Question

Twenty pounds of a salt-water solution contains fifteen percent salt. How much water must be added to weaken the solution to twelve percent salt?

Answer 1

Answer:

the answer is B. 5

Step-by-step explanation:

Answer 2

Using the formula

P₁V₁ = P₂V₂    (where P is percentage and V is volume in pounds)

∴ $(\frac{15}{100}) * (20) = ( \frac{12}{100}) * (x)$  (where x is the unknown volume being found)
$\frac{15}{100} * 20 * \frac{100}{12} = x$
x  = 25

∴ In order for only 12 % of salt to be percent you need to have 25 pounds of water but since you already have 20 pounds of water and 15 % salt, you only need to find how much more pounds of water to add which is 5 (25 - 20 = 5)