 ## Population Growth and Compound depreciation

Subject: Compulsory Mathematics

#### Overview

The growth number of the population does not remain constant. So, it is calculated in compounded way.

##### Population Growth and Compound depreciation

According to Lexmark," the population of the world remains constant that means the total number of animals including social animals in each year is same. But the social animals are increased year by year and rare animals and wild animals are decreased year by year". So, the number of people are increased yearly. In case of Nepal, the population of 2058 and 2064 are different. The population of 2064 BS is more than that of 2058 BS. The growth of the population indicates the additional population than the previous year. The growth number of the population does not remain constant. So, it is calculated in compounded way.

Let 'P' be the population at the beginning of certain year and Pbe the population after 'T' years, then

$P_T = P \left(1 + \frac {R_1} {100}\right)\left(1 + \frac {R_2} {100}\right)\left(1 + \frac {R_3} {100}\right) ................... \left(1 + \frac {R_T} {100}\right)$ where $R_1, R_2, R_3, ....... R_T$ are the rates of 1st, 2nd, 3rd, ............ tth years respectively.

If R1=R2= R3=............., RT = R then $P_T = P \left(1 + \frac {R} {100}\right)^T$

$P_T = P \left(1 + \frac {R} {100}\right)^T$

In case of decrease in population, $P_T = P \left(1 - \frac {R} {100}\right)^T$.

#### Compound depreciation

The value of the machine is decreased yearly due to various reasons like wear and tear, inefficiency etc. The reduced value of the machine is called the depreciated value. The amount of depression is different in successive years even the rate of depreciation is same. Such depression is called compound depreciation. Therefore, the formula for depreciated value or scrap value which is used to solve the problems of compound depreciation can be written as:-

$P_T = P \left (1 - \frac {R} {100} \right ) ^ T$

$or, S = V\left (1 - \frac {R} {100} \right ) ^ T$

Where S, V, R, T indicates amount after compound depreciation, principal (original value), rate of depreciation in percentage per annum and number of years for which goods are used respectively. The amount of compound depreciation determined by subtracting scrap value 'S' from the original value V i.e. Compound depreciation = V - S or P - PT.

##### Things to remember

When the number of periods is not an integer, compound interest for the integral  period is calculated first and the simple interest at the given rate for the fractional period is calculated:

a. Simple interest and compound interest of the first period are equal.

b. The compound interest of more than one simple interest.

c. The difference of amounts of two consecutive periods is equal to the interest  on the amount of first conversion period.

d. The difference of compound interest of the two consecutive periods is equal to the interest of the first conversion period.

• It includes every relationship which established among the people.
• There can be more than one community in a society. Community smaller than society.
• It is a network of social relationships which cannot see or touched.
• common interests and common objectives are not necessary for society.
##### Videos for Population Growth and Compound depreciation ##### Population Explanation and Type 1

Solution:

Present value (PT) = Rs 14580
Previous value (P) = Rs 18000
Time (T) = 2 years
Depreciate rate (R) = ?

\begin{align*} P_T &= P \left( 1 - \frac{R}{100} \right)^T \\ or, 14580 &= 18000 \left( 1 - \frac{R}{100} \right)^2\\ or, 0.81 &=\left( 1 - \frac{R}{100} \right)^2\\ or, (0.9)^2 &= \left(1 - \frac{R}{100}\right)^2\\ or, 0.9 &= 1 - \frac{R}{100}\\ or, \frac{R}{100} &= 1- 0.9 \\ or, R&= 0.1\times100\\ \therefore R &= 10\% \:\:\:_{Ans.} \end{align*}

Solution:

The price before 2 years (P) = Rs 6000
Present price (PT) = Rs 5415
Time(T) = 2 years
Depreciation Rate (R) = ?

\begin{align*} P_T &= P \left( 1 - \frac{R}{100} \right)^T \\ or, 5415 &= 6000 \left( 1 - \frac{R}{100} \right)^2\\ or, 0.9025 &=\left( 1 - \frac{R}{100} \right)^2\\ or, (0.95)^2 &= \left(1 - \frac{R}{100}\right)^2\\ or, 0.95 &= 1 - \frac{R}{100}\\ or, \frac{R}{100} &= 1- 0.95 \\ or, R&= 0.05\times100\\ \therefore R &= 5\% \:\:\:_{Ans.} \end{align*}

Solution:

Present Population (P) = 96000
Growth rate (R) = 5%
population after T year (PT) = 105840
Time (T) = ?

