Question 1092146: How many liters of 80% alcohol solution and 40% alcohol solution must be mixed to obtain 12 liters of 50% alcohol solution? Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
3 parts of the 40% and 1 part of the 80%, i.e. 9 liters of the 40% and 3 liters of the 80%, IF TO SOLVE THE PROBLEM MENTALLY
Let's check: = = = 0.5 = 50%.
Yes, it works.
Now I'll show you how to solve it using equation.
Let x be the amount (the volume in liters) of the 40% solution.
Then the amount of the 80% solution is (12-x) liters.
The "alcohol contents/concentration equation" is
= 0.5.
Multiply by 12 both parts:
0.4x + 0.8*(12-x) = 6 ====>
0.4x + 9.6 - 0.8x = 6 ====>
9.6 - 6 = 0.8x - 0.4x ====> 0.4x = 3.6 ===> x = = 9.
So, we get the same answer: 9 liters of the 40% solution and (12-9) = 3 liters of the 80% solution.
