You purchased concentrated cleaner that must be diluted before use. Container a is the cleaner, which is 20% solution. Container b is pure water. You are mixing together the contents of container a and container b to produce 40 gallons of 9% solution. How many gallons from each container are needed to create the 40 gallons of 9% cleaning solution?

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Ritz01 Ritz01 · Answer 1 · 2020-11-12T07:27:27+01:00

Answer:

18 gallons from container [tex]a[/tex] and 22 gallons from the container [tex]b[/tex] to be mixed to get the required mixture.

Step-by-step explanation:

Given that the container [tex]a[/tex] has 20% concentrated cleaner

and container [tex]b[/tex] has pure water which has 0% cleaner.

Let x gallons from container a and y gallons from the container b are mixed to get a 9% concentrated solution.

So,

[tex]9\%\; \text{of}\; (x+y) = (20\% \; \text{of}\; x )+(0\% \; \text{of}\; y)\cdots(i)[/tex]

As the total amount of the final solution= 40 gallons

So, [tex]x+y=40\cdots(ii)[/tex]

Using the value of equation (ii) in equation (i), we have

9% of 40 = 20% of x

[tex]\Rightarrow 0.09\times 40=0.2 \times x[/tex]

[tex]\Rightarrow x=\frac{0.09\times 40}{0.2}[/tex]

[tex]\Rightarrow x=18[/tex] gallons.

From equation (ii),

18+y=40

[tex]\Rightarrow y=40-18=22[/tex] gallons.

Hence, 18 gallons from container [tex]a[/tex] and 22 gallons from the container [tex]b[/tex] to be mixed to get the required mixture.