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 $r$
$\frac{ dr }{ dt } \propto r \Rightarrow \frac{ dr }{ dt }=- k r$
where $k$ is the constant of proportionality and $k >0$
We have $\frac{ dr }{ r }=- k dt$
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
$\Rightarrow r = Ce ^{- kt } \dots$(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 ^{-10 k } $
$\Rightarrow e ^{10 k }=\frac{49}{50} \times \frac{25}{23}=\frac{49}{46}$
or $e ^{-10 k }=\frac{46}{49}$
Hence, $r =\frac{49}{50} e ^{-10 k \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 $\simeq 864: 1000$