Math Problem Statement

A new car is purchased for 18000 dollars. The value of the car depreciates at 13.5% per year. What will the value of the car be, to the nearest cent, after 14 years?

Solution

The problem asks for the value of a car after 14 years, given an initial purchase price of 18,000 dollars and an annual depreciation rate of 13.5%. This is a typical example of exponential decay in value.

We can use the formula for depreciation:

V(t)=P×(1r)tV(t) = P \times (1 - r)^t

Where:

  • V(t)V(t) is the value of the car after tt years,
  • PP is the initial value of the car (18,000 dollars),
  • rr is the depreciation rate (13.5% or 0.135),
  • tt is the number of years (14 years).

Plugging in the values:

V(14)=18000×(10.135)14V(14) = 18000 \times (1 - 0.135)^{14} V(14)=18000×(0.865)14V(14) = 18000 \times (0.865)^{14} V(14)=18000×0.165303V(14) = 18000 \times 0.165303 V(14)2975.45V(14) \approx 2975.45

Therefore, the value of the car after 14 years, to the nearest cent, is $2,975.45.

Would you like further details or have any questions?

Here are 5 related questions to think about:

  1. How would the car's value change if the depreciation rate were 10% instead of 13.5%?
  2. How much would the car be worth after 10 years instead of 14?
  3. What would happen to the car's value after 20 years if the depreciation continues at the same rate?
  4. How does the concept of exponential decay apply to other assets, like electronics or appliances?
  5. What is the difference between linear and exponential depreciation?

Tip: Remember that exponential decay rapidly reduces the value of an object over time, especially with high depreciation rates.

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Math Problem Analysis

Mathematical Concepts

Exponential Decay
Depreciation
Algebra

Formulas

V(t) = P × (1 - r)^t
V(t) = 18000 × (1 - 0.135)^14

Theorems

Exponential Decay Formula

Suitable Grade Level

Grades 9-12