Subject: Compulsory Mathematics

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

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 P_{T }be 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 1^{st}, 2^{nd}, 3^{rd}, ............ t^{th} years respectively.

If R_{1}=R_{2}= R_{3}=............., R_{T }= 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, \(P_T = P \left(1 - \frac {R} {100}\right)^T\).

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 - P_{T}.

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.

The value of a machine is depreciated from Rs 18000 to Rs 14580 in 2 years. Find the rate of depreciation in present.

**Solution:**

Present value (P_{T}) = 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*}

A radio costing Rs 6,000 is depreciated per year and 2 years later its price becomes to Rs 5,415. Find the rate of depreciation in percent.

**Solution:**

The price before 2 years (P) = Rs 6000

Present price (P_{T}) = 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*}

The population of a town is 96,000. In how many years lateral would it be 1,05,840 if the population increased at the rate of 5% every year.

**Solution:**

Present Population (P) = 96000

Growth rate (R) = 5%

population after T year (P_{T}) = 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

The population of a village was 10,000 one year ago. The population at present is 10210. Find the population growth rate?

**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*}

The population of a village was 7200. 5% of the population was migrated and 2% died due to different cause within a year. What would be the population of the village after a year?

**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*}

The population of a place was 2000, within a year the population is increased by 3% by birth rate and 2% by migration. How much population was there now?

**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*}

The population of a town is increased by 10% each year. If the present population of that town is 242000, find the population of the town 2 years ago?

**Solution:**

Present population (P_{T}) = 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*}

The present population of a town is 64,000 and it increases at the rate of 5% per annum. What will be the population of the town after two years?

**Solution:**

The present population (P) = 64000

Growth rate (R) = 5%

Time (T) = 2 years

Population of a town after 2 years (P_{T}) = ?

\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*}

The present price of motorcycle is Rs 140000. If it depreciate at 7% per year. What will be the price of motorcycle after 2 years?

**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 (P_{T}) = ?

\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*}

The present population of a town is 40,000. If the population increase by 2% by birth and 3% by immigration, what will be the population of the town after 2 years? Find it.

**Solution:**

Present population (P_{T}) = 40000

Growth rate (R) = 2% + 3% =5%

Time (T) = 2 yrs

Population of the town after 2 years (P_{2})= ?

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*}

The population of a town increased every year by 3% . If the present population is 500000.

(i) What was the population of the town 3 years ago?

(ii) What will be the population after one year ago?

**Solution:**

Present population (P_{T}) = 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 \(_{Ans.}\)**

3 year ago the population of a village was 25000. The rate of growth of population is 3% one year ago, 500 people died because of the earthquake. What is the population of village.

**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 \(\:\:_{Ans}\)

The population of a town increases every year by 10% . At the end of two years the total population of the town was 30000. If 5800 people were added by migration. What was the population of the town in the beginning?

**Solution:**

Total population = 30000

Number of people by migration = 5800

\(\therefore\) Present population (P_{T}) = 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 \(\:\:_{Ans.}\)

A man bought a computer for Rs 44100 and after using it for 2 years sold it for Rs 44100 and after using it for 2 years sold it for Rs 40000. Find the rate of compound depreciation of the computer.

**Solution:**

The price of a computer before 2 years (P) = Rs 44100

Present price of computer (P_{T}) = 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% \(\:\:\: _{Ans.}\)

4 ropani of land was bought for Rs 12,50,000 three year ago. If one ropani of land be sold for Rs 160000 at present. At what percentage its value depreciated?

**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 (P_{T}) = 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%

The population of Jhapa increase every year by 2.5%. At the and of two years, the total population of the village was 24895. If 320 people were migrated to the other village. What was the population of the village in the beginning? Find it?

**Solution:**

Present population (P_{T}) = 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 \(\:_{Ans.}\)

In the beginning of 2070 B.S. The population of a two was 1,00,000 and the rate of population growth is 2% every year. In the beginning of 2071 B.S. 8,000 people migrated from different places. What will be the population of the town in 2072 B.S?

**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 (P_{T}) = ?

\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 (P_{T}) =?

\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 \(\:\:_{Ans.}\)

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