Math, asked by jatin27462, 6 months ago

The population of a place increased to 54,000 in 2003 at a rate of 5% per annum .(i) find the population in 2001.(ii) what would be its population in 2006?​

Answers

Answered by nikunjc971
0

Answer:

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.

Step-by-step explanation:

i)

New population =54,000

Time (n)=2 yrs

r=5%

Original population = ?

54000=

(1−

100

5

)

(2)

P

54000=

0.9025

P

P=48735

Population in 2001 was 48,735.

ii)

A=54000(1+

100

5

)

(2)

=54000(1.1025)=59,535

Population in 2005 will be 59,535.

Answered by gauravkumar2424
0

Answer:

i)  

New population =54,000

Time (n)=2 yrs

r=5%

Original population = ?

54000=  

(1−  

100

5

​  

)  

(2)

 

P

​  

 

54000=  

0.9025

P

​  

 

P=48735

Population in 2001 was 48,735.

ii)  

A=54000(1+  

100

5

​  

)  

(2)

 

=54000(1.1025)=59,535

Population in 2005 will be 59,535.

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