1
Question
The population of a place increased to 54000 in 2003 at a rate of 5% per annum. Find the population in 2001?
Open in App
Solution
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) Read more on Brainly.in - https://brainly.in/question/1449114#readmore
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)
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)
Read more on Brainly.in - https://brainly.in/question/1449114#readmore
Suggest Corrections
14
Similar questions