Question 63246: How many grams of salt must be added to 50 grams of a 20% solution in order to increase the content to 25%? Found 2 solutions by ptaylor, 303795:Answer by ptaylor(2198) (Show Source):
You can put this solution on YOUR website! Let x=amount of salt to be added
We know that the amount of salt added (x) plus the amount of salt in the original 50 gms (50)(.20) must equal the amount of salt in the final solution (50+x)(.25). Thus, our equation to solve is:
x+(50)(.20)=(50+x)(.25) simplifying, we get:
x+10=12.5+.25x or
.75x=2.5
x=3.34 gms of salt
A slightly easier way to solve this problem would be to realize that the amount of the "otherstuff" in the solution does not change. Thus, the amount of "otherstuff" in the original 50 gms (50)(.80) equals the amount of "otherstuff" in the final solution (50+x)(.75). So our equation would be:
(50)(.80)=(50+x)(.75) simplifying, we get:
40=37.5+.75x or
.75x=2.5
x=3.34 gms of salt
Hope this helps----ptaylor
You can put this solution on YOUR website! If we have 50 grams of a 20% solution then 20% of the 50 grams is salt.
20/100 x 50 = 10g of salt. This means we must have 40 grams of water.
The amount of salt in the mixture is x. So the amount of solution will therefore be x + 40
The fraction of salt in the mixture will be x/(x+40) which must equal 25%. 25% is one quarter Multiply both sides by (x + 40) Multiply both sides by 4
The original mixture contained 10 g so an extra 3.3g of salt must be added.
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