Growth and Depreciation: Important Questions
Mathematics • Grade 10 • Theory and numerical questions with answersA. Theory (short answer)
2 marks1. Define population growth rate and growth population.
The rate at which the population of a place increases every year when no outside factor interferes is the population growth rate. The increased population is called the growth population. Such problems are solved like compound interest.
3 marks2. Write the formula for the population after T years (a) at a constant rate R%, (b) at different yearly rates R₁, R₂, R₃, …
(a) PT = P(1 + R/100)T
(b) PT = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)…
2 marks3. How is population growth similar to compound interest? Name two factors that affect population growth.
In both, each year’s change is calculated on the previous year’s total, so the same formula (1 + R/100)T is used. Population growth is also affected by migration and deaths.
2 marks4. What is depreciation? What is compound depreciation?
The decrease in the price of a good because of use, age or lower efficiency is depreciation. When the price falls by a fixed percentage each year on the reduced price of the previous year, it is compound depreciation.
3 marks5. Derive the formula VT = V₀(1 − R/100)T.
V₁ = V₀ − V₀R/100 = V₀(1 − R/100). V₂ = V₁(1 − R/100) = V₀(1 − R/100)². V₃ = V₀(1 − R/100)³. Continuing, VT = V₀(1 − R/100)T.
2 marks6. Write two differences between growth and depreciation.
- Growth multiplies by (1 + R/100) each year, so the value increases; depreciation multiplies by (1 − R/100), so it decreases.
- Growth gives PT − P as the increase; depreciation gives V₀ − VT as the loss in value.
B. Numerical questions
3 marks7. A city had 50,000 people in 2078 B.S. with 2% annual growth. Find the population in 2080 B.S.
50,000(1.02)² = 52,020
3 marks8. Two years ago a sack of rice cost Rs. 1,300. If inflation was 5% p.a., find the increase in price.
1,300[(1.05)² − 1] = Rs. 133.25
3 marks9. The price of a photocopy machine rose from Rs. 1,00,000 to Rs. 1,21,000 in 2 years. Find the yearly rate.
1.21 = (1.1)² ⇒ R = 10%
4 marks10. A school has 500 students. In how many years will it have 720 if the number rises 20% every year?
720/500 = 36/25 = (6/5)² ⇒ T = 2 years
4 marks11. Land costs Rs. 15,97,200 per aana now and prices rise 10% p.a. Find the price per ropani 3 years ago (1 ropani = 16 aana).
P = 15,97,200 ÷ 1.331 = 12,00,000 per aana. Per ropani = Rs. 1,92,00,000
4 marks12. 50,000 students appeared in SEE in 2076 B.S. The number rose by 5%, 6% and 4% in the next three years. Find the number in 2079 B.S.
50,000 × 1.05 × 1.06 × 1.04 = 57,876
5 marks13. A municipality had 1,00,000 people in 2078 B.S. In 2079 B.S., 8,000 migrated in and 500 died. At 2% yearly growth, find the population in 2081 B.S.
2079: 1,00,000 × 1.02 + 8,000 − 500 = 1,09,500. 2081: 1,09,500 × (1.02)² = ≈ 1,13,924
3 marks14. The population of a city was 5,18,452 and grows 4.5% p.a. Find the population after 3 years.
5,18,452 × (1.045)³ = ≈ 5,91,640
3 marks15. A university had 20,000 students 2 years ago and has 21,632 now. Find the yearly rate of increase.
1.0816 = (1.04)² ⇒ R = 4%
3 marks16. A papaya plant 4 m tall grows 4% a month. In how many months will it be 4.3264 m tall?
4.3264/4 = 1.0816 = (1.04)² ⇒ 2 months
4 marks17. Bacteria number 1 × 10⁵ at 6 a.m. The hourly increases at 7, 8 and 9 a.m. are 3%, 4% and 5%. Find the number at 9 a.m.
10⁵ × 1.03 × 1.04 × 1.05 = 1.12476 × 10⁵
4 marks18. An insurance company trained 1,000 agents 3 years ago. Every year each group of 5 agents adds 1 more agent. How many agents are there now?
Growth = 1/5 = 20% a year. 1,000(1.2)³ = 1,728
4 marks19. A rural municipality’s population grows 10% a year. At the end of 2 years it reaches 30,000, and then 5,800 people migrate in. Find the initial population.
(30,000 − 5,800) ÷ 1.21 = 24,200 ÷ 1.21 = 20,000
3 marks20. After 5% yearly inflation of the dollar for 2 years, $1 = Rs. 120. How many rupees equalled $1 two years ago?
