Question

"Question 10 The population of a place increased to 54000 in 2003 at a rate of 5% per annum (i) find the population in 2001 (ii) what would be its population in 2005?

Class 8 Comparing Quantities Page 134"

Answer 1

Given:

we use formula of amount of compound interest to find population 

A( 2003)= 54,000,

R = 5%, 

n = 2 years

 

In 2001 Population would be less than 2003 in two years.

A(2003)= P(2001)(1+R/100)^n

54000 = P (2001)(1+5/100)²

54000 = P(2001)(1+1/20)²

54000= P(2001)(21/20)²

54000 = P(2001)((21×21)/(20×20))

P(2001)=( 54000×20×20)/21×21

Population in 2001 = 48979.59= 48980(approx)

(ii) ATQ, population is increasing. Therefore population in 2005,

 

A(2005)= P(1+R/100)^n

= 54000(1+5/100)²

= 54000(1+1/20)²

= 54000( 21/20)²

= 54000(21/20)× (21/20)

=( 54000× 441)/ 400

=( 540×441)/4

= 238140/4

= 59535

Population in 2005= 59535

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Hope this will help you....

Answer 2

Given that population in 2003 = 54000. Population of a place increased at the rate of 5%per annum. Population in 2002 at the rate of 5%per annum = 54000 x ( 5 / 100) = 2700 = 54000 - 2700 = 51300. Population in 2001 at the rate of 5%per annum = 51300 x ( 5 / 100) = 2565. = 51300 - 2565 = 48735. Now population in 2003 = 54000. Population of a place increased at the rate of 5%per annum. Population in 2004 at the rate of 5%per annum = 54000 x ( 5 / 100) = 2700 = 54000 + 2700 = 56700. Population in 2005 at the rate of 5%per annum = 56700 x ( 5 / 100) = 2835 = 56700 + 2835 = 59,535.