水量+10小时进水量
所以,(10-3)小时内的进水量为 1×5×10-1×12×3=14 因此,每小时的进水量为
14÷(10-3)=2
(2)求淘水前原有水量 原有水量=1×12×3-3小时进水量=36-2×3=30
(3)求17人几小时淘完
17人每小时淘水量为17,因为每小时漏进水为2,所以实际上船中每小时减少的水量为(17-2),所以17人淘完水的时间是 30÷(17-2)=2(小时)
答:17人2小时可以淘完水。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"3818c16f10e7d3226139cf31e1036cb2"},"img_del":0,"img_height":720,"img_size":99,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin5d4700985b45549d162f772196279eed__jpg.heic","img_width":960,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"10 ,鸡兔同笼问题
【含义】 这是古典的算术问题。已知笼子里鸡、兔共有多少只和多少只脚,求鸡、兔各有多少只的问题,叫做第一鸡兔同笼问题。已知鸡兔的总数和鸡脚与兔脚的差,求鸡、兔各是多少的问题叫做第二鸡兔同笼问题。 【数量关系】第一鸡兔同笼问题:
假设全都是鸡,则有 兔数=(实际脚数-2×鸡兔总数)÷(4-2)
假设全都是兔,则有 鸡数=(4×鸡兔总数-实际脚数)÷(4-2)
第二鸡兔同笼问题:
假设全都是鸡,则有
兔数=(2×鸡兔总数-鸡与兔脚之差)÷(4+2)
假设全都是兔,则有
鸡数=(4×鸡兔总数+鸡与兔脚之差)÷(4+2)
【解题思路和方法】 解答此类题目一般都用假设法,可以先假设都是鸡,也可以假设都是兔。如果先假设都是鸡,然后以兔换鸡;如果先假设都是兔,然后以鸡换兔。这类问题也叫置换问题。通过先假设,再置换,使问题得到解
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"e7924c5fd1b2e04367356ff93d7f8ea4"},"img_del":0,"img_height":376,"img_size":20,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin4dcb00c6389e90b8e1c3562c46813934__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例1 长毛兔子芦花鸡,鸡兔圈在一笼里。数数头有三十五,脚数共有九十四。请你仔细算一算,多少兔子多少鸡?
解 假设35只全为兔,
则 鸡数=(4×35-94)÷(4-2)=23(只)
兔数7=35-23=12(只)
也可以先假设35只全为鸡,则
兔数=(94-2×35)÷(4-2)=12(只)
鸡数=35-12=23(只)
答:有鸡23只,有兔12只。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"48b8887e966b2a5806a9bf5193bd8efa"},"img_del":0,"img_height":376,"img_size":31,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originac8eac4c0d9006da651ffd78a962510f__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2(第二鸡兔同笼问题)鸡兔共有100只,鸡的脚比兔的脚多80只,问鸡与兔各多少只?
解 假设100只全都是鸡,则有
兔数=(2×100-80)÷(4+2)=20(只)
鸡数=100-20=80(只)
答:有鸡80只,有兔20只。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"93687ca362a467a6556fc10166aba772"},"img_del":0,"img_height":1114,"img_url":"https://ss-mpvolc.meipian.me/users/686154/1615788412387__jpg.heic","img_width":834,"is_origin":false,"location_del":0,"sectionPublished":true,"sell_del":0,"source":1,"splicing_template_id":201,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"497be9aab94b24684a8bbfae85d1532d"},"img_del":0,"img_height":484,"img_size":41,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originbe8784244681503c4daa7ade26df0fcc__jpg.heic","img_width":328,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例3, 有100个馍100个和尚吃,大和尚一人吃3个馍,小和尚3人吃1个馍,问大小和尚各多少人?
