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Depreciation

Question 1

3 years ago a book was published costing Rs. 200, is being sold in an exhibition at 5% rate of yearly depreciation. What is the price of the book this year?

Solution:

Given information:

  • The initial price of the book (V0)=Rs. 200(V_0) = \text{Rs. } 200
  • Rate of depreciation (R)=5%(R) = 5\% p.a.
  • Time (T)=3(T) = 3 years
  • Present price of the book (VT)=?(V_T) = ?

Finding the present price of the book:

We know that for depreciation:

VT=V0(1−R100)TV_T = V_0\left(1 - \frac{R}{100}\right)^T

Substitute the given values:

VT=200(1−5100)3V_T = 200\left(1 - \frac{5}{100}\right)^3

VT=200(95100)3V_T = 200\left(\frac{95}{100}\right)^3

VT=200(0.95)3V_T = 200(0.95)^3

VT=200×0.857375V_T = 200 \times 0.857375

VT=Rs. 171.47V_T = \text{Rs. } 171.47

Final Answer:

Therefore, the book costs Rs. 171.47 this year.

Verification:

  • Original price: Rs. 200
  • After 1 year: Rs. 200 × 0.95 = Rs. 190
  • After 2 years: Rs. 190 × 0.95 = Rs. 180.50
  • After 3 years: Rs. 180.50 × 0.95 = Rs. 171.47 ✓
  • Total depreciation: Rs. 200 - Rs. 171.47 = Rs. 28.53
  • Percentage depreciation: (28.53/200) × 100% = 14.27%

Note: Depreciation follows the compound decay formula, where the value decreases by a fixed percentage each year based on the remaining value. This is the opposite of compound interest growth - instead of (1+R/100)(1 + R/100), we use (1−R/100)(1 - R/100).

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Question 2

Seema admitted in BBA. She purchased a computer for Rs. 40,000 for her study. After using it for 2 years, if the price of the computer depreciates by Rs. 7,600, then find the rate of depreciation.

Solution:

Given information:

  • The initial price of the computer (V0)=Rs. 40,000(V_0) = \text{Rs. } 40,000
  • Rate of depreciation (R)=?(R) = ?
  • Time (T)=2(T) = 2 years
  • Depreciated price = Rs. 7,600

Finding the price after 2 years:

Price after 2 years (VT)=V0−Depreciated amount(V_T) = V_0 - \text{Depreciated amount}

VT=Rs. 40,000−Rs. 7,600=Rs. 32,400V_T = \text{Rs. } 40,000 - \text{Rs. } 7,600 = \text{Rs. } 32,400

Finding the rate of depreciation:

We know that for depreciation:

VT=V0(1−R100)TV_T = V_0\left(1 - \frac{R}{100}\right)^T

Substitute the given values:

32,400=40,000(1−R100)232,400 = 40,000\left(1 - \frac{R}{100}\right)^2

32,40040,000=(1−R100)2\frac{32,400}{40,000} = \left(1 - \frac{R}{100}\right)^2

324400=(1−R100)2\frac{324}{400} = \left(1 - \frac{R}{100}\right)^2

(1820)2=(1−R100)2\left(\frac{18}{20}\right)^2 = \left(1 - \frac{R}{100}\right)^2

Taking square root of both sides:

1820=1−R100\frac{18}{20} = 1 - \frac{R}{100}

R100=1−1820\frac{R}{100} = 1 - \frac{18}{20}

R100=20−1820=220\frac{R}{100} = \frac{20 - 18}{20} = \frac{2}{20}

R=220×100=10%R = \frac{2}{20} \times 100 = 10%

Final Answer:

Therefore, the rate of depreciation of the computer is 10% p.a.

Verification:

  • Original price: Rs. 40,000
  • After 1 year at 10% depreciation: Rs. 40,000 × 0.90 = Rs. 36,000
  • After 2 years: Rs. 36,000 × 0.90 = Rs. 32,400 ✓
  • Total depreciation: Rs. 40,000 - Rs. 32,400 = Rs. 7,600 ✓
  • Alternative check: Rs. 40,000 × (0.90)² = Rs. 40,000 × 0.81 = Rs. 32,400 ✓

Note: This problem demonstrates finding an unknown depreciation rate when we know the initial value, final value, and time period. The key is to work backwards from the compound depreciation formula by setting up the equation and solving for the rate.

