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The population of a town increases every year by 4%. If its present population is 50,000, then after 2 years it will be

153,900
254,000
354,080
454,900

Answer:- 1

Explanation:-

Solution:

\begin{aligned} Here we can use the compound interest based formula, \\ Population after n years \\ = P \times {[1 + (\frac{r}{100})]}^{n} \\ Population after 2 years \\ = 50000 \times {[1 + (\frac{4}{100})]}^{2} \\ Population after 2 years \\ = 54080 \end{aligned}

$\eqalign{ & {\text{Here we can use the compound interest based formula,}} \cr & {\text{Population after }}n{\text{ years}} \cr & = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & {\text{Population after 2 years}} \cr & = 50000 \times {\left[ {1 + \left( {\frac{4}{{100}}} \right)} \right]^2} \cr & {\text{Population after 2 years}} \cr & = 54080 \cr}$
Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
Then, population after 2 years= 54,080.
In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100.

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Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
Then, population after 2 years= 54,080.
In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100. ", "dateCreated": "7/24/2019 10:09:12 AM", "author": { "@type": "Person", "name": "Nitin Sir" } }, "suggestedAnswer": { "@type": "Answer", "text": "

Solution:

\begin{aligned} Here we can use the compound interest based formula, \\ Population after n years \\ = P \times {[1 + (\frac{r}{100})]}^{n} \\ Population after 2 years \\ = 50000 \times {[1 + (\frac{4}{100})]}^{2} \\ Population after 2 years \\ = 54080 \end{aligned}

$\eqalign{ & {\text{Here we can use the compound interest based formula,}} \cr & {\text{Population after }}n{\text{ years}} \cr & = P \times {\left[ {1 + \left( {\frac{r}{{100}}} \right)} \right]^n} \cr & {\text{Population after 2 years}} \cr & = 50000 \times {\left[ {1 + \left( {\frac{4}{{100}}} \right)} \right]^2} \cr & {\text{Population after 2 years}} \cr & = 54080 \cr}$
Alternatively,
we can use, net percentage change graphic as well,
50,000------4%↑---→ 52,000---- 4%↑---→ 54,080.
Then, population after 2 years= 54,080.
In this calculation, we need to find 1% of 50,000 first, which is easily calculated by dividing 50,000 by 100. ", "dateCreated": "7/24/2019 10:09:12 AM" } }

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