解 假设全为大和尚,则共吃馍(3×100)个,比实际多吃(3×100-100)个,这是因为把小和尚也算成了大和尚,因此我们在保证和尚总数100不变的情况下,以“小”换“大”,一个小和尚换掉一个大和尚可减少馍(3-1/3)个。因此,共有小和尚 (3×100-100)÷(3-1/3)=75(人)
共有大和尚 100-75=25(人)
答:共有大和尚25人,有小和尚75人。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"9c8840045e50ab16f41550100941b147"},"img_del":0,"img_height":500,"img_size":264,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin8dc38bc032767879abc140f472a889e9.gif","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"11, 幻方问题
【含义】 把n×n个自然数排在正方形的格子中,使各行、各列以及对角线上的各数之和都相等,这样的图叫做幻方。最简单的幻方是三级幻方。
【数量关系】 每行、每列、每条对角线上各数的和都相等,这个“和”叫做“幻和”。
三级幻方的幻和=45÷3=15
五级幻方的幻和=325÷5=65
【解题思路和方法】首先要确定每行、每列以及每条对角线上各数的和(即幻和),其次是确定正中间方格的数,然后再确定其它方格中的数。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cc74861993706968831495416e8a661a"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例1
把1,2,3,4,5,6,7,8,9这九个数填入九个方格中,使每行、每列、每条对角线上三个数的和相等。
解
幻和的3倍正好等于这九个数的和,所以幻和为
(1+2+3+4+5+6+7+8+9)÷3=45÷3=15
九个数在这八条线上反复出现构成幻和时,每个数用到的次数不全相同,最中心的那个数要用到四次(即出现在中行、中列、和两条对角线这四条线上),四角的四个数各用到三次,其余的四个数各用到两次。看来,用到四次的“中心数”地位重要,宜优先考虑。
设“中心数”为Χ,因为Χ出现在四条线上,而每条线上三个数之和等于15,所以
(1+2+3+4+5+6+7+8+9)+(4-1)Χ=15×4
即 45+3Χ=60 所以 Χ=5 接着用奇偶分析法寻找其余四个偶数的位置,它们分别在四个角,再确定其余四个奇数的位置,它们分别在中行、中列,进一步尝试,容易得到正确的结果。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"fe93f4fc4296b0a22fd44ea8d821fa70"},"img_del":0,"img_height":1892,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin3aac8227b941d618bf33979c188a2b38__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2
把2,3,4,5,6,7,8,9,10这九个数填到九个方格中, 使每行、每列、以及对角线上的各数之和都相等。
解
只有三行,三行用完了所给的个数,所以每行三数之和为
(2+3+4+5+6+7+8+9+10)÷3=18
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"edcaea1fd3d35d0072a001611072070c"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"假设符合要求的数都已经填好,那么三行、三列、两条对角线共8行上的三个数之和都等于18,我们看18能写成哪三个数之和:
最大数是10:
18=10+6+2=10+5+3
最大数是9:
18=9+7+2=9+6+3=9+5+4
最大数是8:
18=8+7+3=8+6+4 最大数是7: 18=7+6+5
刚好写成8个算式。 首先确定正中间方格的数。第二横行、第二竖行、两个斜行都用到正中间方格的数,共用了四次。观察上述8个算式,只有6被用了4次,所以正中间方格中应填6。
9 4 5 2 6 7 8 然后确定四个角的数。四个角的数都用了三次,而上述8个算式中只有9、7、5、3被用了三次,所以9、7、5、3应填在四个角上。
但还应兼顾两条对角线上三个数的和都为18。
10 3 最后确定其它方格中的数。如图。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"6784eb975243b7beb9437ad028146570"},"img_del":0,"img_height":1892,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin8e66cbf3d9949dd85cc2a1f96cf41655__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️
以下另补充4阶幻方的二种制作方法
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"eac2709890f788e03e747e4e597a3e5f"},"img_del":0,"img_height":3264,"img_size":834,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin47953e66f37d5506810b270acddd6ecb__jpg.heic","img_width":1892,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"643b0480bb10843de573fe1e06457e73"},"img_del":0,"img_height":3264,"img_size":789,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin52cf5e743f50c7739ab977913edd8498__jpg.heic","img_width":1878,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"930716ea05c670a3c242b22173d4ba8e"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"12,构图布数问题
