Question

Population of the village in 2013 was 500000.if in 2014 there is an increment of 25%,in 2015 there is an decrement of 15% and in 2016 there is an increment of 12%,then find the population of village at the end of year 2016​

Answer 1

AnswEr :

$\frak{Given}\begin{cases}\sf{Current \:Population \:(2013)=500,000}\\\sf{2014=25\% \: Increase}\\\sf{2015=15\% \:Decrease}\\\sf{2016=12\% \:Increase} \end{cases}$

☯⠀$\underline{\textsf{Let's Head to the Question Now :}}$

$\dashrightarrow\tt\:\:{\scriptsize New\:Population=Old\:Population\times (2014 \:Effect) \times(2015 \:Effect) \times(2016\:Effect)}\\\\\\\dashrightarrow\tt\:\:New\:Population=500000 \times (100 + 25)\% \times (100 - 15)\% \times (100 + 12)\%\\\\\\\dashrightarrow\tt\:\:New\:Population=500000 \times 125\% \times 85\% \times 112\%\\\\\\\dashrightarrow\tt\:\:New\:Population=500000 \times \dfrac{125}{100} \times \dfrac{85}{100} \times \dfrac{112}{100}\\\\\\\dashrightarrow\tt\:\:New\:Population=5 \times 125 \times 85 \times \dfrac{112}{10}\\\\\\\dashrightarrow\tt\:\:New\:Population=125 \times 85 \times 56\\\\\\\dashrightarrow\:\:\underline{\boxed{\red{\tt New\:Population=595000}}}$

⠀

$\therefore\:\underline{\textsf{Population of village at end of 2016 is \textbf{595,000.}}}$

$\rule{220}{2}$

$\boxed{\begin{minipage}{7 cm}\underline{\text{Some Important Formulae Related to it :}}\\ \\\sf Increase = Old Population\times(100+\sf Increase)\%\\ \\Decrease=Old \:Population\times(100-Decrease\%)\end{minipage}}$

Answer 2

Answer:

$\large{\underline{\boxed{\mathfrak\blue{Population\: at \: the \: end \: of \: 2016 =595,000}}}}$

$\rule{100}{2}$

Given:-

• Population of village in 2013 = 500000

• Increase of population in 2014 = 25%

• Decrease of population in 2015 = 15%

• Increase in population in 2016 = 12%

To find:-

• Population at the end of 2016 = ?

Solution:-

As we know that, population increases in the form of compound way.But if it's increasing in just 1 year,then we can calculate population in simple way.

At first,

We know formula:-

$\star\: \: {\boxed{\mathfrak\green{SI = P \times r \times n }}}$

⇒ Population increases=500000×25% × 1

⇒ Population increases = 500000×1/4

⇒Population increases = 125000

$\therefore$ Population in 2014=500000+125000

$\therefore\large\sf\red{Population \: in \: 2014 = 625000}$

$\rule{100}{2}$

Using same formula:-

⇒Population decreases=6250000×15% ×1

⇒Population decreases =625000 ×15/100

⇒Population decreases = 93750

$\therefore$ Population in 2015 = 625000 - 93750

$\therefore\large\sf\red{Population \: in \: 2015 = 531250}$

$\rule{100}{2}$

Using same formula:-

⇒Population increases = 531250×12% × 1

⇒Population increases = 63750

$\therefore$ Population in 2016 = 531250 + 63750

$\therefore\large{\underline{\boxed{\sf\purple{Population\: in \:2016 = 595,000}}}}$

$\therefore{\underline{\text{Population \: at \: the \: end \: of \: 2016 = 595,000 }}}$