有种奖金有三种,一等奖1000元,二等奖250元,三等奖50元,共100人中奖,奖金总额为9500元。

二等奖的有几人?小学鸡兔同笼问题。要解答过程!!!!!
2024-11-15 18:35:51
推荐回答(4个)
回答1:

设一等奖为x人,二等奖为y人,三等奖为z人
x + y + z =100…………①

1000x + 250y + 50z =9500
20x + 5y + z = 190…………②

②-①高高得到:
19x + 4y =90
x = (90-4y)/19
所以,(90-4y)必须为19的倍数

(1)当(90-4y)=19时
y = 71/4 不符合条件

(2)当(90-4y)=38时
y = 13
x = 2
z = 100-x-y = 85
符合条件

(3)当(90-4y)= 57时
y= 33/4 不符合条件

(4)当 90-4y = 76时
y = 14/4 不符合条件

(5)当90-4y = 95时,不符合条件

所以,由上可知,只有(和念宴2)符合条件:
x = 2
y = 13
z = 85
一等奖2人,二等奖唤银13人 ,三等奖85人

回答2:

鸡兔同笼思想:
假设100人全中3等奖,奖金是50*100=5000元
而实际奖金是9500元,多了4500元

寻找多出的部分(4500元)所代表的含义:
若仔早1人中了1等奖就多出了950元
若1人凯戚碰中了2等奖就多出了200元

如果有a个人中1等奖,b个人中2等奖
就盯谈多出了950a+200b元
所以950a+200b=4500
化简19a+4b=90 90-19a=4b 4b是4的倍数,所以90-19a是4的倍数
又因为90除以4余2,所以19a除以4也余2,且19a<90
可以试验发现a=2
所以4b=90-19*2=52
b=13

答:二等奖有13人

回答3:

一等奖A人 二等奖B人 三等奖C人

A+B+C=100
1000A+250B+50C=9500

B=100-A-C
20A+5B+C=190

20A+5(100-A-C)+C=190

20A+500-5A-5C+C=190

15A-4C=-310

4C=310+15A

这时候发现A必然大于1小于5,因为A大于5时 9500-5000=4500 4500/50=90 90+5不满100人了

而C又必定是帆巧物个整数,所以A带宽销入2和4

当A=2时 C=85 B=13带入式子,绝态液对成立!!!!!!!!!!

那么不算A=4了 二等奖13个人

回答4:

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