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*16=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* 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.