## Note on Population Growth and Compound depreciation

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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.

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.

.

### Very Short Questions

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:

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*}

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