Class 10 Mathematics Unit 3 – Growth and Depreciation | Nepal eNotes

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CLASS 10 MATHEMATICS • UNIT 3

Unit 3: Growth and Depreciation

Exercise 3.1 and Exercise 3.2 — WordPress-ready solutions reconstructed from the supplied 39-page handwritten notes.

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Important Formulae

Growth

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

Increase

Increase = P_T − P

Depreciation

V_T = V₀(1 − R/100)^T

Depreciated Amount

V.D. = V₀ − V_T

Different Annual Growth Rates

P_T = P(1 + R₁/100)(1 + R₂/100)(1 + R₃/100) …

Different Annual Depreciation Rates

V_T = V₀(1 − R₁/100)(1 − R₂/100)(1 − R₃/100) …

Notation used in the source: P = present value/population, P_T = value after T years, V₀ = original value, V_T = depreciated value, R = rate percent per year, T = time in years.

PDF Pages 1–20

Exercise 3.1 – Growth

PDF Pages 1–6

Basic Growth Problems

1. Stock price increased by 10% in one year

P = Rs 1,500, R = 10%, T = 1 year

P_T = 1500(1 + 10/100)¹

= 1500 × 1.1

∴ Current price = Rs 1,650

2(a). Population growth problem

P = 5,18,452, R = 4.5%, T = 3 years

P_T = 5,18,452(1.045)³

The handwritten solution gives a future population of approximately 5,91,5xx.

The final population digits on page 2 are partly unclear in the scan, so the source setup is preserved without forcing an exact last-digit transcription.

2(b). Rent increases by 5% every year

P = Rs 10,000, R = 5%, T = 3 years

P_T = 10,000(1.05)³

∴ Rent after 3 years = Rs 11,576.25

2(c). Bacteria increase by 10% per hour

Initial number = 4 × 10¹¹, R = 10%, T = 2 hours

P_T = 4 × 10¹¹(1.1)²

= 4 × 10¹¹ × 1.21

∴ Number after 2 hours = 4.84 × 10¹¹

3(a). Rural-municipality population

P = 28,500, R = 2%, T = 5 years

P_T = 28,500(1.02)⁵

∴ Future population ≈ 31,466

Increase ≈ 31,466 − 28,500 = 2,966

3(b). School monthly fee increased by 10% every year

Fee 3 years ago = Rs 6,500

P_T = 6,500(1.10)³

= Rs 8,651.50

Increase = 8,651.50 − 6,500

∴ Increase = Rs 2,151.50

3(c). Land price grows by 10% per year

Present price = Rs 6,00,000, R = 10%, T = 2 years

P_T = 6,00,000(1.1)²

= Rs 7,26,000

Increase = 7,26,000 − 6,00,000

∴ Increase = Rs 1,26,000

PDF Pages 6–8

Finding Growth Rate

(a) P = 20,000, P_T = 21,682, T = 2 years

21,682 = 20,000(1 + R/100)²

21,682 / 20,000 = (1 + R/100)²

The source concludes the annual growth rate is approximately 4%.

(b) P = Rs 4,00,000, P_T = Rs 4,41,000, T = 2 years

4,41,000 = 4,00,000(1 + R/100)²

1.1025 = (1 + R/100)²

1.05 = 1 + R/100

∴ R = 5%

(c) P = 125, P_T = 216, T = 3 years

216 = 125(1 + R/100)³

216/125 = (1 + R/100)³

1.728 = (1 + R/100)³

1.2 = 1 + R/100

∴ R = 20%

PDF Pages 9–10

Finding Time

(a) P = 10,000, P_T = 13,310, R = 10%

13,310 = 10,000(1.1)^T

1.331 = (1.1)^T

(1.1)³ = (1.1)^T

∴ T = 3 years

(b) Growth at 4% p.a.

