Question

The population of a town is 6,85,625 and is increasing at the rate of 4% per annum. What will be the population of the town after two years?

• A. 7,51,672
• B. 7,48,764
• C. 7,44,6426
• D. 7,41,572

Answer 1

The correct option is D 7,41,572

Present population of the town = $P$ = $6,85,625$
Rate of increase = $R$ = $4$% per annum
Population of the town after two years = $A$
Time = $t$ = $2$ years

Using the formula for compound interest, we get:
$A=P[1+\frac{R}{100}{]}^2$

On substituting the values, we get:

⇒ $A=6,85,625[1+\frac{4}{100}{]}^2$

⇒ $A=6,85,625[\frac{26}{25}{]}^2$

⇒$A=\frac{6,85,625\times 26\times 26}{25\times 25}$

⇒ $A=7,41,572$

Hence, the population of the town after two years will be $7,41,572$.