The population of a district increases by $4\%$ during the first year and decreases by $5\%$ during the second year. If the present population of the district is $4,94,000$, then two years ago it was:

Let the population of the district two years ago be $P$.
1. After the first year, the population increases by $4\%$:

$$P_1=P\times \left(1+\frac{4}{100}\right)=P\times \frac{104}{100}$$

2. After the second year, the population decreases by $5\%$, which is equal to the present population ($4,94,000$):

$$P_2=P_1\times \left(1-\frac{5}{100}\right)=4,94,000$$

$$P\times \frac{104}{100}\times \frac{95}{100}=4,94,000$$

3. Solve for $P$:

$$P=\frac{4,94,000\times 100\times 100}{104\times 95}$$

Simplify the calculation step-by-step:
Notice that $\frac{4,94,000}{95}=5,200$.

$$P=\frac{5,200\times 10,000}{104}$$

Next, $\frac{5,200}{104}=50$.

$$P=50\times 10,000=5,00,000$$