Chapter-3 (Growth and Depreciation) Exercise Solutions

Growth and Depreciation: Exercise Solutions

Mathematics • Grade 10 • Exercise 3.1 and 3.2 with step-by-step solutions

Formulas used in this exercise

• Growth: P_T = P(1 + R/100)^T  |  Increase = P[(1 + R/100)^T − 1]
• Varying growth rates: P_T = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)…
• Depreciation: V_T = V₀(1 − R/100)^T  |  Depreciated amount = V₀ − V_T
• Varying depreciation rates: V_T = V₀(1 − R₁/100)(1 − R₂/100)…
• Unknown rate / time: find V_T ÷ V₀ (or P_T ÷ P) and compare it with powers of (1 ± R/100).
• Two values after n and n + 1 years: the ratio of the later value to the earlier one gives (1 ± R/100).

Money values are in Rs. If a question is garbled in print, the assumption I made is stated in the solution.

Exercise 3.1 — Growth

Q1. (a) Formula for P_T. (b) Number of tigers after N years if it is x now and the growth rate is R%. (c) Number of employees after 3 years with rates R₁, R₂, R₃.

(a) P_T = P(1 + R/100)^T

(b) Tigers after N years = x(1 + R/100)^N

(c) If the number of employees at the start is N₀: N₃ = N₀(1 + R₁/100)(1 + R₂/100)(1 + R₃/100)

Q2. (a) If the death rate is less than the birth rate, does the population increase or decrease? (b) A 25 kg sack of rice cost Rs. 1,500 a year ago. If the price rose 10% p.a., what does it cost now?

(a) The population increases, because more people are being born than are dying.

(b) Price now = 1,500(1 + 10/100)¹ = 1,500 × 1.1 = Rs. 1,650

Q3. (a) Population 5,18,452, growth 4.5% p.a. — population after 3 years? (b) Rent Rs. 10,000 rising 5% p.a. — rent after 3 years? (c) Bacteria 4 × 10¹¹ growing 10% per hour — number after 2 hours?

(a) P₃ = 5,18,452(1.045)³ = 5,18,452 × 1.141166… = 5,91,639.86 ⇒ ≈ 5,91,640

(b) Rent = 10,000(1.05)³ = 10,000 × 1.157625 = Rs. 11,576.25

The book’s answer key prints Rs. 15,208.75 here. With Rs. 10,000 at 5% for 3 years the formula gives Rs. 11,576.25, so the key looks like a misprint (or the question has different data in your edition). Please confirm with your teacher.

(c) Number = 4 × 10¹¹ × (1.1)² = 4 × 10¹¹ × 1.21 = 4.84 × 10¹¹

Q4. (a) Population 28,500 two years ago, growth 2% p.a. — increase in 2 years? (b) Fee Rs. 6,500 three years ago, rising 10% p.a. — increase in 3 years? (c) Land worth Rs. 6,00,000, price rising 10% p.a. — increase in 2 years?

(a) Increase = 28,500[(1.02)² − 1] = 28,500 × 0.0404 = 1,151.4 ⇒ ≈ 1,151 people

(The book gives 1,152 because it rounds the new population 29,651.4 up to 29,652.)

(b) Increase = 6,500[(1.1)³ − 1] = 6,500 × 0.331 = Rs. 2,151.50

(c) Increase = 6,00,000[(1.1)² − 1] = 6,00,000 × 0.21 = Rs. 1,26,000

Q5. Find the yearly rate: (a) 20,000 → 21,632 in 2 years (b) 40,000 → 44,100 in 2 years (c) Rs. 125 → Rs. 216 in 3 years (inflation)

(a) 21,632/20,000 = 1.0816 = (1.04)² ⇒ 1 + R/100 = 1.04 ⇒ R = 4%

(b) 44,100/40,000 = 1.1025 = (1.05)² ⇒ R = 5%

(c) 216/125 = 1.728 = (1.2)³ ⇒ 1 + R/100 = 1.2 ⇒ inflation rate = 20%

Q6. (a) A village’s population is 13,310 now, growing 10% p.a. How many years ago was it 10,000? (b) A papaya plant is 4 m high and grows 4% per month. In how many months will it be 4.3264 m?

