Question

The population of a town increases by 10% annually. If its present population is 60,000 what will be its population after 2 years?

Answer 1

Hint: In this question, we need to evaluate the population of the town after 2 years such that it increases annually by 10% with the present population of 60000. For this, we will first calculate the population of the town after one year and then apply a 10% increase again to determine the population of the town at the end of the second year.

Complete step-by-step answer:
According to the question, the annual increase in the population of the town is 10%. At present, the population is 60000. So, the population of the town at the end of the first year is given as:
 $
\Rightarrow {P_1} = 60000 + 10\% {\text{ of }}60000 \\
   = 60000 + 6000 \\
   = 66000 \;
  $
Again, for the population of the town at the end of the second year, there is an increment of 10% on the population of the town at the end of the first year.
So, the population of the town at the end of the second year will be given as:
 $
\Rightarrow {P_2} = 66000 + 10\% {\text{ of }}66000 \\
   = 66000 + 6600 \\
   = 72600 \;
  $
Hence, the population of the town at the end of the second year is given as 72600.
So, the correct answer is “72600”.

Note: Alternatively, the problem can also be solved by calculating the net increase in the population of the town by using the formula $ \% = x \pm y \pm \dfrac{{xy}}{{100}} $ where ‘x’ and ‘y’ are the increase of decrease in the percentage. Here the value of ‘x’ and ‘y’ is 10. So,
 $
  net\% = 10 + 10 + \dfrac{{10 \times 10}}{{100}} \\
   = 21\% \;
  $
Now, the population is calculated as:
 $
  P = 60000 + 21\% {\text{ of }}60000 \\
   = 60000 + 12600 \\
   = 72600 \;
  $
Hence, the population of the town at the end of the second year is 72600.