A 10 L mixture is made up of 40% antifreeze. How much of the mixture needs to be removed and filled with pure antifreeze to reach a concentration of 60% antifreeze

3.3 liters of the mixture is needed to be removed and filled with 3.33 liters pure antifreeze to reach a concentration of 60% antifreeze.

Explanation:

A 10 L mixture is made up of 40% antifreeze.

Initial Volume of antifreeze in 40% mixture = 40% of 10 L = 4L

Let volume of pure antifreeze added = x

Volume of antifreeze removed = 40% of x = 0.4 x

Volume of antifreeze in 60% mixture = 60% of 10 L = 6 L

Volume of antifreeze left after removal of 0.4 x L of antifreeze and addition of x L of pure antifreeze will be equal to 6 L of antifreeze in the final solution.

4L - (0.4 x) + x = 6L

x = 3.33 L

3.3 liters of the mixture is needed to be removed and filled with 3.3 liter pure antifreeze to reach a concentration of 60% antifreeze.