Solution A is 20% salt and solution B is 80% salt. If you have 30 oz of Solution A and 60 ounces of Solution B, in what ratio could you mix Solution A with solution B to produce 50 oz of a 50% salt solution?

A) 6:4
B) 6:14
C) 4:4
D) 4:6
E) 3:7

2 Methods of Solving:

1. How can I solve this problem with weighted averages?
2. I tried to make the following table:

Sol A 30 oz 20% 60*0.2
Sol B 80 oz 80% 80*0.8
Tot 30+60=90 15% ???

How can I solve it this way? where: ??? what should I put?

Thank you!

Good question, Arturo. The Weighted Average method, from the Problem Solving lesson (lesson 10) works really well here. If solution A is 20 parts salt and solution B is 80 parts salt, but the resulting mixture is 50 parts salt, you could draw out a number line with:

20 (A) 50 (combined) 80 (B)
The distance from A to the average is 30, and from B to the average is 30, so the distances would be in the ratio of 30:30, or 1:1.

So, we know that we'll need equal parts A and B, so the correct answer choice is 4:4.

Now, one last step - we need to make sure that it's possible to have equal amounts of each add up to 50. Because we have 30 of A and 60 of B, it does work - we could have 25 of each.

If you did this algebraically, you could say that:

.2a + .8b = .5(50) (20% of A + 80% of B = 50% of 50, using the amount of salt as the percentage)
a + b = 50 (the combination of a and b will add up to 50)

a = 50 - b (to substitute for a in the first equation)
so...
.2 (50 - b) + .8b = .5(50)
I'd multiply everything by 10 to get rid of the decimals and make it cleaner:

2(50 - b) + 8b = 250
100 - 2b + 8b = 250
100 + 6b = 250
6b = 150
b = 25

So there are 25 ounces of b, and 50 ounces total, so a must also be 25 ounces, giving us a ratio of 1:1, which also works out to 4:4.