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.

Question

Guest

Mathematics

4 November, 19:38

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.

+4

Answers (1)

Know the Answer?

Douglas Smith · 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.