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2019年六年级数学广角专题-4.和倍问题

   日期:2024-05-16     来源:www.danmuer.com    浏览:365    
文章简介:和倍问题   夯实基础 1. 一部手机和一部电话共2240元,手机的单价是电话机的7倍,手机和电话机的单价各是多少元? 【答案】手机和电话机的单价各是1960元、280元 【分析】 考试试题剖析:依据题意,可得一部手机和一部电话共2240元...

和倍问题

 

夯实基础

1. 一部手机和一部电话共2240元,手机的单价是电话机的7倍,手机和电话机的单价各是多少元?

【答案】手机和电话机的单价各是1960元、280元

【分析】

考试试题剖析:依据题意,可得一部手机和一部电话共2240元,手机的单价是电话机的7倍,由和倍公式进一步解答。

解:电话机:2240÷(7+1)=280(元)

手机:280×7=1960(元)

答:手机和电话机的单价各是1960元、280元。

 

 

2. 李大伯家的猪场里有母猪和小猪共84头,其中小猪的头数是母猪的3倍。母猪和小猪各有多少头?

【答案】母猪有21头,小猪有63头

【分析】

考试试题剖析:由于小猪的头数是母猪的3倍,所以母猪和小猪共84头,是母猪的4倍,用除法即可得母猪的头数,再求小猪的头数即可。

解:84÷(3+1)

=84÷4

=21(头)

21×3=63(头)

答:母猪有21头,小猪有63头。

 

 

3. 据信息产业部统计,到现在为止,国内电话用户达3.6亿户,其中移动电话用户是固定电话用户的2倍.求国内移动电话用户和固定电话用户各是多少亿户?

【答案】国内移动电话用户和固定电话用户各是2.4亿户、1.2亿户

【分析】

考试试题剖析:依据题意国内电话用户达3.6亿户,其中移动电话用户是固定电话用户的2倍,可知移动电话用户与固定电话用户的和就是3.6亿,依据和倍公式,和÷(倍数+1)=较小数,就能求出结果。[来源:Z|xx|k.Com]

解:依据题意,由和倍公式可得,

固定电话是:3.6÷(2+1)=1.2(亿户),

移动电话是:1.2×2=2.4(亿户)。

答:国内移动电话用户和固定电话用户各是2.4亿户、1.2亿户。

 

 

4. 育才小学有教师108人,其中女教师人数是男教师的3倍。男教师有多少人?

【答案】男教师有27人

【分析】

考试试题剖析:依据题意了解女教师和男教师的人数的和是108,女教师人数是男教师的3倍,由此借助和倍公式解决问题。

解:男教师的人数:108÷(3+1),

=108÷4,

=27(人),

答:男教师有27人。

 

 

5. 水果店运来苹果和梨共840千克,苹果的水平是梨的3倍,苹果和梨各重多少千克?

【答案】苹果和梨各重630千克、210千克

【分析】

考试试题剖析:依据水果店运来苹果的重量是梨的3倍,把运来梨的重量看作1倍,则运来苹果的重量就是3倍,可知运来苹果的重量和梨的重量共有3+1=4倍,正好运来苹果和梨共840千克,用除法求出梨的重量,再用梨的重量乘以3就是苹果的重量。

解:梨的重量:840÷(3+1)=210(千克),

苹果的重量:210×3=630(千克),

答:苹果和梨各重630千克、210千克。

 

 

6. 一所学校共有810人,其它年级的学生是小学六年级的5倍,小学六年级学生多少人?其它年级一共多少人?

【答案】小学六年级学生135人,其它年级一共675人

【分析】

考试试题剖析:其它年级的学生是小学六年级的5倍,那样学校共有810人,就等于1+5=6个小学六年级人数的和,依据除法意义,求出小学六年级人数,再依据乘法意义即可求出其它年级的人数。

解:810÷(1+5)

=810÷6

=135(人)

其它年级:135×5=675(人)

答:小学六年级学生135人,其它年级一共675人。

 

 

7. 父亲花180元钱给我买了一套服饰,上衣的价钱是裤子的2倍,上衣和裤子各花了多少钱?

