Question

two years ago the population of the town was 10000 during first year it increased at the rate of 5% per annum and during second year it increased at the rate of 6% per annum what i the present population

Answer 1

Answer:

Two years ago population =10000

After one year its population will be at the rate of 5%

=105/100*10000=10500

Current year population at the rate 6%=

106/100*10500=11130

The present population =11130

Answer 2

The present population was 11130

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 = $10000+\frac{5}{100}(10000)=10500$

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

So, Population after 2nd year = $10500+6\%(10500)=10500+\frac{6}{100}(10500)=11130$

So, Present population = 11130

Hence the present population was 11130

#Learn more:

The present population of a town is 176400. If the rate of growth in its population is 5%per

annum, find : (i) its population 2 years hence, (i) its population one year ago​

https://brainly.in/question/10511434