The source simplifies the amount/principal ratio to:

1.0816 = (1.04)^T

(1.04)² = (1.04)^T

∴ T = 2 years

PDF Pages 10–11

Finding Present Population / Present Value

(a) Future total = 2,55,55,200, R = 10%, T = 3 years

2,55,55,200 = P(1.1)³

P = 2,55,55,200 / 1.331

∴ P = 1,92,00,000

(b) Future total = 42,59,200, R = 10%, T = 3 years

42,59,200 = P(1.1)³

P = 42,59,200 / 1.331

∴ P = Rs 32,00,000

PDF Pages 12–14

Growth at Different Rates in Different Years

(a) P = Rs 1,00,000; rates 3%, 4% and 5%

P_T = 1,00,000(1.03)(1.04)(1.05)

= 1,00,000 × 1.12476

∴ P_T = Rs 1,12,476

(b) Population = 1,50,000; rates 2%, 4% and 5%

P_T = 1,50,000(1.02)(1.04)(1.05)

= 1,50,000 × 1.11384

∴ P_T = 1,67,076

Direct growth examples

P = 1,000, T = 3 years, R = 20%

P_T = 1,000(1.2)³ = 1,728

P = 200, T = 2 years, R = 20%

P_T = 200(1.2)² = 288

PDF Pages 15–20

Growth with Migration / Addition / Deduction

Problem: final population/value 30,000 after 2 years, 10% growth and an added 5,800

30,000 = P(1.1)² + 5,800

30,000 − 5,800 = 1.21P

24,200 = 1.21P

∴ P = 20,000

Problem: P_T = 120, R = 5%, T = 2 years

120 = P(1.05)²

P = 120 / 1.1025

∴ P ≈ 108.84

Problem: P = 31,250, R = 6%, with an adjustment of 625

After the first year, the source writes:

P₁ = 31,250(1.06) − 625 = 32,500

For the next year:

P₂ = 32,500(1.06)

∴ P₂ = 34,450

Problem: population 3,75,000, growth 2%, migration 1,980 and deaths 2,750

After 2 years:

3,75,000(1.02)² + 1,980 − 2,750

= 3,88,880

After one more year at 2%:

3,88,880(1.02) = 3,96,657.6

Rounded population ≈ 3,96,658

Population growth from 2075 B.S. to 2079 B.S.

Population in 2075 = 5,00,000

Population after 3 years = 6,65,500

6,65,500 = 5,00,000(1 + R/100)³

1.331 = (1 + R/100)³

∴ R = 10%

Population after another 2 years:

6,65,500(1.1)²

∴ Population = 8,05,255

PDF Pages 21–39

Exercise 3.2 – Depreciation

PDF Pages 21–25

Basic Depreciation Problems

1. Watch costing Rs 5,000, depreciation 7% for 1 year

V_T = 5,000(1 − 7/100)

= 5,000 × 0.93

= Rs 4,650

Depreciated value = 5,000 − 4,650 = Rs 350

2(a). Motorcycle sold for Rs 57,000 after 1 year at 5% depreciation

57,000 = V₀(0.95)

