A chemist needs to mix an 18% acid solution with a 45% acid solution to obtain 12 liters of a 36% solution. How many liters of each of the acid solutions must be used?
.18 a + .45 b = .36 (a+b)
a + b = 12
so b = (12-a)
then
.18 a + .45 (12-a) = .36 [ a + (12-a) ]
To solve this problem, we can use the method of solving a system of linear equations. Let's assume that the chemist needs to mix x liters of the 18% acid solution and y liters of the 45% acid solution.
First, let's set up the equation based on the amount of acid in the solution:
The amount of acid in the 18% solution is 0.18x (18% can be written as 0.18 because percentages are divided by 100).
The amount of acid in the 45% solution is 0.45y.
Since we want to obtain a 12-liter solution that is 36% acid, the equation for the total amount of acid can be written as:
0.18x + 0.45y = 0.36 * 12
Simplifying the equation further, we get:
0.18x + 0.45y = 4.32
We also know that the total volume of the solution is 12 liters, so we can set up another equation:
x + y = 12
Now we have a system of linear equations:
0.18x + 0.45y = 4.32
x + y = 12
We can solve this system of equations using various methods, such as substitution or elimination. Let's use the substitution method:
From the second equation, we can isolate x: x = 12 - y
Substituting this value of x into the first equation, we get:
0.18(12 - y) + 0.45y = 4.32
Simplifying further:
2.16 - 0.18y + 0.45y = 4.32
0.27y = 2.16
y = 2.16 / 0.27
y = 8
So, we need 8 liters of the 45% acid solution.
To find x, we substitute the value of y into the second equation:
x + 8 = 12
x = 12 - 8
x = 4
Therefore, the chemist needs to mix 4 liters of the 18% acid solution and 8 liters of the 45% acid solution to obtain 12 liters of a 36% acid solution.