math word problem.. more!?

Question

how many liters of a -10% solution of a chemical should mike mix with 30 liters of a 50% soution to obtain a mixture that is 40% chemical?


help..

Answer 1

I assume you mean 10%, not -10%, because a solution can't have a negative percent of anything.

Let N be the amount of 10% solution.
That is added to 30 liters of 50% solution

The final N + 30 liter solution should be a 40% solution.

Writing that as an equation:
10%(N) + 50%(30) = 40%(N + 30)

Now you can multiply by 100 on both sides to get rid of the % signs:

10N + 50*30 = 40(N + 30)

Distribute the 40 through the parentheses:
10N + 1500 = 40N + 1200

Subtract 10N from both sides:
1500 = 30N + 1200

Subtract 1200 from both sides:
300 = 30N

Divide both sides by 30:
300/30 = N
N = 10

So you need to add 10 liters of a 10% solution.

Answer 2

Let p = the amount of the 10% solution
Let q = amount of the 50% solution.

The amount of the chemical in the combined solution is
0.10 p + 30 * 0.50.
The quantity of the combined solution is 30 + p
The amount of the chemical in the solution is 40% of the solution.
So...

(0.10p + 30 * 0.50) / (30 + p) = 0.40
(0.10p + 15) / (30 + p) = 0.40
0.10p + 15 = 0.40 (30 + p)
0.10p + 15 = 12 + 0.40p
3 = 0.30p
p = 3/0.30
p = 10.

We end up with 16 liters of chemical in the 40 liters of solution.
16 / 40 = 0.40 = 40%
That's what we're looking for!