SOLUTION: he needs to make 60 liters of a 36% acid solution to test a new product. His supplier only ships a 10% and 40% solution. he decides to make the 36% solution by mixing the 10% solut

Question 1157405: he needs to make 60 liters of a 36% acid solution to test a new product. His supplier only ships a 10% and 40% solution. he decides to make the 36% solution by mixing the 10% solution with the 40% solution. how much of the 10% solution will he need to use
Found 2 solutions by Alan3354, greenestamps:
Answer by Alan3354(69443) (Show Source): You can put this solution on YOUR website!
he needs to make 60 liters of a 36% acid solution to test a new product. His supplier only ships a 10% and 40% solution. he decides to make the 36% solution by mixing the 10% solution with the 40% solution. how much of the 10% solution will he need to use
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Similar problem:
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Birdseed costs $0.56 a pound and sunflower seeds cost $0.86 a pound. Angela Leinenbachs' pet store wishes to make a 40-pound mixture of birdseed and sunflower seeds that sells for $0.75 per a pound. How many pounds of each type of seed should she use?
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b + s = 40 --- total mix
56b + 86s = 40*75 ---- cost in cents
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

A quick and easy non-algebraic way to solve 2-part mixture problems like this:

(1) 36% is 26/30 of the way from 10% to 40% (from 10 to 40 is 30; from 10 to 36 is 26...)
(2) That means 26/30 of the mixture has to be the 40% solution

Since he is making 60 liters of the mixture, he needs 26/30 of 60 liters, or 52 liters, of the 40% solution; the other 8 liters are the 10% solution.

CHECK:
.40(52)+.10(8) = 20.8+0.8 = 21.6
.36(60) = 21.6

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