Two alloys A and B contain copper and zinc in the ratio $7:2$ and $5:3$ respectively. How many $\text{kg}$ of A and B must be melted in order to get an alloy of $44\text{kg}$ containing copper and zinc in the ratio $3:1$?

Let's focus on the fraction of copper in each mixture to apply the alligation method:Copper concentration in Alloy A = $\frac{7}{7+2}=\frac{7}{9}$ Copper concentration in Alloy B = $\frac{5}{5+3}=\frac{5}{8}$ Copper concentration in final mixture = $\frac{3}{3+1}=\frac{3}{4}$ By alligation:
$$\text{Ratio of A and B}=\left(\frac{3}{4}-\frac{5}{8}\right):\left(\frac{7}{9}-\frac{3}{4}\right)$$
$$\text{Ratio}=\left(\frac{6-5}{8}\right):\left(\frac{28-27}{36}\right)=\frac{1}{8}:\frac{1}{36}$$
Simplifying the ratio by multiplying both sides by $72$ (LCM of $8$ and $36$):
$$\text{Ratio}=9:2$$
The total mass of the final alloy is $44\text{kg}$.
$$\text{Quantity of Alloy A}=\frac{9}{9+2}\times 44=\frac{9}{11}\times 44=36\text{kg}$$
$$\text{Quantity of Alloy B}=\frac{2}{11}\times 44=8\text{kg}$$