题目
There are two alloys, A and B, of copper and zinc. The ratio (by weight) of copper and zinc in alloy A is 8 : 1 and that in alloy B is 2 : 7. It is found that, if alloy A and alloy B are mixed in a certain ratio, the weights of copper and zinc in the resultant alloy are also in that ratio. What is that ratio?
解析
有两种铜锌合金A和B。合金A中铜和锌的(重量)比是8:1,合金B中铜和锌的(重量)比是2:7。发现如果合金A和合金B按一定比例混合,所得合金中铜和锌的重量也成该比例。这个比例是多少?设合金\(A\)的重量为\(x\),合金\(B\)的重量为\(y\)。
对于合金\(A\),铜的重量为\(\frac{8}{9}x\),锌的重量为\(\frac{1}{9}x\);对于合金\(B\),铜的重量为\(\frac{2}{9}y\),锌的重量为\(\frac{7}{9}y\)。
混合后铜的总重量为\(\frac{8}{9}x+\frac{2}{9}y\),锌的总重量为\(\frac{1}{9}x+\frac{7}{9}y\)。
已知混合后铜锌重量比等于合金\(A\)和合金\(B\)的混合比例,即\((\frac{8}{9}x+\frac{2}{9}y):(\frac{1}{9}x+\frac{7}{9}y)=x:y\)。
由\((\frac{8}{9}x+\frac{2}{9}y):(\frac{1}{9}x+\frac{7}{9}y)=x:y\)可得:
\[
\begin{align*}
y(\frac{8}{9}x+\frac{2}{9}y)&=x(\frac{1}{9}x+\frac{7}{9}y)\\
\frac{8}{9}xy+\frac{2}{9}y^{2}&=\frac{1}{9}x^{2}+\frac{7}{9}xy\\
\frac{2}{9}y^{2}-\frac{1}{9}x^{2}&=\frac{7}{9}xy-\frac{8}{9}xy\\
2y^{2}-x^{2}&=-xy\\
x^{2}-xy - 2y^{2}&=0\\
(x - 2y)(x + y)&=0
\end{align*}
\]
解得\(x = 2y\)或\(x=-y\)(重量不能为负舍去),所以\(x:y = 2:1\)。
综上,答案是D选项。