The population of a city increases at the rat | vedclass

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(A) Let the population $3$ years ago be $P_0$.Given,the present population $P = 90.51$ lacs,rate $r = 10 \%$,and time $n = 3$ years.The formula for po

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$68$

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(A) Let the population $3$ years ago be $P_0$.Given,the present population $P = 90.51$ lacs,rate $r = 10 \%$,and time $n = 3$ years.The formula for population growth is $P = P_0 \left(1 + \frac{r}{100}ight)^n$.Substituting the values: $90.51 = P_0 \left(1 + \frac{10}{100}ight)^3$.$90.51 = P_0 \left(1.1ight)^3$.$90.51 = P_0 	imes 1.331$.$P_0 = \frac{90.51}{1.331} = 68$.Therefore,the population $3$ years ago was $68$ lacs.

The population of a city increases at the rate of $10 \%$ annually. Its present population is $90.51$ lacs. The population $3$ years ago was nearly:

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