【含义】 这是一种数学游戏,也是现实生活中常用的数学问题。所谓“构图”,就是设计出一种图形;所谓“布数”,就是把一定的数字填入图中。“构图布数”问题的关键是要符合所给的条件。
【数量关系】 根据不同题目的要求而定。
【解题思路和方法】 通常多从三角形、正方形、圆形和五角星等图形方面考虑。按照题意来构图布数,符合题目所给的条件。
例1 :十棵树苗子,要栽五行子,每行四棵子,请你想法子。
解 符合题目要求的图形应是一个五角星。
4×5÷2=10
因为五角星的5条边交叉重复,应减去一半。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"1c8439afeba51fb9567b5e99e4c2bf5e"},"img_del":0,"img_height":768,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originc11c1710b7c46f2f0f2116cbcc773c75__jpg.heic","img_width":1000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2 九棵树苗子,要栽十行子,每行三棵子,请你想法子。
解 :符合题目要求的图形是两个倒立交叉的等腰三角形, 一个三角形的顶点在另一个三角形底边的中线上。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"450d3dbd3b85af06bfeb9b330e6613d8"},"img_del":0,"img_height":2718,"img_size":597,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin0ea2cf0824b3e36ec2850b39cacb50c8__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例3 九棵树苗子,要栽三行子,每行四棵子,请你想法子。
解 :符合题目要求的图形是一个三角形,每边栽4棵树,三个顶点上重复应减去,正好9棵。 4×3-3=9
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01归一问题 02归总问题
03和差问题 04和倍问题
05差倍问题 06相遇问题
07追及问题 08年龄问题
09牛吃草问题 10鸡兔同笼题
11幻方问题 12构图布数问题
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"f8bc456f89cf9070c743332d980f4d59"},"img_del":0,"img_height":420,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin689109a8bd29ee2d6a77822bfee73a00.gif","img_width":600,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"1, 归一问题
【含义】 在解题时,先求出一份是多少(即单一量),然后以单一量为标准,求出所要求的数量。这类应用题叫做归一问题。
【数量关系】 总量÷份数=1份数量
1份数量×所占份数=所求几份的数量
另一总量÷(总量÷份数)=所求份数
【解题思路和方法】 先求出单一量,以单一量为标准,求出所要求的数量。
例1 ,买5支铅笔要0.6元钱,买同样的铅笔16支,需要多少钱?
解
(1)买1支铅笔多少钱?
0.6÷5=0.12(元)
(2)买16支铅笔需要多少钱?
0.12×16=1.92(元)
列成综合算式
0.6÷5×16=0.12×16=1.92(元)
答:需要1.92元。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"8b4f7b6d958a593bd1ff09c4cace433b"},"img_del":0,"img_height":818,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin9d4e78c85002d2de81883a91a967e869__jpg.heic","img_width":624,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"2 , 归总问题
【含义】 解题时,常常先找出“总数量”,然后再根据其它条件算出所求的问题,叫归总问题。所谓“总数量”是指货物的总价、几小时(几天)的总工作量、几公亩地上的总产量、几小时行的总路程等。
【数量关系】
1份数量×份数=总量
总量÷1份数量=份数
总量÷另一份数=另一每份数量
【解题思路和方法】
先求出总数量,再根据题意得出所求的数量。
例1 服装厂原来做一套衣服用布3.2米,改进裁剪方法后,每套衣服用布2.8米。原来做791套衣服的布,现在可以做多少套?
解:
(1)这批布总共有多少米?
3.2×791=2531.2(米)
(2)现在可以做多少套?
2531.2÷2.8=904(套)
列成综合算式 :
3.2×791÷2.8=904(套)
答:现在可以做904套。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"d2d46de1079ad43aa5ef7737595062bf"},"img_del":0,"img_height":346,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin3e36bf910601bf214909f36b9388123d__jpg.heic","img_width":242,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"3 , 和差问题
【含义】 已知两个数量的和与差,求这两个数量各是多少,这类应用题叫和差问题。
【数量关系】
大数=(和+差)÷ 2
小数=(和-差)÷ 2
【解题思路和方法】 简单的题目可以直接套用公式;复杂的题目变通后再用公式。
例1 甲乙两班共有学生98人,甲班比乙班多6人,求两班各有多少人?
