谁知道这两道奥数题怎么做(寻找不变量)?

2025-01-19 02:57:29
推荐回答(5个)
回答1:

3. 假设弟弟带了x元,若弟弟出钱买卷笔刀,剩余x-6元,哥哥带了5(x-6)
若哥哥出钱,哥哥剩余5(x-6)-6,因此(5(x-6)-6)/x=7/5
解得x=10, 哥哥带了5(10-6)=20元
4. 设共有盐x,水y,每次加水k。 则x/(y+k)=3/20, x/(y+2k)=1/10
联立方程组,得y=k,x=0.3k
则x/(y+3k)=0.3k/4k=3/40

回答2:

3.
6元是不变量,用比例解最直接。
7:5=14:10=(20-6):10
5:1=20:4=20:(10-6)
原来哥哥带了20元,弟弟带了10元。
4.
盐是不变量
加入一次水后,盐与盐水的比为3:20
第二次加水后,盐与盐水的比为1:10=3:30
加入水的量为10
第三次加水后,盐与盐水的比为3:(30+10)=3:40
第三次加入同样多的水,盐水中盐占盐水的3/40。

两题都是考察比例的。

回答3:

根据哥哥剩下的钱与弟弟的钱比数是7:5可以知道哥哥剩下的钱数是哥哥和弟弟剩下的总钱数7/(7+5)。根据哥哥的钱与弟弟剩下的钱比数是5:1可以知道哥哥的钱数是哥哥和弟弟剩下的总钱数5/(5+1).用6除以(5/6-7/12)的结果24元就是单位“1”的数量。哥哥的钱数用24乘以5/6结果是20元。弟弟的钱数用24乘以5/12结果是10元。

回答4:

3.解:设共有x元
(7/12)x+6=(5/6)x
(1/4)x =6
x =24
24*7/12+6=20元······哥哥
24*5/12 =10元······弟弟

回答5:

用代数不就行了,假设,哥哥有x弟弟有y

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