\begin{align*} P_T &= P \left( 1 + \frac{R}{100} \right)^T \\ 105840 &= 96000\left( 1 + \frac{5}{100} \right)^T\\ or, \frac{105840}{96000} &= \left( \frac{105}{100}\right)^T \\ or, 1.1025 &= (1.05)^T \\ or, (1.05)^2 &= (1.05)^T\\ T &= 2 \end{align*}

$\therefore$ Time = 2 years

Solution:

The population of a village one year ago (P) = 10,000
Present population (P_T) = 10210
Growth rate (R) = ?
Time (T) = 1 year

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^2 \\ or, 10210 &= 10000 \left(1 + \frac{R}{100}\right)^T \\ or, \frac{10210}{10000} &= 1 + \frac{R}{100}\\ or, 1.021 &= 1 + \frac{R}{100}\\ or, 1.021 - 1 &= \frac{R}{100}\\ or, R &= 0.021 \times 100\% \\ \therefore R &= 2.1\% \end{align*}

Solution:

Population of the last year = 7200

Here,

\begin{align*} Population \: after \: one \: year &= 7200 - 7200 \times \frac{5}{100} - 7200 \times \frac{2}{100}\\ &= 7200 - 360 -144 \\ &= 7200 - 504\\ &= 6996 \:\:\: _{Ans.} \end{align*}

Solution:

The population of the village last year = 2000

\begin{align*} \text{Population of a place after one year } &= 2000 + 2000 \times \frac{3}{100} + 2000 \times \frac{2}{100}\\ &= 2000+60+40\\ &= 2100 \:\:\:\:\: _{Ans} \end{align*}

Solution:

Present population (PT) = 242000
Growth rate (R) = 10%
Time (T) = 2 years
Population of a town 2 years ago (P) = ?
Now,

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 242000 &= P \left( 1 + \frac{10}{100}\right)^2\\ 242000 &= P(1.1)^2 \\ or, P &= \frac{242000}{1.21} \\ \therefore P &= 200,000 \end{align*}

Solution:

The present population (P) = 64000
Growth rate (R) = 5%
Time (T) = 2 years
Population of a town after 2 years (PT) = ?

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ P_T &= 64000 \left( 1 + \frac{1}{100}\right)^2\\ &= 64000 (1.05)^2\\ &=64000 \times 1.1025 \\ &= 70560 \:\:\:\: _{Ans.} \end{align*}

Solution:

Present price of motor cycle (P) = Rs 140,000
Rate of depreciation (R) = 7%
Time (T) = 2 years after
The price of motorcycle 2 years (PT) = ?

\begin{align*} P_T &= P \left( 1 - \frac{R}{100}\right)^T \\ &= 140000 \left( 1 - \frac{7}{100}\right)^2 \\ &= 140000[1 - 0.07]^2 \\ &= 140000 \times (0.93)^2\\ &= 140000 \times 0.8649 \\ &= Rs \: 121086 \: \: \: _{Ans.} \end{align*}

Solution:

Present population (PT) = 40000
Growth rate (R) = 2% + 3% =5%
Time (T) = 2 yrs
Population of the town after 2 years (P2)= ?
Now,

\begin{align*} P_2 &= P_t \left( 1 + \frac{R}{100} \right)^T \\ &= 40000 \left( 1 + \frac{5}{100} \right)^2\\ &=40000\left( \frac{105}{100} \right)^2\\ &= 40000 \times \frac{105 \times 105}{100 \times 100} \\ &= 44,100 _{Ans.} \end{align*}

Solution:

Present population (PT) = 500000
Time (T) = 3 years ago
Population of a town 3 years ago = P
We know that,

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 500000 &= P\left( 1 + \frac{3}{100}\right)^3 \\ or, 500000&= P(1.03)^3\\ or, P &= \frac{500000}{1.092727}\\P&= 457570.83\end{align*}

The population before 3 years = 457571 (Approx)

After one year ago

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ or, 500000 &= P\left( 1 + \frac{3}{100}\right)^1\\ or, P \times 1.03 &= 500000\\ or, P &= \frac{500000}{1.03}\\ P&=515000 \end{align*}

The population before 1 year = 515000 $_{Ans.}$

Solution:

The population of a village before 2 years (P) = 25000
Time (T) = 2 years
Rate of growth (R) = 3%
\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\&= 25000 \left( 1 + \frac{3}{100}\right)^2\\ &= 25000 \times (1.03)^3\\ &= 25000 \times 1.0609\\ &= 26522.5 \approx 26522 \end{align*}

