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

Last updated at Nov. 12, 2018 by Teachoo

Last updated at Nov. 12, 2018 by Teachoo

Transcript

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+๐ /100)^๐ Here, A = Population in year 2003 = 54000 P = Population in year 2001 R = 5% N = Number of years = 2003 โ 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 ร (441/400) (54000 ร 400)/441 = P P = (54 ร 4 ร100000)/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 โ 2003 = 2 Since 5% is compounded rate, We use the formula A = P (1+๐ /100)^๐ Population in year 2005 = 54000 (1+5/100)^2 = 54000 ร (1+1/20)^2 = 54000 ร ((20 + 1)/20)^2 = 54000 ร (21/20)^2 = 54000 ร ((21 ร 21)/(20 ร 20)) = 54000 ร 441/400 = 540/4 ร 441 = 270/2 ร 441 = 135 ร 441 = 59535 โด Population in year 2005 = 59,535

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.