Question

How many ounces of a 35% alcohol solution must be mixed with 10ounces of 40% alcohol solution to make a 37% alcohol solution?

Answer 1

first off, let's change the format of the percentage to decimal.. so 35% is just 35/100 or 0.35 and 40% is 40/100 or 0.4 and so on.


`\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}\\ &-----&-------&-------\\ \textit{35\% sol'n}&x&0.35&0.35x\\ \textit{40\% sol'n}&10&0.40&4.00\\ -----&-----&-------&-------\\ mixture&y&0.37&0.37y \end{array}`

now, whatever "x" amount is, we know that x + 10 must add up to "y". x + 10 = y, because both quantities added will give the mixture amount.

and whatever 0.35x + 4 = 0.37y, because both concentrated amounts must give the 37% of the mixture.


`\bf \begin{cases} x+10=\boxed{y}\\ 0.35x+4=0.37y\\ ----------\\ 0.35x+4=0.37\left( \boxed{x+10} \right) \end{cases}`

solve for "x", to see how much of the 35% solution will be needed.

what about "y"? well, x + 10 = y.