Question

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?

Answer 1

Answer:

18 gallons from container $a$ and 22 gallons from the container $b$ to be mixed to get the required mixture.

Step-by-step explanation:

Given that the container $a$ has 20% concentrated cleaner

and container $b$ 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,

$9\%\; \text{of}\; (x+y) = (20\% \; \text{of}\; x )+(0\% \; \text{of}\; y)\cdots(i)$

As the total amount of the final solution= 40 gallons

So, $x+y=40\cdots(ii)$

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

9% of 40 = 20% of x

$\Rightarrow 0.09\times 40=0.2 \times x$

$\Rightarrow x=\frac{0.09\times 40}{0.2}$

$\Rightarrow x=18$ gallons.

From equation (ii),

18+y=40

$\Rightarrow y=40-18=22$ gallons.

Hence, 18 gallons from container $a$ and 22 gallons from the container $b$ to be mixed to get the required mixture.