Question

If the population of a town is 64,000 and its annual increase is 10%, then its correct population at the end of 3 years will be

• A. 80,000
• B. 85,000
• C. 85,100
• D. 85,184

Answer 1

The correct option is D 85,184

Population (P) $=64,000$
Annual increase rate $\left(r\right)=10\%$
Hence, population after $n=3$ years
$=P(1+\frac{r}{100}{)}^n$
$=64,000(1+\frac{10}{100}{)}^3$
$=64,000\times (\frac{11}{10}{)}^3$
$=64,000\times \frac{11\times 11\times 11}{1000}$
$=85,184$