Two barrels hold mixtures of ethanol and water in the ratios 2 : 1 and 4 : 1 respectively. In what ratio should the mixture from the first barrel and the second barrel be combined to obtain a new mixture with ethanol and water in the ratio 3 : 1?

The correct option is 3 : 5.

Step 1: Determine the concentration of ethanol in each mixture.

In the first barrel, the ratio of ethanol to water is 2 : 1.
Therefore, the fraction of ethanol in the first barrel is:

$\text{Fraction of Ethanol in Barrel 1}=\frac{2}{2+1}=\frac{2}{3}$

In the second barrel, the ratio of ethanol to water is 4 : 1.
Therefore, the fraction of ethanol in the second barrel is:

$\text{Fraction of Ethanol in Barrel 2}=\frac{4}{4+1}=\frac{4}{5}$

In the required final mixture, the ratio of ethanol to water is 3 : 1.
Therefore, the fraction of ethanol in the final mixture is:

$\text{Fraction of Ethanol in Final Mixture}=\frac{3}{3+1}=\frac{3}{4}$

Step 2: Apply the Rule of Alligation.

By the rule of alligation, the ratio of the mixture from Barrel 1 to Barrel 2 is given by:

$\text{Ratio}=\frac{\text{Fraction in Barrel 2}-\text{Fraction in Final Mixture}}{\text{Fraction in Final Mixture}-\text{Fraction in Barrel 1}}$

Step 3: Calculate the individual differences.

Difference for the quantity of Barrel 1:

$\frac{4}{5}-\frac{3}{4}=\frac{16-15}{20}=\frac{1}{20}$

Difference for the quantity of Barrel 2:

$\frac{3}{4}-\frac{2}{3}=\frac{9-8}{12}=\frac{1}{12}$

Step 4: Find the required ratio.

The required ratio of the mixture taken from the first barrel to the second barrel is:

$\text{Required Ratio}=\frac{1}{20}:\frac{1}{12}$

Multiplying both parts by 60 (the least common multiple of 20 and 12):

$\text{Required Ratio}=\left(\frac{1}{20}\times 60\right):\left(\frac{1}{12}\times 60\right)=3:5$

Thus, the mixture from the first barrel and the second barrel should be combined in the ratio 3 : 5.