Question

（8分）某手機(jī)專營(yíng)店代理銷售A、B兩種型號(hào)手機(jī)．手機(jī)的進(jìn)價(jià)、售價(jià)如下
表：

（1）第一季度：用36000元購(gòu)進(jìn)A、B兩種型號(hào)的手機(jī)，全部售完后獲利6300元，求
第一季度購(gòu)進(jìn)A、B兩種型號(hào)手機(jī)的數(shù)量；
（2）第二季度：計(jì)劃購(gòu)進(jìn)A、B兩種型號(hào)手機(jī)共34部，且不超出第一季度的購(gòu)機(jī)總費(fèi)用，則A型號(hào)手機(jī)最多能購(gòu)多少部？

Answer 1

（1）解：設(shè)該專營(yíng)店第一季度購(gòu)進(jìn)A、B兩種型號(hào)手機(jī)的數(shù)量分別為x部和y部． 1分
答：該專營(yíng)店本次購(gòu)進(jìn)A、B兩種型號(hào)手機(jī)的數(shù)分別為15部和18部．······························································ 4分
（2）解：設(shè)第二季度購(gòu)進(jìn)A型號(hào)手機(jī)a部．···················································· 5分
由題意可知：1200a+1000(34－a)≤36000，····················································· 6分
解得：a≤10··························································································· 7分
不等式的最大整數(shù)解為10
答：第二季度最多能購(gòu)A型號(hào)手機(jī)10部．······················································ 8分

解析