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Depreciation Problems

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Introduction

Depreciation problems यह पूछते हैं कि किसी asset की value समय के साथ कैसे घटती है। Aptitude tests में सबसे common model compound (declining-balance) depreciation होता है, जिसमें asset हर period में एक fixed percentage से कम होता है। इस pattern को समझने से resale value, effective life और rate या time को reverse-find करने वाले questions आसानी से solve होते हैं।

Pattern: Depreciation Problems

Pattern: Depreciation Problems

Key concept: Depreciation के बाद value = Principal × (1 - R/100)T.

Notation:
P = original value (cost)
R = depreciation rate per period (%)
T = periods की संख्या (years, half-years, आदि)
A = T periods के बाद asset की value

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

Reverse formulas (जब R या T find करना हो):
1 - R/100 = (A/P)^(1/T) → R = [1 - (A/P)^(1/T)] × 100.
T = ln(A/P) / ln(1 - R/100) (useful जब A/P और R दिए हों)。

Step-by-Step Example

Question

₹50,000 का एक item 10% प्रति वर्ष की दर से depreciate होता है। 3 साल बाद उसकी value find करें।

Solution

  1. Step 1: Values identify करें

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

  2. Step 2: Compound depreciation formula लागू करें

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

  3. Step 3: Powers compute करें

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

  4. Final Answer:

    3 साल बाद value = ₹36,450.

  5. Quick Check:

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

Quick Variations

1. Straight-line depreciation: हर साल equal amount से value घटती है (compound formula से अलग)।

2. Compound depreciation with fractional periods: per-period rate convert करें और fractional exponent use करें (या whole periods + remaining fraction को question के अनुसार handle करें)।

3. Find rate: जब P, A और T दिए हों तो R = [1 - (A/P)^{1/T}] × 100.

4. Find time: जब P, A और R दिए हों तो T = ln(A/P) / ln(1 - R/100).

Trick to Always Use

  • Step 1: जब compounding frequency annual ≠ हो, तो annual rate को per-period rate में convert करें (r = R / periods_per_year).
  • Step 2: Time को उसी units में convert करें जिस units में periods हों (total_periods = periods_per_year × years).
  • Step 3: Decline factor के लिए (1 - r)^{periods} use करें; और arithmetic mistakes पकड़ने के लिए period-by-period quick check ज़रूर करें।

Summary

  • Compound depreciation का core formula: A = P × (1 - R/100)T.
  • Non-annual compounding पर per-period rate use करें: r = R / periods_per_year और total periods = n = periods_per_year × T_years.
  • Rate find करने के लिए: R = [1 - (A/P)^{1/T}] × 100; time find करने के लिए: T = ln(A/P) / ln(1 - R/100).
  • हमेशा year-by-year या period-by-period sequence निकालकर quick sanity check करें।

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.