Question

Chapter - compound interest

the population of a town increased by 4% in a first year and decreased by 4% in the second year . if the population at the end of the second year is 499200 , find the population of the town at the beginning of the first year .

Answer 1

Answer:

Refer to the attachment...

Answer 2

${\large{\bold{\sf{\underline{LET'S \: UNDERSTAND \: THE \: QUESTIONS \: FIRST:-}}}}}$

it is the question of compound interest in which it is given that the population of a town increased by 4% in a first year and decreased by 4% in the second year and then the situation is given that if the population at the end of the second year is 499200 we need to find the population of the town at the beginning of the first year .

Let's solve it !!

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${\large{\bold{\sf{\underline{SOLUTION:-}}}}}$

Given,

» Increasing in the population of town in first year = 4%

»Decreasing in the population of town in 2nd year =4%

» population at the end of second year = 499200

To find,

» population of the town at the beginning of the first year

Suppose,

» the population at the beginning of the first year = P

» time = 1 year

» rate of growth in the 1st year = 4% per annum

Therefore, population of the 1st year

$\begin{gathered}\\\;\;\displaystyle{\sf{:\rightarrow\;\;\bf{P\;\times\;\bigg(1\;+\;\dfrac{4}{100}\bigg)}}}\end{gathered}$

$\begin{gathered}\\\;\;\displaystyle{\sf{:\rightarrow\;\;\bf{P\;\bigg(\dfrac{26}{25}\bigg)}}}\end{gathered}$

rate of decrease in the 2nd year = 4% per annum

Therefore, population of the 2ndyear

$\begin{gathered}\\\;\;\displaystyle{\sf{:\rightarrow\;\;\bf{P\;\bigg(\dfrac{26}{25}\bigg)\bigg(1 - \dfrac{4}{100}\bigg)}}}\end{gathered}$

$\begin{gathered}\\\;\;\displaystyle{\sf{\;\;\ \: { \bigg[decreased \: value = \bf P\;\bigg(1 - \dfrac{r}{100} \bigg) ^{n} \bigg] }}}\end{gathered}$

$\begin{gathered}\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;499200\;=\;\bf{ P\;\bigg(\dfrac{26}{25}\bigg)\bigg(\dfrac{24}{25}\bigg)}}}\end{gathered}$

$\begin{gathered}\\\;\;\displaystyle{\sf{:\Longrightarrow\;\; \frac{499200 \times 25 \times 25}{26 \times 24} \;=\;\bf{ P}}}\end{gathered}$

$\begin{gathered}\\\;\;\displaystyle{\sf{:\Longrightarrow\;\; P \;=\;\bf{ 500000}}}\end{gathered}$

$\begin{gathered}\\\;\underline{\boxed{\tt{\odot\;\;Hence,\;\; Population\;=\;\bf{\blue{500000}}}}}\end{gathered}$

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${\large{\bold{\sf{\underline{MORE \: TO \: KNOW :-}}}}}$

APPLICATIONS OF COMPOUND INTEREST FORMULA

The use of CI formula in some situations helps to simplify calculations.

It can be used to compute the following.

1. Increase (or decrease) in population.

2. The growth of bacteria if the rate of growth is known.

3. The value of an item, if its price increases or decreases in immediate years.