(a) 13,310 = 10,000(1.1)^T ⇒ (1.1)^T = 1.331 = (1.1)³ ⇒ T = 3 years

(b) 4.3264 = 4(1.04)^T ⇒ (1.04)^T = 1.0816 = (1.04)² ⇒ T = 2 months

Q7. Price of land per aana is (a) Rs. 15,97,200 (b) Rs. 2,66,200 now, rising 10% p.a. What was the price per ropani 3 years ago? (1 ropani = 16 aana)

(a) 15,97,200 = P(1.1)³ = 1.331P ⇒ P = Rs. 12,00,000 per aana.

Per ropani = 16 × 12,00,000 = Rs. 1,92,00,000

(b) 2,66,200 = 1.331P ⇒ P = Rs. 2,00,000 per aana.

Per ropani = 16 × 2,00,000 = Rs. 32,00,000

Q8. (a) Bacteria 1 × 10⁵ at 6 a.m.; hourly increases of 3%, 4%, 5% at 7, 8 and 9 a.m. — number at 9 a.m.? (b) City population 1,50,000 three years ago; increases of 2%, 4%, 5% in the three years — present population?

(a) N = 1 × 10⁵ × 1.03 × 1.04 × 1.05 = 1 × 10⁵ × 1.12476 = 1.12476 × 10⁵ ≈ 1.1248 × 10⁵

(b) P = 1,50,000 × 1.02 × 1.04 × 1.05 = 1,50,000 × 1.11384 = 1,67,076

The key prints 1,67,070. The exact product 1.02 × 1.04 × 1.05 = 1.11384 gives 1,67,076, so the key appears to have a small misprint.

Q9. (a) An insurance company had 1,000 agents 3 years ago. Every year each group of 5 agents adds 1 new agent. How many agents are there now? (b) A finance company starts 2078 with 200 marketers, with the same policy. How many are there at the end of 2079?

Every 5 agents add 1 more, so the number grows by 1/5 = 20% each year.

(a) N = 1,000(1.2)³ = 1,000 × 1.728 = 1,728 agents

(b) From the beginning of 2078 to the end of 2079 is 2 years: N = 200(1.2)² = 200 × 1.44 = 288 marketers

Q10. A rural municipality’s population grows 10% every year. At the end of 2 years it reaches 30,000, and then 5,800 people migrate in. Find the initial population.

The final figure includes the 5,800 migrants, so the population before they arrived = 30,000 − 5,800 = 24,200.

24,200 = P(1.1)² = 1.21P ⇒ P = 24,200 ÷ 1.21

Initial population = 20,000

Q11. After 5% p.a. inflation of the American dollar for 2 years, $1 = Rs. 120. How many rupees equalled $1 two years ago?

120 = V(1.05)² = 1.1025V ⇒ V = 120 ÷ 1.1025

$1 = Rs. 108.84

Q12. A village had 31,250 people 2 years ago, growing 6% p.a. One year ago 625 migrated out. Find the present population.

After year 1: 31,250 × 1.06 = 33,125. Then 625 left: 33,125 − 625 = 32,500.

After year 2: 32,500 × 1.06 = 34,450

Q13. A district had 3,75,000 people 3 years ago. At the end of 2 years 1,480 people migrated in and 2,750 died in a disaster. The yearly growth rate is 2%. Find the present population.

After 2 years: 3,75,000(1.02)² = 3,75,000 × 1.0404 = 3,90,150.

Adjust: 3,90,150 + 1,480 − 2,750 = 3,88,880.

