Question

the population of a town increase d by 10 percent annually. if the present population is 22000 find its population a year ago

Answer 1

20000
Now population is 22000
so take initial population as x=100%
each year increased by 10% then now population = 110%
so by equating 110x=22000
then x=200 so current population is 20000

Answer 2

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,,

By applying the formula of compound interest we can make the following assumptions and calculate them for the same. You'll come across this formulae.........

Let us take it as "x" years and Total population to find be "Y" and let "PP" be as present population , we'll get this ,,,

$\boxed{\bf{PP = Y \times [1 - \frac{r}{100}]^x}}$

Now, for population changes before "1" years ago, substitute this into "x" years ,,,

$\bf{22000 = Y \times [1 + \frac{10}{100}]^1}$

$\bf{22000 = Y \times [1 + 0.1]^1}$

$\bf{22000 = Y \times [1.1]^1}$

$\bf{22000 = Y \times [1.1]}$

$\boxed{\bf{\therefore \: \: Y = 20000}}$

Therefore the answer is ,,,

$\boxed{Y = 20000}$

HOPE THIS HELPS YOU AND CLEARS THE DOUBTS FOR CALCULATING THE TIMES OF YEAR BY THIS FORMULAE!!!!!!!