【答案】上衣和裤子各花了120元、60元

【分析】

考试试题剖析:由上衣的价钱是裤子的2倍,把裤子的价格看作1倍,上衣的价钱就是2倍,一共是3倍正好花了180元,用除法求出1倍的,也就是裤子的价格,再用裤子的价格乘2就是上衣的价格。

解:裤子的价格:180÷(1+2)=60(元),

上衣的价钱:60×2=120(元),

答:上衣和裤子各花了120元、60元。

 

 

8. 图书馆有文静书和故事书共960本,其中文静书的本数是故事书的3倍,买来故事书多少本?

【答案】买来故事书240本

【分析】

考试试题剖析:因为文静书的本数是故事书的3倍,把故事书的本数看作单位“1”,则中文静书的本数等于故事书的3倍。如此,两种书的本数和就是故事书的4倍,那样故事书有960÷4本,解答即可。

解:960÷(1+3),

=960÷4,

=240(本);

答:买来故事书240本。

 

 

9. 水果店运来苹果和梨共310千克,运来的苹果是梨的61倍,运来苹果多少千克?

【答案】305千克

【分析】

考试试题剖析:依据水果店运来苹果的重量是梨的61倍,把运来梨的重量看作1倍,则运来苹果的重量就是61倍,可知运来苹果的重量和梨的重量共有61+1=62倍,正好运来苹果和梨共310千克,用除法求出梨的重量,再用梨的重量乘以61就是苹果的重量。

解:梨的重量:

310÷(61+1)

=310÷62

=5(千克)

苹果的重量:

5×61=305(千克)

答:运来苹果305千克。

[来源:学科网]

 

10. 在地震灾害捐款中,参加捐款的成每人数是儿童的3倍,假如在华诚超市一共有652人参加捐款,儿童有多少人?

【答案】儿童有163人

【分析】

解:652÷(3+1)=163(人)

答:儿童有163人。

 

 

11. 甲除以乙的商是10,甲乙的和是77,甲、乙各是多少?

【答案】70,7

【分析】

考试试题剖析:依据题意,甲除以乙的商是10,也就是甲是乙的10倍,又知甲乙的和是77,依据和倍公式进一步解答即可。

解:依据题意可得:甲是乙的10倍;

由和倍公式可得:

乙是:77÷(10+1)=7;

甲是:7×10=70。

答:甲是70,乙是7。

故答案为:70,7。[来源:学*科*网]

 

 

12. 小明和小强共有画片200张,小明的张数比小强的张数的2倍还多20张,则小强有多少张画片?

【答案】60

【分析】

设小强的画片数为1份,

小强有些画片数=÷3=60

[来源:Zxxk.Com]

 

拓展提升

1. 大山羊的年龄比小山羊的年龄多2倍。它们的年龄和是8岁,大山羊、小山羊各多大岁数?

【答案】6;2

【分析】

解:8÷

=8÷4

=2

8-2=6

答:大山羊6岁,大山羊2岁。

 

 

2. 在一个减法算式里,被减数、减数与差的和等于120,而减数是差的3倍,那样差等于多少?

【答案】15

【分析】

解:被减数-减数=差,则被减数=差+减数,于是差与减数的和为120÷2=60,而减数是差的3倍,则视差为“1”,那样减数为“3”,和为“4”。

于是差为60÷=15。

 

 

3. 一个长方形操场,周长是78米,已知长是宽的2倍,这个操场长、宽分别是多少分米?

