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Depreciation Problems

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Introduction

Depreciation problems ask how the value of an asset falls over time. The most common model used in aptitude tests is compound (declining-balance) depreciation, where the asset loses a fixed percentage each period. Mastering this pattern helps with questions on resale value, effective life, and reverse-finding of rate or time.

Pattern: Depreciation Problems

Pattern: Depreciation Problems

Key concept: Value after depreciation = Principal × (1 - R/100)T.

Notation:
P = original value (cost)
R = depreciation rate per period (%)
T = number of periods (years, half-years, etc.)
A = value after T periods.

Core formula (compound depreciation):
A = P × (1 - R/100)T

Reverse formulas (when solving for R or T):
1 - R/100 = (A/P)^(1/T) → R = [1 - (A/P)^(1/T)] × 100.
T = ln(A/P) / ln(1 - R/100) (useful when A/P & R known).

Step-by-Step Example

Question

An item costing ₹50,000 depreciates at 10% per annum. Find its value after 3 years.

Solution

  1. Step 1: Identify values

    P = ₹50,000; R = 10% p.a.; T = 3 years.

  2. Step 2: Apply compound depreciation formula

    A = P × (1 - R/100)T = 50,000 × (1 - 0.10)^3 = 50,000 × (0.90)^3.

  3. Step 3: Compute powers

    (0.90)^2 = 0.81; (0.90)^3 = 0.729 → A = 50,000 × 0.729 = ₹36,450.

  4. Final Answer:

    Value after 3 years = ₹36,450.

  5. Quick Check:

    Yearwise: 50,000 → 45,000 (after 1y) → 40,500 (after 2y) → 36,450 (after 3y) ✅

Quick Variations

1. Straight-line depreciation: equal absolute loss each year (not covered by compound formula).

2. Compound depreciation with fractional periods: convert to per-period rate and use fractional exponents (or handle whole periods + simple depreciation for remaining fraction if stated by the question).

3. Find rate: given P, A and T use R = [1 - (A/P)^{1/T}] × 100.

4. Find time: given P, A and R use T = ln(A/P) / ln(1 - R/100).

Trick to Always Use

  • Step 1: Convert annual rate to per-period rate when compounding frequency ≠ annual (r = R / periods_per_year).
  • Step 2: Convert time into same units as periods (total_periods = periods_per_year × years).
  • Step 3: Use (1 - r)^{periods} for decline factor; do period-by-period quick check to catch arithmetic mistakes.

Summary

  • Compound depreciation core formula: A = P × (1 - R/100)T.
  • For non-annual compounding, use per-period rate: r = R / periods_per_year and total periods = n = periods_per_year × T_years.
  • To find rate from values: R = [1 - (A/P)^{1/T}] × 100; to find time: T = ln(A/P) / ln(1 - R/100).
  • Always verify by computing a stepwise sequence (year-by-year or period-by-period) as a quick sanity check.

Practice

(1/5)
1. An equipment worth ₹40,000 depreciates at 10% per annum. Find its value after 2 years.
easy
A. ₹32,400
B. ₹36,000
C. ₹34,000
D. ₹35,500

Solution

  1. Step 1: Identify values

    P = ₹40,000; R = 10% p.a.; T = 2 years.

  2. Step 2: Apply compound depreciation formula

    A = P × (1 - R/100)^T = 40,000 × (0.9)^2.

  3. Step 3: Compute depreciated value

    (0.9)^2 = 0.81 → A = 40,000 × 0.81 = ₹32,400.

  4. Final Answer:

    Value after 2 years = ₹32,400 → Option A.

  5. Quick Check:

    Yearwise: 40,000 → 36,000 → 32,400 ✅
Hint: Multiply by (1 - R/100)^T directly to get depreciated value.
Common Mistakes: Subtracting a fixed amount each year instead of applying percentage decline.
2. A car purchased for ₹6,00,000 depreciates at 15% per annum. Find its value after 3 years.
easy
A. ₹3,90,000
B. ₹3,50,000
C. ₹3,68,475
D. ₹4,00,000

Solution

  1. Step 1: Identify values

    P = ₹6,00,000; R = 15% p.a.; T = 3 years.

