Question

how many gallons of 80% antifreeze solution must be mixed with 100 gallons of 10% antifreeze to get a mixture that is 70% antifreeze? use the six step method

Answer 1

Let's say we'll mix "x" amount of the 80% solution, how much antifreeze is in it? well is just 80% of antifreeze, and the rest is something else.... what's 80% of "x"? so is (80/100)*x, or 0.8x.

likewise, the mixture will be a 70% solution, and let's say it adds up to "y" amount, so how much antifreeze is in it? well (70/100) * y or 0.7y.

$\bf \begin{array}{lccclll} &\stackrel{gallons}{amount}&\stackrel{\%}{quantity}&\stackrel{antifreeze}{quantity}\\ &------&------&------\\ \textit{80\% sol'n}&x&0.80&0.8x\\ \textit{10\% sol'n}&100&0.10&10\\ ------&------&------&------\\ mixture&y&0.70&0.7y \end{array}$

so whatever "x" is, we know that x + 100 = y.

and we know that 0.8x + 10 = 0.7y.

$\bf \begin{cases} x+100=\boxed{y}\\ 0.8x+10=0.7y\\ ----------\\ 0.8x+10=0.7\left( \boxed{x+100} \right) \end{cases} \\\\\\ 0.8x+10=0.7x+70\implies 0.8x-0.7x=70-10\implies 0.1x=60 \\\\\\ x=\cfrac{60}{0.1}\implies x=600$