After the 3rd year: 3,88,880 × 1.02 = 3,96,657.6 ⇒ ≈ 3,96,658

Q14. 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. if growth continues the same way.

(a) Beginning of 2075 to end of 2077 = 3 years. 6,65,500/5,00,000 = 1.331 = (1.1)³ ⇒ R = 10%

(b) End of 2077 to end of 2079 = 2 years: 6,65,500 × (1.1)² = 6,65,500 × 1.21 = 8,05,255

Project work (3.1): collect the present population of your ward from the ward office, compare it with the census 2078 figure, and report how many people moved in, moved out or died, with the growth rate you find.

Exercise 3.2 — Depreciation

Q1. (a) Formula for the price after T years if the initial price is P and the depreciation rate is R% p.a. (b) What does R represent in V_T = V₀(1 − R/100)^T?

(a) V_T = P(1 − R/100)^T

(b) R is the yearly rate of depreciation (in %). V₀ is the initial price and V_T is the value after T years.

Q2. (a) A watch costing Rs. 5,000 is sold after 1 year at 7% depreciation. Find the depreciated price. (b) A motorcycle sells for Rs. 57,000 after 1 year at 5% p.a. depreciation. What was its purchase price?

(a) Depreciation = 5,000 × 7/100 = Rs. 350 (so the watch sells for 5,000 − 350 = Rs. 4,650).

(b) 57,000 = V₀(1 − 5/100) = 0.95V₀ ⇒ V₀ = 57,000 ÷ 0.95 = Rs. 60,000

Q3. (a) Cupboard Rs. 16,800, depreciating 15% p.a. — price after 2 years? (b) Motorcycle Rs. 2,50,000, depreciating 4% p.a. — price after 3 years?

(a) V₂ = 16,800(0.85)² = 16,800 × 0.7225 = Rs. 12,138

(b) V₃ = 2,50,000(0.96)³ = 2,50,000 × 0.884736 = Rs. 2,21,184

Q4. (a) A phone bought for Rs. 30,000 is sold after 2 years at 30% depreciation — find the depreciated amount. (b) A laptop bought for Rs. 96,000 depreciates 15% a year and is sold after 3 years — find the depreciated amount.

(a) V₂ = 30,000(0.7)² = 30,000 × 0.49 = 14,700. Depreciated amount = 30,000 − 14,700 = Rs. 15,300

(b) V₃ = 96,000(0.85)³ = 96,000 × 0.614125 = 58,956. Depreciated amount = 96,000 − 58,956 = Rs. 37,044

Q5. (a) 4 ropani of land were bought 3 years ago for Rs. 12,50,000 and can be sold now at Rs. 1,60,000 per ropani. Find the yearly depreciation rate. (b) A watch bought for Rs. 5,000 is sold after 3 years for Rs. 625. Find the yearly rate.

(a) Present value = 4 × 1,60,000 = 6,40,000.

6,40,000/12,50,000 = 0.512 = (0.8)³ ⇒ 1 − R/100 = 0.8

R = 20%

(b) 625/5,000 = 0.125 = (0.5)³ ⇒ 1 − R/100 = 0.5 ⇒ R = 50%

Q6. (a) A press machine bought for Rs. 4,00,000 is now worth Rs. 1,96,000 after 30% yearly depreciation. How many years ago was it bought? (b) A car costs Rs. 8,00,000 and loses 10% every year. In how many years will its price be Rs. 5,83,200?

(a) 1,96,000/4,00,000 = 0.49 = (0.7)² ⇒ T = 2 years

(b) 5,83,200/8,00,000 = 0.729 = (0.9)³ ⇒ T = 3 years

Q7. (a) After the dollar loses 5% a year for 2 years, $1 = Rs. 125. How many rupees equalled $1 two years ago? (b) A cow farm bought for Rs. 2,80,000 three years ago depreciates 5% p.a. What does it cost now?

