SOLUTION: A chemist needs 70 milliliters of a 41% solution but has only 21% and 56% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.

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Question 9187: A chemist needs 70 milliliters of a 41% solution but has only 21% and 56% solutions available. Find how many milliliters of each that should be mixed to get the desired solution.
Answer by rapaljer(4551) About Me  (Show Source):
You can put this solution on YOUR website!
Let x = amount of liquid at 56%
70-x = amount of liquid at 21%
70 = amount of liquid at 41%

(The above can be reversed, and it works almost as well!!)

.56x + .21(70-x) = .41(70)
.56x + 14.7 - .21x = 28.7
.35x + 14.7 = 28.7
.35x + 14.7 - 14.7 = 28.7 - 14.7
.35x = 14
x+=+14%2F.35 = 1400%2F35 = 40 milliliters of 56% solution
70 - x = 30 milliliters of 21% solution

Check:
.56(40) + .21(30) = .41(70)
22.4 + 6.3 = 28.7

Praise the Lord!! It checks!!

R^2 at SCC