SOLUTION: 2 solutions are to be mixed to make 50 ml. of a solution that is 16% Bromine. One solution is 10% Bromine and the other solution is 40% Bromine. How much of each solution should

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: 2 solutions are to be mixed to make 50 ml. of a solution that is 16% Bromine. One solution is 10% Bromine and the other solution is 40% Bromine. How much of each solution should       Log On

Ad: Over 600 Algebra Word Problems at edhelper.com


    

Question 14264: 2 solutions are to be mixed to make 50 ml. of a solution that is 16% Bromine. One solution is 10% Bromine and the other solution is 40% Bromine. How much of each solution should be used?
Tried it but don't have the correct answer. Thanks for your help.
Answer by Alwayscheerful(414) About Me  (Show Source):
You can put this solution on YOUR website!
You should start with 2 variables. "D" for dumped and "P" for poured (for the two liquids)
To solve, you need to use either substitution or elimination.
Set up your first equation.
D%2BP=50
Because you are pouring two solutions to make one of 50 mL, you can say that equation is true.
The second equation is a little tricky.
You take the percent and change it to decimals to make the second equation.
.1D%2B.4P=.16%2850%29
How did I get that?
Let's say the 10% bromine solution is D and the 40% solution is P
You want to mix 10% and 40% to get 16% and that 16% solution is 50 mL.
That's how I got the .16%2850%29 part
Then you put them together
D%2BP=50
.1D%2B.4P=.16%2850%29
You can either use substitution or elimination. For this problem, I personally think substitution is much easier.
Isolate one variable.
D=50-P
.1%2850-P%29%2B.4P=8
5-.1P%2B.4P=8
Move isolate the P
.3P=3
P=10
Plug it back in the original equation.
D%2B10=50
D=40
So your answer should be that you should use 10 mL of the 40% solution and 40 mL of the 10% solution
Hope this helps!