Question

David needs a 50% solution of antifreeze and water in his radiator. The radiator currently has a 20% solution of antifreeze and water. If the radiator holds 16 quarts, how much of the 20% solution must be drained and replaced with pure antifreeze to create the desired solution?

Answer 1

Let the amount to be drained and replaced be $x$ quarts.

We know that the volume of the radiator is 16 quarts.

We also know that the pure antifreeze in the final solution is $50%$. Therefore, the volume of antifreeze in the final solution will be $0.5\times16=8$

Thus our equation will be:

$0.2(16-x)+x=8$

This is because the x amount of the 20% solution is removed from 16 quarts and replaced with "pure" antifreeze. And we know that by pure we mean a 100% or 1 antifreeze. Thus, $1\times x =x$.

Therefore, solving the above equation, we get:

$3.2-0.2x+x=8$

$0.8x=4.8$

$x=6$

Thus, $x=6$ quarts is the required answer.