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?
Answers
Answer:
Step-by-step explanation:
(i) Here, A_{2003}=\ Rs.\ 54,000A
2003
= Rs. 54,000, R = 5%, n= 2 years
Population would be less in 2001 than 2003 in two years.
Here the population is increasing.
\therefore A_{2003}=P_{2001}\left(1+\frac{R}{100}\right)^n∴A
2003
=P
2001
(1+
100
R
)
n
\Rightarrow54000=P_{2001}\left(1+\frac{5}{100}\right)^2⇒54000=P
2001
(1+
100
5
)
2
\Rightarrow54000=P_{2001}\left(1+\frac{1}{20}\right)^2⇒54000=P
2001
(1+
20
1
)
2
\Rightarrow54000=P_{2001_{ }}\left(1+\frac{1}{20}\right)^2⇒54000=P
2001
(1+
20
1
)
2
\Rightarrow54000=P_{2001}\left(\frac{21}{20}\right)^2⇒54000=P
2001
(
20
21
)
2
\Rightarrow54000=P_{2001}\times\frac{21}{20}\times\frac{21}{20}⇒54000=P
2001
×
20
21
×
20
21
\Rightarrow P_{2001}=\frac{54000\times20\times20}{21\times21}⇒P
2001
=
21×21
54000×20×20
\Rightarrow P_{2001}=48980⇒P
2001
=48980 (Approx)
(ii) According to question, population is increasing. Therefore population in 2005,
A_{2005}=P\left(1+\frac{R}{100}\right)^nA
2005
=P(1+
100
R
)
n
= 54000\left(1+\frac{5}{100}\right)^254000(1+
100
5
)
2
= 54000\left(1+\frac{1}{20}\right)^254000(1+
20
1
)
2
= 54000\left(\frac{21}{20}\right)^254000(
20
21
)
2
= 54000\times\frac{21}{20}\times\frac{21}{20}54000×
20
21
×
20
21
= 59,535
Hence population in 2005 would be 59,535.
Answer:
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