Question

The population of a town increases by 10% annually. if the present population is 22000,find the population a year ago.

Answer 1

Solution :-

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

The population of the town increases by the rate of 10 % annually and presently the population of the town is 22000

Then, according to the question.

⇒ x + (x*10)/100 = 22000

⇒ x/1 + x/10 = 22000

Taking L.C.M. of the denominators and then solving it.

⇒ (10x + x)/10 = 22000

⇒ 11x/10 = 22000

⇒ 11x = 22000*10

⇒ 11x = 220000

⇒ x = 220000/11

⇒ x = 20000

So, the population of the town one year ago was 20000.

Answer.

Answer 2

Let P the population of the town one year ago

Given:

Present population = 22000
Rate increase= 10%

Present population= P( 1+ 10/100)¹

22000 = P(11/10)¹

22000= 11P/10

22000× 10 = 11P

P= 220000/11

P = 20000

Hence, the population of the town a year ago was 20000.


======================================°===========================

Hope this will help you....