120 ÷ 1.1025 = Rs. 108.84
4 marks21. A village had 31,250 people 2 years ago, growing 6% p.a. One year ago 625 people migrated out. Find the present population.
31,250 × 1.06 − 625 = 32,500; 32,500 × 1.06 = 34,450
5 marks22. A district had 3,75,000 people 3 years ago. At the end of 2 years 1,480 migrated in and 2,750 died. With 2% yearly growth, find the present population.
3,75,000 × 1.0404 = 3,90,150; + 1,480 − 2,750 = 3,88,880; × 1.02 = ≈ 3,96,658
5 marks23. A city had 5,00,000 people at the beginning of 2075 B.S. and 6,65,500 at the end of 2077 B.S. (a) Find the growth rate. (b) Find the population at the end of 2079 B.S.
(a) 1.331 = (1.1)³ ⇒ R = 10%
(b) 6,65,500 × 1.21 = 8,05,255
3 marks24. A book published 3 years ago at Rs. 200 depreciates 5% a year. Find its price now.
200(0.95)³ = ≈ Rs. 171.47
4 marks25. A computer bought for Rs. 40,000 depreciates by Rs. 7,600 in 2 years. Find the yearly rate.
V₂ = 32,400; 32,400/40,000 = 0.81 = (0.9)² ⇒ R = 10%
5 marks26. A factory built with Rs. 4 crore earned Rs. 75 lakh in 3 years and depreciated 2.5% a year. It is sold after 3 years. Find the profit or loss.
V₃ = 4,00,00,000(0.975)³ = 3,70,74,375. Total = 4,45,74,375. Profit = Rs. 45,74,375
4 marks27. A finance company’s share price fell 10% a year for 2 years and the shares sold for Rs. 25,920. How many shares of Rs. 100 each were sold?
V₀ = 25,920 ÷ 0.81 = 32,000. 320 shares
4 marks28. A laptop bought for Rs. 96,000 depreciates 15% a year. Find the amount it depreciates in 3 years.
V₃ = 96,000(0.85)³ = 58,956. Depreciation = Rs. 37,044
4 marks29. A photocopy machine costs Rs. 5,00,000 now. It depreciates 15%, 10% and 5% in the next three years. Find its price after 3 years.
5,00,000 × 0.85 × 0.90 × 0.95 = Rs. 3,63,375
5 marks30. A truck bought for Rs. 48,00,000 earned Rs. 6,80,000 in 2 years and is sold after 2 years with 10% yearly depreciation. Find the profit or loss.
Selling price = 48,00,000 × 0.81 = 38,88,000. Total = 45,68,000. Loss = Rs. 2,32,000
5 marks31. A bus costing Rs. 16,00,000 earned Rs. 5,10,000 in 3 years and depreciates 5% a year. Find the profit or loss when it is sold after 3 years.
Selling price = 16,00,000(0.95)³ = 13,71,800. Total = 18,81,800. Profit = Rs. 2,81,800
5 marks32. A lamination machine sells for Rs. 24,168 after depreciating 4% and 5% in two successive years. Find its price 2 years ago.
24,168 ÷ (0.96 × 0.95) = 24,168 ÷ 0.912 = Rs. 26,500
5 marks33. At a certain yearly depreciation rate a good is worth Rs. 10,240 after 2 years and Rs. 8,192 after 3 years. Find the rate and the initial price.
8,192/10,240 = 0.8 ⇒ R = 20%. V₀ = 10,240 ÷ 0.64 = Rs. 16,000
5 marks34. An article is worth Rs. 5,41,500 after 2 years and Rs. 5,14,425 after 3 years at a certain yearly depreciation rate. Find the rate and the initial price.
5,14,425/5,41,500 = 0.95 ⇒ R = 5%. V₀ = 5,41,500 ÷ 0.9025 = Rs. 6,00,000
4 marks35. A hydropower company’s share price falls 10% a year. Sashi sells all his shares for Rs. 7,10,775. Find the value of his shares 2 years ago and the number bought at Rs. 100 each.
V₀ = 7,10,775 ÷ 0.81 = Rs. 8,77,500; shares = 8,775
4 marks36. After the dollar loses 5% a year for 2 years, $1 = Rs. 125. How many rupees equalled $1 two years ago?
125 ÷ 0.9025 = Rs. 138.50
Growth and Depreciation • Grade 10 Mathematics • Important Questions
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