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The population of a town increased from 1,75,000 to 2,62,500 in a decade. The average percent increase of population per year is:

14.37%
25%
36%
48.75%

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Increase in 10 years \\ = (262500 - 175000) \\ = 87500. \\ Increase % \\ = (\frac{87500}{175000} \times 100) % = 50 % . \\ ∴ Required average \\ = (\frac{50}{10}) % = 5 % . \end{aligned}

$\eqalign{ & {\text{Increase}}\,{\text{in}}\,{\text{10}}\,{\text{years}} \cr & = \left( {262500 - 175000} \right) \cr & = 87500. \cr & {\text{Increase}}\% \cr & = \left( {\frac{{87500}}{{175000}} \times 100} \right)\% = 50\% . \cr & \therefore {\text{Required}}\,{\text{average}} \cr & = \left( {\frac{{50}}{{10}}} \right)\% = 5\% . \cr}$

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", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Increase in 10 years \\ = (262500 - 175000) \\ = 87500. \\ Increase % \\ = (\frac{87500}{175000} \times 100) % = 50 % . \\ ∴ Required average \\ = (\frac{50}{10}) % = 5 % . \end{aligned}

$\eqalign{ & {\text{Increase}}\,{\text{in}}\,{\text{10}}\,{\text{years}} \cr & = \left( {262500 - 175000} \right) \cr & = 87500. \cr & {\text{Increase}}\% \cr & = \left( {\frac{{87500}}{{175000}} \times 100} \right)\% = 50\% . \cr & \therefore {\text{Required}}\,{\text{average}} \cr & = \left( {\frac{{50}}{{10}}} \right)\% = 5\% . \cr}$ ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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