How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 80% antifreeze?

The amount of water to be added is ¹/₄ gallon

Step-by-step explanation:

Given;

amount of pure antifreeze at 100% = 1 gallon

let the amount of water to be added with 0% antifreeze = (x) gallon

then, the amount of mixture at 80% antifreeze = (x + 1) gallon

For conservation of mass, we will have the following equation;

(1 x 1) + (0)(x) = 0.8(x +1)

1 = 0.8x + 0.8

0.8x = 1 - 0.8

0.8x = 0.2

x = 0.2 / 0.8

x = ²/₈

x = ¹/₄ gallon

Therefore, the amount of water to be added is ¹/₄ gallon

step-by-step explanation:

what grade level is this? need to know so i can look in my old textbooks!