A mixture of 13% disinfectant solution is to be made from 11 % and 19 % disinfectant solutions. How muchof each solution should be used if 16 gallons of the 13% solution are needed?

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28-10-2022
Mathematics

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A mixture of 13% disinfectant solution is to be made from 11 % and 19 % disinfectant solutions. How muchof each solution should be used if 16 gallons of the 13% solution are needed?

Answer :

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OleJ299159 OleJ299159 · Answer 1 · 2022-10-28T17:15:14-04:00

We have to find the volume of each solution to make 16 gallons of a blend of 13% disinfectant solution.

Solution A has 11% disinfectant, while Solution B has 19% disinfectant.

We can call A and B to the volume of each respective solution.

The total volume (16 gallons) is made only from solutions A and B, so the sum of their volumes is equal to the total volume.

Then, we can write:

[tex]A+B=16[/tex]

We can repeat this analyisis but only with the volume of pure disinfectant.

The total volume will be 0.13*16 = 2.08 gallons.

This will be made of the volume of disinfectant of solution A, which is 0.11*A, and the volume of disinfectant of solution B, which is 0.19*B.

Then, we can write:

[tex]0.11A+0.19B=2.08[/tex]

We can now use the first equation to express B in function of A:

[tex]\begin{gathered} A+B=16 \\ B=16-A \end{gathered}[/tex]

We can then replace B in the second equation and solve for A as:

[tex]\begin{gathered} 0.11A+0.19B=2.08 \\ 0.11A+0.19(16-A)=2.08 \\ 0.11A+3.04-0.19A=2.08 \\ (0.11-0.19)A=2.08-3.04 \\ -0.08A=-0.96 \\ A=\frac{0.96}{0.08} \\ A=12 \end{gathered}[/tex]

Now, we can calculate B as:

[tex]\begin{gathered} B=16-A \\ B=16-12 \\ B=4 \end{gathered}[/tex]

Answer:

Volume of 11% disinfectant solution = 12 gallons.

Volume of 19% disinfectant solution = 4 gallons.