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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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