Question 952093
 A solution of iodine and alcohol contains 3 ounces of iodine and 21 ounces of alcohol.
 My question is how much be added to produce a solution that is 25% iodine?
:
Find the initial percent of iodine (in decimal form): {{{3/((21+3))}}} = .125
:
Let x = amt of iodine to be added to make it 25%
.125(24) + x = .25(x + 24)
3 + x = .25x + 6
3 + x - .25x = 6
.75x = 6 - 3
.75x = 3
x = 3/.75
x = 4 oz of pure iodine added to make the mixture 25% iodine