Question

In the $2001$ census, the population of India was found to be $8.7\times {10}^7$. If the population increases at the rate of $2.5$% every year, what would be population in $2011?$ The population of India in $2011$ is $k\times {10}^7$, then what is $k$?

Answer 1

Here, $P_0=8.7\times {10}^7$, $r=2.5$, and $n=10$
Let $P$ be the population in $2011$. Then,
$P=P_0{(1+\frac{r}{100})}^n$
$=8.7\times {10}^7{(1+\frac{2.5}{100})}^{10}$
$=8.7\times {10}^7{\left(1.025\right)}^{10}$
Taking log on both side, we get
$\log P=\log [8.7\times {10}^7{\left(1.025\right)}^{10}]$
$=\log 8.7+\log {10}^7+\log {\left(1.025\right)}^{10}$
$=\log 8.7+7\log 10+10\log \left(1.025\right)$
$=0.9395+7+0.1070$
$=8.0465$
$\Rightarrow$ $P=antilog\left(8.0465\right)=1.113\times {10}^8$ (using antilog table)
To compare with given population , writing it in that form $11.13\times {10}^7$

So, $k=11.13$