Question 548035
A chemist mixed a 20% alcohol solution with a 12% alcohol solution to obtain 200mL of a 15% alcohol solution. 
How many mL of the 20% solution did he use?"
:
Let x = amt of 20% solutions required
It states the results should be 200 ml, therefore
(200-x) = amt of 12% required
:
Write a decimal equiv mixture equation
:
.20x + .12(200-x) = .15(200)
.20x + 24 - .12x = 30
.20x - .12x = 30 - 24
.08x = 6
x = {{{6/.08}}}
x = 75 ml of 20% solution
:
:
Check this (200-75=125 ml of 12%)
.20(75) + .12(125) = .15(200)
15 + 15 = 30; confirms our solution