SOLUTION: how many liters of 40% silver must be added to 50 liters of 90% silver to make an alloy that is 60% silver

Question 815232: how many liters of 40% silver must be added to 50 liters of 90% silver to make an alloy that is 60% silver
Answer by FightinBlueHens(27) (Show Source): You can put this solution on YOUR website!
Let's start with what you know.
You need to mix 2 solutions, we'll call them A and B, to get a third solution of a different concentration.
For solution A you know that it is 50 L, and that it contains 90% silver, which means that it has 45 L of silver (50*.9=45)
For solution B, you don't know how much you need, so let's designate that as x, but it you know it has 40% silver, so the amount of silver that's in that solution is .4x.
The total (solution C) will result in the amount of 50 L + x liters (the amount of the 40% solution that you will add). This solution will have 60% silver.
Before you solve the problem, draw a table to help put this informaiton together. Write "Solution, Amount of solution, % of solution made up of silver, Amount of silver in solution" across the top, and underneath the column solution, write, A, B, and C that represnt. To get a better idea of how to complete the table see the link at the bottom of this answer.
Fill in the table.
The amount of silver you will get by mixing the two solutions will be 60% of 50+x, which you can represent by .6(50+x). It will also be equal to the amount of silver in the first solution (45 L), plus the amount of silver in the second solution (.4x, because we don't know how much of the second solution we want to add yet). Because these two amounts are equal, you can set them equal to each other to solve for x.
That is .6(50+x)=45+.4x
First use the distribution property on the left hand side of the equation, and you get 30+.6x=45+.4x
Next subtract 30 from both sides, and then .4x from both sides. These two steps are:
.6x=15+.4x and
.2x=15
Now all that's left to do is to divide both sides by .2 to solve for x, and x=75.
So you have to add 75 L of the 40% solution to get to a solution that had 60% silver.

The Khan Academy's "Mixture Problems 2" page: https://www.khanacademy.org/math/algebra/solving-linear-equations-and-inequalities/More-equation-practice/v/mixture-problems-2.
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