Question 1113358: How many ounces of a 35% acid solution should be mixed with a 60% acid solution to get 40 ounces of a 50% acid solution? Found 2 solutions by mananth, ikleyn:Answer by mananth(16946) (Show Source):
You can put this solution on YOUR website! component percent ---------------- quantity
Acid type I 0.35 ---------------- x ounce
Acid type II 0.6 ---------------- 40 - x ounce
Mixture 0.50 ---------------- 40
0.35 x + 0.6 ( 40 - x ) = 40.00 * 0.50
0.35 x + 24 - 0.6 x = 20.00
0.35 x - 0.6 x = 20 - 24
-0.25 x = -4
/ -0.25
x = 16
Let x = the amount of the 35% acid solution needed (in ounces).
The the amount of the 60% acid solution needed is (40-x) ounces.
The amount of the "pure" acid in these solutions is 0.35*x + 0.6*(40-x).
It is exactly 50%, or 0.5 of 40 ounces, i.e. 0.5*40.
So, your "pure acid amount" equation is
0.35*x + 0.6*(40-x) = 0.5*40.
Simplify and solve it for x:
0.35x + 24 - 0.6x = 20,
-0.25x = 20 - 24,
-0.25x = -4,
x = = = 16.
Answer. 16 ounces of the 35% acid solution is needed.
Check. 16*0.35 + (40-16)*0.60 = 20 = 0.5*40 ounces. ! Correct !
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