Question

The population of a town increase by 5% every year if the preasent population of the town is 44100 what was the population 2 years ago

Answer 1

Given :

Present population of the town : 44100

Rate of appreciation : 5%

To find :

The population of the town, 2 years ago

Solution :

First, we should find the value of the town one year ago.

Let the population of the town one year ago be m.

Population of town one year ago :

$\sf \dashrightarrow m + 5\% \: of \: m = 44100$

$\sf \dashrightarrow m + \dfrac{5}{100} \times m = 44100$

$\sf \dashrightarrow m + \dfrac{1}{20} \times m = 44100$

$\sf \dashrightarrow m + \dfrac{1m}{20} = 44100$

$\sf \dashrightarrow \dfrac{20m + 1m}{20} = 44100$

$\sf \dashrightarrow \dfrac{21m}{20} = 44100$

$\sf \dashrightarrow 21m = 44100 \times 20$

$\sf \dashrightarrow 21m = 882000$

$\sf \dashrightarrow m = \dfrac{882000}{19}$

$\sf \dashrightarrow m = 42000$

Now, let's find the population of the town two years ago.

Let the population of the town two years ago be n.

Population of the town two years ago :

$\sf \dashrightarrow n + 5\% \: of \: n = 42000$

$\sf \dashrightarrow n + \dfrac{5}{100} \times n = 42000$

$\sf \dashrightarrow n + \dfrac{1}{20} \times n = 42000$

$\sf \dashrightarrow n + \dfrac{1n}{20} = 42000$

$\sf \dashrightarrow \dfrac{20n + 1n}{20} = 42000$

$\sf \dashrightarrow \dfrac{21n}{20} = 42000$

$\sf \dashrightarrow 21n = 42000 \times 20$

$\sf \dashrightarrow 21n = 840000$

$\sf \dashrightarrow n = \dfrac{840000}{21}$

$\sf \dashrightarrow n = 40000$

Hence, the population of the town, two years ago was 40000.