Summary
Judiciary is the independent organ of the government whose function is to provide justice to the innocent, punishment to the culprit and safeguard the rights and duties of the citizen.
This note provides us the information about judiciary and its functions.
Subjective Questions
Q1:
The value of a machine is depreciated from Rs 18000 to Rs 14580 in 2 years. Find the rate of depreciation in present.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present value (P<sub>T</sub>) = Rs 14580<br>Previous value (P) = Rs 18000<br>Time (T) = 2 years<br>Depreciate rate (R) = ?</p> <p>\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*}</p>
Q2:
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.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The price before 2 years (P) = Rs 6000<br>Present price (P<sub>T</sub>) = Rs 5415<br>Time(T) = 2 years<br>Depreciation Rate (R) = ?</p> <p>\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*}</p>
Q3:
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.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present Population (P) = 96000<br>Growth rate (R) = 5%<br>population after T year (P<sub>T</sub>) = 105840<br>Time (T) = ?</p> <p>\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*}</p> <p>\(\therefore \) Time = 2 years</p>
Q4:
The population of a village was 10,000 one year ago. The population at present is 10210. Find the population growth rate?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The population of a village one year ago (P) = 10,000<br>Present population (P_T) = 10210<br>Growth rate (R) = ?<br>Time (T) = 1 year</p> <p>\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*}</p>
Q5:
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?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Population of the last year = 7200</p> <p>Here,</p> <p>\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*}</p>
Q6:
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?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The population of the village last year = 2000</p> <p>\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*}</p>
Q7:
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?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present population (P<sub>T</sub>) = 242000<br>Growth rate (R) = 10%<br>Time (T) = 2 years<br>Population of a town 2 years ago (P) = ?<br>Now,</p> <p>\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*}</p>
Q8:
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?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The present population (P) = 64000<br>Growth rate (R) = 5%<br>Time (T) = 2 years<br>Population of a town after 2 years (P<sub>T</sub>) = ?</p> <p>\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*}</p>
Q9:
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?
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present price of motor cycle (P) = Rs 140,000<br>Rate of depreciation (R) = 7%<br>Time (T) = 2 years after<br>The price of motorcycle 2 years (P<sub>T</sub>) = ?</p> <p>\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*}</p>
Q10:
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.
Type: Short
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present population (P<sub>T</sub>) = 40000<br>Growth rate (R) = 2% + 3% =5%<br>Time (T) = 2 yrs<br>Population of the town after 2 years (P<sub>2</sub>)= ?<br>Now,</p> <p>\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*}</p> <p></p> <p></p>
Q11:
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?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present population (P<sub>T</sub>) = 500000<br>Time (T) = 3 years ago<br>Population of a town 3 years ago = P<br>We know that,</p> <p>\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*}</p> <p><strong>The population before 3 years = 457571 (Approx)</strong></p> <p>After one year ago</p> <p>\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*}</p> <p><strong>The population before 1 year = 515000 \(_{Ans.}\)</strong></p>
Q12:
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.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The population of a village before 2 years (P) = 25000<br>Time (T) = 2 years<br>Rate of growth (R) = 3%<br>\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*}</p> <p>No. of died people = 500<br>Population after 2 years = 26522 - 500 = 26022<br>Population before 1 year (P) =26022<br>Time (T) =1 year<br>Population on growth rate (R) = 3%</p> <p>\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*}</p> <p>\(\therefore \) The present population of village = 26803 \(\:\:_{Ans}\)</p>
Q13:
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?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Total population = 30000<br>Number of people by migration = 5800<br>\(\therefore\) Present population (P<sub>T</sub>) = 30000 - 5800 = 24200<br>Growth rate (R) = 10%<br>Time (T) = 2 years<br>We know that,</p> <p>\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*}</p> <p>\(\therefore\) The population of the town at first = 20,000 \(\:\:_{Ans.}\)</p>
Q14:
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.
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The price of a computer before 2 years (P) = Rs 44100<br>Present price of computer (P<sub>T</sub>) = Rs 40000<br>Time (T) = 2 years<br>Rate of depreciation (R) = ?</p> <p>We know that,</p> <p>\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*}</p> <p>\(\therefore\) the rate of depreciation = 4.76% \(\:\:\: _{Ans.}\)</p>
Q15:
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?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>\begin{align*} \text{Price of the one ropani of land 3 year ago (P)} &= \frac{1250000}{4}\\ & = Rs\: 312500 \end{align*}</p> <p>Price of one ropaniof land at present (P<sub>T</sub>) = Rs 160000<br>Rate of depreciated (R) = ?<br>Time (T) = 3 years<br>We know that,</p> <p>\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*}</p> <p>The depreciated rate (R) = 20%</p>
Q16:
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?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>Present population (P<sub>T</sub>) = 24895 + 320 = 25215<br>Time (T) = 2 years<br>Growth rate (R) = 2.5%<br>Population before 2 years (P) =?</p> <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*}</p> <p>\(\therefore\) the population of a village 2 years before = 24000 \(\:_{Ans.}\)</p>
Q17:
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?
Type: Long
Difficulty: Easy
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Answer: <p><strong>Solution:</strong></p> <p>The population of a town in 2070 (P) = 1,00,000<br>Population growth (R) = 2%<br>Time (T) = 1 year<br>The population of a town in 2071 (P<sub>T</sub>) = ?</p> <p>\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*}</p> <p>Migrated people = 8,000<br>The population of a town after migrated people in 2071 (P) = 102000 + 8000 =1,10,000<br>Time (T) = 2 years<br>Rate (R) = 2%<br>The population of a town in the beginning of 2071 (P<sub>T</sub>) =?</p> <p>\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*}</p> <p>\(\therefore\) The population of a town in the beginning of 2071 = 1,14,444 \(\:\:_{Ans.}\)</p>
