**In what ratio should a 20% methyl alcohol solution be mixed with a 50% methyl alcohol solution so that the resultant solution has 40% methyl alcohol in it?**

To determine the ratio in which a 20% methyl alcohol solution should be mixed with a 50% methyl alcohol solution to obtain a resultant solution with 40% methyl alcohol, we can use a mathematical approach.

Let's assume we need to mix x units of the 20% methyl alcohol solution with y units of the 50% methyl alcohol solution. The resultant solution will have a total of (x + y) units.

To calculate the amount of methyl alcohol in the resultant solution, we can multiply the concentration of each solution by the corresponding volume and sum them up. Then we equate it to the concentration of the desired mixture.

The equation can be set up as follows:

0.20x + 0.50y = 0.40(x + y)

Now we can solve for the ratio by rearranging the equation:

0.20x + 0.50y = 0.40x + 0.40y 0.10y = 0.20x y = 2x

The ratio of the 20% methyl alcohol solution to the 50% methyl alcohol solution is 1:2. This means that for every unit of the 20% solution, you need to mix 2 units of the 50% solution to obtain a resultant solution with 40% methyl alcohol.

For example, if you mix 1 unit (e.g., liter) of the 20% solution with 2 units (e.g., liters) of the 50% solution, you will have a total of 3 units of the resultant solution, and it will contain 40% methyl alcohol.

Thank You