Class 10 Compulsory Mathematics – जनसङ्ख्या वृद्धि र मिश्रहास (Population Growth and Compound Depreciation)
Chapter Information
Chapter: जनसङ्ख्या वृद्धि र मिश्रहास (Population Growth and Compound Depreciation)
Subject: Compulsory Mathematics, Class 10 SEE
Writer: D.R. Simkhada
Publisher: Readmore Publishers & Distributors
Description: Complete notes on Population Growth and Compound Depreciation covering formulas, examples, and solved problems as per SEE Nepal curriculum.
अभ्यास (Exercises)
१. एउटा सहरको सुरुको जनसङ्ख्या ८०,००० थियो। यदि जनसङ्ख्या वृद्धि दर १०% छ भने, दुई वर्षपछि सहरको जनसङ्ख्या कति होला? (The population of a town at the beginning was 80,000. If the population growth rate is 10%, what will be the population of the town after two years?)
Solution:
\( P = 80,000, R = 10\%, T = 2 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T = 80000 \left(1 + \frac{10}{100}\right)^2 \)
\( = 80000 \times (1.1)^2 = 80000 \times 1.21 = 96,800 \)
तसर्थ, दुई वर्षपछि सहरको जनसङ्ख्या ९६,८०० हुनेछ। (Hence, the population after two years will be 96,800.)
२. एउटा गाउँको जनसङ्ख्या हरेक वर्ष ५% ले बढ्छ। यदि त्यो गाउँको हालको जनसङ्ख्या १६,००० छ भने, ३ वर्षपछि गाउँको जनसङ्ख्या पत्ता लगाउनुहोस्। (The population of a village increases every year by 5%. If the present population of that village is 16,000, find the population of the village after 3 years.)
Solution:
\( P = 16,000, R = 5\%, T = 3 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T = 16000 \left(1 + \frac{5}{100}\right)^3 \)
\( = 16000 \times (1.05)^3 = 16000 \times 1.157625 = 18,522 \)
तसर्थ, ३ वर्षपछि जनसङ्ख्या १८,५२२ हुनेछ। (Hence, the population after 3 years will be 18,522.)
३. दुई वर्षअघि एउटा नगरपालिकाको जनसङ्ख्या ५०,००० थियो। यदि जनसङ्ख्या वृद्धि दर प्रतिवर्ष २% छ भने, त्यो नगरपालिकाको हालको जनसङ्ख्या कति छ? (The population of a municipality was 50,000, two years ago. If the rate of increase of population is 2% per year, what is the present population of that municipality?)
Solution:
\( P = 50,000, R = 2\%, T = 2 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T = 50000 \left(1 + \frac{2}{100}\right)^2 \)
\( = 50000 \times (1.02)^2 = 50000 \times 1.0404 = 52,020 \)
तसर्थ, हालको जनसङ्ख्या ५२,०२० हो। (Hence, the present population is 52,020.)
४. एउटा सहरको जनसङ्ख्या १,३३,१०० छ। यदि जनसङ्ख्या वृद्धि दर १०% छ भने, ३ वर्षअघि जनसङ्ख्या कति थियो? (The population of a city is 133,100. If the population growth rate is 10%, what was the population before 3 years?)
Solution:
\( P_T = 133,100, R = 10\%, T = 3 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T \)
\( 133100 = P \times (1.1)^3 = P \times 1.331 \)
\( P = \frac{133100}{1.331} = 100,000 \)
तसर्थ, ३ वर्षअघि जनसङ्ख्या १,००,००० थियो। (Hence, the population before 3 years was 100,000.)
५. तीन वर्षअघि एउटा सहरको जनसङ्ख्या २०,००० थियो। पछिल्ला तीन वर्षमा वृद्धि दर क्रमशः २%, २.५% र ४% थियो। सहरको हालको जनसङ्ख्या पत्ता लगाउनुहोस्। (Three years ago the population of a town was 20,000. The rates of growth in the last three years were 2%, 2.5%, and 4% respectively. Find the present population of the town.)
