Question

The population of a town $3$ years ago was Rs. $50,000$.If the population in these three years has increased at the rate of $10\%,15\%$ and $8\%$ respectively, find the present population.

• A. Rs. $68,210$
• B. Rs. $68,310$
• C. Rs. $68,410$
• D. Rs. $68,510$

Answer 1

The correct option is B Rs. $68,310$

Let $P=.50,000$ and

$R_1=10\%,R_2=15\%,R_3=8\%$.

Present population = $P\times (1+\frac{R_1}{100})(1+\frac{R_2}{100})(1+\frac{R_3}{100})$
On substituiting the given values,

Present population $=$ $50000\times (1+\frac{10}{100})\times (1+\frac{15}{100})\times (1+\frac{8}{100})$

$=$ $50000\times \frac{11}{10}\times \frac{23}{20}\times \frac{27}{25}=68310$

Therefore, present population $=$ $68310$.