【答案】260,130

【分析】

考试试题剖析:依据长方形的周长公式,先求出一条长与一条宽的和是:780÷2=390分米,由于“长是宽的2倍”把长与宽的和平均分成3份,则宽就是其中1份,由此即可求出宽,从而求出长。

解:78米=780分米,

780÷2÷3=130(分米),

130×2=260(分米),

答:长是260分米,宽是130分米。

故答案为:260,130。

 

 

4. 两个数的和是616,其中一个加数的个位上是0,若把0去掉,就与另一个加数相同,这两个数分别是多少?

【答案】560和56

【分析】

考试试题剖析:由“一个加数的个位是0,若把0去掉,就与另一个加数相同”可知,其中一个加数是另一个加数的10倍。那样它们的和就是另一个加数的(10+1)倍。即另一个加数的11倍是616,求另一个加数列式为616÷(10+1)。再依据另一个加数求出其中的一个加数。

解:616÷(10+1)

=616÷11

=56

56×10=560

答:这两个数分别是560和56。

 

 

5. 学校买来2张桌子和5把椅子,共用去110元,已知每张桌子价钱是椅子的3倍,问:每张桌子多少钱?

【答案】每张桌子30元

【分析】

解:1张桌子=3把椅子,所以2张桌子=6把椅子;

所以2张桌子+5把椅子=6把椅子+5把椅子=11把椅子=110元;

所以椅子单价为110÷11=10(元);

所以桌子单价为10×3=30(元)。

 

 

6. 三块布共长220米,第二块布长是第一块的3倍,第三块布长是第二块的2倍,第一块布长多少米?

【答案】22

【分析】

解:设第一块布长为1份,

第一块布长=220÷=22

 

 

7. 果园里有桃树和梨树共340棵,梨树的棵数比桃树的3倍还多20棵,果园里有桃树、梨树各多少棵?

【答案】桃树有80棵,梨树有260棵

【分析】

解:设桃树的棵数为1份,由题意得:

(340-20)÷(3+1)=80(棵)

3×80+20=260(棵);

答:桃树有80棵,梨树有260棵。

 

 

8. 一张桌子、一张椅子和一个熨斗共540元。已知一张椅子的价格比一个熨斗多60元,桌子单价是椅子的2倍。请问一张椅子多少元?

【答案】一张椅子150元

【分析】

考试试题剖析:桌子单价是椅子的2倍,也就是说一张桌子的价格等于2把椅子的价格,一张椅子的价格比一个熨斗多60元,也就是说一个熨斗加60元就等于一张椅子的价格,据此可得:540元钱加上60元,就等于2+1+1=4把椅子的价格,依据除法意义即可解答。

解:(540+60)÷(2+1+1)

=600÷4

=150(元)

答:一张椅子150元。

 

 

9. 某校共有学生560人,其中男孩比女孩的3倍少40人。则男、女孩各有多少人?

【答案】410 150

【分析】

解:把女孩人数看作1份,其中男孩人数不够女孩人数的3倍,假如把男孩人数的和560人加上40人就等于女孩人数的4倍。

所以女孩人数=÷=150

男孩人数=150×3-40=410

 

 

10. 甲、乙、丙三个数的和是120,其中甲、乙两个数的和是丙的3倍,甲比乙多10。三个数各是多少?

【答案】甲是50、乙是40、丙是30

【分析】

考试试题剖析:算出丙是解题的重点,由题意“甲、乙、丙三个数的和是120,甲、乙丙个数的和是丙的3倍”能算出丙:再依据丙算出甲乙丙数的和:又因“甲比乙多10”依据和差关系算出甲乙。

解:丙:120÷(3+1)=30   30×3=90

甲:(90+10)÷2=50

乙:(90-10)÷2=40

答:三个数是,甲是50,乙是40,丙是30。

 

 

11. 两个数的和是194.5,较大数除以较小数商是4,余数是12,这两个数分别是__________

多少?

