Solving a percent mixture problem

Question

a scientist has two solutions, which she has labeled solution A and Solution B. Each contains salt. She knows that Solutuin A is 60% salt and Solution B is 85% salt. She wants to obtain 130 ounces of a mixture that is 80% salt. How many ounces of each solution should she use? 

Answer 1

60%x+85%(130-x)=80%(130)

0.6x+0.85(130-x)=0.8(130)

0.6x+110.5-0.85x=104

-0.25x+110.5=104

-0.25x=104-110.5

-0.25x=-6.5

0.25x=6.5

25x=650

x=650/25

x=26 ounces of the 60% solution

130-26=104 ounces of the 85 % solution

 

Answer 2

You can set this up as a system of two equations, using the two things you know about weight, in oz, of Solution A you need (which we can label w_A) and the weight of Solution B you need (w_B).

The first thing you know is that the total weight is 183 oz, which we can represent in equation form as:

w_A + w_B = 130

Then, we know that 60% of Solution A and 85% of Solution B must equal 80% of 130 oz, to give the correct amount of salt in the resultant solution.  This can be written as

0.6w_A + 0.85w_B = 0.8*130 = 104

We can solve this system in several ways, of which I will choose substitution: using the first equation above to write the second equation in terms of one variable, and then solve for that variable, like this:

w_A = 130 - w_B (this is just a rearrangement of the first equation)

Then the second equation can be written as: 0.6(130-w_B) + 0.85w_B = 104
This can now be solved algebraically for w_B:

78 - 0.6w_B + 0.85w_B = 104 -->78 + 0.25w_B = 104 --> 0.25w_B = 104 - 78 = 26 --> w_B = 26/.25 = 104 oz

So 104 oz of Solution B is required. Putting this value back into the first equation lets us solve for w_A, which is 26 oz.