| Exam Board | Edexcel |
|---|---|
| Module | PMT Mocks (PMT Mocks) |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Functions |
| Type | Interpret model constants |
| Difficulty | Moderate -0.8 This is a straightforward exponential model question requiring basic substitution to find A, simple algebra to find k (solving 50000 = 2500k^20), interpretation of constants (standard textbook exercise), and logarithmic calculation to find when P exceeds £100000. All steps are routine applications of exponential function techniques with no novel problem-solving required. |
| Spec | 1.06i Exponential growth/decay: in modelling context |
| Year | 1990 | 2010 |
| Average weekly pay | \(\pounds 2500\) | \(\pounds 50000\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2500 = Ak^0 \Rightarrow A = 2500\) | B1 | Substitutes \(t=0, P=2500\) into \(P=Ak^t\) to reach a value of \(A\) |
| \(50000 = 2500k^{20} \Rightarrow 20 = k^{20} \Rightarrow k = 20^{\frac{1}{20}} = 1.16159\) | M1, A1 | Substitutes \(t=20, A=2500, P=50000\) into \(P=Ak^t\); \(k=1.16159\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(A\) is the weekly pay of a footballer on 1st August 1990 | B1 | Must reference the footballer, weekly pay/wage and time \(t=0\) |
| \(k\) is the rate at which the weekly pay rises each year | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(100000 = 2500 \times 1.16159^t\) | M1 | Using the model and proceeding to \(1.16159^t = k\) |
| \(40 = 1.16159^t \Rightarrow \log 40 = t\log 1.16159\) | M1 | Correct method to find \(t\) |
| \(t = 24.627\) | A1 | \(t = 24.627...\) or \(t = \log_{1.16159}40\) |
| Year 2015 | A1 | Start date is August 1990, so \(t=24.6\) years first exceeds £100000 in 2015 |
## Question 12(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2500 = Ak^0 \Rightarrow A = 2500$ | B1 | Substitutes $t=0, P=2500$ into $P=Ak^t$ to reach a value of $A$ |
| $50000 = 2500k^{20} \Rightarrow 20 = k^{20} \Rightarrow k = 20^{\frac{1}{20}} = 1.16159$ | M1, A1 | Substitutes $t=20, A=2500, P=50000$ into $P=Ak^t$; $k=1.16159$ |
## Question 12(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $A$ is the weekly pay of a footballer on 1st August 1990 | B1 | Must reference the footballer, weekly pay/wage and time $t=0$ |
| $k$ is the rate at which the weekly pay rises each year | B1 | |
## Question 12(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $100000 = 2500 \times 1.16159^t$ | M1 | Using the model and proceeding to $1.16159^t = k$ |
| $40 = 1.16159^t \Rightarrow \log 40 = t\log 1.16159$ | M1 | Correct method to find $t$ |
| $t = 24.627$ | A1 | $t = 24.627...$ or $t = \log_{1.16159}40$ |
| Year **2015** | A1 | Start date is August 1990, so $t=24.6$ years first exceeds £100000 in 2015 |
---12. The table shows the average weekly pay of a footballer at a certain club on 1 August 1990 and 1 August 2010.
\begin{center}
\begin{tabular}{ | l | l | l | }
\hline
Year & 1990 & 2010 \\
\hline
Average weekly pay & $\pounds 2500$ & $\pounds 50000$ \\
\hline
\end{tabular}
\end{center}
The average weekly pay of a footballer at this club can be modelled by the equation
$$P = A k ^ { t }$$
where $\pounds P$ is the average weekly pay $t$ years after 1 August 1990, and $A$ and $k$ are constants.\\
a. i. Write down the value of $A$.\\
ii. Show that the value of $k$ is 1.16159 , correct to five decimal places.\\
b. With reference to the model, interpret\\
i. the value of the constant $A$,\\
ii. the value of the constant $k$,
Using the model,\\
c. find the year in which, on 1 August, the average weekly pay of a footballer at this club will first exceed $\pounds 100000$.\\
\hfill \mbox{\textit{Edexcel PMT Mocks Q12 [9]}}