Question

The population of a country increased by an average of $2$% per year from $2000$ to $2003$. If the population of this country was $2,000,000$ on December $31$, $2003$, then the population of this country on January $1$, $2000$, to the nearest thousand would have been

• A. $1846000$
• B. $1852000$
• C. $1000000$
• D. $1500000$

Answer 1

The correct option is A $1846000$

A population of size $P$ increasing at the rate of $2\%$ may be modelled as follows:

$P=P_0{e}^{0.02t}$, where $t$ is the number of years after $t=0$ and $P_0$ is the population at $t=0$.

Let $t=0$ corresponds to January $1$, $2000$ and therefore, $t=4$ corresponds to December $31$.

But $P$ is $2,000,000$ when $t=4$.

Hence, $2,000,000=P_0{e}^{0.02\times 4}$

Solve the above for $P_0$ as:

$P_0=\frac{2,000,000}{e^{0.02\times 4}}=1846000$