Question

Liquid X is composed of 30% alcohol and 70% water, while Liquid Y is composed of 18% alcohol and 82% water. The two liquids are combined to form Mixture Z, which is composed of 21% alcohol and 79% water. What is the ratio in Mixture Z of Liquid X to Liquid Y?

Answer 1

Let \(a\) be the amount of solution \(X\) in \(Z\) and \(b\) be the amount of solution \(Y\) in \(Z\). \(A\) is Alcohol and \(W\) is water.
Then
$$
X=0.3A+.7W\\
Y=.18A+.82W\\
Z=.2A+.79W=aX+bY\\
=a(0.3A+.7W)+b(.18A+.82W)\\
=A(.3a+.18b)+W(.7a+.82b)
$$
This means that
$$
Z=.2A+.79W=A(.3a+.18b)+W(.7a+.82b)\\
\implies \\
.2=(.3a+.18b)\\
.79=(.7a+.82b)
$$
Two equations. Two unknowns. Solvable.

When you get a and b you can then determine the ratio of X to Y in Z:
$$
\frac{a}{b}
$$

Does this make sense?

Answer 2

what is the Y?

Answer 3

oh nvm i see it

Answer 4

*Above \(Z=.2\color{red}1A+.79W\) -- notice I missed the "1"

Answer 5

hmmmm I'm going to have to look at this several times ....

Answer 6

OK

Answer 7

Hey this is a hard problem right? It's not just me right?