解:
甲班人数=(98+6)÷2=52(人)
乙班人数=(98-6)÷2=46(人)
答:甲班有52人,乙班有46人。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"923000ba1cb0626de1a4490ac11c5ded"},"img_del":0,"img_height":376,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originaaf74ce071a4a62d285f4fcbb7789de0__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"4,和倍问题
【含义】已知两个数的和及大数是小数的几倍(或小数是大数的几分之几),要求这两个数各是多少,这类应用题叫做和倍问题。
【数量关系】
总和÷(几倍+1)较小的数
总和-较小的数 = 较大的数
较小的数 ×几倍 = 较大的数
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 果园里有杏树和桃树共248棵,桃树的棵数是杏树的3倍,求杏树、桃树各多少棵?
解 :
(1)杏树有多少棵?
248÷(3+1)=62(棵)
(2)桃树有多少棵?
62×3=186(棵)
答:杏树有62棵,桃树有186棵。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"f5cb7dd056ee30943d5503fb971a42fd"},"img_del":0,"img_height":300,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originfcb2315ecbdf0cd2b72a5ad90b90f70d__jpg.heic","img_width":600,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"5 ,差倍问题
【含义】 已知两个数的差及大数是小数的几倍(或小数是大数的几分之几),要求这两个数各是多少,这类应用题叫做差倍问题。
【数量关系】
两个数的差÷(几倍-1)=较小的数
较小的数×几倍=较大的数
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 果园里桃树的棵数是杏树的3倍,而且桃树比杏树多124棵。求杏树、桃树各多少棵? 解 :
(1)杏树有多少棵?
124÷(3-1)=62(棵)
(2)桃树有多少棵?
62×3=186(棵)
答:果园里杏树是62棵,桃树是186棵。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"b20c4efce44d5d0375776b37d06b5f1c"},"img_del":0,"img_height":1070,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origincd1756cdd380435109e9a98afbc8e89d__jpg.heic","img_width":1080,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"6,相遇问题
【含义】两个运动的物体同时由两地出发相向而行,在途中相遇。这类应用题叫做相遇问题。
【数量关系】相遇时间=总路程÷(甲速+乙速)
总路程=(甲速+乙速)×相遇时间
【解题思路和方法】 简单的题目可直接利用公式,复杂的题目变通后再利用公式。
………………
例(由于该题比较典型故选此题): 甲乙二人同时从两地骑自行车相对而行,甲每小时行15千米,乙每小时行13千米,两人在距中点3千米处相遇,求两地的距离。
解 : “两人在距中点3千米处相遇”是正确理解本题题意的关键。从题中可知甲骑得快,乙骑得慢,甲过了中点3千米,乙距中点3千米,就是说甲比乙多走的路程是(3×2)千米,因此,相遇时间=(3×2)÷(15-13)=3(小时)
两地距离=(15+13)×3=84(千米)
答:两地距离是84千米。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"d42287306c55cb8eb13bbbb6ba3e4547"},"img_del":0,"img_height":818,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin9d4e78c85002d2de81883a91a967e869__jpg.heic","img_width":624,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"7, 追及问题
【含义】 两个运动物体在不同地点同时出发(或者在同一地点而不是同时出发,或者在不同地点又不是同时出发)作同向运动,在后面的,行进速度要快些,在前面的,行进速度较慢些,在一定时间之内,后面的追上前面的物体。这类应用题就叫做追及问题。
【数量关系】
追及时间=追及路程÷(快速-慢速)
追及路程=(快速-慢速)×追及时间
【解题思路和方法】 简单的题目直接利用公式,复杂的题目变通后利用公式。
例1 好马每天走120千米,劣马每天走75千米,劣马先走12天,好马几天能追上劣马?
解 :
(1)劣马先走12天能走多少千米?
75×12=900(千米)
(2)好马几天追上劣马?
