Question 1190166
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The fact that a saline solution can have a concentration of no more than 26% is not universal knowledge.  This is a math website, not a physics/chemistry website.  The intent of the question is clearly to give the student practice in learning how to solve mixture problems.  Ignore the fact that the given situation is physically impossible and help the student learn.<br>
(1) Using formal algebra....<br>
x = mL of water to be added<br>
Adding water adds no salt to the mixture, so the 30% of the (150+x) mL of the final mixture that is salt is still only the 80% of the 150mL of the original solution.  So<br>
{{{.30(150+x)=.80(150)}}}
{{{45+.30x=120}}}
{{{.30x=75}}}
{{{x=75/.30=250}}}<br>
ANSWER (algebraically): 250mL<br>
(2) If a formal algebraic solution is not required....<br>
You are starting with a solution that is 80% salt, adding water (0% salt), stopping when the solution reaches 30%.<br>
Look at the three percentages 80, 30, and 0 on a number line and observe/calculate that 30 is 5/8 of the way from 80 to 0.
That means 5/8 of the final mixture is what you are adding (water); and that makes the 150mL you started with 3/8 of the final mixture.
A simple proportion then shows the amount of water to be added is 250mL.<br>
ANSWER (informally): 250mL<br>