Question

the present population of the town is 20000 its population increases by 10% in first year and 15% in the second find the population of the term at the end of 2 years​

Answer 1

➤ Given :-

Present population in a town :- 20000

Increased percent in first year :- 10%

Increased percent in second year :- 15%

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➤ To Find :-

Population of the town in second year...

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★ How to do :-

Here, we are given with the population of a town and it increases by 10% in the first year and 15% in the second year respectively. We are asked to find the population of the same after two years. So, first we will find the population of the town after first year by using the present population and the increased population in first year. Then, we find the population after second year by using the value after first year and the second increased population. So, let's solve!!

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➤ Solution :-

Increased population in first year :-

${\tt \leadsto 10 \bf\% \: \: \tt of \: \: 20000}$

Write the percentage as fraction and 'of' as multiplication sign.

${\tt \leadsto \dfrac{10}{100} \times 20000}$

Cancel the zeros in the fraction.

${\tt \leadsto \dfrac{1 \not{0}}{10 \not{0}} \times 20000 = \dfrac{1}{10} \times 20000}$

Multiply the remaining numbers.

${\tt \leadsto \dfrac{1 \times 20000}{10} = \dfrac{20000}{10}}$

Write the fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{20000}{10} = 2000}$

Population after first year :-

${\tt \leadsto 20000 + 2000}$

Add the values now.

${\tt \leadsto 22000}$

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Increased population in second year :-

${\tt \leadsto 15 \bf\% \: \: \tt of \: \: 22000}$

Write the percentage as fraction and 'of' as multiplication sign.

${\tt \leadsto \dfrac{15}{100} \times 22000}$

Write the fraction in lowest form by cancellation method.

${\tt \leadsto \dfrac{\cancel{15}}{\cancel{100}} \times 22000 = \dfrac{3}{20} \times 22000}$

Multiply the remaining numbers.

${\tt \leadsto \dfrac{3 \times 22000}{20} = \dfrac{66000}{20}}$

Write the fraction in lowest form by cancellation method.

${\tt \leadsto \cancel \dfrac{66000}{20} = 3300}$

Population after second year :-

${\tt \leadsto 22000 + 3300}$

Add the values to get the answer.

${\tt \leadsto 25300}$

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${\red{\underline{\boxed{\bf So, \: the \: population \: after \: two \: years \: is \: 25300}}}}$