Question 1092040
<br>First a traditional algebraic solution; and then I will show you the unorthodox method I use to solve mixture problems.  If you understand my method, it will get you to the answer much faster and with far less work.<br>
You are draining x liters of the 50% salt solution and replacing it with distilled water (0% salt) to obtain 40 liters of 40% salt solution.  Don't get confused with the "draining and replacing"; in the end you are simply mixing two ingredients.<br>
  let x = liters of distilled water (0% salt)
  then 40-x = liters of 50% salt solution<br>
You want the 40 liters you end up with to be 40% salt.  Equating the amounts of salt in the two ingredients to the amount in the final mixture gives you the algebraic equation:<br>
{{{(0)x+.50(40-x) = .40(40)}}}
{{{20-.5x = 16}}}
{{{4 = .5x}}}
{{{x = 8}}}<br>
You need to drain 8 liters of the 50% salt solution and replace it with distilled water to get 40 liters of 40% salt solution.<br>
Now my method, based on the ratios in which the two ingredients are mixed.<br>
The difference between the percentages of salt in the original salt solution and in the desired mixture is 50-40=10; the difference between the percentages of salt in the desired mixture and in the distilled water is 40-0 = 0.<br>
Those differences of 40 and 10 mean the two ingredients must be mixed in the ratio 40:10, or 4:1.  And since 40% is closer to 50% than to 0%, the larger portion must be the 50% salt solution.<br>
But 40 liters in the ratio 4:1 means 32 liters of one ingredient and 8 of the other.<br>
So you need 32 liters of the original 50% salt solution and 8 liters of distilled water.<br>
No equations to solve; just a couple of subtractions and finding a ratio....