Let:
• A ,be the number of millilitres (mL) of solution A used.
,• B ,be the number of mL of solution B used.
We know that Dale uses four times as much solution A as solution B, meaning
[tex]A=4B[/tex]Now, we know that we will end up with 570 mL of pure alcohol in the final solution. Using the dilution of both A and B (20% means 0.2 and 15% is 0.15) we would have that:
[tex]0.2A+0.15B=570[/tex]We would have the following system of equations:
[tex]\begin{cases}A=4B \\ 0.2A+0.15B=570\end{cases}[/tex]Substituting equation 1 in equation 2 and solving for B :
[tex]\begin{gathered} 0.2A+0.15B=570 \\ \rightarrow0.2(4B)+0.15B=570 \\ \rightarrow0.8B+0.15B=570 \\ \rightarrow0.95B=570\rightarrow B=\frac{570}{0.95} \\ \Rightarrow B=600 \end{gathered}[/tex]Substituting in equation 1 and solving for A:
[tex]\begin{gathered} A=4B \\ \rightarrow A=4(600) \\ \Rightarrow A=2400 \end{gathered}[/tex]This way, we can conclude that 2400 mL of solution A and 600mL of solution B were used.
