Question

Suppose a lab need to need to make 400 liters of 39% acid solution, but the only solutions available to the lab are 20% acid and 50% acid. What system of equation can be used to find the number of liters of each solution the should be mixed to make the 39% solution? Let c represent the number of liters of 20% acid solution and let d represent the number of liters of 50% acid solution.

Answer 1

c + d = 400

0.2c + 0.5d = 156

Explanation:

Let

c = number of liters of 20% acid solution

d = number of liters of 50% acid solution.

c + d = 400 (1)

0.2c + 0.5d = 400(0.39)

0.2c + 0.5d = 156 (2)

From (1)

c = 400 - d

Substitute c = 400 - d into (2)

0.2c + 0.5d = 156 (2)

0.2(400 - d) + 0.5d = 156

80 - 0.2d + 0.5d = 156

- 0.2d + 0.5d = 156 - 80

0.3d = 76

d = 76/0.3

d = 253.33 liters

Substitute d = 253.33 into (1)

c + d = 400

c + 253.33 = 400

c = 400 - 253.33

c = 146.67 liters