Question

The population of a city increases 10% every year. In 2019, if the population
is 1,33, 100, what was the population of the city in 2017?​

Answer 1

Answer:

The population of the city in 2017 was 1,10,000.

Step-by-step explanation:

This is really easy sum. You just have to apply theformula of increase or decrease.

Here's how it should be done :

Time (t) = (2019 – 2017) = 2 years

Present Population (A) = 1,33,100

Rate of increase of population (r) = 10%

Let, the population in 2017 be P.

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

[Tip: I'm applying the formula with the plus because the population is increasing. If it's decreasing, use minus]

→ 133100 = P ( 1 + 10/100)^2

→ 133100 = P ( 1 + 1/10)^2

→ 133100 = P ( 11/10)^2

→ 133100 × 100/121 = P

→ 110000 = P

[Do the calculation yourself, lol :P]

I hope this helps. I actually learnt this last year and my memory is kinda hazy about this stuff. But I hope this helps ☺️.

Answer 2

Let the population before two years ago be x

Given the population increases by 10%

So increase in population after one year is

100

10

×x=0.1x

So population after one year is x+0.1x=1.1x

Increase in population after another one year is

100

10

×1.1x=0.11x

So population at present is 1.1x+0.11x=1.21x

But given population at present is 3,02,500

⟹1.21x=302500

⟹x=

1.21

302500

=2,50,000

Population of city two years ago is 2,50,000