Question

9. The population of a place increased to 54,000 in 2003 at a rate of 5% per annum
(1) find the population in 2001.
(ii) what would be its population in 2005?


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Answer 1

Step-by-step explanation:

Solution:

(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+ 100R ) n

$\Rightarrow54000=P_{2001}\left(1+\frac{5}{100}\right)^2⇒54000=P$

2001

(1+ 10052

$\Rightarrow54000=P_{2001}\left(1+\frac{1}{20}\right)^2⇒54000=P$

2001

(1+ 201 ) 2

$\Rightarrow54000=P_{2001_{ }}\left(1+\frac{1}{20}\right)^2⇒54000=P$

2001

(1+ 201 ) 2

$\Rightarrow54000=P_{2001}\left(\frac{21}{20}\right)^2⇒54000=P$

2001 ( 2021 ) 2

$\Rightarrow54000=P_{2001}\times\frac{21}{20}\times\frac{21}{20}⇒54000=P$

2001

2021 × 2021

$\Rightarrow P_{2001}=\frac{54000\times20\times20}{21\times21}⇒P </p><p>2001$

=

21×21

54000×20×20

$\Rightarrow P_{2001}=48980⇒P$

2001

=48980 (Approx)

(ii) According to question, population is increasing. Therefore population in 2005,

$A_{2005}=P\left(1+\frac{R}{100}\right)^nA </p><p>2005$

=P(1+ 100R ) n

$= 54000\left(1+\frac{5}{100}\right)^254000(1+ </p><p>100$

5 ) 2

$= 54000\left(1+\frac{1}{20}\right)^$254000(1+

20 ) 2

$= 54000\left(\frac{21}{20}\right)$^254000(

2021 ) 2

$= 54000\times\frac{21}{20}\times\frac{21}{20$}54000× 20

21 × 2021

= 59,535

Hence population in 2005 would be 59,535.

Answer 2

Answer:

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Step-by-step explanation:

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