Question

A chemist needs 120 milliliters of a 71% solution but has only 55% and 79% solutions available. Find how many milliliters of each that should be mixed to get the desired solution

Answer 1

x = amount of 55% solution in mL
y= amount of 79% solution in mL


`\bf \begin{array}{lccclll} &amount&concentration&\textit{concentration amount}\\ &-----&-----&--------\\ \textit{55\% sol'n}&x&0.55&0.55x\\ \textit{79\% sol'n}&y&0.79&0.79y\\ -----&-----&-----&--------\\ mixture&120&0.71&(120)(0.71) \end{array}`

now.. notice, we use the decimal format for the percents, thus 55% is just 55/100, and 79% is just 79/100 and 71% is well, you guessed it, 71/100 or 0.71 anyway

whatever those amounts of x and y are, they will end up as 120mL total
so, we can say that

x + y = 120

now, the concentration in each, they'll add up to (120)(0.71)

thus
`\bf \begin{cases} x+y=120\to \boxed{y}=120-x\\\\ 0.55x+0.79y=(120)(0.71)\\ ----------\\ 0.55x+0.79\left( \boxed{120-x} \right)=(120)(0.71) \end{cases}`

solve for "x" to see how much 55% solution will go into the 71% mix

what about the 79% solution? well y = 120 - x