Question

A scientist needs 70 liters of a 40% solution of alcohol. He has a 30% and a 60% solution available. How many liters of the 30% and how many liters of the 60% solutions should he mix to make the 40% solution?

Answer 1

Hello!

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.

Why?

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:

$30(Percent)SolutionVolume=X\\60(Percent)SolutionVolume=Y$

If both solutions will make a 70 liters solution, so the first equation will be:

$x+y=70L$

Let's work with real numbers converting the percent values into real numbers by dividing it into 100, so:

$30(Percent)=\frac{30}{100}=0.3\\\\40(Percent)=\frac{40}{100}=0.4\\\\60(Percent)=\frac{60}{100}=0.6$

Then, we know that both solutions will make 70 liters of a 40% solution, so the second equation will be:

$0.30x+0.60y=0.4*70$

Therefore, from the first equation we have:

$x=70-y$

Then, substituting x into the second equation to find y, we have:

$0.30(70-y)+0.60y=28\\21-0.30y+0.60y=28\\0.3y=28-21\\y=\frac{7}{0.3}=23.33$

Hence, substituting y into the first equation to find x, we have:

$x=70-y=70-23.33=46.67$

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!