In a mixture, the ratio of the first and second liquid is $2:3$, and in another mixture, the ratio of those two liquids is $5:4$. The ratio in which the two mixtures must be mixed so that the quantity of the two liquids in the new mixture is equal is:

Let us solve this using the method of Alligation on the proportion of the first liquid.
1. In the first mixture ($M_1$), the ratio is $2:3$.

$$\text{Fraction of the first liquid }(x_1)=\frac{2}{2+3}=\frac{2}{5}$$

2. In the second mixture ($M_2$), the ratio is $5:4$.

$$\text{Fraction of the first liquid }(x_2)=\frac{5}{5+4}=\frac{5}{9}$$

3. In the final target mixture, the two liquids must be equal, meaning the ratio is $1:1$.

$$\text{Mean fraction of the first liquid }(x_m)=\frac{1}{1+1}=\frac{1}{2}$$

Applying the Alligation rule:

* Difference between the second fraction and the mean fraction:

$$x_2-x_m=\frac{5}{9}-\frac{1}{2}=\frac{10-9}{18}=\frac{1}{18}$$

* Difference between the mean fraction and the first fraction:

$$x_m-x_1=\frac{1}{2}-\frac{2}{5}=\frac{5-4}{10}=\frac{1}{10}$$

The required mixing ratio of $M_1:M_2$ is:

$$\text{Ratio}=\frac{1}{18}:\frac{1}{10}=10:18=5:9$$