Question

How many ounces of 6% and 15% alcohol solutions must be mixed to obtain 100 ounces of a 10% solution

Answer 1

so hmmm first off, let's use the decimal format of a percentage, thus, 6% is rather just 6/100 or 0.06, and 15% is just 15/100 or just 0.15 and so on

now
`\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentrated\\ amount \end{array}\\ &-----&-------&-------\\ \textit{6\% solution}&x&0.06&0.06x\\ \textit{15\% solution}&y&0.15&0.15y\\ -----&-----&-------&-------\\ mixture&100&0.10&10 \end{array}`

whatever "x" and "y" may be, we know they must add up to 100 ounces, thus

x + y = 100

and whatever the concentrated amounts in the solution are for each, they must add up to 10 oz, thus

0.06x + 0.15y = 10

thus
`\bf \begin{cases} x+y=100\implies \boxed{y}=100-x\\ 0.06x+0.15y=10\\ ----------\\ 0.06x+0.15\left( \boxed{100-x} \right)=10 \end{cases}`

solve for "x", to see how much of the 6% solution will be needed

what about "y"? well, y = 100 - x