The population of a place increased to 54,000 in 2003 at a rate of 5% per am
per annum, interest being
1) find the population in 2001.
(1) what would be its population in 2005?
count of bacteria in a certain experiment was increasing
Answers
Step-by-step explanation:
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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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.
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