七年级不等式应用题专项练习
一元一次不等式(组)之
笔记本5元 ·············································································· (2)设买a支钢笔,则买笔记本(48-a)本
依题意得:
应用题 姓名
1.(2009年凉山州)我国沪深股市交易中,如果买、卖一次股票均需付交易金额的0.5%
作费用.张先生以每股5元的价格买入―西昌电力‖股票1000股,若他期望获利不低于1000元,问他至少要等到该股票涨到每股多少元时才能卖出?(精确到0.01元)
3a5(48a)200
··························································
48aa
解得:20a24 ···································
所以,一共有5种方案.
································································································
即购买钢笔、笔记本的数
解:设至少涨到每股x元时才能卖出. ·············量分别为:·················· ·········································· 1分 根
1
据
x0
0
题
x0
意
≥
得
(
20,28; 21,27; 22,
26; 23,25; 24,24. ·························································5
···································································································································· 4分 ·解这个不等式得x≥
1205199
,即x≥6.06.3.(2009年株洲市)初中毕业了,孔明同学
准备利用暑假卖报纸赚取140~200元钱,····································································································································· 6分 答:至少涨到每股6.06元时才能卖出. ········································································ 7分 买一份礼物送给父母.已知:在暑假期间,
2.(2009年益阳市)开学初,小芳和小亮去学校商店购买学习用品,小芳用18元钱买了1支钢笔和3本笔记本;小亮用31元买了同样的钢笔2支和笔记本5本.
(1)求每支钢笔和每本笔记本的价格; (2)校运会后,班主任拿出200元学校奖
励基金交给班长,购买上述价格的钢笔和笔记本共48件作为奖品,奖给校
运会中表现突出的同学,要求笔记本数不少于钢笔数,共有多少种购买方案?请你一一写出.
如果卖出的报纸不超过1000份,则每卖出一份报纸可得0.1元;如果卖出的报纸超过1000份,则超过部分每份可得0.2元. ....(1)请说明:孔明同学要达到目的,卖出报纸的份数必须超过1000份.
(2)孔明同学要通过卖报纸赚取140~200元,请计算他卖出报纸的份数在哪个范围内.
解:(1)如果孔明同学卖出1000份报纸,
解:(1)设每支钢笔x元,每本笔记本y
则可获得:10000.1100元,没有超过
元 ········································································································································ 1分
140元,从而不能达到目的.(注:其它说理
依题意得:
x3y182x5y31
正确、合理即可.)
(2)设孔明同学暑假期间卖出报纸x份,
············································································································································ 3分
由(1)可知x1000,依题意得:
x3
解得: ··························································································4分 10000.10.2(x1000)140
y5
10000.10.2(x1000)200
答:每支钢笔3元,每本
x解得 1200
1
15 0
答:孔明同学暑假期间卖出报纸的份数在1200~1500份之间.
4.(2009年桂林市、百色市)(本题满分8分)在保护地球爱护家园活动中,校团委把一批树苗分给初三(1)班同学去栽种.如果每人分2棵,还剩42棵;如果前面每人分3棵,那么最后一人得到的树苗少于5棵(但至少分得一棵). (1)设初三(1)班有x名同学,则这批树苗有多少棵?(用含x的代数式表示). (2) 初三(1)班至少有多少名同学?最多有多少名?
15x2020x365
…………………
18x3020x492
………………………3分
解得:7≤
x ≤
9 ………………………………………………………………4分
∵ x为整数 ∴ x = 7,8 ,9 ,∴满足条件的方案有三种.. ……………5分 (2)设建造A型沼气池 x 个时,总费用为y万元,则:
y = 2x + 3( 20-x) = -x+ 60 ………………………………………………6分
∵-1
当x·=9 时,y的值最小,此时y= 51( 解(1)这批树苗有(2x42)棵 ·············································································· 1万分
(2)根据题意,得元 ) …………………………………7分
∴此时方案为:建造A型沼气池9个,建造
2x423(x1)5
·····································B····型沼气池···············11····个.······· ··……………8····················分··· ····· 5分
2x423(x1)≥1
解这个不等式组,得40
某体育用品商场预测某品牌运动服能够畅
销,就用32000元购进了一批这种运动服,
5.(2009年湖北十堰市)为执行中央―节能上市后很快脱销,商场又用68000元购进第减排,美化环境,建设美丽新农村‖的国策,我市某村计划建造A、B两种型号的沼气池共20个,以解决该村所有农户的燃料问题.两种型号沼气池的占地面积、使用农户数及造价见下表:
二批这种运动服,所购数量是第一批购进数量的2倍,但每套进价多了10元. (1)该商场两次共购进这种运动服多少套?
