Question

A radiator contains 11 quarts of fluid, 10% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 40% antifreeze?

Answer 1

Amount of antifreeze needed in new mixture = 11 * 0.4 quarts
Let amount on orignal mixture to be drained and amount of pure antifreeze to be added = x quarts.
Amount of antifreeze left in original mixture after draining = (11 - x) * 0.1 quarts.
Adding amounts of remaining antifreeze after draining and amount of added pure antifreeze to get required 40% antifreeze in new mixture gives:
(11 - x) * 0.1 quarts + x = 11 * 0.4
1.1 - 0.1x + x = 4.4

Can you solve the last equation to find x ?

Answer 2

of original*