Question 1094740
<br>First, here is an easy way to solve mixture problems like this, if you can understand it....<br>
(1) Look to see how far (or close) the percentage of the mixture is to the percentages of the two ingredients:
{{{100-40 = 60}}}
{{{40-10 = 30}}}<br>
The percentage of the mixture, 40%, is "twice as close" to 10% as it is to 100%.  That means there must be twice as much of the 10% ingredient as the 100% ingredient.  So 2/3 of the mixture must be the 10% acid solution, and 1/3 must be the pure (100%) acid.<br>
2/3 of the 21 liters is 14 liters; so you need 14 liters of the 10% acid solution and 7 liters of pure acid.<br>
If you want to use the slow traditional algebraic solution method...<br>
let x = liters of 10% acid solution
then 21-x = liters of pure (100%) acid<br>
The total mixture is 21 liters; the amount of acid is 10% of the x, plus 100% of the (21-x).  You want the amount of acid to be 40% of the total mixture, so
{{{.10(x)+1(21-x) = .40(21)}}}
{{{.1x + 21-x = 8.4}}}
{{{12.6 = .9x}}}
{{{14 = x}}}<br>
liters of the 10% acid solution: x = 14
liters of pure (100%) acid: 21-x = 7