(2)如果这两批运动服每套的售价相同,
(每列对一个不等式给2分)
20%,那么每?(利润率
解:(1)设商场第一次购进x套运动服,由题意得:
680002x
32000x
365m2,该村农户共有492户.
(1)满足条件的方案共有几种?写出解答过
程.
(2)通过计算判断,哪种建造方案最省钱.
解: (1) 设建造A型沼气池 x 个,则建造B 型沼气池(20-x )个………1分 依题意得
:
10, ···························································
解这个方程,得x200.
经检验,x200是所列方程的根.
2xx2200200600.
所以商场两次共购进这种运动服600套. ·································
(2)设每套运动服的售价为y元,由题意
2
得:
600y3200068000
3200068000
≥20%,
20≤a≤
1507
,
解这个不等式,得y≥200,
a是正整数,a=20或21,
当a20时b50,当a21时
b45.……………2分
所以每套运动服的售价至少是200元. ·················方案一:建室内车位·································20····个,露天车位··················· 850分
个;方案二:室内车位21个,露天车位45
7.(2009年湖州)随着人民生活水平的不断提高,我市家庭轿车的拥有量逐年增加.据统计,某小区2006年底拥有家庭轿车64辆,2008年底家庭轿车的拥有量达到100辆. (1) 若该小区2006年底到2009年底家庭
轿车拥有量的年平均增长率都相
同,求该小区到2009年底家庭轿车将达到多少辆?
(2) 为了缓解停车矛盾,该小区决定投资
15万元再建造若干个停车位.据测算,建造费用分别为室内车位5000
元/个,露天车位1000元/个,考虑到实际因素,计划露天车位的数量不少于室内车位的2倍,但不超过室内车位的2.5倍,求该小区最多可建两种车位各多少个?试写出所有可能的方案.
解:
(1) 设家庭轿车拥有量的年平均增长
率为x,则:
641x100,……………2分
2
个.
8.(2009年哈尔滨)跃壮五金商店准备从宁云机械厂购进甲、乙两种零件进行销售.若每个甲种零件的进价比每个乙种零件的进价少2元,且用80元购进甲种零件的数量与用100元购进乙种零件的数量相同. (1)求每个甲种零件、每个乙种零件的进价分别为多少元? (2)若该五金商店本次购进甲种零件的数量比购进乙种零件的数量的3倍还少5个,购进两种零件的总数量不超过95个,该五金商店每个甲种零件的销售价格为12元,每个乙种零件的销售价格为15元,则将本次购进的甲、乙两种零件全部售出后,可使销售两种零件的总利润(利润=售价-进价)超过371元,通过计算求出跃壮五金商店本次从宁云机械厂购进甲、乙两种零件有几种方案?请你设计出来.
解得:x1
1
9
25%,x244
舍去),……………2分
100125%125.……………1分
答:该小区到2009年底家庭轿车将达到125辆.……………1分
(2) 设该小区可建室内车位a个,露天车位b个,则: 0.5a0.1b15①
……………2分
2a≤b≤2.5a②
9.(2009年漳州)为了防控甲型H1N1流感,某校积极进行校园环境消毒,购买了甲、乙
由①得:b=150-5a代入②得:
3
两种消毒液共100瓶,其中甲种6元/瓶,乙种9元/瓶.
(1)如果购买这两种消毒液共用780元,求甲、乙两种消毒液各购买多少瓶? (2)该校准备再次购买这两种消毒液(不..包括已购买的100瓶),使乙种瓶数是甲种瓶数的2倍,且所需费用不多于(不...1200元包括780元),求甲种消毒液最多能再购买多少瓶?
