If the population of a town is 64000 and its annual increase is 10%, then its population at the end of 3 years will be :

A. 80000

B. 85184

C. 85000

D. 85100

Answer: Option B

Solution(By Examveda Team)

Present population = 64000
1^st year = 66400
2^nd year = 6400 + 640
3^rd year = 6400 + 2 × 640 + 64

Total population after 3 years
= 64000 + 3 × 6400 + 3 × 640 + 64
= 85184

Alternate:
Population after n years
$$\eqalign{ & \Rightarrow {\text{p}}' = {\text{p}}{\left( {1 \pm \frac{{\text{R}}}{{100}}} \right)^{ \pm {\text{n}}}} \cr & {\text{p}}' = 64000{\left( {1 \pm \frac{{10}}{{100}}} \right)^3} \cr & \,\,\,\,\,\,\,\, = 85184 \cr} $$