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 $t$. (4) 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. (2) 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. (1)

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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 2

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 - it... 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 2

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 depreciation of car A's value, we can use the exponential decay formula:

$V = A e^{kt}$

where:

$A$ is the initial value of the car, which is £20000,
$k$ is the rate of depreciation,
$t$ is the age of the car in years,

We know from the problem that the value after one year is £16000. Substituting these values into the equation gives:

$16000 = 20000 e^{k imes 1}$

To isolate $k$ , we divide both sides by 20000:

$e^{k} = rac{16000}{20000} = 0.8$

Taking the natural logarithm of both sides:

$k = ext{ln}(0.8)$

Thus, the possible equation linking $V$ and $t$ is:

$V = 20000 e^{ ext{ln}(0.8)t}$

Step 2

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

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Answer

After 10 years, we need to evaluate the model's prediction and compare it with the known value of car A, which is also £20000. Substituting $t = 10$ into our model:

$V = 20000 e^{ ext{ln}(0.8) imes 10}$

Calculating this gives:

eq £20000$$ Since the model suggests a value lower than £20000 (as it suggests a decrease in value due to depreciation), the reliability is questionable. Hence, we can state: - The model is reliable if the value approaches or exceeds £20000 after the full monitoring period, which, in this case, it does not.

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 model car B, which depreciates more slowly, we can adapt the original model by using a smaller value for $k$ . We could express this as:

$V = 20000 e^{kt}$

where $k$ is less negative than the original $k$ used for car A, reflecting the slower rate of depreciation. For example, if $k$ were to be changed to something like $-0.1$ , car B would retain its value better over time.

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 2

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 2

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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