Question

I found this problem on line and I think the answer is wrong.
http://www.softmath.com/algebra-word-problems/show.php?id=10442
A car's 4.3 liter radiator contains 60% antifreeze and 40% water. How much has to be drained and replaced with water to make the solution 50% antifreeze and 50% water?
Their answer is .43 liters but I think the correct answer is .71666... liters.

Answer 1

Why did you think that?

Answer 2

Just using my answer shows I'm right.
Draining .716666... liters of the solution leaves 3.5833333333 liters of the 60%/40% solution. This means that it contains 3.5833333 * .6 liters (or 2.15 liters)of antifreeze.
2.15 liters is exactly one half of the radiator volume and filliing the remainder with water makes a 50/50 solution.

Answer 3

the numerator is pure antifreeze, the denominator is the total liquid. you want 50% antifreeze

Answer 4

Isn't the easiest thing to do is just use my answer?

Answer 5

yes

Answer 6

(see above)

Answer 7

woops i made a mistake , one sec checking

Answer 8

I get .71666 as well

Answer 9

okay perl!! I'm surprised they have that error there - it is an algebra software company.