900÷(120-75)=20(天)
列成综合算式
75×12÷(120-75)
=900÷45=20(天) 答:好马20天能追上劣马。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"5fc9ea4c93b07440930fa7a29a68496f"},"img_del":0,"img_height":898,"img_size":519,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin28f1be94949187103eda55e393910dc5__jpg.heic","img_width":1078,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"……………………………………
例2 小明和小亮在200米环形跑道上跑步,小明跑一圈用40秒,他们从同一地点同时出发,同向而跑。小明第一次追上小亮时跑了500米,求小亮的速度是每秒多少米。
解 :
小明第一次追上小亮时比小亮多跑一圈,即200米,此时小亮跑了(500-200)米,要知小亮的速度,须知追及时间,即小明跑500米所用的时间。又知小明跑200米用40秒,则跑500米用[40×(500÷200)]秒,
所以小亮的速度是
(500-200)÷[40×(500÷200)]
=300÷100=3米/秒
答:小亮的速度是每秒3米。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"01b1bc736816b300832ddd1888f84a13"},"img_del":0,"img_height":372,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originfc24f7f997ca5abce21841924df56143__jpg.heic","img_width":514,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"8, 年龄问题
【含义】这类问题是根据题目的内容而得名,它的主要特点是两人的年龄差不变,但是,两人年龄之间的倍数关系随着年龄的增长在发生变化。
【数量关系】年龄问题往往与和差、和倍、差倍问题有着密切联系,尤其与差倍问题的解题思路是一致的,要紧紧抓住“年龄差不变”这个特点。
【解题思路和方法】 可以利用“差倍问题”的解题思路和方法。
例1 爸爸今年35岁,亮亮今年5岁,今年爸爸的年龄是亮亮的几倍?明年呢?
解 35÷5=7(倍)
(35+1)÷(5+1)=6(倍) 答:今年爸爸的年龄是亮亮的7倍,明年爸爸的年龄是亮亮的6倍。
例2 母亲今年37岁,女儿今年7岁,几年后母亲的年龄是女儿的4倍?
解 (1)母亲比女儿的年龄大多少岁?
37-7=30(岁)
(2)几年后母亲的年龄是女儿的4倍?
30÷(4-1)-7=3(年)
列成综合算式
(37-7)÷(4-1)-7=3(年)
答:3年后母亲的年龄是女儿的4倍
例3 3年前父子的年龄和是49岁,今年父亲的年龄是儿子年龄的4倍,父子今年各多少岁?
解 今年父子的年龄和应该比3年前增加(3×2)岁,今年二人的年龄和为 :
49+3×2=55(岁) 把今年儿子年龄作为1倍量,则今年父子年龄和相当于(4+1)倍,因此,今年儿子年龄为
55÷(4+1)=11(岁)
今年父亲年龄为
11×4=44(岁)
答:今年父亲年龄是44岁,儿子年龄是11岁。
例4 甲对乙说:“当我的岁数曾经是你现在的岁数时,你才4岁”。
乙对甲说:“当我的岁数将来是你现在的岁数时,你将61岁”。求甲乙现在的岁数各是多少? 解 :这里涉及到三个年份:过去某一年、今年、将来某一年。列表分析:
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"7dddae1bee9196fc7d1a5e18ce46be18"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"这里涉及到三个年份:过去某一年、今年、将来某一年。列表分析:
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"dda846f53437a54bd3f088f730a0f268"},"img_del":0,"img_height":470,"img_size":58,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin81cea68393f06eca20807ecb99cbdd10__jpg.heic","img_width":1080,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"表中二个"□"表示同一个数,表中二个"△"表示同一个数。
因为两个人的年龄差总相等:
□-4=△-□=61-△,也就是4,□,△,61成等差数列,所以,61应该比4大3个年龄差,因此二人年龄差为
(61-4)÷3=19(岁)
甲今年的岁数为
△=61-19=42(岁)
乙今年的岁数为
□=42-19=23(岁)
答:甲今年的岁数是42岁,乙今年的岁数是23岁
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cf029721afd011c72013f47454768240"},"img_del":0,"img_height":346,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin2c467601e0e422fea3164d6f9eb9950c__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"9,“牛吃草”问题
【含义】 “牛吃草”问题是大科学家牛顿提出的问题,也叫“牛顿问题”。这类问题的特点在于要考虑草边吃边长这个因素。
【数量关系】 草总量=原有草量+草每天生长量×天数 【解题思路和方法】 解这类题的关键是求出草每天的生长量。
例1 一块草地,10头牛20天可以把草吃完,15头牛10天可以把草吃完。问多少头牛5天可以把草吃完?
