SOLUTION: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal
Algebra ->
Customizable Word Problem Solvers
-> Mixtures
-> SOLUTION: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal
Log On
Question 190158: A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must be drained from the radiator and replaced by pure antifreeze so that the radiator will contain 15 gal of a 40% antifreeze solution?
You can put solution on your website.
Thanks. Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! A radiator contains 15 gal of a 20% antifreeze solution. How many gallons must
be drained from the radiator and replaced by pure antifreeze so that the
radiator will contain 15 gal of a 40% antifreeze solution?
:
Let x = amt of 20% to be drained, and amt of pure antifreeze to be added
:
Pure antifreeze = 1.0x
:
.20(15-x) + 1.0x = .40(15)
:
3 - .2x + 1x = 6
:
-.2x + 1x = 6 - 3
:
.8x = 3
x =
x = 3.75 gal removed and 3.75 gal of pure antifreeze added.
:
:
See if that's true:
.2(15-3.75) + 3.75 = .4(15)
3 - .75 + 3.75 = 6
