Question

The population of a town increases every year by 10% If the present population is 180000 what will be the population the town after 2 years fast

Answer 1

$\sf\large\underline{Given:}$

$\tt{\implies Population\:_{(increase\:every\: year)}=10\%}$

$\tt{\implies Present\: Population\:_{(town)}=180000}$

$\sf\large\underline{To\: Find:}$

$\tt{\implies Population\:_{(after\:2\: years\:of\:town)}=?}$

$\sf\large\underline{Solution:}$

• Here ,to calculate population we have to find out the population of the separately each year]

$\sf\large\underline{Population\:of\:1st\: year:}$

$\tt{\implies Population\:_{(1st\: year)}=\dfrac{100+10}{100}\times\:180000}$

$\tt{\implies Population\:_{(1st\: year)}=\dfrac{110}{100}\times\:180000}$

$\tt{\implies Population\:_{(1st\: year)}=110\times\:1800}$

$\tt{\implies Population\:_{(1st\: year)}=198000}$

$\sf\large\underline{Population\:of\:2nd\: year:}$

$\tt{\implies Population\:_{(2nd\: year)}=\dfrac{100+10}{100}\times\:198000}$

$\tt{\implies Population\:_{(2nd\: year)}=\dfrac{110}{100}\times\:198000}$

$\tt{\implies Population\:_{(2nd\: year)}=110\times\:1980}$

$\tt{\implies Population\:_{(2nd\: year)}=217800}$

$\sf\large{Hence,}$

$\bf{\implies Population\:_{(after\:2\: years\:of\:town)}=217800}$

Answer 2

Answer:

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