Suppose that you have a supply of a 30% solution of alcohol and a 70% solution of alcohol. How many quarts of each should be mixed to produce 40 quarts that is 40% alcohol?
.3*A+.7B=.4*40
A+B=40
Does that help?
To determine the amounts of the 30% and 70% alcohol solutions needed to produce a 40-quart mixture that is 40% alcohol, we can use the concept of the equation.
Let's assume you need "x" quarts of the 30% solution and "y" quarts of the 70% solution. Since you want to produce a total of 40 quarts, we have the equation:
x + y = 40 -- Equation 1
Next, we consider the amount of alcohol in the mixture. The 30% solution contains 30% alcohol, which means that out of each quart, 0.30 quarts are alcohol. Similarly, the 70% solution contains 70% alcohol, which is 0.70 quarts per quart. Therefore, the equation for the alcohol content in the mixture is:
0.30x + 0.70y = 0.40(40) -- Equation 2
Simplifying equation 2, we have:
0.30x + 0.70y = 16
Now, we can solve this system of equations (equation 1 and equation 2) to find the values of x and y.
There are multiple ways to solve these equations, but let's use the method of substitution. Solving equation 1 for x, we get:
x = 40 - y
Substituting x in equation 2, we have:
0.30(40 - y) + 0.70y = 16
Expanding the equation:
12 - 0.30y + 0.70y = 16
Combining like terms:
0.40y = 4
Divide both sides by 0.40:
y = 10
Substituting this value of y back into equation 1:
x + 10 = 40
x = 30
Therefore, you would need 30 quarts of the 30% alcohol solution and 10 quarts of the 70% alcohol solution to produce 40 quarts of a 40% alcohol mixture.