Your automobile radiator contains 15 quarts of an antifreeze mixture.

A test shows the mixture is 40% antifreeze and 60% water. The owner's manual says the mixture should be 50% antifreeze and 50% water.

How much of the mixture should be removed and replaced with pure antifreeze so the final mixture is 50% antifreeze.

there should be 7.5 qt each of water and antifreeze

there are 9 qt of water ... .60 * 15 = 9

removing 1.5 qt of water from a 60% mix
... 1.5 = .60 * x ... 2.5 = x

remove 2.5 gal of the mixture and refill with pure antifreeze

To find out how much of the mixture should be removed and replaced with pure antifreeze, we need to follow these steps:

Step 1: Determine the amount of antifreeze currently in the radiator.
Since the mixture in the radiator contains 15 quarts of liquid and it is 40% antifreeze, we can calculate the amount of antifreeze present as follows:
Antifreeze = 40/100 * 15 quarts
Antifreeze = 6 quarts

Step 2: Determine the amount of water currently in the radiator.
Since the mixture in the radiator is 60% water, we can calculate the amount of water present as follows:
Water = 60/100 * 15 quarts
Water = 9 quarts

Step 3: Determine the total volume of the mixture that needs to be replaced.
Let's assume we need to remove 'x' quarts of the mixture and replace it with pure antifreeze. The total volume of the mixture that needs to be replaced is x quarts.

Step 4: Set up an equation based on the desired final ratio of 50% antifreeze and 50% water.
The equation can be set up as follows:
6 quarts (current antifreeze) - x quarts (removed antifreeze) + x quarts (added antifreeze) = 0.5 * (15 quarts)

Step 5: Solve the equation to find the value of 'x'.
Let's solve the equation:
6 - x + x = 0.5 * 15
6 = 7.5 - 0.5x
0.5x = 7.5 - 6
0.5x = 1.5
x = 1.5 / 0.5
x = 3

So, 3 quarts of the mixture should be removed, and 3 quarts of pure antifreeze should be added to achieve a final mixture of 50% antifreeze.