Question

A pharmacist has 40% and 80% iodine solutions on hand .How many liters of each iodine solutions will be required to produce 6 liters of a 50% iodine mixture?

Answer 1

So... let's change the concentration percentage to decimal format. .. so 40% is just 4/100 or 0.4  and so on.

$\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{40\% sol'n}&x&0.4&0.4x\\ \textit{80\% sol'n}&y&0.8&0.8y\\ -----&-----&-------&-------\\ mixture&6&0.5&3 \end{array}$

so... whatever "x" and "y" may be, we know the must add up to 6 liters.

and whatever 0.4x and 0.8y are, we also know, they must add up to a 3 of concentrated amount.

thus    $\bf \begin{cases} x+y=6\implies \boxed{y}=6-x\\ 0.4x+0.8y=3\\ ----------\\ 0.4x+0.8\left( \boxed{6-x} \right)=3 \end{cases}$

solve for "x", to see how much of the 40% solution will be needed.

what about "y"?  well, y = 6 - x.