A district has 64000 inhabitants. If the popu | vedclass

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(B) The present population of the district is $P = 64000$.The rate of increase is $R = 2 \frac{1}{2} \% = 2.5 \% = \frac{5}{2} \%$.The time period is

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$68921$

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(B) The present population of the district is $P = 64000$.The rate of increase is $R = 2 \frac{1}{2} \% = 2.5 \% = \frac{5}{2} \%$.The time period is $n = 3$ years.Using the formula for population growth: $A = P \left(1 + \frac{R}{100}ight)^n$.Substituting the values: $A = 64000 \left(1 + \frac{2.5}{100}ight)^3$.$A = 64000 \left(1 + \frac{25}{1000}ight)^3 = 64000 \left(1 + \frac{1}{40}ight)^3$.$A = 64000 \left(\frac{41}{40}ight)^3$.$A = 64000 	imes \frac{41 	imes 41 	imes 41}{40 	imes 40 	imes 40}$.$A = 64000 	imes \frac{68921}{64000}$.$A = 68921$.Thus,the population after $3$ years will be $68921$.

$A$ district has $64000$ inhabitants. If the population increases at the rate of $2 \frac{1}{2} \%$ per annum,then the number of inhabitants at the end of $3$ years will be...

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