Question

population of a village increased by 54000 in 2003 by 5%p.a find the population in 2001

Answer 1

Given that population in 2003 = 54000. Population of a place increased at the rate of 5%per annum. Population in 2002 at the rate of 5%per annum = 54000 x ( 5 / 100) = 2700 = 54000 - 2700 = 51300. Population in 2001 at the rate of 5%per annum = 51300 x ( 5 / 100) = 2565. = 51300 - 2565 = 48735. Now population in 2003 = 54000. Population of a place increased at the rate of 5%per annum. Population in 2004 at the rate of 5%per annum = 54000 x ( 5 / 100) = 2700 = 54000 + 2700 = 56700. Population in 2005 at the rate of 5%per annum = 56700 x ( 5 / 100) = 2835 = 56700 + 2835 = 59,535.

Answer 2

Answer:

${\huge{\underline{\underline{\sf{\blue{Solution :-}}}}}}$

Hare, Population in 2003 = 44100

Rate of growth of population (R) = 5% per annum.

Number of years (n) = 2

Let P be the population in 2001.

Then , population after n years

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \: p \: (1 \: \frac{R}{100})^{n}$

$i.e., \: population \: after \: 2 \: years = P ( \: 1 + \frac{5}{100})^{2}$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 44100 \: p \: \frac{21}{20} ^{2}$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \frac{20^{2} \times 44100 }{21^{2} } = p$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p \: = \: \frac{44100 \times 400}{441}$

$or \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: p \: = 4000$

Thus, the population of the village in 2001 was 40000.