Answer:

49,680

Step-by-step explanation:

present population=54000

percent in which the population increase =8

so population that was a year ago =8percent decrease in 54000

8/100*54000=4320

=54000-4320

=49,680

Answer:

Given :-

• The population of a town increases by 8% annually.
• The present population is 54000.

To Find :-

• What is the population of a town a year ago ?

Solution :-

Let,

[tex]\small \mapsto \bf Population\: Of\: Town\: A\: Year\: Ago =\: x\\[/tex]

Given :

• Present Population Of Town (A) = 54000
• Rate of Increase (r%) = 8%
• Time Period (n) = 1 year

According to the question by using the formula we get,

[tex]\implies \sf\boxed{\bold{A =\: P\bigg(1 + \dfrac{r}{100}\bigg)^n}}\\[/tex]

[tex]\implies \sf 54000 =\: x\bigg(1 + \dfrac{8}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 54000 =\: x\bigg(\dfrac{108}{100}\bigg)^1\\[/tex]

[tex]\implies \sf 54000 =\: x \times \dfrac{108}{100}\\[/tex]

[tex]\implies \sf 54000 =\: \dfrac{108x}{100}\\[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 108x =\: 54000(100)\\[/tex]

[tex]\implies \sf 108x =\: 5400000\\[/tex]

[tex]\implies \sf x =\: \dfrac{5400000}{108}\\[/tex]

[tex]\implies \sf\bold{x =\: 50000}\\[/tex]

[tex]\therefore[/tex] The population of a town a year ago is 50000 .