Question

The female-male ratio of a village decreases continuously at the rate proportional to their ratio at any time. If the ratio of female : male of the villages was $980:1000$ in $2001$ and $920:1000$ in $2011$ . What will be the ratio in $2021$ ?

• A. 864 : 1000
• B. 864 : 100
• C. 1000 : 864
• D. 100 : 864

Answer 1

Answer: (A)

Let female-male ratio at any time be $r$
$\frac{dr}{dt}\propto r\Rightarrow \frac{dr}{dt}=-kr$
where $k$ is the constant of proportionality and $k>0$
We have $\frac{dr}{r}=-kdt$
Integrating both sides, we have
$\int \frac{dr}{r}=-k\int dt$
$\log r=-kt+\log C$
$\log r-\log C=-kt$
$\Rightarrow \log \left(\frac{r}{C}\right)=-kt$
where $\log C$ is the constant of integration
$\Rightarrow r=C{e}^{-kt}\dots$(i)
Let us start time from the year $2001$,
So in $2001,t=0,r=\frac{980}{1000}=\frac{49}{50}$
Putting $t=0$ in (i), we have
$\frac{49}{50}=C$
$\Rightarrow r=\frac{49}{50}{e}^{-kt}\dots$(ii)
Also in the year $2011,t=10$ and
$r=\frac{920}{1000}=\frac{23}{25}$
Putting in (ii), we have
$\frac{23}{25}=\frac{49}{50}{e}^{-10k}$
$\Rightarrow e^{10k}=\frac{49}{50}\times \frac{25}{23}=\frac{49}{46}$
or $e^{-10k}=\frac{46}{49}$
Hence, $r=\frac{49}{50}{e}^{-10k\times \frac{t}{10}}$
$\Rightarrow r=\frac{49}{50}{\left(\frac{46}{49}\right)}^{\frac{t}{10}}$
In the year 2021, $t=20$
$\therefore r=\frac{49}{50}{\left(\frac{46}{49}\right)}^{\frac{20}{10}}$
$=\frac{49}{50}\times \frac{46}{49}\times \frac{46}{49}=0.864$
Thus, at this trend female $:$ male $\simeq 864:1000$