<PRE><B>Question 622587</B>
A solution containing 45% water is mixed with another solution containing 10% bromine. 
If the chemist needs 30 liters of the mixture, how many liters of each solution are needed to make a 34.5% water/bromine solution?
:
The 45% water solution must be a 55% bromine solution
let x = amt of 55% bromine solution
then
30-x = amt of 10% bromine
:
A typical mixture equation
.55x + .10(30-x) = .345(30)
.55x + 3 - .1x = 10.35
.55x - .10x = 10.35 - 3
.45x = 7.35
x = 7.35/.45
x = 16{{{1/3}}} liters of the 55% bromine solution (45% water)
and
30 - 16{{{1/3}}} = 13{{{2/3}}} liters of 10% solution</PRE>