(1)解法一:设甲种消毒液购买x瓶,则
乙、丙三种电冰箱的出厂价格分别为:1 200元/台、1 600元/台、2 000元/台.(1)至少购进乙种电冰箱多少台?
(2)若要求甲种电冰箱的台数不超过丙种电冰箱的台数,则有哪些购买方案? 解:(1)设购买乙种电冰箱x台,则购买甲种电冰箱2x台,
丙种电冰箱(803x)台,根据题意,列不
等式: ·····················································································
12002x1600x(803x)2000≤132000
乙种消毒液购买(100x)瓶. ······················································································ 1分 依题意,得6x9(100x)780.
. ····························································································
解这个不等式,得x≥14. ····················································
···············································至少购进乙种电冰箱14台. ·
解得:x40. ··········································································································· 3分
(2)根据题意,得2x≤803x. ·········································
. ··············································································· 4分 100x1004060(瓶)
解这个不等式,得x≤16. ····················································
答:甲种消毒液购买40瓶,乙种消毒液购
由(1)知x≥14.
买60瓶. ····················································································································· 5分
14≤x≤16.
解法二:设甲种消毒液购买x瓶,乙种消毒
又x为正整数,
液购买y瓶. ··············································································································· 1分
x14,15,16. ·····································································
xy100,所以,有三种购买方案:
依题意,得 ························································································· 3分
6x9y780.方案一:甲种电冰箱为28台,乙种电冰箱
为14台,丙种电冰箱为38台; 解得:
x40,
方案二:甲种电冰箱为30台,乙种电冰箱
············································································································ 4分
为15台,丙种电冰箱为35台; y60.
方案三:甲种电冰箱为32台,乙种电冰箱
答:甲种消毒液购买40瓶,乙种消毒液购
为16台,丙种电冰箱为32台. ···············································
买60瓶. ····················································································································· 5分
(2)设再次购买甲种消毒液y瓶,刚购买
乙种消毒液2y瓶. ······································································································ 6分 11.市政公司为绿化一段沿江风光带,计划购买甲、乙两种树苗共500株,甲种树苗依题意,得6y92y≤1200. ················································································ 8分 每株50 元,乙种树苗每株80元.有关统计表明:解得:y≤50. ·········································································································· 9分 甲、乙两种树苗的成活率分别为90%和95%. 答:甲种消毒液最多再购买50瓶. ·············································································10分
(1)若购买树苗共用了28000元,求甲、乙
两种树苗各多少株?
(2)若购买树苗的钱不超过34000元,应如
10.(2009威海)响应―家电下乡‖的惠农政
何选购树苗?
策,某商场决定从厂家购进甲、乙、丙三种
(3)若希望这批树苗的成活率不低于
不同型号的电冰箱80台,其中甲种电冰箱
92%,且购买树苗的费用最低,应如何选
的台数是乙种电冰箱台数的2倍,购买三种
购树苗?
电冰箱的总金额不超过132 000元.已知甲、...
4
解:(1)设购甲种树苗x株,则乙种树苗为(500-x)株.依题意得 S0x+80(500—x)=28000 解之得:x=400
∴500-x=500-400=100 答:购买甲种树苗400株,乙种树苗100株
(2)由题意得 50x+80(500-x)≤34000 解之得x≥200
答:购买甲种树苗不小于200株 (3)由题意可得
90%x+95%(500—x)≥92%·500 ∴x≤300
设购买两种树苗的费用之和为y元,则 y=50x+80 (500-x)=40000-30x
函数y=40000-3x的值随x的增大而减小
x=300时
y最小值=40000-30╳300=31000 12.(2009襄樊市)为实现区域教育均衡发展,我市计划对某县A、B两类薄弱学校全部进行改造.根据预算,共需资金1575万元.改造一所A类学校和两所B类学校共需资金230万元;改造两所A类学校和一所B类学校共需资金205万元.
(1)改造一所A类学校和一所B类学校所需的资金分别是多少万元?
