A Chemist needs 200 liters of 30% acid solution. she has a 20% and 50% acid solution. How much of each must she mix?

amount of the 20% stuff --- x L

amount of the 50% stuff --- 200-x L

.2x + .5(200-x) = .3(200)
times 10
2x + 5(200-x) = 3(200)
2x + 1000 - 5x = 600
-3x = -400
x = 400/3

The chemist needs 400/3 L of the 20% solution and 200/3 L of the 50% solution

To find out how much of each solution the chemist must mix, we can set up a system of equations based on the information given.

Let's assume x represents the amount (in liters) of the 20% acid solution, and y represents the amount (in liters) of the 50% acid solution.

The first equation is based on the total volume of the solution:
x + y = 200

The second equation is based on the concentration of the acid in the solution:
(0.20x + 0.50y) / (x + y) = 0.30

Now, we can solve this system of equations to find the values of x and y.

1. Solve the first equation for x:
x = 200 - y

2. Substitute this value of x into the second equation:
(0.20(200-y) + 0.50y) / (200-y + y) = 0.30

Simplifying the equation:
(40 - 0.20y + 0.50y) / 200 = 0.30
(40 + 0.30y) / 200 = 0.30

Multiply both sides of the equation by 200 to remove the fraction:
40 + 0.30y = 0.30 * 200
40 + 0.30y = 60

Subtract 40 from both sides:
0.30y = 60 - 40
0.30y = 20

Divide both sides by 0.30 to solve for y:
y = 20 / 0.30

Calculating:
y ≈ 66.67

So, the chemist needs to mix approximately 66.67 liters of the 50% acid solution.

3. Substitute this value of y back into the first equation to find x:
x = 200 - y
x = 200 - 66.67

Calculating:
x ≈ 133.33

Therefore, the chemist needs to mix approximately 133.33 liters of the 20% acid solution.

To summarize, the chemist should mix approximately 133.33 liters of the 20% acid solution with approximately 66.67 liters of the 50% acid solution to obtain 200 liters of a 30% acid solution.