Question

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.  Find the population in 2001 if you give this answer I will make you brilliant ​

Answer 1

Answer:

Population in 2001 at the rate of 5%per annum = 51300 x ( 5 / 100) = 2565. = 51300 - 2565 = 48735. Now population in 2003 = 54000.

Step-by-step explanation:

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Answer 2

Answer:

48,980 approx.

Step-by-step explanation:

Given,

Population of a place in 2003 = 54000

It has increased at the rate of 5% P.A.

Here 5% is the compounded rate

Therefore formula used

$a = p(1 + \frac{r}{100} )$

Here,

A = population in year 2003

P = population in year 2001

R = 5%

N = number of years

= 2003 - 2001 = 2 years

Hence,

$54000 = p(1 \frac{5}{100} ) ^{2}$

$54000 = p(1 + \frac{1}{20})^{2}$

$54000 =( p \frac{20 + 1}{20} ) ^{2}$

$54000 = p( \frac{21}{20} ) ^{2}$

$54000 = p \times ( \frac{441}{400})$

$\frac{54000 \times 400}{441} = p$

$p = \frac{54 \times 4 \times 100000}{441}$

$p = \frac{21600000}{441}$

$p = 48979.59$

Since, population cannot be decimal

Thus, population in year 2001 is around 48,980