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Number and Algebra - Depreciation

Grade 12IB_AI

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Depreciation is the systematic reduction in the value of an asset over time, primarily due to wear and tear, age, or obsolescence.

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The Reducing Balance Method is the most common form of depreciation in the IB AI syllabus, where the asset's value decreases by a fixed percentage rr of its value at the beginning of each period.

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The Salvage Value or Scrap Value is the estimated value of an asset at the end of its useful life.

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The formula for reducing balance depreciation is a variation of the compound interest formula, where the interest rate is negative: FV=PV(1−r100)nFV = PV(1 - \frac{r}{100})^n.

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In the formula, PVPV represents the initial value (Present Value), FVFV represents the future value after depreciation, rr is the annual percentage rate of depreciation, and nn is the number of years.

📐Formulae

FV=PV(1−r100)nFV = PV \left(1 - \frac{r}{100}\right)^n

Total Depreciation=PV−FVTotal \ Depreciation = PV - FV

i=r100i = \frac{r}{100}

💡Examples

Problem 1:

A logistics company purchases a delivery truck for 45000. The value of the truck depreciates at a rate of 12%12\% per year. Calculate the value of the truck after 55 years.

Solution:

Given: PV=45000PV = 45000, r=12r = 12, n=5n = 5. Using the formula: FV=45000×(1−12100)5FV = 45000 \times \left(1 - \frac{12}{100}\right)^5 FV=45000×(0.88)5FV = 45000 \times (0.88)^5 FV≈45000×0.5277319FV \approx 45000 \times 0.5277319 FV≈23747.94FV \approx 23747.94

Explanation:

Substitute the initial purchase price, the depreciation rate, and the time period into the reducing balance formula to find the future value.

Problem 2:

A piece of industrial machinery is bought for 12000. After 33 years, its value is 7680. Find the annual percentage rate of depreciation.

Solution:

Given: PV=12000PV = 12000, FV=7680FV = 7680, n=3n = 3. We need to find rr: 7680=12000×(1−r100)37680 = 12000 \times \left(1 - \frac{r}{100}\right)^3 768012000=(1−r100)3\frac{7680}{12000} = \left(1 - \frac{r}{100}\right)^3 0.64=(1−r100)30.64 = \left(1 - \frac{r}{100}\right)^3 Take the cube root of both sides: 0.643=1−r100\sqrt[3]{0.64} = 1 - \frac{r}{100} 0.86177≈1−r1000.86177 \approx 1 - \frac{r}{100} r100=1−0.86177\frac{r}{100} = 1 - 0.86177 r≈13.8%r \approx 13.8\%

Explanation:

Rearrange the depreciation formula to solve for the rate rr. This involves dividing by the initial value and then taking the nn-th root.

Problem 3:

A computer system costs 8000 and depreciates at 25%25\% per annum. Calculate the total value lost through depreciation over 22 years.

Solution:

First, find the value after 22 years: FV=8000×(1−0.25)2FV = 8000 \times (1 - 0.25)^2 FV=8000×(0.75)2FV = 8000 \times (0.75)^2 FV=8000×0.5625=4500FV = 8000 \times 0.5625 = 4500 To find the total depreciation, subtract the final value from the initial value: 8000−45003500\begin{array}{r} 8000 \\ - 4500 \\ \hline 3500 \end{array} The total depreciation is 3500.

Explanation:

First, calculate the future value after the specified time. Then, find the difference between the original price and the depreciated price to determine the total loss in value.

Depreciation Grade 12 Notes & Examples | IB AI Maths