9514 1404 393
Answer:
- 10
- 12%: 4.8 L; 32%: 3.2 L
Step-by-step explanation:
For mixture problems, I like to use a single variable to represent the quantity of the greatest contributor (greatest cost or greatest concentration). Let that variable be x.
1. In this problem, the amount of the greatest contributor is already fixed, so we'll use x for the amount of walnuts. The cost of the mix is ...
0.80x + 1.25(8) = 1.00(x +8)
2 = 0.2x . . . . . . . . . . . . . . . . . subtract 0.8x+8, simplify
10 = x . . . . . . . . . divide by 0.2
10 pounds of walnuts must be used.
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2. The amount of acid in the mix is ...
12%(8 -x) +32%(x) = 20%(8)
20%(x) = 8%(8) . . . . . . . . . . . . subtract 12%(8)
x = 8(8)/20 = 3.2 . . . . . . . . . . divide by 20%
8-x = 4.8
3.2 liters of 32% acid must be mixed with 4.8 liters of 12% acid
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<em>Comment on mixture problems</em>
The general solution for mixture problems that have a greater contributor (G), a lesser contributor (L), and a mix value (M) is ...
g/l = (M -L)/(G -M) . . . . . . ratio of quantities of G and L
In the first problem, this becomes ...
8/l = (1.00 -.80)/(1.25 -1.00) = .2/.25 = 4/5 ⇒ l = 8(5/4) = 10
In the second problem, this becomes ...
g/l = (20-12)/(32-20) = 8/12 = 2/3 ⇒ g = (2/5)(8) = 3.2; l = (3/5)(8) = 4.8
In this case, we recognize that the ratio of 2 parts to 3 parts means that the greater contributor is 2 of 5 total parts.