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Question 3

The price of a house is Rs. 20,00,000 now. If its price decreases by 10% every year, then in how many years will its price be Rs. 14,58,000?

Solution:

Given information:

  • The present price of the house (V0)=Rs. 20,00,000(V_0) = \text{Rs. } 20,00,000
  • Rate of depreciation (R)=10%(R) = 10\% p.a.
  • Price after T years (VT)=Rs. 14,58,000(V_T) = \text{Rs. } 14,58,000
  • Time (T)=?(T) = ?

Finding the time required:

We know that for depreciation:

VT=V0(1−R100)TV_T = V_0\left(1 - \frac{R}{100}\right)^T

Substitute the given values:

14,58,000=20,00,000(1−10100)T14,58,000 = 20,00,000\left(1 - \frac{10}{100}\right)^T

14580002000000=(90100)T\frac{1458000}{2000000} = \left(\frac{90}{100}\right)^T

14582000=(0.9)T\frac{1458}{2000} = (0.9)^T

0.729=(0.9)T0.729 = (0.9)^T

Now, we need to find T such that (0.9)T=0.729(0.9)^T = 0.729

Let's check: (0.9)3=0.9×0.9×0.9=0.729(0.9)^3 = 0.9 \times 0.9 \times 0.9 = 0.729 ✓

Therefore: T=3T = 3 years

Final Answer:

Thus, 3 years later, the price of the house will be Rs. 14,58,000.

Step-by-step verification:

  • Year 0: Rs. 20,00,000 (initial price)
  • Year 1: Rs. 20,00,000 × 0.90 = Rs. 18,00,000
  • Year 2: Rs. 18,00,000 × 0.90 = Rs. 16,20,000
  • Year 3: Rs. 16,20,000 × 0.90 = Rs. 14,58,000 ✓

Alternative verification:

  • Direct calculation: Rs. 20,00,000 × (0.90)³ = Rs. 20,00,000 × 0.729 = Rs. 14,58,000 ✓
  • Total depreciation: Rs. 20,00,000 - Rs. 14,58,000 = Rs. 5,42,000
  • Percentage decrease: (5,42,000/20,00,000) × 100% = 27.1%

Note: This problem demonstrates solving for time in compound depreciation problems. The key insight is recognizing when we can express the decimal result as a power of the depreciation factor, allowing us to determine the number of years directly.

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Question 4

A factory established with the capital of Rs. 4 crore gained Rs. 75 Lakhs in 3 years but its cost depreciated by 2.5% p.a. Then company was sold after 3 years. Now, calculate whether the factory is in profit or loss.

Solution:

Given information:

  • Initial price of investment (V0)=Rs. 4,00,00,000(V_0) = \text{Rs. } 4,00,00,000 (4 crore)
  • Rate of depreciation (R)=2.5%(R) = 2.5\% p.a.
  • Time (T)=3(T) = 3 years
  • Profit gained in 3 years = Rs. 75,00,000 (75 Lakhs)
  • Price after 3 years (V3)=?(V_3) = ?

Finding the depreciated value after 3 years:

We know that for depreciation:

VT=V0(1−R100)TV_T = V_0\left(1 - \frac{R}{100}\right)^T

Substitute the given values:

V3=4,00,00,000(1−2.5100)3V_3 = 4,00,00,000\left(1 - \frac{2.5}{100}\right)^3

V3=4,00,00,000(97.5100)3V_3 = 4,00,00,000\left(\frac{97.5}{100}\right)^3

V3=4,00,00,000(0.975)3V_3 = 4,00,00,000(0.975)^3

V3=4,00,00,000×0.926859V_3 = 4,00,00,000 \times 0.926859

V3=Rs. 3,70,74,375V_3 = \text{Rs. } 3,70,74,375

Calculating total returns and profit/loss:

