The population of a place increased to 54,000 in 2003 at a rate of 5% per annum. (i) Find the population in 2001. (ii) What would be its population in 2005?

(i) Here, A₂₀₀₃ = 54,000, R = 5%, n = 2 years
Population would be less in 2001 than 2003 in two years.
Here population is increasing.

∴ A₂₀₀₃ = P₂₀₀₁ (1+R/100)ⁿ ⇒ 54000 = P₂₀₀₁(1+5/100)²
⇒ 54000 = P₂₀₀₁ (1+ 1/20)² ⇒ 54000 = P₂₀₀₁(21/20)²
⇒ 54000 = P₂₀₀₁ x 21/20 x 21/20 ⇒ P₂₀₀₁= 54000x20X20/21×21
⇒ P₂₀₀₁ = 48,980 (approx.)

(ii) According to question, population is increasing.
Therefore population in 2005,
A₂₀₀₅ = P(1+R/100)ⁿ = 54000 (1+5/100)² = 54000(1+ 1/20)²
= 54000(21/20)² = 54000 x 21/20 x 21/20 = 59,535
= Hence population in 2005 would be 59,535.

Class 8 Maths Chapter 8 Exercise 8.3 Solution in Video

for more answers vist to:
https://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-8/