Suppose that you have a supply of a 25% solution of alcohol and a 75% solution of alcohol. How many quarts of each should be mixed to produce 90 quarts that is 60% alcohol?
Thank you,and how did you get come up with the equation?
Thank you,and how did set up the equation?
If you let x be the amount of 25% solution, then .25x is the amount of alcohol in that solutions.
Since there are 90 qts in all, 90-x is the amount of the 75% solution. .75(90-x) is the amount of alcohol in that solution.
Add up the two amounts of alcohol, and it must equal the amount of alcohol in 90 qts of 60% solution.
Ta-daa
Thanks
To solve this problem, we can use a method called a mixture problem.
Let's denote the amount of 25% alcohol solution as x (in quarts) and the amount of 75% alcohol solution as y (in quarts).
Since the total amount of the mixture is 90 quarts, we have the equation:
x + y = 90
We also know that the resulting mixture should be 60% alcohol. The amount of alcohol in the 25% solution is 0.25x, and the amount of alcohol in the 75% solution is 0.75y.
Therefore, the equation for the total amount of alcohol in the mixture is:
0.25x + 0.75y = 0.6 * 90
Simplifying the equation:
0.25x + 0.75y = 54
Now we have a system of two equations:
x + y = 90
0.25x + 0.75y = 54
To solve this system, we can use substitution or elimination.
Let's use the substitution method:
From the first equation, solve for x:
x = 90 - y
Substitute this value of x into the second equation:
0.25(90-y) + 0.75y = 54
Distribute:
22.5 - 0.25y + 0.75y = 54
Combine like terms:
0.5y = 31.5
Divide by 0.5:
y = 63
Now substitute this value of y back into the first equation to find x:
x + 63 = 90
x = 90 - 63
x = 27
Therefore, you would need 27 quarts of the 25% alcohol solution and 63 quarts of the 75% alcohol solution to produce 90 quarts of a 60% alcohol mixture.
work with the amount of alcohol in each part. It has to add up to the total alcohol.
.25x + .75(90-x) = .60(90)
x = 27
so, 27 qt @ 25% and 63 qt @ 75%