题目内容

(8分)某手机专营店代理销售A、B两种型号手机.手机的进价、售价如下

表:

(1)第一季度:用36000元购进 A、B两种型号的手机,全部售完后获利6300元,求

第一季度购进A、B两种型号手机的数量;

(2)第二季度:计划购进A、B两种型号手机共34部,且不超出第一季度的购机总费用,则A型号手机最多能购多少部?

 

(1)解:设该专营店第一季度购进A、B两种型号手机的数量分别为x部和y部.····· 1分

答:该专营店本次购进A、B两种型号手机的数分别为15部和18部.······························································ 4分

(2)解:设第二季度购进A型号手机a部.···················································· 5分

由题意可知:1200a+1000(34-a)≤36000,····················································· 6分

解得:a≤10··························································································· 7分

不等式的最大整数解为10

答:第二季度最多能购A型号手机10部.······················································ 8分

解析:略

 

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