Question 1156222
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I am capable of forgiving the author of the problem for not knowing that saline solutions of those percentages are not possible.  Clearly the purpose of the problem is to learn how to solve mixture problems.  So let's change the saline solutions to antifreeze or acid and work the problem.<br>
Algebraically....<br>
20% of x ounces, plus 50% of 200 ounces, makes 30% of (x+200) ounces:<br>
{{{.20(x)+.50(200) = .30(x+200)}}}
{{{.2x+100 = .3x+60}}}
{{{40 = .1x}}}
{{{x = 400}}}<br>
ANSWER: 400 ounces<br>
There is a MUCH easier way to solve problems like this, if a formal algebraic solution is not required.<br>
Here is the complete solution, in two sentences:<br>
30% is "twice as close to 20% as it is to 50%"; therefore the amount of the 20% ingredient is twice the amount of the 50% ingredient".
200 ounces of 50% means 2*200 = 400 ounces of 20%.<br>