Question

A druggist wants to reduce a 20-ounce quantity of a 7% solution of iodine to a 4% solution by adding pure alcohol. How much alcohol must he add to accomplish this?

Answer 1

So hmmm 7% of anything is (7/100) * anything.

now... how much iodine is in the 20oz solution? well (7/100) * 20, or 1.4 is iodine, the rest is some other substances, but 1.4oz is iodine.

now, he wants to drop it to a mixture of 4% iodine, let's say the mixture "m" in ounces, has 4% iodine, how much is that? well (4/100) * m, or 0.04m.

well, he wants to do so by adding say "a" oz of pure alcohol... how much iodine in the alcohol?  none or 0%, or (0/100) * a, or 0.00a or just 0a or 0 anyway.

$\bf \begin{array}{lccclll} &\stackrel{oz}{amount}&\stackrel{iodine~\%}{quantity}&\stackrel{iodine~oz}{quantity}\\ &------&------&------\\ \textit{20oz sol'n}&20&0.07&1.4\\ \textit{pure alcohol}&a&0.00&0a\\ ------&------&------&------\\ mixture&m&0.04&0.04m \end{array}$

now, whatever "m" may be, we know that 20 + a = m, and we also know likewise that 1.4+0 = 0.04m.

$\bf \begin{cases} 20+a=\boxed{m}\\ 1.4+0=0.04m\\ ----------\\ 1.4=0.04\left( \boxed{20+a} \right) \end{cases} \\\\\\ 1.4=0.04a+0.8\implies 1.4-0.8=0.04a\implies 0.6=0.04a \\\\\\ \cfrac{0.6}{0.04}=a\implies \stackrel{oz}{15}=a$