How much of 20% alcohol solution and 50% alcohol solution must be mixted to get 9 liters of 30% alcohol solution?
To find out how much of each solution is needed, we can set up an algebraic equation based on the amount of pure alcohol in each solution.
Let's assume that x liters of the 20% alcohol solution are mixed with (9 - x) liters of the 50% alcohol solution to get a total of 9 liters of the 30% alcohol solution.
Now, let's calculate the amount of pure alcohol in each solution:
Amount of pure alcohol in the 20% solution:
0.20x
Amount of pure alcohol in the 50% solution:
0.50(9 - x)
Therefore, the amount of pure alcohol in the 30% solution will be:
0.30 * 9 = 0.20x + 0.50(9 - x)
Now, we can solve this equation to find the value of x, which represents the amount of the 20% alcohol solution needed:
0.30 * 9 = 0.20x + 0.50(9 - x)
2.7 = 0.20x + 4.5 - 0.50x
2.7 - 4.5 = -0.30x
-1.8 = -0.30x
x = -1.8 / -0.30
x = 6
So, 6 liters of the 20% alcohol solution are needed, and the remaining (9 - 6 = 3) liters will be the 50% alcohol solution.