Question

The population of a city is 131000. If it increases by \(6\%\) in the first year and \(5\%\) in the second year, then the population of the city at the end of two years is

• A. 145800
• B. 145803
• C. 145600
• D. 145900

Hint:

For successive percentage increase: \[ \boxed{\text{Final Value}=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)} \] Do not add the percentages directly; multiply the successive growth factors.

Answer 1

The Correct Option is B

Concept: When the population increases every year, the increase is calculated using the principle of successive percentage increase (compound growth). The formula is: \[ \boxed{\text{Final Population}=P\left(1+\frac{r_1}{100}\right)\left(1+\frac{r_2}{100}\right)} \] where \(P\) is the initial population.

Step 1: Increase the population by \(6\%\) in the first year.
Initial population \[ P=131000. \] After the first year, \[ 131000\times\frac{106}{100} =138860. \]

Step 2: Increase the new population by \(5\%\).
After the second year, \[ 138860\times\frac{105}{100} =145803. \]

Step 3: Final conclusion.
Therefore, the population at the end of two years is \[ \boxed{145803.} \]