V₀ = 57,000 / 0.95

∴ Purchase price = Rs 60,000

3(a). Cupboard costing Rs 16,800, depreciated by 15% for 2 years

V_T = 16,800(0.85)²

= Rs 12,138

3(b). Motorcycle price Rs 2,50,000, depreciation 4% for 3 years

V_T = 2,50,000(0.96)³

∴ V_T = Rs 2,21,184

4(a). V₀ = Rs 30,000, R = 30%, T = 2 years

V_T = 30,000(0.70)²

= Rs 14,700

Depreciated amount = 30,000 − 14,700

∴ Rs 15,300

4(b). V₀ = Rs 96,000, R = 15%, T = 3 years

V_T = 96,000(0.85)³

= Rs 58,956

Depreciated amount = 96,000 − 58,956

∴ Rs 37,044

PDF Pages 25–27

Finding Depreciation Rate

(a) V₀ = Rs 3,12,500 and V_T = Rs 1,60,000 after 3 years

1,60,000 = 3,12,500(1 − R/100)³

1,60,000 / 3,12,500 = (1 − R/100)³

0.512 = (1 − R/100)³

(0.8)³ = (1 − R/100)³

0.8 = 1 − R/100

∴ R = 20%

(b) V₀ = Rs 5,000 and V_T = Rs 625 after 3 years

625 = 5,000(1 − R/100)³

0.125 = (1 − R/100)³

(0.5)³ = (1 − R/100)³

0.5 = 1 − R/100

∴ R = 50%

PDF Pages 27–30

Finding Time and Earlier Value

(a) V₀ = Rs 4,00,000, V_T = Rs 1,96,000, R = 30%

1,96,000 = 4,00,000(0.70)^T

0.49 = (0.70)^T

(0.70)² = (0.70)^T

∴ T = 2 years

(b) V₀ = Rs 8,00,000, V_T = Rs 5,83,200, R = 10%

5,83,200 = 8,00,000(0.9)^T

0.729 = (0.9)^T

(0.9)³ = (0.9)^T

∴ T = 3 years

Earlier value: V_T = Rs 125, R = 5%, T = 2 years

125 = V₀(0.95)²

V₀ = 125 / 0.9025

∴ V₀ ≈ Rs 138.50

Present value Rs 2,80,000 depreciated by 5% for 3 years

V_T = 2,80,000(0.95)³

∴ V_T = Rs 2,40,065

PDF Pages 30–31

Different Depreciation Rates in Different Years

(a) V₀ = Rs 5,00,000; rates 15%, 10% and 5%

V_T = 5,00,000(0.85)(0.90)(0.95)

∴ V_T = Rs 3,63,375

(b) Two different annual depreciation rates

The source uses:

V_T = V₀(1 − R₁/100)(1 − R₂/100)

The amount and final original-value digits on page 31 are partly overwritten. The method is clear, but the individual numerical values are not copied as definite facts.

PDF Pages 32–33

Depreciation with Earnings – Profit and Loss

9. Vehicle value Rs 48,00,000, earnings Rs 6,80,000, depreciation 10% for 2 years

V_T = 48,00,000(0.9)²

= Rs 38,88,000

Depreciated amount = 48,00,000 − 38,88,000

= Rs 9,12,000

Loss = 9,12,000 − 6,80,000

∴ Loss = Rs 2,32,000

10. Asset value Rs 16,00,000, earnings Rs 5,10,000, depreciation 5% for 3 years

V_T = 16,00,000(0.95)³

= Rs 13,71,800

Depreciated amount = 16,00,000 − 13,71,800

= Rs 2,28,200

Profit = 5,10,000 − 2,28,200

∴ Profit = Rs 2,81,800

PDF Pages 34–35

Share-Value Problems

11. Value after 2 years = Rs 28,350, depreciation 10%

28,350 = V₀(0.9)²

28,350 = 0.81V₀

∴ V₀ = Rs 35,000

If each share is Rs 100:

Number of shares = 35,000 / 100

∴ 350 shares

12. Value after 2 years = Rs 7,10,775, depreciation 10%

7,10,775 = V₀(0.9)²

V₀ = 7,10,775 / 0.81

∴ V₀ = Rs 8,77,500

At Rs 100 per share:

Number of shares = 8,77,500 / 100

∴ 8,775 shares

PDF Pages 36–39

Finding Depreciation Rate and Original Value from Two Future Values

13. Value after 2 years = Rs 10,240 and after 3 years = Rs 8,192

For 2 years:

V₀(1 − R/100)² = 10,240 … (i)

For 3 years:

V₀(1 − R/100)³ = 8,192 … (ii)

Dividing (ii) by (i):

1 − R/100 = 8,192 / 10,240 = 0.8

R/100 = 0.2

∴ R = 20%

Using equation (i):

V₀(0.8)² = 10,240

0.64V₀ = 10,240

∴ V₀ = Rs 16,000

14. Value after 2 years = Rs 5,41,500 and after 3 years = Rs 5,14,425

For 2 years:

V₀(1 − R/100)² = 5,41,500 … (i)

For 3 years:

V₀(1 − R/100)³ = 5,14,425 … (ii)

Dividing (ii) by (i):

1 − R/100 = 5,14,425 / 5,41,500 = 0.95

R/100 = 0.05

∴ R = 5%

Putting R = 5% in equation (i):

V₀(0.95)² = 5,41,500

0.9025V₀ = 5,41,500

∴ V₀ = Rs 6,00,000

Original Scanned PDF – View Notes

Important Formulae

Exercise 3.1 – Growth

Basic Growth Problems

1. Stock price increased by 10% in one year

2(a). Population growth problem

2(b). Rent increases by 5% every year

2(c). Bacteria increase by 10% per hour

3(a). Rural-municipality population

3(b). School monthly fee increased by 10% every year

3(c). Land price grows by 10% per year

Finding Growth Rate

(a) P = 20,000, PT = 21,682, T = 2 years

(b) P = Rs 4,00,000, PT = Rs 4,41,000, T = 2 years

(c) P = 125, PT = 216, T = 3 years

Finding Time

(a) P = 10,000, PT = 13,310, R = 10%

(b) Growth at 4% p.a.