Solution:
\( P = 20,000, R_1 = 2\%, R_2 = 2.5\%, R_3 = 4\% \)
\( P_T = P \left(1 + \frac{R_1}{100}\right) \left(1 + \frac{R_2}{100}\right) \left(1 + \frac{R_3}{100}\right) \)
\( = 20000 \times 1.02 \times 1.025 \times 1.04 \)
\( = 20000 \times 1.08784 = 21,757 \)
तसर्थ, हालको जनसङ्ख्या २१,७५७ हो। (Hence, the present population is 21,757.)
६. २०७० सालको अन्त्यमा एउटा सहरको जनसङ्ख्या २०,००० थियो। जन्म दर ५% र मृत्यु दर प्रतिवर्ष २% थियो। २०७२ सालको अन्त्यमा सहरको जनसङ्ख्या कति हुनेछ? (The population of a city at the end of the year 2070 B.S. was 20,000. The birth rate was 5% and the death rate was 2% per year. What will be the population of the city at the end of the year 2072 B.S.?)
Solution:
\( P = 20,000, \text{Birth rate} = 5\%, \text{Death rate} = 2\% \)
\( \text{Actual growth rate } R = 5\% – 2\% = 3\%, T = 2 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T = 20000 \left(1 + \frac{3}{100}\right)^2 \)
\( = 20000 \times (1.03)^2 = 20000 \times 1.0609 = 21,218 \)
तसर्थ, २०७२ सालको अन्त्यमा जनसङ्ख्या २१,२१८ हुनेछ। (Hence, the population at the end of 2072 B.S. will be 21,218.)
७. एउटा मेसिनको मूल्य रु. १,००,००० छ। यदि यसको मूल्य प्रतिवर्ष १०% ले घट्छ भने, ३ वर्षपछि मेसिनको मूल्य पत्ता लगाउनुहोस्। (The price of a machine is Rs. 100,000. If its price depreciates by 10% per year, find the price of the machine after 3 years.)
Solution:
\( V_0 = 100,000, R = 10\%, T = 3 \)
\( V_T = V_0 \left(1 – \frac{R}{100}\right)^T = 100000 \left(1 – \frac{10}{100}\right)^3 \)
\( = 100000 \times (0.9)^3 = 100000 \times 0.729 = 72,900 \)
तसर्थ, ३ वर्षपछि मेसिनको मूल्य रु. ७२,९०० हुनेछ। (Hence, the price after 3 years will be Rs. 72,900.)
८. एक व्यक्तिले एउटा मोटरसाइकल रु. ८१,००० मा किन्यो। यदि यसको मूल्य प्रतिवर्ष १०% ले घट्छ भने, कति वर्षमा यसको मूल्य रु. ६५,६१० हुनेछ? (A man bought a motorcycle for Rs. 81,000. If its price depreciates by 10% per year, in how many years will its price be Rs. 65,610?)
Solution:
\( V_0 = 81,000, V_T = 65,610, R = 10\% \)
\( V_T = V_0 \left(1 – \frac{R}{100}\right)^T \)
\( 65610 = 81000 \times (0.9)^T \)
\( \frac{65610}{81000} = (0.9)^T \implies 0.81 = (0.9)^T \)
\( (0.9)^2 = (0.9)^T \implies T = 2 \)
तसर्थ, २ वर्षमा मूल्य रु. ६५,६१० हुनेछ। (Hence, in 2 years the price will be Rs. 65,610.)
९. एउटा सामानको मूल्य २ वर्षमा रु. २५,००० बाट घटेर रु. २०,२५० भयो। ह्रासको दर पत्ता लगाउनुहोस्। (The price of an article is depreciated from Rs. 25,000 to Rs. 20,250 in 2 years. Find the rate of depreciation.)
Solution:
\( V_0 = 25,000, V_T = 20,250, T = 2 \)
\( V_T = V_0 \left(1 – \frac{R}{100}\right)^T \)
\( 20250 = 25000 \times \left(1 – \frac{R}{100}\right)^2 \)
\( \frac{20250}{25000} = \left(1 – \frac{R}{100}\right)^2 \implies 0.81 = \left(1 – \frac{R}{100}\right)^2 \)
\( \sqrt{0.81} = 1 – \frac{R}{100} \implies 0.9 = 1 – \frac{R}{100} \)
\( \frac{R}{100} = 0.1 \implies R = 10\% \)
तसर्थ, ह्रासको दर १०% हो। (Hence, the depreciation rate is 10%.)