(2)若该县的A类学校不超过5所,则
B类学校至少有多少所? (3)我市计划今年对该县A、B两类
学校共6所进行改造,改造资金由
国家财政和地方财政共同承担.若今年国家财政拨付的改造资金不超过400万元;地方财政投入的改造资金不少于70万元,其中地方财政投入到A、B两类学校的改造资金分别为每所10万元和15万元.请你通过计算求出有几种改造方案?
解:(1)设改造一所A类学校和一所B类
学校所需的改造资金分别为a万
元和b万元.依题意得:
a2b230
···························································
2ab205
解之得
a60b85
··························································
答:改造一所A类学校和一所B类学
校所需的改造资金分别为60万元和85万元.
(2)设该县有A、B两类学校分
别为m所和n所.则
1712
31512
60m85n1575 ·····················································m
n
······················································
∵A类学校不超过5所 ∴
1712n
31515
≤5
∴n≥15
即:B类学校至少有15所.······································(3)设今年改造A类学校x所,则
改造B类学校为6x所,依题意得:
50x706x≤400
··············································
10x156x≥70
解之得1≤x≤4 ··························································∵x取整数
23,4 ∴x1,,
即:共有4种方案. ······················································
13.(2009襄樊市)星期天,小明和七名同学共8人去郊游,途中,他用20元钱去买饮料,商店只有可乐和奶茶,已知可乐2元一杯,奶茶3元一杯,如果20元钱刚好用完.
(1)有几种购买方式?每种方式可乐和奶茶各多少杯?
(2)每人至少一杯饮料且奶茶至少二杯时,有几种购买方式?
5
解:(1)设买可乐、奶茶分别为x、y杯,根据题意得
2x+3y=20(且x、y均为自然数) …………………………………………………………2分 ∴x=
203y2
中所有方案获利相同,a值应是多少?此时,哪种方案对公司更有利?
(1)解:设今年三月份甲种电脑每台售价x元
100000x1000
80000x
···································································
≥0 解得y≤
20
3
∴y=0,1,2,3,4,5,6.代入2x+3y=20 并检验得
x10,x7,x4,x1,
…………
y0;y2;y4;y6.
解得:x4000 ······································································
经检验:x4000是原方程的根, ··········································所以甲种电脑今年每台售价4000元. (2)设购进甲种电脑x台,
······························48000≤3500x3000(15x)≤50000 ·
…………………………………………………6分
所以有四种购买方式,每种方式可乐和奶茶的杯数分别为:(亦可直接列举法求得) 10,0;7,2;4,4;1,6.………………………………………………………………7分 (2)根据题意:每人至少一杯饮料且奶茶至少二杯时,即y≥2且x+y≥8
由(1)可知,有二种购买方式.……………………………………………………………10分
14.(2009年齐齐哈尔市)某电脑公司经销甲种型号电脑,受经济危机影响,电脑价格不断下降.今年三月份的电脑售价比去年同期每台降价1000元,如果卖出相同数量的电脑,去年销售额为10万元,今年销售额只有8万元.
(1)今年三月份甲种电脑每台售价多少元?
(2)为了增加收入,电脑公司决定再经销乙种型号电脑,已知甲种电脑每台进价为3500元,乙种电脑每台进价为3000元,公司预计用不多于5万元且不少于4.8万元的资金购进这两种电脑共15台,有几种进货方案?
(3)如果乙种电脑每台售价为3800元,为打开乙种电脑的销路,公司决定每售出一台乙种电脑,返还顾客现金a元,要使(2)
6
解得6≤x≤10 ······································································
因为x的正整数解为6,7,8,9,10,所以共有5种进货方案 ····································································(3)设总获利为W元,
W(40003500)x(38003000a)(15x)(a300)x1200015a
································································································
当a300时,(2)中所有方案获利相同. ································································································此时,购买甲种电脑6台,乙种电脑9台时对公司更有利. ······································································· 15.(2009贺州)已知一件文化衫价格为18元,一个书包的价格是一件文化衫的2倍还少6元. (1)求一个书包的价格是多少元?