  2. Step 2: Apply compound depreciation formula

    A = P × (1 - R/100)^T = 6,00,000 × (0.85)^3.

  3. Step 3: Compute depreciated value

    (0.85)^3 = 0.614125 → A = 6,00,000 × 0.614125 = ₹3,68,475.

  4. Final Answer:

    Value after 3 years = ₹3,68,475 → Option C.

  5. Quick Check:

    Yearwise: 6,00,000 → 5,10,000 → 4,33,500 → 3,68,475 ✅
Hint: Multiply by 0.85 each year for 15% depreciation.
Common Mistakes: Using simple subtraction of percentages instead of compounding.
3. A machine was bought for ₹1,00,000 and its value after 2 years is ₹81,000. Find the annual rate of depreciation.
medium
A. 9%
B. 8%
C. 11%
D. 10%

Solution

  1. Step 1: Identify values

    P = ₹1,00,000; A = ₹81,000; T = 2 years.

  2. Step 2: Form the depreciation equation

    (1 - R/100)^T = A/P → (1 - R/100)^2 = 81,000 / 1,00,000 = 0.81.

  3. Step 3: Solve for R

    Take square root: 1 - R/100 = √0.81 = 0.9 → R/100 = 0.1 → R = 10%.

  4. Final Answer:

    Depreciation rate = 10% → Option D.

  5. Quick Check:

    (0.9)^2 = 0.81 → 1,00,000 × 0.81 = 81,000 ✅
Hint: Take the T-th root of A/P to find (1 - R/100), then subtract from 1.
Common Mistakes: Using difference A - P instead of ratio A/P to find rate.
4. A computer is bought for ₹80,000 and depreciates at 12% per annum. Find its value after 1.5 years (annual compounding).
medium
A. ₹66,040.96
B. ₹70,400.96
C. ₹68,288.96
D. ₹69,000.96

Solution

  1. Step 1: Identify values

    P = ₹80,000; R = 12% p.a.; T = 1.5 years.

  2. Step 2: Apply fractional-year depreciation

    A = P × (1 - R/100)^T = 80,000 × (0.88)^1.5.

  3. Step 3: Compute fractional exponent

    √0.88 ≈ 0.938083153 → (0.88)^1.5 = 0.88 × 0.938083153 ≈ 0.825512 → A ≈ 80,000 × 0.825512 = ₹66,040.96.

  4. Final Answer:

    Value ≈ ₹66,040.96 → Option A.

  5. Quick Check:

    After 1 year: 80,000 × 0.88 = 70,400; half-year factor ≈ 0.938 → 70,400 × 0.938 ≈ 66,041 ✅
Hint: For 1.5 years multiply by (1 - R/100) and then by the square root of (1 - R/100).
Common Mistakes: Applying the 12% decline twice instead of using the 1.5 exponent.
5. A bus worth ₹9,00,000 depreciates at 20% per annum. Find how many years it will take for its value to become ₹4,60,800.
medium
A. 4 years
B. 3 years
C. 2.5 years
D. 5 years

Solution

  1. Step 1: Identify values

    P = ₹9,00,000; A = ₹4,60,800; R = 20% p.a.

  2. Step 2: Compute ratio A/P

    A/P = 4,60,800 / 9,00,000 = 0.512.

  3. Step 3: Match with powers of decline

    Set (1 - R/100)^T = 0.512 → (0.8)^T = 0.512. Observe (0.8)^3 = 0.512 → T = 3 years.

  4. Final Answer:

    Time = 3 years → Option B.

  5. Quick Check:

    Yearwise: 9,00,000 → 7,20,000 → 5,76,000 → 4,60,800 ✅
Hint: Compare A/P with powers of (1 - R/100) to find T when it matches an integer power.
Common Mistakes: Using simple percentage reduction (R×T) instead of compounding decline.