SOLUTION: A radiator contains 3 gallons of 40% antifreeze solution. How many gallons of pure antifreeze must be added to obtain 50% solution? Please Help!!!

Algebra ->  Customizable Word Problem Solvers  -> Mixtures -> SOLUTION: A radiator contains 3 gallons of 40% antifreeze solution. How many gallons of pure antifreeze must be added to obtain 50% solution? Please Help!!!      Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 751428: A radiator contains 3 gallons of 40% antifreeze solution. How many gallons of pure antifreeze must be added to obtain 50% solution? Please Help!!!
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3*0.4 = 1.2 gallons of pure antifreeze.

Let x = amount (in gallons) of pure antifreeze needed to add

If you add x gallons of pure antifreeze to 1.2 gallons of pure antifreeze, you'll have 1.2+x gallons of pure antifreeze

You are also adding x gallons of pure antifreeze to the 3 gallons of solution already there, so the total amount of solution is now 3+x gallons.

We can now say this

(1.2+x)/(3+x) = 0.5

since we want the percentage of pure antifreeze to total solution must be 50%

Now solve for x

(1.2+x)/(3+x) = 0.5

1.2+x = 0.5(3+x)

1.2+x = 1.5+0.5x

x = 1.5+0.5x-1.2

x-0.5x = 1.5-1.2

0.5x = 0.3

x = 0.3/0.5

x = 0.6

So you need to add 0.6 gallons of pure antifreeze to obtain 50% solution.