A building has been purchased by a person at a cost of Rs. 25,000. The useful life of the building is 40 years and the scrap value of the building is Rs. 3,000. Calculate the annual sinking fund (Rs.) at the rate of 5% interest. (Take 1.0540 = 7.04)
182
Calculating Annual Sinking Fund Deposit
A sinking fund is established to accumulate a certain sum of money over a period of time by making regular deposits. This fund is typically used to replace an asset at the end of its useful life or to repay a debt. In this problem, we need to calculate the annual deposit required for a sinking fund to replace a building.
Understanding the Sinking Fund Requirement
The total cost of the building is Rs. 25,000. At the end of its useful life of 40 years, the building is expected to have a scrap value of Rs. 3,000. This means the amount that needs to be accumulated in the sinking fund to replace the building is the initial cost minus the scrap value.
Amount needed (A) = Cost of Building - Scrap Value
\( A = \text{Rs. } 25,000 - \text{Rs. } 3,000 \)
\( A = \text{Rs. } 22,000 \)
So, Rs. 22,000 needs to be accumulated over 40 years.
Sinking Fund Formula and Calculation
The annual sinking fund deposit (P) can be calculated using the formula:
\( P = \frac{A \times i}{(1 + i)^n - 1} \)
Where:
- \( A \) = Total amount to be accumulated (Rs. 22,000)
- \( i \) = Annual interest rate (5% or 0.05)
- \( n \) = Useful life in years (40 years)
- \( (1 + i)^n \) = Future value factor of a single sum. We are given \( 1.05^{40} = 7.04 \).
Now, let's substitute the given values into the formula:
\( P = \frac{22000 \times 0.05}{(1 + 0.05)^{40} - 1} \)
\( P = \frac{22000 \times 0.05}{1.05^{40} - 1} \)
Using the given value \( 1.05^{40} = 7.04 \):
\( P = \frac{1100}{7.04 - 1} \)
\( P = \frac{1100}{6.04} \)
Calculating the value of P:
\( P \approx 182.119 \)
Matching with Options
The calculated annual sinking fund deposit is approximately Rs. 182.119. We now compare this value with the given options:
- Option 1: 136
- Option 2: 155
- Option 3: 182
- Option 4: 207
The calculated value of 182.119 is closest to Rs. 182.
| Parameter | Value |
|---|---|
| Cost of Building | Rs. 25,000 |
| Scrap Value | Rs. 3,000 |
| Amount Needed (A) | Rs. 22,000 |
| Useful Life (n) | 40 years |
| Interest Rate (i) | 5% (0.05) |
| \( (1+i)^n \) | 7.04 |
| Annual Deposit (P) | Rs. 182.119 ≈ Rs. 182 |
Revision Table: Key Concepts
| Term | Explanation |
|---|---|
| Sinking Fund | A fund created by setting aside money periodically to replace an asset or pay off a liability at a future date. |
| Useful Life | The estimated period over which an asset is expected to be usable. |
| Scrap Value (Salvage Value) | The estimated residual value of an asset at the end of its useful life. |
| Interest Rate | The rate at which the deposited funds grow over time. |
Additional Information: Applications of Sinking Funds
Sinking funds are commonly used in various financial and engineering contexts. Some typical applications include:
- Funding the replacement of machinery or equipment.
- Providing for the redemption of bonds or loans.
- Saving for future major expenses, like building renovations.
- Ensuring funds are available for dismantling costs at the end of a project's life.
The concept relies on the principle of compound interest, where each periodic deposit and the accumulated interest earn further interest, allowing a target sum to be reached efficiently over time.
Important Questions from Valuation
Calculate the year’s purchase for a property of useful life of 30 years and rate of interest of 5% per annum.
The junk or demolition value of a structure, calculated at the end of its utility span, that has lost all of its structural strength and is near to its demolition is called:
X is the measure and adjustment of price levels for goods and services across a broad sector of the economy, where X is:
In which of the following cases, valuation is not required?
Method used to make an estimate is
