Question

The population of a place increased to 1,21,000 in 2003 at a rate of 10% per annum. Find the population in 2001. *​

Answer 1

$\bold{ANSWER≈}$

(i) Population in the year 2001

Let the population in the year 2001 be 'P' and the population in 2003 is 'A' = 54000

Also, R = 5%, n = 2

A = P[1 + (R/100)]n

54000 = P[1 + (5/100)]2

54000 = P[1 + (1/20)]2

54000 = P × (21/20)2

54000 = P × (21/20) × (21/20)

P = 54000 × (400/441)

P = 48979.6

The population in 2001 = 48980 (approx.)

(ii) Population in the year 2005

Now, the population in 2003 is considered as 'P' = 540000 and the population in 2005 is 'A'

R = 5%, n = 2

A = P[1 + (R/100)]n

A = 54000[1 + (5/100)]2

A = 54000[1 + (1/20)]2

A = 54000 × (21/20)2

A = 54000 × (21/20) × (21/20)

A = 54000 × (441/400)

A = 135 × 441

A = 59535

The population in 2005 = 59535

Answer 2

Answer:

100000

Step-by-step explanation:

At first you should know the formula of population growth and depreciation then See the given clues and start to solve.