解 草是均匀生长的,所以,草总量=原有草量+草每天生长量×天数。求“多少头牛5天可以把草吃完”,就是说5 天内的草总量要5 天吃完的话,得有多少头牛?
设每头牛每天吃草量为1,按以下步骤解答:
(1)求草每天的生长量
因为,一方面20天内的草总量就是10头牛20天所吃的草,即(1×10×20);另一方面,20天内的草总量又等于原有草量加上20天内的生长量,所以
1x10x20=原有草量÷20天内生长量
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"c4f7a65e18455d056d7bf7b6d8bd12c9"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"同理
1×15×10=原有草量+10天内生长量
由此可知 (20-10)天内草的生长量为 1×10×20-1×15×10=50
因此,草每天的生长量为
50÷(20-10)=5
(2)求原有草量
原有草量=10天内总草量-10内生长量=1×15×10-5×10=100
(3)求5 天内草总量
5 天内草总量=原有草量+5天内生长量=100+5×5=125
(4)求多少头牛5 天吃完草 因为每头牛每天吃草量为1,所以每头牛5天吃草量为5。因此5天吃完草需要牛的头数
125÷5=25(头)
答:需要5头牛5天可以把草吃完。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"ce0ed9cbef67a7cf3c3d08ea7a708f20"},"img_del":0,"img_height":500,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originb356fce34d77c711c3da5dd2d0001d8e__jpg.heic","img_width":750,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2 一只船有一个漏洞,水以均匀速度进入船内,发现漏洞时已经进了一些水。如果有12个人淘水,3小时可以淘完;如果只有5人淘水,要10小时才能淘完。求17人几小时可以淘完?
解 这是一道变相的“牛吃草”问题。与上题不同的是,最后一问给出了人数(相当于“牛数”),求时间。设每人每小时淘水量为1,按以下步骤计算:
(1)求每小时进水量
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cb9ffa1e0c00e501be5e64222518b865"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"因为,3小时内的总水量=1×12×3=原有水量+3小时进水量
10小时内的总水量=1×5×10=原有水量+10小时进水量
所以,(10-3)小时内的进水量为 1×5×10-1×12×3=14 因此,每小时的进水量为
14÷(10-3)=2
(2)求淘水前原有水量 原有水量=1×12×3-3小时进水量=36-2×3=30
(3)求17人几小时淘完
17人每小时淘水量为17,因为每小时漏进水为2,所以实际上船中每小时减少的水量为(17-2),所以17人淘完水的时间是 30÷(17-2)=2(小时)
答:17人2小时可以淘完水。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"3818c16f10e7d3226139cf31e1036cb2"},"img_del":0,"img_height":720,"img_size":99,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin5d4700985b45549d162f772196279eed__jpg.heic","img_width":960,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"10 ,鸡兔同笼问题
【含义】 这是古典的算术问题。已知笼子里鸡、兔共有多少只和多少只脚,求鸡、兔各有多少只的问题,叫做第一鸡兔同笼问题。已知鸡兔的总数和鸡脚与兔脚的差,求鸡、兔各是多少的问题叫做第二鸡兔同笼问题。 【数量关系】第一鸡兔同笼问题:
假设全都是鸡,则有 兔数=(实际脚数-2×鸡兔总数)÷(4-2)
假设全都是兔,则有 鸡数=(4×鸡兔总数-实际脚数)÷(4-2)
第二鸡兔同笼问题:
假设全都是鸡,则有
兔数=(2×鸡兔总数-鸡与兔脚之差)÷(4+2)
假设全都是兔,则有
鸡数=(4×鸡兔总数+鸡与兔脚之差)÷(4+2)
【解题思路和方法】 解答此类题目一般都用假设法,可以先假设都是鸡,也可以假设都是兔。如果先假设都是鸡,然后以兔换鸡;如果先假设都是兔,然后以鸡换兔。这类问题也叫置换问题。通过先假设,再置换,使问题得到解
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"e7924c5fd1b2e04367356ff93d7f8ea4"},"img_del":0,"img_height":376,"img_size":20,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin4dcb00c6389e90b8e1c3562c46813934__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例1 长毛兔子芦花鸡,鸡兔圈在一笼里。数数头有三十五,脚数共有九十四。请你仔细算一算,多少兔子多少鸡?
