Solution A is 25% and solution B is 40% acid. How many gallons of each solution must be used in order to create a 50 gallon mixture that is 34% acid?
Solution A___
Solution B___
A+B=50
0.25A+0.40B=0.34(50)
Solve for A and B.
I don't know... i was born in the 1940's ..
i chose to reborn later in 2000 cause i died when i was thriteen in 1940's you know...
To solve this problem, we can use a mixture equation. Let's assign variables to the unknowns:
Let x be the number of gallons of solution A.
Let y be the number of gallons of solution B.
We know that the total number of gallons in the mixture is 50, so we can set up the equation:
x + y = 50 (Equation 1)
We also know that the concentration of acid in solution A is 25% and in solution B is 40%. The concentration of acid in the resulting mixture is 34%. We can use this information to set up a second equation based on the acid content:
0.25x + 0.40y = 0.34(50) (Equation 2)
Now we have a system of two equations (Equation 1 and Equation 2) with two unknowns (x and y). We can solve this system of equations to find the values of x and y.
To solve the system of equations, we can use the method of substitution or elimination.
Let's solve it using the method of substitution:
1. Solve Equation 1 for x:
x = 50 - y
2. Substitute the value of x into Equation 2:
0.25(50 - y) + 0.40y = 0.34(50)
3. Simplify and solve for y:
12.5 - 0.25y + 0.40y = 17
0.15y = 4.5
y = 4.5 / 0.15
y = 30
Now we know that we need 30 gallons of solution B.
4. Substitute the value of y back into Equation 1 to solve for x:
x + 30 = 50
x = 50 - 30
x = 20
Therefore, we need 20 gallons of solution A.
To summarize:
Solution A: 20 gallons
Solution B: 30 gallons