Get live Maths 1-on-1 Classs - Class 6 to 12
Example 14 - Chapter 8 Class 8 Comparing Quantities
Last updated at March 23, 2023 by Teachoo
Examples
Example 1 (i)
Example 1 (ii)
Example 1 (iii) Important
Example 2
Example 3 Important
Example 4
Example 5
Example 6 Important
Example 7 Important
Example 8
Example 9
Example 10 – GST (Page Number - 124)
Example 10 (Page no. 126) Important
Example 11
Example 12
Example 13 Important
Example 14 You are here
Example 15 Important
Transcript
Example 14 The population of a city was 20,000 in the year 1997. It increased at the rate of 5% p.a. Find the population at the end of the year 2000. Population of city in 1997 = 20,000 It Increases at the Rate of 5% Per Annum Here, 5% is compounded Rate So we use the formula A = P (1+𝑅/100)^𝑛 Here, A = Population at end of 2000 P = Population in 1997 = 20,000 R = 5 % N = 2000 − 1997 = 3 Putting Values in Formula Population at end of 2000 = 20000 (1+5/100)^3 = 20000 (1+1/20)^3 = 20000 ((20 + 1)/20)^3 = 20000 (21/20)^3 = 20000 × (21 × 21 × 21)/(20 × 20 × 20) = (5 × 21 × 21 × 21)/2 = (441 × 21 × 5)/2 = (441 × 105)/2 = 46305/2 = 23152.5 Since population cannot be in decimals ∴ Estimated Population = 23,153
