Algebra Word Problem

Question

A chemist is mixing one solution of 32% sodium and another solution that is 12% sodium. How many liters of each type of solution should be mixed to make 50 liters of a 20% sodium solution?

Answer 1

Hi, William

So 32% of one solution and 12% of another .

Okay, so let's call X the amount of 32% solution needed.

Then 50 - X is the amount of 32% solution needed

.32X + .12(50 - X) = .20(50)

.32X + 6 - .12X = 10

.20X = 4

X = 4/.20 = 20

Therefore, :

You need 20 L of the 32% solution

and 30 L of the 12% solution

Answer 2

Let a represent liters of 32% sodium solution.

Let b represent liters of 12% sodium solution.

We wish to find a and b such that:

a+b = 50

0.32a + .12b = 0.20*50 = 10

Solve for a by first eliminating b:

b = 50 - a

0.32a + 0.12(50 - a) = 10

0.32a + 6 - 0.12a = 10

0.20a + 6 = 10

0.20a = 4

a = 20 liters

b = 50 - a = 30 liters

Check:

0.32(20) + 0.12(30) = 6.4 + 3.6 = 10, Correct!