Question

A pharmacist has an 18% alcohol solution and a 40% alcohol solution. How much of each should he mix together to make 10L of a 20% alcohol solution? Pls help

Answer 1

`x=\textit{Liters of solution at 18\%}\\\\ ~~~~~~ 18\%~of~x\implies \cfrac{18}{100}(x)\implies 0.18 (x) \\\\\\ y=\textit{Liters of solution at 40\%}\\\\ ~~~~~~ 40\%~of~y\implies \cfrac{40}{100}(y)\implies 0.4 (y) \\\\\\ \textit{10 Liters of solution at 20\%}\\\\ ~~~~~~ 20\%~of~10\implies \cfrac{20}{100}(10)\implies 2 \\\\[-0.35em] ~\dotfill`

`\begin{array}{lcccl} &\stackrel{Liters}{quantity}&\stackrel{\textit{\% of Liters that is}}{\textit{alcohol only}}&\stackrel{\textit{Liters of}}{\textit{alcohol only}}\\ \cline{2-4}&\\ \textit{1st Sol'n}&x&0.18&0.18x\\ \textit{2nd Sol'n}&y&0.4&0.4y\\ \cline{2-4}&\\ mixture&10&0.2&2 \end{array}~\hfill \begin{cases} x + y = 10\\\\ 0.18x+0.4y=2 \end{cases} \\\\[-0.35em] ~\dotfill`

`\stackrel{\textit{using the 1st equation}}{x+y=10\implies y=10-x} \\\\\\ \stackrel{\textit{substituting on the 2nd equation from above}}{0.18x+0.4(10-x)=2}\implies 0.18x+4-0.40x=2 \\\\\\ -0.22x+4=2\implies -0.22x=-2\implies x=\cfrac{-2}{-0.22}\implies \boxed{x\approx 9.09} \\\\\\ \stackrel{\textit{since we know that}}{y=10-x}\implies y\approx 10-9.09\implies \boxed{y\approx 0.91}`

Answer 2

0.18x + 4 - 0.40x = 2

-0.22x = -2

x = 9.09

Therefore, the pharmacist should mix 9.09 liters of 18% alcohol solution and 0.91 liters of 40% alcohol solution to make 10 liters of 20% alcohol solution.