Question

David needs to have a 20% solution of antifreeze in his truck's radiator. If the radiator is now full and contains 8 quarts of 10% antifreeze solution, how much of that solution must be drained and replaced with pure antifreeze to create the desired solution?

Answer 1

Approximately 0.89 quarts solution must be drained and replaced with pure antifreeze to create the desired solution.

Step-by-step explanation

Suppose,
`x` quarts solution must be drained and replaced with pure antifreeze.

So, the amount of solution in the radiator after draining out
`=(8-x)` quarts, which is 10% antifreeze solution.

That drained out solution is replaced with
`x` quarts 100% antifreeze solution.

David needs to have a 20% solution of antifreeze in the radiator, which can contain 8 quarts solution. So, the equation will be.......

`10\%*(8-x)+100\%*(x)=20\%*(8)\\ \\ 0.10(8-x)+1.00(x)=0.20(8)\\ \\ 0.80-0.10x+1.00x=1.60\\ \\ 0.90x=1.60-0.80\\ \\ 0.90x=0.80\\ \\ x=(0.80)/(0.90)=0.888... \approx 0.89`

So, the amount of solution that must be drained and replaced with pure antifreeze will be approximately 0.89 quarts.