# Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities

Last updated at Nov. 12, 2018 by Teachoo

Ex 8.3, 10 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum (i) find the population in 2001. Given, Population of place in 2003 = 54000 It has increased at a rate of 5% P.A. Here 5 % is compounded rate So we use the formula A = P (1+π /100)^π Here, A = Population in year 2003 = 54000 P = Population in year 2001 R = 5% N = Number of years = 2003 β 2001 = 2 Putting Values in formula, 54000 = P (1+5/100)^2 54000 = P (1+1/20)^2 54000 = P ((20 + 1)/20)^2 54000 = P (21/20)^2 54000 = P Γ (441/400) (54000 Γ 400)/441 = P P = (54 Γ 4 Γ100000)/441 P = 21600000/441 P = 48979.59 Since population cannot be decimal Thus, Population in year 2001 is around 48,980 Ex 8.3, 10 The population of a place increased to 54,000 in 2003 at a rate of 5% per annum (ii) what would be its population in 2005Given, Population in year 2003 (P) = 54000 Rate (R) = 5% p.a n = Number of Years = 2005 β 2003 = 2 Since 5% is compounded rate, We use the formula A = P (1+π /100)^π Population in year 2005 = 54000 (1+5/100)^2 = 54000 Γ (1+1/20)^2 = 54000 Γ ((20 + 1)/20)^2 = 54000 Γ (21/20)^2 = 54000 Γ ((21 Γ 21)/(20 Γ 20)) = 54000 Γ 441/400 = 540/4 Γ 441 = 270/2 Γ 441 = 135 Γ 441 = 59535 β΄ Population in year 2005 = 59,535