【答案】158.1,36.4

【分析】

考试试题剖析:依据大数除以小数,商4余数是12,所以大数减去12后是小数的4倍,则和194.5减去12就是小数的(4+1)倍,因此,依据除法的意义,小数可求得,然后进一步可以求出大数。

解:(194.5﹣12)÷(4+1)

=182.5÷5

=36.4

194.5﹣36.4=158.1

故答案为:158.1,36.4。

 

 

12. 年前父亲的年龄是儿子的5倍,年后父子二人年龄和是98岁,父子二人今年分别多大了?

【答案】爸爸49岁,儿子21岁

【分析】

解:年后父子二人年龄和是岁,那样年前父子二人年龄和是,所以年前儿子的年龄是:,父亲的年龄是:,今年父亲的年龄是:,儿子年龄是

 

 

巅峰突破

1. 甲、乙、丙三数之和是100,甲数除以乙数,丙数除以甲数,商都是5,余数都是1,乙数是多少?

【答案】3

【分析】

解:把乙数看作1份,那样甲数是5份加1;丙数是5×再加1,即25份加6。所以每份是:

 ÷=93÷31=3

即乙数是3。

 

 

2. 三堆苹果共有130个,第二堆的苹果数是第一堆的3倍,第三堆的苹果数是第二堆的2倍多10个,问三堆苹果各有多少个?

【答案】12;36;82

【分析】

考试试题剖析:

 

由于第二堆是第一堆的3倍,第三堆又是第二堆的2倍多10个,所以减去10个后,第三堆就等于第一堆的3×2=6。总数变为130-10=120,等于第一堆的1+3+6=10,可以求出第一堆的个数,依据有关条件再求第二堆和第三堆的个数。

解:130-10=120

1+3+3×2=10

120÷10=12

12×3=36

36×2+10=82

答:第一堆有12个,第二堆有36个,第三堆有82个。

 

 

3. 少先队1、2、三中队共植树200棵,二中队植树的棵数是一中队的2倍多5棵,三中队植树的棵数比1、二中队之和多4棵,三个中队各植树多少棵?

【答案】31;67;102

【分析】

考试试题剖析:

 

二中队比一中队的2倍多5棵,假如减去5就正好是一中队的2倍,三中队比1、二中队的和多4棵,如减去4就是1、二中队的和,由于二中队比一中队的2倍多5棵,所以还要减去一个5才符合倍数关系。如此,总数就变为200-5-4-5=186,等于一中队的1+2+1+2=6,如此就能求出一中队植树的棵数,相应也就能求出2、三中队植树的棵树了。

解:200-5-4-5=186

1+2+1+2=6

186÷6=31

31×2+5=67

31+67+4=102

答:一中队植树31棵,二中队植树67棵,三中队植树102棵。

 

 

4. 甲、乙两个粮仓存粮320吨,后来从甲仓运出40吨,给乙仓运进20吨,这个时候甲仓存粮是乙仓的2倍,两个粮仓原来各存粮分别为多少吨?

【答案】240,80

【分析】

考试试题剖析:

 

如图,甲、乙两仓原来共存粮320吨,“后来从甲仓运出40吨,给乙仓运进20吨,”甲、乙两仓目前共存粮=300吨,这个时候甲仓存粮是乙仓的2倍,可以先求出在乙仓存粮多少吨,然后再减去运进的20吨就是乙仓原来存粮的吨数。如此甲仓原吨数就好求了。

解:现乙仓存粮=÷=100

乙仓原存粮=100-20=80

甲仓原存粮=320-80=240

 

 

5. 有两层书架,共有书173本。从第一层拿走38本书后,第二层的书是第一层的2倍还多6本,则第二层有多少本书?

【答案】92

【分析】

解:

 

设把第一层剩下的书算作“1”份,由图可知:

每一份=÷3=43

第二层的书共有:43×2+6=92

 

 

6. 甲乙两书架共有118本书,如果从甲书架上拿20本到乙书架上,乙书架上的书就比甲书架上的书的2倍还多10本,两书架原来各有多少本书?

