Question

The population of a town increases by 10% every year. If the present population is 60,500, find its population (i) after 2 years (ii) 2 years ago.

Answer 1

Given:

• Present population, P = 60,500
• Rate of increase, r = 10% per year
• Time = 2 years

(i) Population after 2 years

The formula for population after n years is:

`P_("future") = P(1 + r/100)^n`

Substitute the values:

`P_("after 2 years") = 60,500(1 + 10/100)^2`

`P_("after 2 years") = 60,500 xx (1.1)^2`

Step by step:

1.1 × 1.1 = 1.21

60,500 × 1.21 = 73,205

Population after 2 years: 73,205

(ii) Population 2 years ago

The formula for population n years ago (backward) is:

`P_("past") = (P_("present"))/(1 + r/100)^n`

Substitute the values:

`P_(2  "years ago") = (60,500)/(1.1)^2`

`P_(2  "years ago") = (60,500)/(1.21)`

`P_(2  "years ago") ≈ 50,000`

Population 2 years ago: 50,000