Question

The population of a town increases at the rate of 7% every year if the present population is 90000 what will it be after two years

Answer 1

Answer:

103,041

Step-by-step explanation:

Population increase = 7%.

Present Population = 90000.

Population after two years = 90000 * (1 + (7/100)) ^2.

                                            = 90000 * ( 1 + 0.07)^2.

                                            = 90000 * (1.07)^2

                                             = 103,041.

Answer 2

Answer:

It will be 103041 population after two years.

Step-by-step explanation:

Given,

population of a town increase at the rate of 7% every year.

So rate of interest $= 7\%$

and also given the present population is 90000 rupees.

So, Principal $= 90000 \: rupees$

Now we want to find what will be the population after two years ?

So,here given time is 2 years.

We know, population always increase by Compound interest.

By Compound interest rule,

Amount $= p {(1 + \frac{r}{100}) }^{t}$

Where p is principal,r is rate of interest and t is time.

Here ,

$p = 90000 \: rupees \\ r = 7\% \\ t = 2 \: years$

So, Population after 2 years

$= 90000 \times {(1 + \frac{7}{100} )}^{2} \\ = 90000 \times \frac{107}{100} \times \frac{107}{100} \\ = 9 \times 107 \times 107 \\ = 103041$

Know more about Compound interest,

1) https://brainly.in/question/1950647

2)https://brainly.in/question/11735147

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