The population of a city in $1960 A . D .$ was $2000$ and in $1970 A . D .$ was $5000$ . if the rate is increasing the population is proportional to the population, then what will be population is $2000 A . D .$ ?

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Solution

Population of a city in $1960 A . D = 2000$
Populaion of a city in $1970 A . D = 5000$
Net difference in population $10$ years $= 5000 - 2000 = 3000$
So $1980$ it will be $= 5000 + 300 = 8000$
$1990$ it will be $= 8000 + 3000 = 11000$
$2000$ it will be $= 11000 + 3000 = 14000$

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The population of a city in 1960A.D. was 2000 and in 1970A.D. was 5000. if the rate is increasing the population is proportional to the population, then what will be population is 2000A.D.?

Population of a city in 1960A.D=2000Populaion of a city in 1970A.D=5000Net difference in population 10 years=5000−2000=3000So 1980 it will be=5000+300=80001990 it will be =8000+3000=110002000 it will be =11000+3000=14000

The population of a city in $1960 A . D .$ was $2000$ and in $1970 A . D .$ was $5000$ . if the rate is increasing the population is proportional to the population, then what will be population is $2000 A . D .$ ?

Population of a city in $1960 A . D = 2000$
Populaion of a city in $1970 A . D = 5000$
Net difference in population $10$ years $= 5000 - 2000 = 3000$
So $1980$ it will be $= 5000 + 300 = 8000$
$1990$ it will be $= 8000 + 3000 = 11000$
$2000$ it will be $= 11000 + 3000 = 14000$