Boris has 10 ounces of a 10% acid solution. How many ounces of 25% acid solution should he add to make a solution that is 20% acid?

Thank you.

10 oz pot has 1 oz acid and 9 oz water

x oz pot has .25 x oz acid and .75 x oz water

so

acid = 1 + .25 x
water = 9 + .75 x
total = 10 + x
and
(1 + .25 x) = .2 ( 10 + x)
1 + .25 x = 2 + .2 x
.05 x = 1
x = 20 oz

Thank you very much for all your great help! Once again, thank.

You are welcome :)

To solve this problem, we need to use a mixture equation, which states that the amount of acid in the final solution is equal to the sum of the amounts of acid in the individual solutions.

Let's assume Boris adds x ounces of the 25% acid solution to his 10 ounces of 10% acid solution.

In the 10-ounce solution, the amount of acid is 10% of 10, which is 10(0.10) = 1 ounce.

In the x-ounce solution of 25% acid, the amount of acid is 25% of x, which is 0.25x ounces.

In the final mixture, the amount of acid is the sum of the acid amounts in the individual solutions:

1 ounce (from the 10% solution) + 0.25x ounces (from the 25% solution).

According to the problem, we want a final mixture that is 20% acid. So, the amount of acid in the final mixture is 20% of the total mixture, which is 0.20 times the sum of the amounts of the individual solutions:

0.20 (1 + 0.25x) = 0.20 + 0.05x

Now we can set up an equation to solve for x:

1 + 0.25x = 0.20 + 0.05x

Subtract 0.20 and 0.05x from both sides:

0.25x - 0.05x = 0.20 - 1

Combine like terms:

0.20x = -0.80

Divide both sides by 0.20:

x = -0.80 / 0.20

x = -4

The solution gives us a negative value for x, which does not make sense in the context of the problem. It is not possible to add a negative amount of 25% acid solution.

Therefore, there is no solution to this problem. Boris cannot make a solution that is 20% acid by adding the given amounts of 10% and 25% acid solutions.