(2)某公司出资1800元,拿出不少于350元但不超过400元的经费奖励山区小学的优秀学生,剩余经费还能为多少名山区小学的学生每人购买一个书包和一件文化衫?
解:(1)182630(元) ····································
所以一个书包的价格是30元. ··········································(注:用其它方法解出正确答案也给予相应的分值) (2)设还能为x 名学生每人购买一个书包
和一件文化衫,根据题意得: ··················································
……
∴B型号彩电的政府补贴要多
些
(2)设购进A型彩电x台,则购进
(1830)x≥1800400
B型彩电(100–x)台,
···················································································· 4分 (1830)x≤1800350·
由题意知,222000≤
x≥29x≤30
16524
解之得:
2916
≤x≤30
2000x+2400(100–x)≤222800 解得:43≤x≤45
∵x取非负整数,∴x的值为43、44、45
∴有三种购进方案:①A型彩
所以不等式组的解集为:
5
电43台,B型彩电57台;
······································································································ 5分 24②A型彩
电44台,B型彩电56台;
∵x为正整数,
③A型彩
∴x=30 ················································································································ 6分
电45台,B型彩电55台。
答:剩余经费还能为30名学生每人购买
设获得的利润为W(元)
一个书包和一件文化衫. ·················································································· 7得分 · · · · · · · · · · · ·由题意, W=500x+600(100–x)= –100x+600000 16.(2009年宜宾)从2008年12月1日起,国家开始实施家电下乡计划,国家按照农民购买家电金额的13%予以政策补贴,某商场计划购进A、B两种型号的彩电共100台,已知该商场所筹购买的资金不少于222000元,但不超过222800元,国家规定这两种型号彩电的进价和售价如下表:
型号 进价
A 2000
B 2400
当x=43时,W最大=55700(元) 即购进A型彩电43台,B型彩电57台,获得利润最大
17.(2009柳州)某校积极推进―阳光体育‖工程,本学期在九年级11个班中开展篮球单循环比赛(每个班与其它班分别进行一场比赛,每班需进行10场比赛).比赛规则规定:每场比赛都要分出胜负,胜一场得3分,负一场得1分.
(1)如果某班在所有的比赛中只得14分,那么该班胜负场数分别是多少? (2)假设比赛结束后,甲班得分是乙班的3倍,甲班获胜的场数不超过5场,且甲班获胜的场数多于乙班,请你求出甲班、乙班各胜了几场.
(元/台)
售价2500 (元/台)
3000
(1)农民购买哪种型号的彩电获得的政府补贴要多些?请说明理由;
(2)该商场购进这两种型号的彩电共有哪些方案?其中哪种购进方案获得的利润最大?请说明理由.(注:利润=售价-进价)。
解:(1)∵A型号彩电政府的补贴为325元, B型号彩电政府的补贴为390元,
7
由题意,
得 ………… 2分
解之,得
…… ………… …… …4分
答:A、B两种纪念品的进价分别为20元、30元… …… …5分
(2)设上点准备购进A种纪念品a件,则购进B种纪念品(40-x)件,
由题意,得
… …… …… ……7分 解
之
,
得
:
30a32… ……………………………
…………………8分
∵总获利w5a7(40a)2a280
是a的一次函数,且w随a的增大而减小 ∴当a=30时,w最大,最大值w=-2×30+280=220.
∴40-a=10
∴应进A种纪念品30件,B种纪念品10件,在能是获得利润最大,最大值是220元。………………………………………………………………………………10分
18.(2009泰安)某旅游商品经销店欲购进A、B两种纪念品,若用380元购进A种纪念品7件,B种纪念品8件;也可以用380元购进A种纪念品10件,B种纪念品6件。
求A、B两种纪念品的进价分别为多少? 若该商店每销售1件A种纪念品可获利5元,每销售1件B种纪念品可获利7元,该商店准备用不超过900元购进A、B两种纪念品40件,且这两种纪念品全部售出候总获利不低于216元,问应该怎样进货,才能使总获利最大,最大为多少? 解:(1)设A、B两种纪念品的进价分别为x元、y元。
8