Question

Suppose a lab needs to make 400 liters of a 39% acid solution, but the only solutions available to the lab are 20% acid and 50% acid. What system of equations can be used to find the number of liters of each solution that should be mixed to make the 39% solution? Let c represent the number of liters of 20% acid solution and let d represent the number of liters of 50% acid solution.

Answer 1

Mixture problem.
Ratio of 20% : 50% acids
=50-39 : 39-20
=11:19
total volume
=400L
vol. of 20% acid=400*(11/(11+19)=400*(11/30)=146.667 L
vol. of 50% acid=400*(19/(11+19)=253.333 L

Alternatively, use c=volume of 20%, d=vol of 50%
then
0.2c+0.5d=400*0.39
Since c+d=400, we have
0.2c+0.5(400-c)=400*0.39=156
solve for c:
0.3c=44
c=146.667 and d=400-c=253.333 as before.