Applications of Compound Interest Formula
Applications of Compound Interest Formula
Last updated at April 16, 2024 by Teachoo
Ex 7.3, 1 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+𝑅/100)^𝑛 Here, A = Population in year 2003 = 54000 P = Population in year 2001 R = 5% N = Number of years = 2003 − 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 × (𝟒𝟒𝟏/𝟒𝟎𝟎) (54000 × 400)/441 = P P = (54 × 4 ×100000)/441 P = 21600000/441 P = 48979.59 Since population cannot be decimal Thus, Population in year 2001 is around 48,980 Ex 7.3, 1 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 − 2003 = 2 Since 5% is compounded rate, We use the formula A = P (1+𝑅/100)^𝑛 Population in year 2005 = 54000 (𝟏+𝟓/𝟏𝟎𝟎)^𝟐 = 54000 × (1+1/20)^2 = 54000 × ((20 + 1)/20)^2 = 54000 × (21/20)^2 = 54000 × ((21 × 21)/(20 × 20)) = 54000 × 𝟒𝟒𝟏/𝟒𝟎𝟎 = 540/4 × 441 = 270/2 × 441 = 135 × 441 = 59535 ∴ Population in year 2005 = 59,535