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Answer:
- 36 liters of 20%
- 18 liters of 35%
- 54 liters of 55%
Step-by-step explanation:
Let a, b, c represent the quantities of 20%, 35%, and 55% solutions, respectively. Then for the given mix, we have ...
a + b + c = 108 . . . . . total number of liters
.20a +.35b +.55c = 0.40·108 . . . . . liters of acid
0a - 3b + c = 0 . . . . . 55% solution was 3 times 35% solution
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Solving these three equations by your favorite method gives ...
(a, b, c) = (36, 18, 54)
The chemist used 36 liters of 20%, 18 liters of 35%, and 54 liters of 55%.
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Comment on alternate approach
Since the ratio of 55% to 35% is 3 : 1, that mix will have an acid content of ...
(3(.55) +1(.35))/4 = 0.50 = 50%
So, the final mix is equivalent to some quantity of 50% mix being added to some quantity of 20% mix. The fraction that is 50% mix* will be (40-20)/(50-20) = 2/3 of the total, or (2/3)(108 L) = 72 L.
Now we know that (1/3)(108 L) = 36 L of 20% solution is needed and (1/4)(72 L) = 18 L of 35% solution is needed. 3 times that is 54 L of 55% solution.
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* After you work a few mixture problems, you can see the pattern. If we start with an equation for x = fraction of 50% to use with 20% to make 40%, we get ...
x(50%) +(1 -x)(20%) = 40%
x(50% -20%) = 40% -20% . . . . . subtract 1(20%) and factor out x
x = (40% -20%)/(50% -20%)
The key here is to see where the numbers 20%, 40%, and 50% show up in the fraction.
