A resturant has 161L of a cleaning solution which consists of 22% water and the remaining amount is soap. How much water is needed to be added to obtain a 55% water mixture.

Answer 1

let the amount of water added be x L

look at it from the view of the constant amount of soap which does not change.
.78(161) + 0x = .55(161+x)
125.58 = 88.55 + .55x

x = 67.33 L of pure water to be added

Answer 2

thanks, but i still dont know why you did 0.78(161)

Answer 3

If 22% of the original solution is water than the other part must be 78%
(22% + 78% = 100%, 100% would be the "whole thing")

and 78% = .78

so the amount of soap in the original 161L is .78(161)L = 125.58 L
This does not change, since we are adding x L of pure water
so our amount of soap is still 125.58
which is the left side of my equation: .78(161) + 0*x
but what is that ????
we now have 161L + xL = 161+x
55% of that is supposed to be soap, thus my right side of the
equation is .55(161+x)

TahDah!!!!