An 80% acid solution is to be mixed with a 30% solution to get 90 gallons of a 50% solution. How many gallons of each solution are needed?
0.8x + 0.3(90-x) = .5(90)
.8x + 27 -.3x = 45
.5x +27 = 45
.5x + 27 - 27 = 45-27
.5x = 18
x = 36
36 at 80%
54 at 30%
Kuai, you are awesome!!! Thank you very much! I really appreciate it!
To solve this problem, we'll use a system of equations.
Let's assume x represents the number of gallons of the 80% acid solution, and y represents the number of gallons of the 30% acid solution.
We know that the total volume of the mixture is 90 gallons, so our first equation is:
x + y = 90 (Equation 1)
We also know that the acid concentration in the mixture is 50%, so we can set up a second equation based on the acid concentration:
(0.80x + 0.30y) / (x + y) = 0.50 (Equation 2)
Now we can solve this system of equations using substitution or elimination.
Let's solve by substitution. From Equation 1, we can write x = 90 - y.
Substituting this value in Equation 2, we have:
(0.80(90 - y) + 0.30y) / (90) = 0.50
Simplifying, we get:
72 - 0.80y + 0.30y = 0.50 * 90
72 - 0.50y = 45
-0.50y = 45 - 72
-0.50y = -27
y = (-27) / (-0.50)
y = 54
Substituting the value of y back into Equation 1, we can find x:
x + 54 = 90
x = 90 - 54
x = 36
Therefore, we need 36 gallons of the 80% acid solution and 54 gallons of the 30% acid solution to get 90 gallons of a 50% acid solution.