How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 50% antifreeze?

Question

A radiator contains 14 quarts of fluid, 35% of which is antifreeze. How much fluid should be drained and replaced with pure antifreeze so that the new mixture is 50% antifreeze? (Round your answer to two decimal places.)

Answer 1

The formula for this solution is C1·V1 + C2·V2= C3·V3, where C = Concentration, and V= Volume

 

let x = the volume to be drained, so C1= 35%, and V1 = 14q - x

then, the values for C2 and V2 will be x·100%

those two expressions will equal C3·V3 which is 14q·50%

 

So, the equation is: (14q - x)·35% + x·100% = 14q·50%

when we expand the equation, we get

490q - x·35% +x·100% = 700q%

 

after we combine the like terms, we get -35%·x + x·100% = 210q%

which will simplify to 65%·x = 210q%

we then divide both sides by 65%, and x = 3.2q which is the amount to be drained, and replaced with pure antifreeze

 

the amount of the old antifreeze is 14q - 3.2q = 10.8q

 

Proof: 10.8·35% + 3.2q·100% = 14q·50%

380q% + 320q% = 700q%