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Given that, the current population of a city is $11,02,500, $ rate $ = 5\%,$ and time $ = 2$ years.

We know that, $A = P\left(1+\dfrac{r}{100}\right)^{t}$

Here, $A = 11,02,500,r = 5\%,t = 2$ years.

Now, $1102500 = P\left(1 + \dfrac{5}{100}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{105}{100}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{21}{20}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{441}{400}\right)$

$\implies P = \dfrac{1102500 \times 400}{441} = 10,00,000$

$\textbf{Short Method:}$ Let’s assume, the two years ago the population was $x.$

Now$,x \times \dfrac{105}{100} \times \dfrac{105}{100} = 1102500$

$\implies x = 10,00,000$

So, the correct answer is $(C).$

We know that, $A = P\left(1+\dfrac{r}{100}\right)^{t}$

Here, $A = 11,02,500,r = 5\%,t = 2$ years.

Now, $1102500 = P\left(1 + \dfrac{5}{100}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{105}{100}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{21}{20}\right)^{2}$

$\implies 1102500 = P\left(\dfrac{441}{400}\right)$

$\implies P = \dfrac{1102500 \times 400}{441} = 10,00,000$

$\textbf{Short Method:}$ Let’s assume, the two years ago the population was $x.$

Now$,x \times \dfrac{105}{100} \times \dfrac{105}{100} = 1102500$

$\implies x = 10,00,000$

So, the correct answer is $(C).$