No. of died people = 500
Population after 2 years = 26522 - 500 = 26022
Population before 1 year (P) =26022
Time (T) =1 year
Population on growth rate (R) = 3%

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T \\ &=26022 \left( 1 + \frac{3}{100}\right)^1 \\ &=26022 \times \frac{103}{100}\\ &= 26803 \:(Approx)\end{align*}

$\therefore$ The present population of village = 26803 $\:\:_{Ans}$

Solution:

Total population = 30000
Number of people by migration = 5800
$\therefore$ Present population (PT) = 30000 - 5800 = 24200
Growth rate (R) = 10%
Time (T) = 2 years
We know that,

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ or, 24200&= P \left( 1 + \frac{10}{100}\right)^2 \\ or, 24200 &= P(1.1)^2 \\ or, P \times 1.21 &= 24200 \\ or, P &= \frac{24200}{1.21} \\ \therefore P &= 20000 \end{align*}

$\therefore$ The population of the town at first = 20,000 $\:\:_{Ans.}$

Solution:

The price of a computer before 2 years (P) = Rs 44100
Present price of computer (PT) = Rs 40000
Time (T) = 2 years
Rate of depreciation (R) = ?

We know that,

\begin{align*} P_T &= P \left( 1 - \frac{R}{100}\right)^T\\ or, 40000 &= 44100 \left( 1 - \frac{R}{100}\right)^2\\ or, \frac{40000}{44100} &= \left( 1 - \frac{R}{100}\right)^2\\ or, \left( 1 - \frac{R}{100}\right)^2 &= \left(\frac{20}{21}\right)^2\\ or, 1 - \frac{R}{100}&= \frac{20}{21}\\ or,1-\frac{20}{21} &= \frac{R}{100} \\ or, \frac{21-20}{21} &= \frac{R}{100}\\ or, \frac{R}{100} &= \frac{1}{21}\\ or, R &= \frac{100}{21}\\ R &= 4.76\% \end{align*}

$\therefore$ the rate of depreciation = 4.76% $\:\:\: _{Ans.}$

Solution:

\begin{align*} \text{Price of the one ropani of land 3 year ago (P)} &= \frac{1250000}{4}\\ & = Rs\: 312500 \end{align*}

Price of one ropaniof land at present (PT) = Rs 160000
Rate of depreciated (R) = ?
Time (T) = 3 years
We know that,

\begin{align*} P_T &= P \left( 1 - \frac{R}{100}\right)^T\\ or, 160000 &= 3125000 \left( 1 - \frac{R}{100}\right)^3\\ or, \frac{160000}{312500}&= \left( 1 - \frac{R}{100}\right)^3\\ or, \frac{64}{125}&= \left( 1 - \frac{R}{100}\right)^3\\ or, \left( \frac{4}{5} \right)^3 &= \left( 1 - \frac{R}{100}\right)^3\\ \frac{4}{5}&=1-\frac{R}{100}\\ or, \frac{R}{100}&= 1-\frac{4}{5} \\ or, R &= \frac{5-4}{5} \times 100\\ R &=20\% \end{align*}

The depreciated rate (R) = 20%

Solution:

Present population (PT) = 24895 + 320 = 25215
Time (T) = 2 years
Growth rate (R) = 2.5%
Population before 2 years (P) =?

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ or, 25215 &= P \left( 1 + \frac{2.5}{100}\right)^2\\or, 25215 &= P(1.021)^2\\ or, P \times 1.050625 &= 25215\\ P &= \frac{25215}{1.050625}\\ P &= 24000 \end{align*}

$\therefore$ the population of a village 2 years before = 24000 $\:_{Ans.}$

Solution:

The population of a town in 2070 (P) = 1,00,000
Population growth (R) = 2%
Time (T) = 1 year
The population of a town in 2071 (PT) = ?

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ &=100000\left( 1 + \frac{2}{100}\right)^1\\ &= 100000 \times \frac{102}{100}\\ &= 1,02,000 \end{align*}

Migrated people = 8,000
The population of a town after migrated people in 2071 (P) = 102000 + 8000 =1,10,000
Time (T) = 2 years
Rate (R) = 2%
The population of a town in the beginning of 2071 (PT) =?

\begin{align*} P_T &= P \left( 1 + \frac{R}{100}\right)^T\\ &=110000\left( 1 + \frac{2}{100}\right)^2\\ &= 110000 \times \frac{102}{100} \times \frac{102}{100}\\ &= 1,14,444 \end{align*}

$\therefore$ The population of a town in the beginning of 2071 = 1,14,444 $\:\:_{Ans.}$