Question 1074549

One solution contains 20% alcohol and another solution contains 60% alcohol. Some of each of the two solutions is mixed to produce 10 liters of a 50% solution. How many liters of the 60% solution should be used?
<pre>Let amount of 60% (needed solution) be S
Then amount of 20% solution = 10 - S
We then get the following MIXTURE equation: .6S + .2(10 - S) = .5(10)
.6S + 2 - .2S = 5
.6S - .2S = 5 - 2
.4S = 3
S, or amount of 60% (needed) solution to mix = {{{highlight_green(matrix(1,4, 3/.4, or, 7.5, L))}}}
That's how SIMPLE this is.....nothing COMPLEX and/or CONFUSING!