The population of a village increases by 5% yearly. If at present the population of the village is 4410, then what was the population before two years?

Let the population of the village two years ago be $P$.The population grows at a compounding rate of $R=5\%$ per year for $n=2$ years. The formula for the final population is:
$$\text{Present Population}=P\times {\left(1+\frac{R}{100}\right)}^n$$
Substitute the given values into the equation:
$$4410=P\times {\left(1+\frac{5}{100}\right)}^2$$
$$4410=P\times {\left(1+\frac{1}{20}\right)}^2$$
$$4410=P\times {\left(\frac{21}{20}\right)}^2$$
$$4410=P\times \frac{441}{400}$$
Solving for $P$:
$$P=\frac{4410\times 400}{441}$$
$$P=10\times 400=4000$$
Thus, the population of the village two years ago was $4000$.