१०. एक दशकमा एउटा सहरको जनसङ्ख्या १,७५,००० बाट बढेर २,६२,५०० पुग्यो। प्रतिवर्ष जनसङ्ख्यामा औसत प्रतिशत वृद्धि पत्ता लगाउनुहोस्। (The population of a city increased from 175,000 to 262,500 in a decade. Find the average percentage increase in population per year.)
Solution:
\( P = 175,000, P_T = 262,500, T = 10 \)
\( P_T = P \left(1 + \frac{R}{100}\right)^T \)
\( 262500 = 175000 \times \left(1 + \frac{R}{100}\right)^{10} \)
\( \left(1 + \frac{R}{100}\right)^{10} = \frac{262500}{175000} = 1.5 \)
\( 1 + \frac{R}{100} = (1.5)^{0.1} \approx 1.041 \)
\( \frac{R}{100} = 0.041 \implies R = 4.1\% \)
तसर्थ, औसत प्रतिशत वृद्धि ४.१% हो। (Hence, the average percentage increase is 4.1%.)
११. एउटा सहरको जनसङ्ख्या वार्षिक ५% ले बढ्छ। सहरको हालको जनसङ्ख्या १,६०,००० छ। दुई वर्षपछि, २% जनसङ्ख्या अर्को सहरमा बसाइँ सर्यो। ३ वर्षमा सहरको जनसङ्ख्या पत्ता लगाउनुहोस्। (The population of a town increases by 5% annually. The present population of the town is 160,000. After two years, 2% of the population migrated to another town. Find the population of the town in 3 years.)
Solution:
\( P = 160,000, R = 5\% \)
Population after 2 years:
\( P_2 = 160000 \times (1.05)^2 = 160000 \times 1.1025 = 176,400 \)
Migration: \( 2\% \text{ of } 176400 = 0.02 \times 176400 = 3,528 \)
Population after migration: \( 176400 – 3528 = 172,872 \)
Population after 3rd year: \( 172872 \times 1.05 = 181,516 \)
तसर्थ, ३ वर्षमा सहरको जनसङ्ख्या १,८१,५१६ हुनेछ। (Hence, the population after 3 years will be 181,516.)
१२. एक व्यक्तिले रु. ५०,००० मा एउटा मेसिन किन्यो। पहिलो दुई वर्षको लागि यसको मूल्य १०% को दरले घट्छ र त्यसपछि तेस्रो वर्षको लागि ८% को दरले बढ्छ। ३ वर्षपछि यसको मूल्य कति हुनेछ? (A man bought a machine for Rs. 50,000. Its value depreciates at the rate of 10% for the first two years and then appreciates at the rate of 8% for the third year. What will be its value after 3 years?)
Solution:
\( V_0 = 50,000 \)
After 2 years depreciation (10%):
\( V_2 = 50000 \times (0.9)^2 = 50000 \times 0.81 = 40,500 \)
After 3rd year appreciation (8%):
\( V_3 = 40500 \times 1.08 = 43,740 \)
तसर्थ, ३ वर्षपछि मेसिनको मूल्य रु. ४३,७४० हुनेछ। (Hence, the value after 3 years will be Rs. 43,740.)
१३. एउटा मेसिनको मूल्य पहिलो २ वर्षको लागि प्रतिवर्ष १०% ले र अर्को २ वर्षको लागि प्रतिवर्ष १५% ले घट्यो। यदि मेसिन रु. ४०,००० मा किनिएको थियो भने, ४ वर्षपछि यसको मूल्य पत्ता लगाउनुहोस्। (The price of a machine depreciated at 10% per year for the first 2 years and at 15% per year for the next 2 years. If the machine was bought for Rs. 40,000, find its value after 4 years.)
Solution:
\( V_0 = 40,000 \)
After first 2 years (10% depreciation):
\( V_2 = 40000 \times (0.9)^2 = 40000 \times 0.81 = 32,400 \)
After next 2 years (15% depreciation):
\( V_4 = 32400 \times (0.85)^2 = 32400 \times 0.7225 = 23,409 \)
तसर्थ, ४ वर्षपछि मेसिनको मूल्य रु. २३,४०९ हुनेछ। (Hence, the value after 4 years will be Rs. 23,409.)
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