A 60% solution is mixed with 100% solution to produce 25L of a 68% solution...?

Question

How much of the 60% solution was there to start and how much 100% solution was added?

Answer 1

Let
x = amount of 60% solution
y = amount of 100% solution

We have

x + y = 25
y = 25 - x

.6x + y = .68*25 = 17
.6x + (25 - x) = 17
8 = .4x
20 = x

y = 25 - x = 25 - 20 = 5

The mixture contains 20L of 60% solution and 5L of 100% solution.

Answer 2

.68*25 = 17L, so the final ingredient in the solution is 17L

.6*x + y = 17, where x is the starting 60% solution and y is 100% solution.
You also know that x+y=25, y=25-x
.6x+25-x=17 ->x=20, y=5

Answer 3

Set up a system of eqns:

Let x=the amount of 60% solution, and y=the amount of 100% solution

x+y=25
60x+100y=(25*.68) >>> 60x+100y=17

Now simply solve for x and y.