How would I go about setting this up?
How many liters of a 30% alcohol solution must be mixed with 80 liters 0f a 90% solution to get a 40% solution ?
Thanks
Hello, Jerry99!
$\displaystyle \text{How many liters of a 30\% alcohol solution must be mixed}$
$\displaystyle \text{with 80 liters 0f a 90\% solution to get a 40\% solution?}$
Here is one way to reason it out . . .
We will use $\displaystyle \,x$ liters of the 30% solution.
. . It contains: .$\displaystyle 0.30x$ liters of alcohol.
We have $\displaystyle 80$ liters of the 90% solution.
. . It contains: .$\displaystyle 0.90 \times 80 \:=\:72$ liters of alcohol.
The mixture will have: .$\displaystyle 0.30x + 72$ liters of alcohol. .[1]
But we know that the mixture will be $\displaystyle x+80$ liters which is 40% alcohol.
. . The mixture will have: .$\displaystyle 0.40(x+80)$ liters of alcohol. .[2]
We just described the final amount of alcohol in two ways.
There is our equation! . . . $\displaystyle 0.30x + 72 \;=\;0.40(x + 80)$