A certain car is known to deprecitate at a rate of 20% per year. The equation V(n)=p`(x)^(n)` can be used to calculate the value of the car, V. after n years where p is the purchase price. If the purchase price of the car is $25,000, to the nearest dollar, howmuch more is the car worth after 2 year than 3 years?

Options

A. 1600
B. 2400
C. 3200
D. 4000

Answer

To solve the problem step by step, we will use the formula provided for calculating the value of the car after a certain number of years, taking into account the depreciation rate.

### Step 1: Understand the formula
The value of the car after n years is given by the formula:
\[ V(n) = P \cdot (1 - r)^{n} \]
where:
- \( V(n) \) is the value of the car after n years,
- \( P \) is the purchase price,
- \( r \) is the rate of depreciation (as a decimal),
- \( n \) is the number of years.

### Step 2: Identify the values
From the problem:
- Purchase price \( P = 25,000 \) dollars,
- Rate of depreciation \( r = 20\% = 0.20 \),
- We need to find the values after 2 years (\( V_1 \)) and after 3 years (\( V_2 \)).

### Step 3: Calculate the value after 2 years (\( V_1 \))
Using the formula:
\[ V_1 = 25,000 \cdot (1 - 0.20)^{2} \]
\[ V_1 = 25,000 \cdot (0.80)^{2} \]
\[ V_1 = 25,000 \cdot 0.64 \]
\[ V_1 = 16,000 \]

### Step 4: Calculate the value after 3 years (\( V_2 \))
Using the same formula:
\[ V_2 = 25,000 \cdot (1 - 0.20)^{3} \]
\[ V_2 = 25,000 \cdot (0.80)^{3} \]
\[ V_2 = 25,000 \cdot 0.512 \]
\[ V_2 = 12,800 \]

### Step 5: Find the difference in value
Now, we need to find how much more the car is worth after 2 years than after 3 years:
\[ V_1 - V_2 = 16,000 - 12,800 \]
\[ V_1 - V_2 = 3,200 \]

### Conclusion
The car is worth $3,200 more after 2 years than after 3 years.

### Final Answer
The answer is $3,200.

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