Finding Present Population / Present Value

(a) Future total = 2,55,55,200, R = 10%, T = 3 years

(b) Future total = 42,59,200, R = 10%, T = 3 years

Growth at Different Rates in Different Years

(a) P = Rs 1,00,000; rates 3%, 4% and 5%

(b) Population = 1,50,000; rates 2%, 4% and 5%

Direct growth examples

Growth with Migration / Addition / Deduction

Problem: final population/value 30,000 after 2 years, 10% growth and an added 5,800

Problem: PT = 120, R = 5%, T = 2 years

Problem: P = 31,250, R = 6%, with an adjustment of 625

Problem: population 3,75,000, growth 2%, migration 1,980 and deaths 2,750

Population growth from 2075 B.S. to 2079 B.S.

Exercise 3.2 – Depreciation

Basic Depreciation Problems

1. Watch costing Rs 5,000, depreciation 7% for 1 year

2(a). Motorcycle sold for Rs 57,000 after 1 year at 5% depreciation

3(a). Cupboard costing Rs 16,800, depreciated by 15% for 2 years

3(b). Motorcycle price Rs 2,50,000, depreciation 4% for 3 years

4(a). V0 = Rs 30,000, R = 30%, T = 2 years

4(b). V0 = Rs 96,000, R = 15%, T = 3 years

Finding Depreciation Rate

(a) V0 = Rs 3,12,500 and VT = Rs 1,60,000 after 3 years

(b) V0 = Rs 5,000 and VT = Rs 625 after 3 years

Finding Time and Earlier Value

(a) V0 = Rs 4,00,000, VT = Rs 1,96,000, R = 30%

(b) V0 = Rs 8,00,000, VT = Rs 5,83,200, R = 10%

Earlier value: VT = Rs 125, R = 5%, T = 2 years

Present value Rs 2,80,000 depreciated by 5% for 3 years

Different Depreciation Rates in Different Years

(a) V0 = Rs 5,00,000; rates 15%, 10% and 5%

(b) Two different annual depreciation rates

Depreciation with Earnings – Profit and Loss

9. Vehicle value Rs 48,00,000, earnings Rs 6,80,000, depreciation 10% for 2 years

10. Asset value Rs 16,00,000, earnings Rs 5,10,000, depreciation 5% for 3 years

Share-Value Problems

11. Value after 2 years = Rs 28,350, depreciation 10%

12. Value after 2 years = Rs 7,10,775, depreciation 10%

Finding Depreciation Rate and Original Value from Two Future Values

13. Value after 2 years = Rs 10,240 and after 3 years = Rs 8,192

14. Value after 2 years = Rs 5,41,500 and after 3 years = Rs 5,14,425

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(a) P = 20,000, P_T = 21,682, T = 2 years

(b) P = Rs 4,00,000, P_T = Rs 4,41,000, T = 2 years

(c) P = 125, P_T = 216, T = 3 years

(a) P = 10,000, P_T = 13,310, R = 10%

Problem: P_T = 120, R = 5%, T = 2 years

4(a). V₀ = Rs 30,000, R = 30%, T = 2 years

4(b). V₀ = Rs 96,000, R = 15%, T = 3 years

(a) V₀ = Rs 3,12,500 and V_T = Rs 1,60,000 after 3 years

(b) V₀ = Rs 5,000 and V_T = Rs 625 after 3 years

(a) V₀ = Rs 4,00,000, V_T = Rs 1,96,000, R = 30%

(b) V₀ = Rs 8,00,000, V_T = Rs 5,83,200, R = 10%

Earlier value: V_T = Rs 125, R = 5%, T = 2 years

(a) V₀ = Rs 5,00,000; rates 15%, 10% and 5%