The answer is:
There are needed 46.67 liters of the 30% solution and 23.33 liters of the 60% solution in order to make 70 liters of a 40% solution.
We can solve this problem creating a system of equations. Let's make the two equations that will help us to solve the problem.
From the statement we know that a solution of 70 liters of a 40% is needed, and there are two differents solutions available, of 30% and 60% and we need to use them to make 70 liters of a 40% solution, so:
So:
[tex]30(Percent)SolutionVolume=X\\60(Percent)SolutionVolume=Y[/tex]
If both solutions will make a 70 liters solution, so the first equation will be:
[tex]x+y=70L[/tex]
Let's work with real numbers converting the percent values into real numbers by dividing it into 100, so:
[tex]30(Percent)=\frac{30}{100}=0.3\\\\40(Percent)=\frac{40}{100}=0.4\\\\60(Percent)=\frac{60}{100}=0.6[/tex]
Then, we know that both solutions will make 70 liters of a 40% solution, so the second equation will be:
[tex]0.30x+0.60y=0.4*70[/tex]
Therefore, from the first equation we have:
[tex]x=70-y[/tex]
Then, substituting x into the second equation to find y, we have:
[tex]0.30(70-y)+0.60y=28\\21-0.30y+0.60y=28\\0.3y=28-21\\y=\frac{7}{0.3}=23.33[/tex]
Hence, substituting y into the first equation to find x, we have:
[tex]x=70-y=70-23.33=46.67[/tex]
So, there are needed 46.67 liters of the 30% solution and 23.33 liters of the 60% solution in order to make 70 liters of a 40% solution.
Have a nice day!
