Question

How many ounces of a 40% silver nitrate solution must be mixed with 8 ounces of a 15% silver nitrate solution to make a 30% solution?

Answer 1

let's say we have to mix "x" ounces of 40% silver nitrate, now the solution is 40% of silver nitrate, so 40% in those "x" oz is silver nitrate, the rest is hmmm water or so. So how much silver nitrate is there in "x" anyway? (40/100) * x, or 0.4x.

we know in 8 oz of 15% silver nitrate, there's only 15% of silver nitrate, the rest is other substances, how much is that? (15/100) * 8, or 1.2.

our mixture will be 30% of silver nitrate, and hmmm say we end up with "y" ounces, how much silver nitrate is in the "y" oz? (30/100) * y, or 0.3y.

we also know that, whatever "x" and "y" are, we know that x + 8 = y and that 0.4x + 1.2 = 0.3y


`\bf \begin{array}{lccclll} &\stackrel{ounces}{amount}&\stackrel{\%~nitrate}{quantity}&\stackrel{oz~nitrate}{quantity}\\ &------&------&------\\ \textit{40\% sol'n}&x&0.40&0.4x\\ \textit{15\% sol'n}&8&0.15&1.2\\ ------&------&------&------\\ mixture&y&0.30&0.3y \end{array} \\\\\\ \begin{cases} x+8=\boxed{y}\\ 0.4x+1.2=0.3y\\ ----------\\ 0.4x+1.2=0.3\left( \boxed{x+8} \right) \end{cases} \\\\\\ 0.4x+1.2=0.3x+2.4\implies 0.1x=1.2\implies x=\cfrac{1.2}{0.1}\implies x=\stackrel{oz}{12}`