7. 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 £20000. (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.

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7. In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

Question 9

7. 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 ... show full transcript

Worked Solution & Example Answer:7. In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

Step 1

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

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Answer

To model the value of car A using an exponential decay formula, we start with the general equation:

$V = A e^{-kt}$

Where:

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

From the information provided:

When $t = 0$ , $V = 20000$ ,
When $t = 1$ , $V = 16000$ .

Substituting the values into the formula:

$16000 = 20000 e^{-k(1)}$

To find $k$ , we first solve for $e^{-k}$ :

$e^{-k} = \frac{16000}{20000} = 0.8$

Taking the natural logarithm of both sides:

$-k = ext{ln}(0.8) \\ k \approx 0.223$

Thus, the equation linking $V$ with $t$ becomes:

$V = 20000 e^{-0.223t}$

Step 2

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

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Answer

After monitoring car A for 10 years, its value is observed to be £20000.

Using the model, we find the theoretical value at $t = 10$ :

$V = 20000 e^{-0.223(10)}$ $V \approx 20000 e^{-2.23} \approx 20000 \cdot 0.1071 \approx 2142.00$

The predicted value is significantly lower than the observed £20000. This discrepancy indicates that the model may not reliably predict the car's value over a longer period, as the actual value remains much higher and does not follow the expected exponential decay. Therefore, the model is not reliable in this context.

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.

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Answer

To adapt the equation for car B, we can modify the decay constant to reflect that it depreciates more slowly than car A. Thus, if we let the new decay constant be $k_B < k_{A}$ , the revised equation would be:

$V_B = 20000 e^{-k_B t}$

Where $k_B$ could be a reduced value, for example, $k_B = 0.15$ , to demonstrate a slower rate of depreciation.

7. In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 2019 - Paper 1

Worked Solution & Example Answer:7. In a simple model, the value, $V$, of a car depends on its age, $t$, in years - Edexcel - A-Level Maths Pure - Question 9 - 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.

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