七年级不等式应用题专项练习

1一元一次不等式(组)之应用题姓名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）解：设至少涨到每股x元时才能卖出．·······························································1分根据题意得1000(50001000)0.5%50001000xx≥···················································································································4分解这个不等式得1205199x≥，即6.06x≥．···················································································································6分答：至少涨到每股6.06元时才能卖出．·······························································7分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.解：(1)设每支钢笔x元，每本笔记本y元······················································································································1分依题意得：3152183yxyx·························································································································3分解得：53yx·············································································４分答：每支钢笔3元，每本笔记本5元·········································································································５分(2)设买a支钢笔，则买笔记本(48－a)本依题意得：aaaa48200)48(53·························································································7分解得：2420a····································································8分所以，一共有５种方案．·························································································································9分即购买钢笔、笔记本的数量分别为：20，28；21，27；22，26；23，25；24，24．······················································································10分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，买一份礼物送给父母．已知：在暑假期间，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，则可获得：10000.1100元，没有超过140元，从而不能达到目的.（注：其它说理正确、合理即可.）（2）设孔明同学暑假期间卖出报纸x份，由（1）可知1000x，依题意得：10000.10.2(1000)14010000.10.2(1000)200xx解得12001500x2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？解（1）这批树苗有（242x）棵·····································································1分（2）根据题意，得2423(1)52423(1)1xxxx≥····················································································5分（每列对一个不等式给2分）解这个不等式组，得40x≤44································································7分5．（2009年湖北十堰市）为执行中央“节能减排，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：型号占地面积(单位:m2/个)使用农户数(单位:户/个)造价(单位:万元/个)A15182B20303已知可供建造沼气池的占地面积不超过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:492203018365202015xxxx…………………………………………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∵－10，∴y随x增大而减小，当x=9时，y的值最小，此时y=51(万元)…………………………………7分∴此时方案为：建造A型沼气池9个，建造B型沼气池11个．……………8分6．（2009年山东青岛市）北京奥运会开幕前，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，上市后很快脱销，商场又用68000元购进第二批这种运动服，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，且全部售完后总利润率不低于20%，那么每套售价至少是多少元？（利润率100%利润成本）解：（1）设商场第一次购进x套运动服，由题意得：6800032000102xx，···················································································3分解这个方程，得200x．经检验，200x是所列方程的根．22200200600xx．所以商场两次共购进这种运动服600套．·····························································5分（2）设每套运动服的售价为y元，由题意3得：600320006800020%3200068000y≥，解这个不等式，得200y≥，所以每套运动服的售价至少是200元．·······························································8分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：2641100x，……………2分解得：11254x%，294x（不合题意，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：0.50.11522.5ababa①≤≤②……………2分由①得：b=150-5a代入②得：20a150≤≤7，a是正整数，a=20或21，当20a时50b，当21a时45b.……………2分方案一：建室内车位20个，露天车位50个；方案二：室内车位21个，露天车位45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．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4两种消毒液共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．．．1200元（不包括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，则乙种消毒液购买(100)x瓶．··················································