Question

two years ago the population of a town was 1000.during first year it increases at the rate at 5%per annum and during second year it increases at the rate of 6% per annum. what is it's present population​

Answer 1

Answer:

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Step-by-step explanation:

We are given that two years ago the population of the town was 10000

During first year it increased at the rate of 5% per annum

So, Population after 1st year = 10000+\frac{5}{100}(10000)=1050010000+

100

5

(10000)=10500

During second year it increased at the rate of 6% per annum

So, Population after 2nd year = 10500+6\%(10500)=10500+\frac{6}{100}(10500)=1113010500+6%(10500)=10500+

100

6

(10500)=11130

So, Present population = 11130

Hence the present population was 11130