Question

The population of a town increases every near by 8%. if the population in the year 2012 is 1,18,800, what was the population in the year 2011?​

Answer 1

$\sf\red{Given:}$

• Percentage of increase population every year=$\frak{8}$%
• Population in year $\frak{2012=1,18,800}$

$\rm\blue{To\:find:}$

• The population in the year $\frak{2011}$

$\sf\red{Solution:}$

Let us take the population of the town in the year $\rm{2011\:\:be\:\:x}$

Increased per cent=$\frak{8}$%

Actual increase in population

$\rm\longrightarrow{8}$%$\sf{\:of\:x=}\dfrac{8}{100}\:of\:x$

$\rm\longrightarrow\dfrac{2x}{25}$

Population in the year $\rm{2012}$

(in the terms of x)$\rm{=x+}\dfrac{2x}{25}$

$\rm\longrightarrow\dfrac{25x+2x}{25}$

$\rm\longrightarrow\dfrac{27x}{25}$

Thus,it is given that

$\rm\longrightarrow\dfrac{27x}{25}{=1,18,800.}$

$\rm\longrightarrow{x=}\dfrac{1,18,800×25}{27}$

$\longrightarrow1,10,000$

Hence,the population of the town in the year $\rm{2011=1,10,000}$

Answer 2

$\huge\star\underline\mathtt\blue{AnsWer}$

Let the population of the town after 2 years be x

Increase percent of population every year = 10%

Present population = 180000

= 10% of 180000

= 10/100 × 180000 = 10×1800 = 18000

Increase in population after 1 year = 180000+18000 = 198000

Increase in population afer 2 years = 10% of population in 1 year + population of 1 year

= 10% of 198000+198000

= 10/100×198000 = 10×1980 = 19800

= 198000+19800 = 217800

Therefore, the population of the town after 2 years is 2,17,800.

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