Question

The population of a town increases by 10% annually if its present population is 60000 what will be its population after 2 years??

Answer 1

The population of a town increases by 10% each year.

Now, the present population of the town is 60000.

After 1 year, the present population of the country will increase by 10%.

10% of 60000 = $60000*10$%

= $60000* \frac{10}{100}$

= $600*10$

= $6000$

After increasing by 10%, the population is = $60000+60000*10$%

= $60000+6000$

= $66000$

So, after 1 year, the population of the town will be 66000.

But,  we still need to find the population of the town 2 years later.

After 2 year later, the population of the town will increase by 10% of 66000.

10% of 66000 = $66000*10$%

= $66000* \frac{10}{100}$

= $660*10$

= $6600$

After increasing by 10%, the population is = $66000+66000*10$%

= $66000+6600$

= $72600$

ANSWER: After 2 years, the population of the town will be 72600.

Answer 2

Solution

Population of town increased by 10%.

Present population is = 60000

Population after two year = ?

According to the question,

     10% of 60000

    ⇒ $\dfrac{10}{100}\times60000$

    ⇒ 6000

Population after 2 years = 6000 × 2

                                        = 12000

                                       = 12000 + 6000

                                      = 18000 answer