(a) 125 = V₀(0.95)² = 0.9025V₀ ⇒ V₀ = 125 ÷ 0.9025 = $1 = Rs. 138.50

(b) V₃ = 2,80,000(0.95)³ = 2,80,000 × 0.857375 = Rs. 2,40,065

Q8. (a) A photocopy machine costs Rs. 5,00,000 now and depreciates 15%, 10% and 5% in the next three years. Find its price after 3 years. (b) A lamination machine sells for Rs. 24,168 after depreciating 4% and 5% in two successive years. Find its price 2 years ago.

(a) V₃ = 5,00,000 × 0.85 × 0.90 × 0.95 = 5,00,000 × 0.72675 = Rs. 3,63,375

(b) 24,168 = V₀ × 0.96 × 0.95 = 0.912V₀ ⇒ V₀ = 24,168 ÷ 0.912 = Rs. 26,500

Q9. An entrepreneur bought a heavy truck for Rs. 48,00,000, earned Rs. 6,80,000 in 2 years, and sold it after 2 years with 10% yearly depreciation. Find the profit or loss.

Selling price = 48,00,000(0.9)² = 48,00,000 × 0.81 = 38,88,000.

Total received = 38,88,000 + 6,80,000 = 45,68,000.

Investment = 48,00,000, so loss = 48,00,000 − 45,68,000 = Rs. 2,32,000 loss

The key prints “Rs. 2,32,000 shares”, which is a misprint; it is a loss of Rs. 2,32,000.

Q10. A bus owner bought a bus for Rs. 16,00,000, ran it for 3 years and earned Rs. 5,10,000. If its value depreciates 5% a year and he sells it after 3 years, find the profit or loss.

Selling price = 16,00,000(0.95)³ = 16,00,000 × 0.857375 = 13,71,800.

Total received = 13,71,800 + 5,10,000 = 18,81,800.

Profit = 18,81,800 − 16,00,000 = Rs. 2,81,800 profit

Q11. A finance company’s share price falls 10% (every year) for 2 years. Your neighbour sold his shares for Rs. 28,350. How many shares had been bought two years ago at Rs. 100 each?

28,350 = V₀(0.9)² = 0.81V₀ ⇒ V₀ = 28,350 ÷ 0.81 = Rs. 35,000.

Number of shares = 35,000 ÷ 100 = 350 shares

I took “10% in 2 years” to mean 10% each year, because that matches the book’s answer of 350 shares.

Q12. A hydropower company’s share price falls 10% p.a. Sashi sells all his shares now for Rs. 7,10,775. (a) What was the value of his shares 2 years ago? (b) How many shares did he buy at Rs. 100 each in the IPO?

(a) 7,10,775 = V₀(0.9)² = 0.81V₀ ⇒ V₀ = Rs. 8,77,500

(b) Shares = 8,77,500 ÷ 100 = 8,775 shares

Q13. At a certain yearly depreciation rate the price of a good is Rs. 10,240 after 2 years and Rs. 8,192 after 3 years. Find (a) the rate, (b) the initial price.

(a) 1 − R/100 = 8,192/10,240 = 0.8 ⇒ R = 20%

(b) 10,240 = V₀(0.8)² = 0.64V₀ ⇒ V₀ = Rs. 16,000

Q14. A yearly depreciation rate makes the price of an article Rs. 5,41,500 after 2 years and Rs. 5,14,425 after 3 years. Find (a) the rate, (b) the initial price.

(a) 1 − R/100 = 5,14,425/5,41,500 = 0.95 ⇒ R = 5%

(b) 5,41,500 = V₀(0.95)² = 0.9025V₀ ⇒ V₀ = Rs. 6,00,000

Project work (3.2): visit a second-hand shop, note the cost price and the present value of items such as a vehicle, photocopy machine or furniture, then find the depreciation rate and the profit or loss percentage using V_T = V₀(1 − R/100)^T.

Growth and Depreciation • Grade 10 Mathematics • Exercise Solutions