Question

the population of a town increases by 8% annually if the present population is 54000 what was it a year ago

Answer 1

let the population of town 1 year ago be x
if the population increases by 8% annually we can say that
x+8x/100=5400
108x/100=5400
x=5400×100/108
x= 5000

Answer 2

Let the last year population = 'x'

Present year population = 54000

Increased percent = g = 8%

Present population = last year population × $\dfrac{(100+g)}{100}$

$54000=x\times\dfrac{(100+8)}{100}\\\\\\\implies54000=x\times\dfrac{108}{100}\\\\\\\implies x=\dfrac{54000\times100}{108}\\\\\\\implies x=\boxed{50000}$

Let us verify ,

$54000=50000\times\dfrac{100+8}{100}\\\\\\\implies54000=50000\times\dfrac{108}{100}\\\\\\\implies54000=500\times108\\\\\\\implies54000=54000$

Here,

   LHS = RHS

Hence,

  it is verified

Therefore last year population is 50000 answer