Solution A is 80% Alcohol and Solution B is 30% Alcohol. How many Liters of each Solutions should be mixed in order to make 200 liters of a Solution that is 62% Alcohol ?

.8x + .3(200-x) = .62(200)

yes

To determine the number of liters of each solution needed to obtain a 200-liter solution that is 62% alcohol, we can set up a system of equations.

Let x represent the volume (in liters) of Solution A.
Let y represent the volume (in liters) of Solution B.

Since the total volume of the final solution is 200 liters, we have the equation:
x + y = 200 -- Eq. 1

To determine the alcohol content of the final solution, we need to consider the alcohol content of each solution. The equation based on alcohol concentration is as follows:
0.8x + 0.3y = 0.62(200) -- Eq. 2

Simplifying Eq. 2, we have:
0.8x + 0.3y = 124 -- Eq. 2

Now, we can solve the system of equations (Eq. 1 and Eq. 2) to find the values of x and y.

Method 1: Substitution method
Solve Eq. 1 for x:
x = 200 - y

Substitute x into Eq. 2:
0.8(200 - y) + 0.3y = 124

Simplify and solve for y:
160 - 0.8y + 0.3y = 124
0.5y = 36
y = 72

Substitute y back into Eq. 1 to solve for x:
x + 72 = 200
x = 200 - 72
x = 128

Therefore, 128 liters of Solution A and 72 liters of Solution B should be mixed to obtain 200 liters of a solution that is 62% alcohol.

Method 2: Elimination method
Multiply Eq. 1 by 0.8 to match the x term in Eq. 2:
0.8x + 0.8y = 160 -- Eq. 1'

Subtract Eq. 1' from Eq. 2:
0.8x + 0.3y - (0.8x + 0.8y) = 124 - 160
0.8x - 0.8x + 0.3y - 0.8y = -36
-0.5y = -36
y = 72

Substitute y back into Eq. 1 to solve for x:
x + 72 = 200
x = 200 - 72
x = 128

Therefore, 128 liters of Solution A and 72 liters of Solution B should be mixed to obtain 200 liters of a solution that is 62% alcohol.