We must mix 4 gallons of a 90% alcohol solution and 8 gallons of a 30% alcohol solution.
We can solve the exercise writing two equations. In these equations let's call x the 30% alcohol solution and y the 90% alcohol solution:
- equation 1: x.(0.3) + y.(0.9) = 12.(0.5)
- equation 2: x+y=12
Now, we can replace x from the equation 2 on equation 1.
First, from equation 2 we resolve x=12-y so we can replace it on the equation 2:
[tex]\begin{gathered} x.\mleft(0.3\mright)+y.\mleft(0.9\mright)=12.\mleft(0.5\mright) \\ (12-y)\cdot(0.3)+y\cdot(0.9)=12\cdot(0.5) \\ 3.6-y\cdot(0.3)+y\cdot(0.9)=6 \\ y\cdot(0.6)=6-3.6 \\ y=\frac{2.4}{0.6} \\ y=4 \end{gathered}[/tex]So, it is needed 4 gallons of 90% alcohol solution.
Finally, we replace the value of y=4 on the equation 2 to find x:
[tex]\begin{gathered} x+y=12 \\ x+4=12 \\ x=12-4 \\ x=8 \end{gathered}[/tex]And It is needed 8 gallons of 30% alcohol solution.
We must mix 4 gallons of a 90% alcohol solution and 8 gallons of a 30% alcohol solution.