Question 12 - A-Level Maths

Edexcel Paper 1 2018 June — Question 12 10 marks

Exam Board	Edexcel
Module	Paper 1 (Paper 1)
Year	2018
Session	June
Marks	10
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Exponential Equations & Modelling
Type	Finding x from given y value
Difficulty	Standard +0.3 This is a straightforward exponential modelling question requiring substitution of two data points to form simultaneous equations, solving by division to find p, back-substitution for A, and using logarithms to solve for t. All steps are standard A-level techniques with no novel insight required, making it slightly easier than average.
Spec	1.06a Exponential function: a^x and e^x graphs and properties 1.06g Equations with exponentials: solve a^x = b 1.06i Exponential growth/decay: in modelling context

The value, $\pounds V$, of a vintage car $t$ years after it was first valued on 1 st January 2001, is modelled by the equation

$$V = A p ^ { t } \quad \text { where } A \text { and } p \text { are constants }$$ Given that the value of the car was $\pounds 32000$ on 1st January 2005 and $\pounds 50000$ on 1st January 2012

1. find $p$ to 4 decimal places,
2. show that $A$ is approximately 24800
With reference to the model, interpret
1. the value of the constant $A$,
2. the value of the constant $p$. Using the model,
find the year during which the value of the car first exceeds $\pounds 100000$

Show mark scheme Show mark scheme source

Question 12:

Part (a)(i):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Method to find $p$: divides $32000 = Ap^4$ by $50000 = Ap^{11}$, giving $p^7 = \frac{50000}{32000} \Rightarrow p = \sqrt[7]{\frac{50000}{32000}}$	M1	Attempts to use both pieces of information within $V = Ap^t$, eliminates $A$ correctly and solves equation of form $p^n = k$. Allow slips on 32000, 50000 and values of $t$
$p = 1.0658$	A1	$p =$ awrt 1.0658

Part (a)(ii):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Substitutes $p = 1.0658$ into either equation: $A = \frac{32000}{1.0658^4}$ or $A = \frac{50000}{1.0658^{11}}$	M1	Substitutes their $p = 1.0658$ into either equation and finds $A$
$A = 24795 \rightarrow 24805 \approx 24800$	A1*	Shows $A$ is between 24795 and 24805 before stating $\approx 24800$. Accept with or without units

Part (b):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
$A$ / (£)24800 is the value of the car on 1st January 2001	B1	Must reference the car, cost/value, and "0" time
$p$/1.0658 is the factor by which the value rises each year. Accept: value rises by 6.6% a year	B1	Must reference a yearly rate and an increase in value or multiplier

Part (c):

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Attempts $100000 = 24800 \times 1.0658^t$	M1	Uses model $100000 = 24800 \times 1.0658^t$ and proceeds to $1.0658^t = k$
$1.0658^t = \frac{100000}{24800}$ leading to $t = \log_{1.0658}\left(\frac{100000}{24800}\right)$	dM1	Complete method using (i) the model equation and (ii) correct method to find $t$
$t = 21.8$ or $21.9$	A1	awrt 21.8 or 21.9 or $\log_{1.0658}\left(\frac{100000}{24800}\right)$
2022	A1	cso

# Question 12:

## Part (a)(i):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Method to find $p$: divides $32000 = Ap^4$ by $50000 = Ap^{11}$, giving $p^7 = \frac{50000}{32000} \Rightarrow p = \sqrt[7]{\frac{50000}{32000}}$ | M1 | Attempts to use both pieces of information within $V = Ap^t$, eliminates $A$ correctly and solves equation of form $p^n = k$. Allow slips on 32000, 50000 and values of $t$ |
| $p = 1.0658$ | A1 | $p =$ awrt 1.0658 |

## Part (a)(ii):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Substitutes $p = 1.0658$ into either equation: $A = \frac{32000}{1.0658^4}$ or $A = \frac{50000}{1.0658^{11}}$ | M1 | Substitutes their $p = 1.0658$ into either equation and finds $A$ |
| $A = 24795 \rightarrow 24805 \approx 24800$ | A1* | Shows $A$ is between 24795 and 24805 before stating $\approx 24800$. Accept with or without units |

## Part (b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ / (£)24800 is the value of the car on 1st January 2001 | B1 | Must reference **the car**, **cost/value**, and **"0" time** |
| $p$/1.0658 is the factor by which the value rises each year. Accept: value rises by 6.6% a year | B1 | Must reference **a yearly rate** and **an increase in value or multiplier** |

## Part (c):

| Answer/Working | Mark | Guidance |
|---|---|---|
| Attempts $100000 = 24800 \times 1.0658^t$ | M1 | Uses model $100000 = 24800 \times 1.0658^t$ and proceeds to $1.0658^t = k$ |
| $1.0658^t = \frac{100000}{24800}$ leading to $t = \log_{1.0658}\left(\frac{100000}{24800}\right)$ | dM1 | Complete method using (i) the model equation and (ii) correct method to find $t$ |
| $t = 21.8$ or $21.9$ | A1 | awrt 21.8 or 21.9 or $\log_{1.0658}\left(\frac{100000}{24800}\right)$ |
| 2022 | A1 | cso |

---

Show LaTeX source

\begin{enumerate}
  \item The value, $\pounds V$, of a vintage car $t$ years after it was first valued on 1 st January 2001, is modelled by the equation
\end{enumerate}

$$V = A p ^ { t } \quad \text { where } A \text { and } p \text { are constants }$$

Given that the value of the car was $\pounds 32000$ on 1st January 2005 and $\pounds 50000$ on 1st January 2012\\
(a) (i) find $p$ to 4 decimal places,\\
(ii) show that $A$ is approximately 24800\\
(b) With reference to the model, interpret\\
(i) the value of the constant $A$,\\
(ii) the value of the constant $p$.

Using the model,\\
(c) find the year during which the value of the car first exceeds $\pounds 100000$

\hfill \mbox{\textit{Edexcel Paper 1 2018 Q12 [10]}}

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