Question 1124678: Terry needs 20 Gallons of 40% acid. The only available acid solutions are 25% and 80%. How much of each should she mix? Answer by ikleyn(52776) (Show Source):
Let x = the amount of the 25% acid solution needed (in gallons).
Then the amount of the 80% acid solution needed is (20-x) gallons.
The amount of the "pure" acid in these solutions is 0.25*x + 0.8*(20-x) gallons.
It should be exactly 40%, or 0.4 of 20 gallons, i.e. 0.4*20 gallons.
So, your "pure acid amount" equation is
0.25*x + 0.8*(20-x) = 0.4*20.
Simplify and solve it for x:
0.25x + 16 - 0.8x = 8,
-0.55x = 8 - 16,
-0.55x = -8,
x = = 14.54(54) gallons.
Answer. 14.54 gallons of the 25% acid solution and 20-14.54 = 5.45 of the 80% acid solution are needed.
Check. 0.25*14.54 + 0.8*5.45 = 8 gallons of pure acid = 0.4*20 ounces. ! Correct !
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