Question

6. A solution containing 80% chicken broth is to be mixed with water to

make 200L containing 60% chicken broth. How much of each should be
mixed?

Answer 1

x = amount of the 80% broth

y = amount of water with 0% broth

keeping in mind that water has no broth in it, so 0% of broth, let's plug in those percentages in decimal format to see how much each one has.

what's 80% of "x"? well (8/100) * x = 0.80x.

what's 0% of "y"? well hell is just 0.

what's 60% of 200? well anyhow is just 120.

`\begin{array}{lcccl} &\stackrel{liters}{quantity}&\stackrel{\textit{\% of liters that is}}{\textit{broth only}}&\stackrel{\textit{liters of}}{\textit{broth only}}\\ \cline{2-4}&\\ \textit{80\% Sol'n}&x&0.80&0.80x\\ water&y&0.00&0\\ \cline{2-4}&\\ mixture&200&0.60&120 \end{array}~\hfill \begin{cases} x + y = 200\\\\ 0.80x+0=120 \end{cases} \\\\[-0.35em] ~\dotfill\\\\`

`\stackrel{\textit{using the 2nd equation}}{0.8x=120}\implies x=\cfrac{120}{0.8}\implies \boxed{x=150~liters} \\\\\\ \stackrel{\textit{using the 1st equation}}{150 + y = 200}\implies y=200-150\implies \boxed{y=50~liters}`