【答案】甲书架原来有56本,乙书架原来有62本

【分析】

考试试题剖析:由题意,若从总数118里减去10本,剩下的本数就是甲书架上的书的(1+2)倍,由此用除法可求得后来甲书架上的书有多少本,再加上20本就是原来的本数,进而求得乙书架原来的本数,解决问题。

解:甲书架:(118﹣10)÷(1+2)+20

=108÷3+20

=36+20

=56(本)

乙书架:118﹣56=62(本)

答:甲书架原来有56本,乙书架原来有62本。

 

 

7. 哥哥5年后的年龄与弟弟3年前的年龄和是29岁,弟弟目前的年龄是两人年龄差的4倍。哥哥今年多大?

【答案】15岁

【分析】

解:兄弟二人目前的年龄和是27岁,两人的年龄差是(岁),哥哥目前(岁)。

 

 

8. 甲、乙、丙三人,甲的年龄是乙的2倍还大3岁,乙的年龄是丙的2倍小2岁,三个人的年龄之和是109岁,分别求出三人的年龄。

【答案】63;30;16

【分析】

考试试题剖析:大家都以丙为1倍量来剖析。乙比丙的2倍小2岁,假如加上2就正好是丙的2倍,甲要想和丙联系起来,需要由乙来搭桥。假如甲去掉大出3岁就正好是乙的2倍,但乙比丙的2倍小2,所以甲要加上两个2才能是丙的2×2=4。所以总数变为109-3+2+2×2=112,等于丙的1+2+2×2=7可以先求出丙的年龄,再相应求出乙和甲的年龄。

 

解:109+2-3+2×2=112

1+2+2×2=7

112÷7=16

16×2-2=30

30×2+3=63

答:甲63岁,乙30岁,丙16岁。

 

 

