Question

How many ounces of pure salt must be added to 16oz of 40% salt to produce a 65% solution (pure salt is 100%)

Answer 1

so we're adding pure salt to a saline solution, how much percent of salt is in pure salt? well, hell is 100% salt. We'll be adding that to 16oz of 40% salt solution to end up with a mixture of "m" oz at 65% of salt.

`x=\textit{oz of solution at 100\%}\\\\ ~~~~~~ 100\%~of~x\implies \cfrac{100}{100}(x)\implies x \\\\\\ 16=\textit{oz of solution at 40\%}\\\\ ~~~~~~ 40\%~of~16\implies \cfrac{40}{100}(16)\implies 0.4 (16)\implies 6.4 \\\\\\ m=\textit{oz of solution at 65\%}\\\\ ~~~~~~ 65\%~of~m\implies \cfrac{65}{100}(m)\implies 0.65 (m) \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{oz}{quantity}&\stackrel{\textit{\% of oz that is}}{\textit{salt only}}&\stackrel{\textit{oz of}}{\textit{salt only}}\\ \cline{2-4}&\\ \textit{1st Sol'n}&x&1.00&x\\ \textit{2nd Sol'n}&16&0.4&6.4\\ \cline{2-4}&\\ mixture&m&0.65&0.65m \end{array}~\hfill \begin{cases} x + 16 = m\\\\ x+6.4=0.65m \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{substituting on the 2nd equation}}{x+6.4~~ = ~~0.65\stackrel{ m }{(x+16)}}\implies x+6.4=0.65x+10.4 \\\\\\ 0.35x+6.4=10.4\implies 0.35x=4\implies x=\cfrac{4}{0.35}\implies \boxed{x\approx 11.43}~oz`