Twice the Percentage of Another Amount

Let $\dfrac x{100} = \alpha$ .

• Then the alcohol concentrations in sample $A$ and sample $B$ are $\alpha$ and $2\alpha$ respectively.
• And the water concentrations in sample $A$ and sample $B$ are $1-\alpha$ and $1-2\alpha$ respectively.

Then the water concentration in the mixture of $A$ and $B$ is:

$\begin{aligned} \frac 32 (1-\alpha) + \frac 83 (1-2\alpha) & = 1.98 & \small \color{#3D99F6} \text{Multiply both sides by }3 \\ 4.5-4.5\alpha + 8 - 16\alpha & = 5.94 \\ 20.5 \alpha & = 6.56 \\ \implies \alpha & = 0.32 \end{aligned}$

Therefore, the water concentration in sample $B$ , $V_{Bw} = \dfrac 83 (1-2\alpha) = \dfrac 83 (1-2\times 0.32) = \boxed{0.96}$ .

The new mixture, which we will call Sample $C$ , is a $\frac{25}{6}$ litre solution since it's the sum of Sample $A$ and Sample $B$ . It was said that Sample $C$ has a water concentration of $1.98$ litres. This is $47.52\%$ of Sample $C$ .

$\small \frac{198}{100}÷\frac{25}{6}=\frac{198}{100}\cdot\frac{6}{25}=\frac{1188}{2500} \text{or} \frac{297}{625} \space \text{or} \space 47.52\%$

So $100\%-47.52\%=52.48\%$ is the alcohol concentration of Sample $C$ . Since the $x\%$ alcohol concentration of Sample $A$ added to the $2x\%$ alcohol concentration of Sample $B$ is the $52.48\%$ alcohol concentration of Sample $C$ , we can figure out the value of $x$ .

$\small \begin{aligned} (\frac{3}{2}\cdot x\%)+(\frac{8}{3}\cdot2x\%)&=\frac{25}{6}\cdot52.48\% \\ (\frac{3}{2}\cdot\frac{x}{100})+(\frac{8}{3}\cdot\frac{2x}{100})&=\frac{25}{6}\cdot\frac{5248}{10000} \\ \frac{3x}{200}+\frac{16x}{300}&=\frac{131200}{60000} \\ \frac{9x}{600}+\frac{32x}{600}&=\frac{1312}{600} \\ \frac{41x}{600}&=\frac{1312}{600} \\ 41x&=\frac{600\cdot1312}{600} \\ 41x&=1312 \\ x&=\frac{1312}{41} \\ x&=32 \end{aligned}$

Since $2x\%$ is $64\%$ , we can get the water concentration of Sample $B$ , which is $100\%-64\%=36\%$ .

$\frac{8}{3}\cdot36\%=\frac{8}{3}\cdot\frac{36}{100}=\frac{288}{300}\text{or}\frac{24}{25}\text{or}\boxed{0.96}$