can anyone help me set up a table for this problem, and then solve it?

a lab technician has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of each solution is needed to make a 100L solution that is 50% chlorine?

Printable View

- Jun 4th 2008, 05:41 PMheather_meaghansolving tables
can anyone help me set up a table for this problem, and then solve it?

a lab technician has one solution that is 60% chlorinated and another that is 40% chlorinated. How much of each solution is needed to make a 100L solution that is 50% chlorine? - Jun 4th 2008, 06:07 PMSoroban
Hello, Heather!

Quote:

A lab tech has a solution that is 60% chlorine and another that is 40% chlorine.

How much of each solution is needed to make a 100L solution that is 50% chlorine?

. . $\displaystyle \begin{array}{|c|c|c|c|}

& \text{Liters} & \text{Percent} & \text{Chlorine} \\ \hline

A & x & 0.60 & 0.6x \\

B & 100-x & 0.40 & 0.4(100-x) \\ \hline

\text{Mix} & 100 & 0.50 & 0.5(100) \end{array}$

The equation is in the last column: . $\displaystyle 0.6x + 0.4(100-x) \;=\;50 $

Got it?