9. 甲、乙、丙、丁四个人一共做了370个零件,假如把甲做的个数加2,乙做的个数减3,丙做的个数乘2,丁做的个数除以2,四个人做的零件个数正好相等,问四个人各做多少个零件?[来源:学base64,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人工智能RAxEAPwD+/iiiigAooooAKKKKACiiigAooooAKKKKACiivyB/4K2fDj/hbeqf8E1/h7/woL9n/wDag/4SD9v/AMQf8WL/AGpNZ/4R/wCBPjj+yv8AgnH/AMFC9b/4rnV/+FK/tEfZP+EZ/s3/人工智能TDwz/xZ3xh9o8Y+H/D1r/xT/nf8JTogB+v1FfxB/HP9ir+yv2bf+C9Wt/8OiP+CQHhn/hUn/C4v+Kw8M/Ev7R4x/Zf/s3/人工智能JPfsp/EL/jFy1/4dg+H/8AhIP+Ef8A+Eg/4Xp4e/4qn9nfzvjt448c6R/xJPsn/C4vGH9vlABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRRRQAUUUUAFFFFABRXn/h34peBPFvjv4i/Dbwzrv9t+LfhJ/wiMHxHttP0zWLjR/CWseONHn8TeH/AAjqfixdP/4RL/hYH/CJf2N42134dW2uXHjvwl4E8cfC3xz4s8O6J4S+Lfwx1nxb8AaZ/wAFhP2Itb0f4W+ItGl/a/1fw/8AHH+z/wDhSmu6Z/wTP/4KV3+j/GD+1/Amu/FLSv8AhVup2v7JEtl8QP7T+GXhfxN8RdP/AOETn1f7b4E8O674utvN8P6RqGoW4B+n9FeAfs9/tO/CL9qDR/Hms/CS7+IH/Fr/人工智能gN8LfiF4f+KXwT+Nn7P/jvwf47/wCEE8CfE6HQtd+HHx++Hvwy+IFl9t+H/wATfAfizTNTk8M/2Pquj+JtPudN1C7/ANIWH3+gAorz/wAf/FLwJ8Lv+EKk8f67/w人工智能1ZfED4geHPhb4c1m90zWJfDi+O/Gf2228FaF4k8TWen3Ph/wT/wm3iC2s/AXgrU/GupeH9H8X/FPxR4A+EnhvUNT+JvxI8AeE/EnoFABRRRQAUV+fS/Hb9qr9oW913Uv2P8Awt8BfDnwa0HxR4o8FWHx3/aB1Hx74ln+I/iPwRr+o+EfGVz8Ofgt8PF8K3svgPQPF+i654as/HHi74r+GLvxdqGjX+qeF/B974HuPDnjXxLN/w人工智能5/wVN/6LJ+wB/wCI1ftFf/RYUAff9eAftCfsq/s4/tZaP4D8M/tM/Bf4f/Hfwl8NfiAvxS8J+CfiloFr4x8CQ+O4/Anjv4bW2u674G1kXXhLxj9g8JfErxhbaZpnjHRtf0fS9Yv9P8Wadp9r4t8OeGdc0f5//wCEc/4Km/8ARZP2AP8AxGr9or/6LCj/人工智能Rr/gqa/wAn/C6v2ALbf8v2j/hmD9oq98jdx532P/hruw+1+Vnf9m+32Xn7fK+122/zkAPjbwZ/wRy/4JgR/FL9sj4S/F/9jz/gn9rP/DS3ja7+KXwQ+GvhX4PfDLw58Vvhv+zPp/7L37LP7PXj0eEp9B8GeEvHHweW1/aA0H4l+M7HxD8HNes7fQPEnxR0jxzpXi3R/iZ4l1qDSv3Ir84H/YL8Xaj4kj/aG179o/Wpv27LLTxoeh/tI6D8OdG0XwF4Y+H5ee6vP2f9E/Z+vvEOv2U37N3iDWLh/E3ivwZ4k+I/iP4q694tt9C8Xv8AHG18U+BvhvrPgvoP+EX/AOCpsP7n/heH7AGo+V+7/tD/人工智能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人工智能J7fDvwbZeEo4jY6THf+B/h3qPiXwLpWg+JtK0/x1d64niXxG/4Y/tGfDD9pn/gmD8BvjD/AMFOv2avGfiDR9Q8C/ALxBD4M+NXhr4AfEjR/wBtL40+N/h9ajwq/wAK/wBnn4E/E/4maD8S73x34v1yS/8Ahz8Jvil8O/Bup6X4wbw74xtfhv4Y1LwTYeIPC37lkBgVYBlYEMp人工智能IIwQQeCCOCDwRXgPxC/ZP8A2Wfi58NfCHwZ+K37NPwA+J3wf+HzaQ3gL4UfEL4N/Drxp8NfBDeH9IuPD+gt4Q8C+JPDmpeF/DTaJoN3daJpB0bS7I6bpFzcabZGGymkhbKUJOljKd4TliZ5Z7CdSnTk8BHDVcS8XXw0Z0qtOeLlTrxqYP6zSq4VYynQqY2nVoYLCUzeNSCjh0lOnKjWxlWrKnJr61CtllbC0cNWmp060MO8TOnLFRw9SnOWE9v9XcMbiJVV8Z/C74J/Dn9nr/glJ8QvhZ+25Y2/h34AaV8Av2m/EH7QXw8sNS8Waxo/wY/Zj8eJ8TvHnin4B+H9U+GUs3jG+8O/AX4IeJJfg/oNx8MHGqPpXgu0Hw5igA0KCO5d/Fj9sn4Ufssf8Eyv+FpfEv8AZA+Gv7TXxK+IH7GPwn/bQ/4a88ZWPg7/人工智能T7xH4x+HTf8NF/DT9mD/hTOpad8P8AxV+1/wCKviBp2of8KV8G6F9v+FmufYPEv9mabdaLa6Xs+tND/ZF/Zi8J+Cvh78MvA/wK+Gnw6+Fvws+JemfGHwV8KPhj4X0/4Z/Cix+JOhXWoav4f8Vat8MPAUfh3wF4su/Dfi2+t/iR4atfF3h7XdN8O/F3w34E+MWiWVh8Uvh14C8X+G/S/H/wn+FnxX/4Qr/haXw0+H/xK/4Vr8QPDnxY+HX/AAn/人工智能N8OeMf+EB+Kfg77b/wiPxL8Ff8JFpuo/8ACK/EDwr/AGjqH/COeMtC+weI9D+33v8AZmpWv2qff14iqq1aVRc9pRoxXtJzqVFGlhsHhoqdWpUq1a81DB0ozxNerUxOKmqmLxU5YvGYuT4qFOVKkozcHNzr1JunCFKnKpXxOLxVSUKNOFOlh4Tq4utOGGoU4YbCwnDCYWKwuFw/N6BRRRWJsfjz+yL+3v8AHn4h/sq/s2/EbSf+CSH7W+nWPxI+BPwo+I6Wvwd8V/8ABNvw38KBN4/8DaH4wvLj4caD47/4KB+B/HWk+Cby81q4vPDVl448GeE/GcOk3Fqvivw3oniD+0dOtvof/hsj9or/AKRO/t//APhxv+CWX/0yyvv+igD4A/4bI/aK/wCkTv7f/wD4cb/gll/9Mso/4bI/aK/6RO/t/wD/人工智能cb/gll/wDTLK+/6KAPgD/h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/> 

