Question

the population of a town increase at the rate of 40 per thousand its population qfter 2 years will be 6084 find iys present population​

Answer 1

Soln:-

Rate =40/1000=4/100=4℅

Population after 2years =6084

let the present population be x

A=P(1+r/100)^t

6084=x(1+4/100)^2

6084=x(1+1/25)^2

6084=x(26/25)^2

6084=x(26/25×26/25)

6084=x×(676/625)

x=6084×625/676

=5625

so the present population of the town is 5625

Answer 2

◆ Answer -

[tex]\sf{Present~ population = 5625 }[/tex]

● Explanation -

$\sf{Let ~'x'~ be~ present~ population~ of ~the ~town.}$

$\sf{Population~ after~ 2 ~years ~is ~given~by:}$

$\tt{P' = P (1 + r)^t}$

[tex]\tt{6084 = P \frac({1 + 40}{1000})^2 }[/tex]

$\tt{6084 = P (\frac{1 + 40}{1000})^2}$

$\tt{6084 = P (\frac{1 + 1}{25})^2}$

$\tt{6084 = P × (\frac{26}{25})^2}$

$\tt{P = 6084 × \frac{625}{676}}$

$\tt{P = 5625}$

Hence, present population of the town is 5625.