解 假设35只全为兔,
则 鸡数=(4×35-94)÷(4-2)=23(只)
兔数7=35-23=12(只)
也可以先假设35只全为鸡,则
兔数=(94-2×35)÷(4-2)=12(只)
鸡数=35-12=23(只)
答:有鸡23只,有兔12只。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"48b8887e966b2a5806a9bf5193bd8efa"},"img_del":0,"img_height":376,"img_size":31,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originac8eac4c0d9006da651ffd78a962510f__jpg.heic","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2(第二鸡兔同笼问题)鸡兔共有100只,鸡的脚比兔的脚多80只,问鸡与兔各多少只?
解 假设100只全都是鸡,则有
兔数=(2×100-80)÷(4+2)=20(只)
鸡数=100-20=80(只)
答:有鸡80只,有兔20只。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"93687ca362a467a6556fc10166aba772"},"img_del":0,"img_height":1114,"img_url":"https://ss-mpvolc.meipian.me/users/686154/1615788412387__jpg.heic","img_width":834,"is_origin":false,"location_del":0,"sectionPublished":true,"sell_del":0,"source":1,"splicing_template_id":201,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"497be9aab94b24684a8bbfae85d1532d"},"img_del":0,"img_height":484,"img_size":41,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originbe8784244681503c4daa7ade26df0fcc__jpg.heic","img_width":328,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例3, 有100个馍100个和尚吃,大和尚一人吃3个馍,小和尚3人吃1个馍,问大小和尚各多少人?
解 假设全为大和尚,则共吃馍(3×100)个,比实际多吃(3×100-100)个,这是因为把小和尚也算成了大和尚,因此我们在保证和尚总数100不变的情况下,以“小”换“大”,一个小和尚换掉一个大和尚可减少馍(3-1/3)个。因此,共有小和尚 (3×100-100)÷(3-1/3)=75(人)
共有大和尚 100-75=25(人)
答:共有大和尚25人,有小和尚75人。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"9c8840045e50ab16f41550100941b147"},"img_del":0,"img_height":500,"img_size":264,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin8dc38bc032767879abc140f472a889e9.gif","img_width":500,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"11, 幻方问题
【含义】 把n×n个自然数排在正方形的格子中,使各行、各列以及对角线上的各数之和都相等,这样的图叫做幻方。最简单的幻方是三级幻方。
【数量关系】 每行、每列、每条对角线上各数的和都相等,这个“和”叫做“幻和”。
三级幻方的幻和=45÷3=15
五级幻方的幻和=325÷5=65
【解题思路和方法】首先要确定每行、每列以及每条对角线上各数的和(即幻和),其次是确定正中间方格的数,然后再确定其它方格中的数。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"cc74861993706968831495416e8a661a"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例1
把1,2,3,4,5,6,7,8,9这九个数填入九个方格中,使每行、每列、每条对角线上三个数的和相等。
解
幻和的3倍正好等于这九个数的和,所以幻和为
(1+2+3+4+5+6+7+8+9)÷3=45÷3=15
九个数在这八条线上反复出现构成幻和时,每个数用到的次数不全相同,最中心的那个数要用到四次(即出现在中行、中列、和两条对角线这四条线上),四角的四个数各用到三次,其余的四个数各用到两次。看来,用到四次的“中心数”地位重要,宜优先考虑。
设“中心数”为Χ,因为Χ出现在四条线上,而每条线上三个数之和等于15,所以
(1+2+3+4+5+6+7+8+9)+(4-1)Χ=15×4
即 45+3Χ=60 所以 Χ=5 接着用奇偶分析法寻找其余四个偶数的位置,它们分别在四个角,再确定其余四个奇数的位置,它们分别在中行、中列,进一步尝试,容易得到正确的结果。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"fe93f4fc4296b0a22fd44ea8d821fa70"},"img_del":0,"img_height":1892,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin3aac8227b941d618bf33979c188a2b38__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2
