Question

Gave needs 60 liters of 40% acid solution he currently has 20% solution and 50% solution how many liters of each does he need to make the needed 60 liters of 40% acid solution

Answer 1

`\bf \begin{array}{lccclll} &amount&concentration& \begin{array}{llll} concentration\\ amount \end{array}\\ &-----&-------&-------\\ \textit{20\% sol'n}&x&0.20&0.20x\\ \textit{50\% sol'n}&y&0.50&0.50y\\ -----&-----&-------&-------\\ mixture&60&0.40&(60)(0.40) \end{array}`

notice above... we use the decimal format for the percentage, thus 20% is really just 20/100, 40% is 40/100 and so on.

so.. whatever "x" and "y" amounts are, we know they have to add up to 60 Liters, that is x+y = 60

and whatever the concentration amount of each is, it must add up to (60)(0.40), that is 0.20x+0.50y=(60)(0.40)

thus
`\bf \begin{cases} x+y=60\implies \boxed{y}=60-x\\\\ 0.20x+0.50y=(60)(0.40)\\ ----------\\ 0.20x+0.50\left( \boxed{60-x} \right)=(60)(0.40) \end{cases}`

solve for "x", to see how much of the 20% solution will be needed

what about "y"? well, y = 60 - x