How much water should be added to 1 gallon of pure antifreeze to obtain a solution that is 65% ​antifreeze?

Answer 1

1.00(1) = 1 = .65(1+x)

.65x = 1-.65 = .35

x = 35/65 = 7/13 gallon of water,

added to 1 gallon pure antifreeze yields 1 7/13 gallons of 65%

100-65=35

65-0= 65

ratio of 65/35 of 100%/0%=13/7

= 13/7 antifreeze/water

= 1/(7/13)= 13/7