In a simple model, the value, $V$, of a car depends on its age, $t$, in years. The following information is available for car A: - its value when new is £20000 - its value after one year is £16000 (a) Use an exponential model to form, for car A, a possible equation linking $V$ with $t$. The value of car A is monitored over a 10-year period. Its value after 10 years is £2000. (b) Evaluate the reliability of your model in light of this information. The following information is available for car B: - it has the same value, when new, as car A - its value depreciates more slowly than that of car A. (c) Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

Photo AI

In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

Question 8

In a simple model, the value, $V$, of a car depends on its age, $t$, in years. The following information is available for car A: - its value when new is £20000 - i... show full transcript

Worked Solution & Example Answer:In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

Step 1

Use an exponential model to form, for car A, a possible equation linking $V$ with $t$.

96%

114 rated

Answer

To model the value of car A using an exponential decay function, we can express it as:

$V = A e^{-kt}$

Where:

$V$ is the value of the car after $t$ years,
$A$ is the initial value (£20000),
$k$ is the decay constant,
$t$ is the time in years.

From the information provided, we can substitute the known values:

$16000 = 20000 e^{-k}$

Solving for $k$ , we rearrange this to:

$e^{-k} = \frac{16000}{20000} = 0.8 \implies -k = \ln(0.8) \implies k = -\ln(0.8).$

Thus, the model becomes:

$V = 20000 e^{-\ln(0.8) t}.$

This can be simplified to:

$V = 20000 (0.8)^t.$

Step 2

Evaluate the reliability of your model in light of this information.

99%

104 rated

Answer

After 10 years, the model predicts:

$V = 20000 (0.8)^{10} \approx 20000 \times 0.1074 \approx 2148.09.$

However, the actual value is stated to be £2000. This indicates that while our model provides close results, the prediction is not fully accurate. The difference between £2000 and £2148.09 suggests that although the exponential model captures the general trend of depreciation, it may not account for all variables affecting the car's value over the long term. Therefore, the model is reasonably reliable but may require further adjustment.

Step 3

Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

96%

101 rated

Answer

To adapt the model for car B, which depreciates more slowly, we would alter the decay constant $k$ to a smaller value. Thus, the equation can be represented as:

$V_B = 20000 e^{-kt} \text{ where } k < -\ln(0.8).$

The exact value of $k$ would depend on the rate at which car B depreciates relative to car A.

In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

Worked Solution & Example Answer:In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 8 - 2019 - Paper 1

Use an exponential model to form, for car A, a possible equation linking $V$ with $t$.

Evaluate the reliability of your model in light of this information.

Explain how you would adapt the equation found in (a) so that it could be used to model the value of car B.

Join the A-Level students using SimpleStudy...