Ex 8.3, 10 - The population of a place increased to 54,000 in 2003

Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities - Part 2
Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities - Part 3
Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities - Part 4
Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities - Part 5
Ex 8.3, 10 - Chapter 8 Class 8 Comparing Quantities - Part 6

  1. Chapter 8 Class 8 Comparing Quantities
  2. Serial order wise

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 โˆ’ 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 ร—100000)/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 โˆ’ 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 โˆด Population in year 2005 = 59,535

About the Author

Davneet Singh's photo - Teacher, Engineer, Marketer
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.