Question

[Will give medal+fan]

A radiator contains 5 quarts of fluid, 20% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 55% antifreeze? (Round your answer to two decimal places.)

Answer 1

Amount of pure antifreeze in 55% antifreeze fluid = 5 * 0.55 = 2.75 quarts.
Let the amount to be drained be x quarts. Therefore the amount of pure antifreeze needed to be added is also x quarts.
The amount of 20% antifreeze remaining after draining is 5 - x quarts.
Now we can write an equation based on amounts of pure antifreeze as follows:
0.2(5 - x) + x = 2.75
which can be simplified to:
x - 0.2x = 2.75 - 1 ..........(1)
Now you just need to solve equation (1) to find how much fluid should be drained and replaced with pure antifreeze, which is the value of x.