Question

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

• A. $8000$
• B. $85000$
• C. $85100$
• D. $85184$

Answer 1

The correct option is D $85184$

Population at the end of $3$ years.

$=64000{(1+\frac{10}{100})}^3$

$=64000\times \frac{11}{10}\times \frac{11}{10}\times \frac{11}{10}$

$=85184$