“山村”老师提出了一个“把含糖5%和含糖8%的两种糖水混合成含糖6%的糖水”的数学问题，原题如下：把含糖5%（百分之五）和含糖8%（百分之八）的两种糖混合制成含糖6%（百分之六）的糖水600克，应取5%（百分之五）的糖水多少克？取8%（百分之八）的糖水多少克？

“闲庭信步”老师数学题解答：此题列方程解答比较容易理解。首先，含糖5%和含糖8%的糖水合起来是600克，设其中一种糖水重量为x克，那另一种就是（600-x）克；其次，糖水混合前两种糖水中含糖的重量的和，与混合后600克糖水中含糖的重量是相等的。据此可以列出方程：

解：设含糖8%的糖水为x克，则含糖5%的糖水为（600-x）克。

8%x + 5%(600-x)=6%×600

8%x + 30 – 5%x  = 36

3%x + 30 = 36

3%x = 6

x = 200

则含糖5%的糖水为：600-200＝400（克）。

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