Profit gained by the factory after 3 years = Rs. 75,00,000

Total amount gained from the factory = Depreciated value + Profit gained

Total returns=Rs. 3,70,74,375+Rs. 75,00,000=Rs. 4,45,74,375\text{Total returns} = \text{Rs. } 3,70,74,375 + \text{Rs. } 75,00,000 = \text{Rs. } 4,45,74,375

Determining profit or loss:

  • Total investment = Rs. 4,00,00,000
  • Total returns = Rs. 4,45,74,375
  • Net result = Rs. 4,45,74,375 - Rs. 4,00,00,000 = Rs. 45,74,375

Final Answer:

Since the total returns (Rs. 4,45,74,375) exceed the initial investment (Rs. 4,00,00,000), the factory is in profit of Rs. 45,74,375.

Detailed analysis:

  • Initial investment: Rs. 4,00,00,000
  • Factory value after depreciation: Rs. 3,70,74,375
  • Loss due to depreciation: Rs. 4,00,00,000 - Rs. 3,70,74,375 = Rs. 29,25,625
  • Operational profit: Rs. 75,00,000
  • Net profit: Rs. 75,00,000 - Rs. 29,25,625 = Rs. 45,74,375
  • Return on investment: (45,74,375/4,00,00,000) × 100% = 11.44%

Note: This problem demonstrates how to analyze business profitability considering both operational gains and asset depreciation. The key is to account for the depreciated asset value plus operational profits against the initial investment.

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Question 5

When the price of a share of a finance company depreciates continuously for 2 years by 10% p.a. and it is sold for Rs. 25,920 then how many shares of Rs. 100 were sold? Find it.

Solution:

Given information:

  • Present value of the shares (VT)=Rs. 25,920(V_T) = \text{Rs. } 25,920
  • Depreciation rate (R)=10%(R) = 10\% p.a.
  • Time (T)=2(T) = 2 years
  • Initial price per share = Rs. 100
  • Initial total value of shares (V0)=?(V_0) = ?
  • Number of shares = ?

Finding the initial value of shares:

We know that for depreciation:

VT=V0(1−R100)TV_T = V_0\left(1 - \frac{R}{100}\right)^T

Substitute the given values:

25,920=V0(1−10100)225,920 = V_0\left(1 - \frac{10}{100}\right)^2

25,920=V0(90100)225,920 = V_0\left(\frac{90}{100}\right)^2

25,920=V0×8110025,920 = V_0 \times \frac{81}{100}

V0=25,920×10081V_0 = \frac{25,920 \times 100}{81}

V0=Rs. 32,000V_0 = \text{Rs. } 32,000

Finding the number of shares:

Initial total value of shares = Rs. 32,000

Price per share 2 years ago = Rs. 100

Number of shares=Total initial valuePrice per share\text{Number of shares} = \frac{\text{Total initial value}}{\text{Price per share}} Number of shares=32,000100=320\text{Number of shares} = \frac{32,000}{100} = 320

Final Answer:

Therefore, 320 shares of Rs. 100 were sold by the finance company.

Verification:

  • Initial situation: 320 shares × Rs. 100 = Rs. 32,000
  • After 1 year: Rs. 32,000 × 0.90 = Rs. 28,800
  • After 2 years: Rs. 28,800 × 0.90 = Rs. 25,920 ✓
  • Alternative check: Rs. 32,000 × (0.90)² = Rs. 32,000 × 0.81 = Rs. 25,920 ✓
  • Per share value after 2 years: Rs. 25,920 ÷ 320 = Rs. 81

Additional analysis:

  • Total depreciation in value: Rs. 32,000 - Rs. 25,920 = Rs. 6,080
  • Depreciation per share: Rs. 100 - Rs. 81 = Rs. 19
  • Percentage depreciation: (6,080/32,000) × 100% = 19%
  • This matches: 1 - (0.90)² = 1 - 0.81 = 0.19 = 19% ✓

Note: This problem demonstrates finding the original principal amount and quantity when we know the final depreciated value. It requires working backwards from the compound depreciation formula and then using the unit price to determine quantity.

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