SOLUTION: Solution A has 50% alcohol wine while solution B has 75% alcohol,how much of each solution must be added to come up with 4 liters of alcohol solution with 60% alcohol?
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Question 1097577: Solution A has 50% alcohol wine while solution B has 75% alcohol,how much of each solution must be added to come up with 4 liters of alcohol solution with 60% alcohol?
Let x = the amount of the 50% alcohol wine needed (in liters).
Then the amount of the 75% alcohol wine needed is (4-x) liters.
The amount of the "pure" alcohol in these solutions is 0.5*x + 0.75*(4-x).
It must be exactly 60%, or 0.6 of 4 liters, i.e. 0.6*4 liters.
So, your "pure alcohol" equation is
0.5*x + 0.75*(4-x) = 0.6*4.
Simplify and solve it for x:
0.5x + 3 - 0.75x = 2.4,
-0.25x = 2.4 - 3,
-0.25x = -0.6,
x = = 2.4.
Answer. 2.4 liters of 50% alcohol wine and 4-2.4 = 1.6 liters of 75% alcohol wine are needed.
Check. 2.4*0.5 + 1.6*0.75 = 2.4 = 4*0.6. ! Correct !
