SOLUTION: how much water should be added to a radiator that contains 10 gallons of an 80% solution antifreeze solution to dilute it to a 50% antifreeze solution?

Question 282968: how much water should be added to a radiator that contains 10 gallons of an 80% solution antifreeze solution to dilute it to a 50% antifreeze solution?
Found 2 solutions by richwmiller, modi.kk91:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
Using water percentages
.2*(10)+x=.5*(10+x)
x=6
using antifreeze percent
10*.8=.5*(10+x)
x=6
If the radiator can hold 16 gallons then add 6 gallons of water.
Otherwise we don't have enough info ie the capacity of the radiator.

Answer by modi.kk91(20) (Show Source): You can put this solution on YOUR website!
allegation mixture

0% 80%

50%
80-50=30 and 50-0=50
ratio=30:50=====3:5

3:5::x:10

X=6 gallons
second method
let the added water be x at 0%
x*0%+10*80%=50%(x+10)
800=50x+500

300=50x
x=6 gallons

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