为了改善办学条件,某中学计划购买部分A品牌电脑和B品牌课桌。

为了改善办学条件,某中学计划购买部分A品牌电脑和B品牌课桌。第一次,用9万元购买了A品牌电脑10台和B品牌课桌200张。第二次,用9万元购买了A品牌电脑12台和B品牌课桌... 为了改善办学条件,某中学计划购买部分A品牌电脑和B品牌课桌。第一次,用9万元购买了A品牌电脑10台和B品牌课桌200张。第二次,用9万元购买了A品牌电脑12台和B品牌课桌120张。
(1)每台A品牌电脑与每张B品牌课桌的价格各是多少元?
(2)第三次购买时,销售商对一次购买量大的客户打折销售。规定:一次购买A品牌电脑35台以上(含35台),按九折销售。一次购买B品牌课桌600张以上(含600张)按八折销售。学校准备用27万元购买电脑和课桌,其中电脑不少于35台,课桌不少于600张,问有几种购买方案?请设计出来。
【注意:考点为《不等式组的应用》,要有详细的过程!】
很急的啊~~~~~~~~~
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x沢田纲吉x
推荐于2016-03-26 · TA获得超过532个赞
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(1)解设电脑X元/台 课桌Y元/ 张 由题意得
10x+200y=90000 (1)
12x+120y=90000 (2)
整理得
x+20y=9000 (3)
x+10y=7500 (4)
(3)-(4) 得
10y=1500
y=150
把 y=150代入(4)得
x+10×150=7500
x=6000
∴x=6000
y=150
答:每台电脑售价6000元,桌子售价150元
第二题。。。。暂时想不起来怎么写了。。如果会写了的话明天给你答案
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有两种方案 : 列方程式 6000 X 0.9 x + 150 X 0.8 y =270000 其中 x>= 35 , y >=600 且x y都是整数
方案一: x= 35 , y= 675 即是买电脑35台 买课桌675 台
方案二:x= 36 , y = 630 即是买电脑36台 买课桌630台
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