Problem 2 - Algebra for College Students

Question

Suppose it is estimated that the value of a car depreciates \(30 \%\) per year for the first 5 years. The equation \(A=P_{0}(0.7)^{t}\) yields the value \((A)\) of a car after \(t\) years if the original price is \(P_{0}\). Find the value (to the nearest dollar) of each of the following cars after the indicated time: (a) \(\$ 16,500\) car after 4 years \(\$ 3962\) (b) \(\$ 22,000\) car after 2 years \(\$ 10,780\) (c) \(\$ 27,000\) car after 5 years \(\$ 4538\) (d) \(\$ 40,000\) car after 3 years \(\$ 13,720\)

Step-by-Step Solution

Answer

(a) $3962, (b) $10,780, (c) $4538, (d) $13,720.

Step 1: Understand the Formula

The formula given for depreciation is: \( A = P_0(0.7)^t \), where \( A \) is the value of the car after \( t \) years, \( P_0 \) is the original price, \( 0.7 \) is the yearly retention factor (because the car retains 70% of its value each year), and \( t \) is the time in years.

Step 2: Calculate Depreciation for Car (a)

For a car with an original price of \( P_0 = \$16,500 \) after 4 years, substitute into the formula: \( A = 16,500 \times (0.7)^4 \). Calculate \( (0.7)^4 = 0.2401 \). Therefore, \( A = 16,500 \times 0.2401 = 3961.65 \). Round to the nearest dollar: \( A \approx 3962 \).

Step 3: Calculate Depreciation for Car (b)

For a car with an original price of \( P_0 = \$22,000 \) after 2 years, substitute into the formula: \( A = 22,000 \times (0.7)^2 \). Calculate \( (0.7)^2 = 0.49 \). Therefore, \( A = 22,000 \times 0.49 = 10,780 \).

Step 4: Calculate Depreciation for Car (c)

For a car with an original price of \( P_0 = \$27,000 \) after 5 years, substitute into the formula: \( A = 27,000 \times (0.7)^5 \). Calculate \( (0.7)^5 = 0.16807 \). Therefore, \( A = 27,000 \times 0.16807 = 4537.89 \). Round to the nearest dollar: \( A \approx 4538 \).

Step 5: Calculate Depreciation for Car (d)

For a car with an original price of \( P_0 = \$40,000 \) after 3 years, substitute into the formula: \( A = 40,000 \times (0.7)^3 \). Calculate \( (0.7)^3 = 0.343 \). Therefore, \( A = 40,000 \times 0.343 = 13,720 \).

Key Concepts

Depreciation

Depreciation refers to the reduction in the value of an asset over time, which is particularly common with tangible assets like cars. This happens because of factors like wear and tear, age, and market conditions. Understanding depreciation is crucial for economic decisions as it affects resale values. For instance, if you buy a car, its immediate value after purchase starts declining as it ages.

• Depreciation calculation is based on percentages; in our exercise, it is 30% per year.
• For vehicles, depreciation can lead to lower insurance costs as well over time.
• Keeps in mind, depreciation doesn't involve cash flow but affects asset book value.

By understanding depreciation, individuals can forecast future values of assets and plan accordingly. In practical terms, for the exercise above, a car valued initially at $16,500 would be expected to decrease in worth by 30% annually for the first five years.

Algebraic Equations

Algebraic equations serve as a fundamental building block in mathematics which represent relationships between different quantities. In the context of our exercise with cars, the equation given (\(A = P_{0}(0.7)^{t}\)) is used to calculate depreciation.

• Here, \(A\) is the resulting value of the car after a certain number of years, \(P_{0}\) represents the initial purchase price, and \(t\) is the time in years.
• The concept of solving algebraic equations typically involves finding the values for the variables involved.
• For depreciation, solving such equations helps predict future values, letting you estimate how much your car or asset will be worth in the future.

Using algebra helps in rearranging formulae and solving for unknowns efficiently. In this situation, it provides a systematic way to handle the effects of time on value as represented numerically.

Exponential Functions

Exponential functions are a type of mathematical function that is pivotal when understanding the rate of change processes like growth and decay. The formula used in the exercise, \( A = P_{0}(0.7)^{t} \), is an example of exponential decay.

• In exponential decay, the value of something decreases by a constant percentage over equal time periods.
• The base of the exponential, in this case 0.7, indicates the portion of the value retained each year (70%).
• It contrasts with linear decay, where the value decreases by a constant absolute amount over time.

These functions are especially useful in modeling scenarios where there is a consistent percentage reduction over time, helping in making predictions about future values. Through understanding exponential functions, we grasp how quickly a quantity like car value diminishes within specific intervals, which is immensely valuable for financial planning and science.