<PRE><B>Question 254975</B>
Cole has a solution that is 20% water and another solution that is 60% water.
 He wants to mix these to make a 40 mL of each solution that is 30% water.
 How many mL of each solution should he mix together?
:
Let x = amt of 60% solution required
It wants the results to be 40 ml, therefore:
(40-x) = amt of 20% solution
:
.60x + .20(40-x) = .30(40)
:
.60x + 8 - .20x = 12
:
.60x - .20x = 12 - 8
:
.40x = 4
x = {{{40/4}}}
x = 10 ml of 60% solution required
and
40 - 10  = 30 ml of 20% solution
:
:
Check solution in original equation
.6(10) = .2(30) = .4(40)
6 + 6 = 12 </PRE>