把2,3,4,5,6,7,8,9,10这九个数填到九个方格中, 使每行、每列、以及对角线上的各数之和都相等。
解
只有三行,三行用完了所给的个数,所以每行三数之和为
(2+3+4+5+6+7+8+9+10)÷3=18
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"edcaea1fd3d35d0072a001611072070c"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"假设符合要求的数都已经填好,那么三行、三列、两条对角线共8行上的三个数之和都等于18,我们看18能写成哪三个数之和:
最大数是10:
18=10+6+2=10+5+3
最大数是9:
18=9+7+2=9+6+3=9+5+4
最大数是8:
18=8+7+3=8+6+4 最大数是7: 18=7+6+5
刚好写成8个算式。 首先确定正中间方格的数。第二横行、第二竖行、两个斜行都用到正中间方格的数,共用了四次。观察上述8个算式,只有6被用了4次,所以正中间方格中应填6。
9 4 5 2 6 7 8 然后确定四个角的数。四个角的数都用了三次,而上述8个算式中只有9、7、5、3被用了三次,所以9、7、5、3应填在四个角上。
但还应兼顾两条对角线上三个数的和都为18。
10 3 最后确定其它方格中的数。如图。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"6784eb975243b7beb9437ad028146570"},"img_del":0,"img_height":1892,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin8e66cbf3d9949dd85cc2a1f96cf41655__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️⬇️
以下另补充4阶幻方的二种制作方法
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"eac2709890f788e03e747e4e597a3e5f"},"img_del":0,"img_height":3264,"img_size":834,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin47953e66f37d5506810b270acddd6ecb__jpg.heic","img_width":1892,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"643b0480bb10843de573fe1e06457e73"},"img_del":0,"img_height":3264,"img_size":789,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin52cf5e743f50c7739ab977913edd8498__jpg.heic","img_width":1878,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"930716ea05c670a3c242b22173d4ba8e"},"img_del":0,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"12,构图布数问题
【含义】 这是一种数学游戏,也是现实生活中常用的数学问题。所谓“构图”,就是设计出一种图形;所谓“布数”,就是把一定的数字填入图中。“构图布数”问题的关键是要符合所给的条件。
【数量关系】 根据不同题目的要求而定。
【解题思路和方法】 通常多从三角形、正方形、圆形和五角星等图形方面考虑。按照题意来构图布数,符合题目所给的条件。
例1 :十棵树苗子,要栽五行子,每行四棵子,请你想法子。
解 符合题目要求的图形应是一个五角星。
4×5÷2=10
因为五角星的5条边交叉重复,应减去一半。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"1c8439afeba51fb9567b5e99e4c2bf5e"},"img_del":0,"img_height":768,"img_url":"https://ss-mpvolc.meipian.me/users/686154/originc11c1710b7c46f2f0f2116cbcc773c75__jpg.heic","img_width":1000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例2 九棵树苗子,要栽十行子,每行三棵子,请你想法子。
解 :符合题目要求的图形是两个倒立交叉的等腰三角形, 一个三角形的顶点在另一个三角形底边的中线上。
","text_del":0,"type":1,"video_del":0,"video_thumbnail_del":0,"vote_del":0},{"audio_del":0,"audio_music_id":0,"ext":{"unique_id":"450d3dbd3b85af06bfeb9b330e6613d8"},"img_del":0,"img_height":2718,"img_size":597,"img_url":"https://ss-mpvolc.meipian.me/users/686154/origin0ea2cf0824b3e36ec2850b39cacb50c8__jpg.heic","img_width":2000,"is_origin":true,"location_del":0,"sectionPublished":true,"sell_del":0,"source":0,"text":"例3 九棵树苗子,要栽三行子,每行四棵子,请你想法子。
解 :符合题目要求的图形是一个三角形,每边栽4棵树,三个顶点上重复应减去,正好9棵。 4×3-3=9
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