The mixture

If a 6-liter solution with a concentration of 20% alcohol is mixed with a 4-liter solution with a 60% concentration of alcohol. Find the percentage of alcohol in the mixture.

Final Answer:

p =  36 %

Step-by-step explanation:

$$\begin{gathered} p_1=20\%=\frac{20}{100}=\frac{1}{5}=0.2 \\ {p}_2=60\%=\frac{60}{100}=\frac{3}{5}=0.6 \\ {V}_1=6\text{l} \\ {V}_2=4\text{l} \\ a=p_1\cdot V_1+p_2\cdot V_2=0.2\cdot 6+0.6\cdot 4=\frac{18}{5}=3.6\text{l} \\ V=V_1+V_2=6+4=10\text{l} \\ p=100\cdot \frac{a}{V}=100\cdot \frac{3.6}{10}=36\% \end{gathered}$$

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