1. Meaning
Depreciation is the reduction in value of assets (machines, cars, computers, etc.) over time.
Important points:
Value decreases by a percentage each year.
It decreases proportionally, not by the same amount every year.
Works similar to reverse compound interest.
2. Reasons for Depreciation
A. Wear and Tear
Continuous use reduces efficiency.
B. Age
Old items lose value even if not used much.
C. Technological Change
New models come out β old models lose value.
D. Market Value
Demand and supply can reduce the value of items.
E. Damage / Usage
Accidents or heavy usage reduce value faster.
3. Depreciation Formulas
A. Constant Rate
VT = V0(1 β R/100)^T
Where:
V0 = Initial Value
R = Depreciation rate
VT = Value after T years
B. Variable Rate
VT = V0(1 β R1/100)(1 β R2/100)β¦(1 β RT/100)
Each year has a different depreciation rate.
C. Amount Depreciated
Depreciation Amount = V0 β VT
This shows how much value is lost.
4. Growth vs Depreciation (Detailed Table)
FeatureGrowthDepreciationMeaningValue increasesValue decreasesMultiplier(1 + R/100)(1 β R/100)NatureWorks like compound interestWorks like reverse compound interestResultFinal value > Initial valueFinal value < Initial valueExamplesPopulation, price rise, followersMachines, vehicles, gadgetsVariable rateMultiply yearlyMultiply yearly5. Detailed Examples
1. Population Growth (Constant rate)
Population = 50,000
Rate = 2%
Time = 2 years
P2 = 50000(1.02)Β² = 52020
2. Variable Population Growth
P3 = 80000 Γ 1.03 Γ 1.04 Γ 1.05 = 89208
3. Growth + Migration
After applying growth:
120000 Γ 1.02 = 122400
Adjust migration + births + deaths:
122400 + 2000 β 800 + 3500 β 1200 = 125900
4. Depreciation (Constant rate)
Value = 60,000
Rate = 12%
V2 = 60000(0.88)Β² = 46464
5. Variable Depreciation
V3 = 50000 Γ 0.90 Γ 0.92 Γ 0.94 = 38988
6. Find Depreciation Rate
Initial = 40,000
Current = 25,000
Time = 4 years
R = 11%
6. Memory Tricks (Expanded)
Population Growth
Use (1 + R/100)
Migration β growth first, migration second
Population always increases unless rate is negative
Depreciation
Use (1 β R/100)
Value never increases
New technology β higher depreciation
Heavy usage β faster depreciation
Variable Rates
Multiply each year separately
Do NOT add the rates
Do NOT multiply the percentages directly
Visit this link for further practice!!
Some Important Questions
1
Q: A town has population 80,000 and grows at 3% per year. What is the population after 2 years?
Solution:
PT = P(1 + R/100)^T = 80,000 Γ (1.03)^2
(1.03)^2 = 1.0609
PT = 80,000 Γ 1.0609 = 80,000 + 80,000Γ0.0609 = 80,000 + 4,872 = 84,872
2
Q: Population = 150,000; increases 4%, 5%, 3% in three years. Find population after 3 years.
Solution:
P3 = 150,000 Γ 1.04 Γ 1.05 Γ 1.03
Stepwise: 150,000 Γ 1.04 = 156,000
156,000 Γ 1.05 = 156,000 + 7,800 = 163,800
163,800 Γ 1.03 = 163,800 + 4,914 = 168,714
3
Q: Village population = 60,000; natural growth 2% for the year; there are 900 births and 500 deaths. What is population at year end?
Solution:
Apply growth first: 60,000 Γ 1.02 = 61,200
Adjust births/deaths: 61,200 + 900 β 500 = 61,200 + 400 = 61,600
4
Q: City had 120,000 people. During the year 3,000 immigrated and 1,200 emigrated. Natural growth = 2%. Find year-end population.
Solution:
Apply growth: 120,000 Γ 1.02 = 122,400
Adjust migration: 122,400 + 3,000 β 1,200 = 122,400 + 1,800 = 124,200
5
Q: A mobile costs Rs. 40,000; depreciation 12% p.a. Find value after 2 years.
Solution:
V2 = 40,000 Γ (1 β 0.12)^2 = 40,000 Γ (0.88)^2
(0.88)^2 = 0.7744
V2 = 40,000 Γ 0.7744 = 40,000 Γ (0.7 + 0.0744) = 28,000 + 2,976 = Rs. 30,976
6
Q: Machine = Rs. 60,000; depreciation 8%, 10%, 5% in three years. Value after 3 years?
Solution:
V3 = 60,000 Γ 0.92 Γ 0.90 Γ 0.95
60,000 Γ 0.92 = 55,200
55,200 Γ 0.90 = 49,680
49,680 Γ 0.95 = 49,680 β 2,484 = Rs. 47,196
7
Q: Value of a car becomes Rs. 300,000 after 3 years. Annual depreciation = 10%. What was original price?
Solution:
Let V0 = original price. After 3 years: V3 = V0 Γ (0.9)^3 = V0 Γ 0.729
So V0 = 300,000 / 0.729 = 300,000 Γ· 0.729 β Rs. 411,522.63
(rounded) β Rs. 411,523 (to nearest rupee)
8
Q: A computer depreciates at 15% p.a. Current value = Rs. 20,000. Find its value after 3 years.
Solution:
V3 = 20,000 Γ (0.85)^3
(0.85)^2 = 0.7225; Γ0.85 = 0.614125
V3 = 20,000 Γ 0.614125 = Rs. 12,282.50
9
Q: Population increased from 100,000 to 120,000 in 2 years. Find annual growth rate.
Solution:
(1 + r)^2 = 120,000 / 100,000 = 1.2
1 + r = β1.2 β 1.095445115
r β 0.095445115 = 9.5445% β 9.545%
10
Q: Machine bought for Rs. 50,000; after 4 years its value is Rs. 30,000. Find annual depreciation rate.
Solution:
30,000 = 50,000 Γ (1 β r)^4 β (1 β r)^4 = 0.6
1 β r = 0.6^(1/4) β 0.880106
r = 1 β 0.880106 β 0.119894 = 11.99% β 12.0%
Visit this link for further practice!!