上图可以看出丙做得最少,因为丙做的个数乘以2和丁做的个数除以2相等,也就是丙做的2倍和丁的一半相等,即丁做的个数是丙的4倍。甲加上2后是丙的2倍,乙减去3后是丙的2倍,依据如此的倍数关系可以先求出丙做的个数,再分别求出甲、乙、丁做的个数。

370+2-3=369

2+2+1+4=9

369÷9=41

41×2-2=80

41×2+3=85

41×4=164

答:甲做80个,乙做85个,丙做41个,丁做164个。

 

 

10. 有货物108件,分成四堆存放在仓库时,第一堆件数的2倍等于第二堆件数的一半,比第三堆的件数少2,比第四堆的件数多2。问每堆各存放多少件?

【答案】12;48;26;22

【分析】

解:第一堆的件数的4倍等于第二堆件数,第三堆的件数比第一堆件数的2倍还多2,第四堆的件数比第一堆的件数的2倍少2。

第一堆件数+4个第一堆件数++=108

所以 9个第一堆件数=108,所以第一堆的件数为108÷9=12件。

则第二堆件数为12×4=48,第三堆件数为12×2+2=26件,第四堆件数为12×2-2=22件。

 

 

11. 小明、小红、小玲共有73块糖.假如小玲吃掉3块,那样小红与小玲的糖就一样多;假如小红给小明2块糖,那样小明的糖就是小红的糖的2倍。问小红有多少块糖?

【答案】19

【分析】

解:小玲比小红多3块糖,小明糖数再增加2就等于小红糖数降低2后2倍,所以小明的糖数是小红的2倍少6颗,

有小红+小玲+小明=小红++ =4小红-3=73。

所以小红有糖÷4=19块。

 

 

12. 今年哥两个的岁数加起来是55岁,过去有一年,哥哥的岁数与今年弟弟的岁数相同,那时哥哥的岁数恰好是弟弟的2倍,哥哥今年多大?

【答案】33

【分析】

解:设那时弟弟的岁数是1份。哥哥的岁数是2份,那样哥哥与弟弟的岁数之差为1份。二人的岁数之差是不会变的,今年他们的年龄仍差1份。

而题目中说:“那时哥哥的岁数与今年弟弟的岁数相同”。因此今年弟弟的岁数也是2份,而哥哥今年的岁数是2+1=3。

今年,哥哥与弟弟的年龄之和是:3+2=5

每份是:55÷5=11

所以今年哥哥是:11×3=33

 

 

 

 

 
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