帮我出一个二元一次方程组与一元一次不等式的综合应用(文字题)

如题。
2025-03-14 22:40:26
推荐回答(2个)
回答1:

光明中学9年级甲、乙两班在为“希望工程”捐款活动中,两班捐款的总数相同,均多于300元且少于400元。已知甲班有一人捐6元,其余都每人捐9元;一班有一人捐13元,其余每人都捐8元。求甲、乙两班学生总人数共是少人。

甲班捐款总数是9x+6=9(x+1)-3;
乙班捐款总数是8x+13=8(y+2)-3;

所以每班捐款总金额-3必是8和9的公倍数.

8和9的公倍数中,在300到400之间的只有360,所以捐款总额必是357.

甲班39人,乙43

小明的妈妈带了100元钱去超市购物,她用了50元买床上用品,30元给小明买书包.如果她再买3千克香蕉,则她所带的钱就不够了;如果她再买2.5千克香蕉,则还有余钱,若香蕉的单价是一个整数,求证香蕉的单价?

因为再买3千克香蕉,则她所带的钱就不够,故3X大于20;又因为再买2.5千克香蕉,则还有余钱,故2.5X小于20.

20/3小于X小于20/2.5
所以整数X=7

10.某饮料厂为开发新产品,用A,B两种果汁原料各19千克甲种新型饮料每千克含量A为0.5,B为0.2
乙种新型饮料每千克含量A为0.3,B为0.4
1.假设甲种饮料需配制x千克,列出满足题意的不等式组,并求出解集.
2.甲种饮料每千克成本为4元,乙种饮料每千克成本为3元,根据1.的运算结果,确定当配制多少千克甲种饮料时,甲乙两种饮料的成本总额最小?

1 A果汁 0.5X+0.3*(50-X)=<19 解得X=<20
B果汁 0.2X+0.4*(50-X)=<19 解得X>=5 所以 5=2 成本为 4X+3(50-X)=X+150 X最小时成本最低 所以当配置5千克甲饮料时成本最小

回答2:

http://www.jysls.com/thread-332547-1-1.html试卷。用快车下载

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