Question

The difference between the population of a city
1 year and 2 year ago is 5000. If increment in
population every year is 10% then find the present population of city.​

Answer 1

Answer:

Population of city two years ago=P

5000 = PX10100PX10100

=>P=50000

Present population of city

= 50000 (1+10100)3(1+10100)3

=66550

Answer 2

Okay so 2016=x1 2015=x2 so that means that x1=x2+5000 since there is a difference of 5000! x1=x2+5000! We know the population grows every year by 10% from the previous year! First 10%=10/100=0.1! So that means x1=x2+(x2*10%)! But we know that x1=x2+5000 so that means that x2+5000=x2+(x2*10%)! So x2–x2–5000+(x2*10%)=0–5000+(x2*10%)=-5000+(x2*10%)! Okay so we need to find x2! So x2=5000/10%! So x2=5000/0.1 people! So x2=50000 people! But we know x2=2015 population so 2015=50000 people! Okay let us find 2016=x1! x1=x2+5000 Okay so that means that 2016=50000+5000 and 2016=55000 or x1=55000 people! So how do we find 2017? 2017=x3 so since we know that the population increases by 10% per year from the previous year we also know that x3=x1+(x1*10%)! So we have x3=55000+(55000*0.1)! So x3=55000+5500 and x3=60500 people! So since 2017=x3 than in 2017 the population is 60500 people!