| Exam Board | Edexcel |
|---|---|
| Module | AS Paper 1 (AS Paper 1) |
| Year | 2018 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Exponential Equations & Modelling |
| Type | log(y) vs x: convert and interpret |
| Difficulty | Moderate -0.3 This is a standard exponential modelling question requiring conversion between logarithmic and exponential forms. Part (a) involves straightforward manipulation of log equations (comparing log₁₀V = 0.05t + 4.8 with log₁₀(pq^t) to find p = 10^4.8 and q = 10^0.05). Parts (b) and (c) are routine interpretation and substitution. While it requires understanding of logarithms and exponential models, it's a textbook exercise with no novel problem-solving, making it slightly easier than average. |
| Spec | 1.06g Equations with exponentials: solve a^x = b1.06h Logarithmic graphs: reduce y=ax^n and y=kb^x to linear form |
| Answer | Marks | Guidance |
|---|---|---|
| For a correct equation in \(p\) or \(q\): \(p = 10^{4.8}\) or \(q = 10^{0.05}\) | M1 | Usually \(p = 10^{4.8}\) or \(q = 10^{0.05}\); may be \(\log q = 0.05\) or \(\log p = 4.8\) |
| For \(p =\) awrt 63100 or \(q =\) awrt 1.122 | A1 | |
| For correct equations in both \(p\) and \(q\): \(p = 10^{4.8}\) and \(q = 10^{0.05}\) | dM1 | Links two equations; both values imply M1 M1 |
| For \(p =\) awrt 63100 and \(q =\) awrt 1.122 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (i) The value of the painting on 1st January 1980 (is £63,100) | B1 | Accept original value/cost of the painting |
| (ii) The proportional increase in value each year | B1 | Value rises 12.2% a year; "1.122 is the decimal multiplier representing year on year increase"; do not accept "the amount by which it is rising" |
| Answer | Marks | Guidance |
|---|---|---|
| Uses \(V = 63100 \times 1.122^{30}\) or \(\log V = 0.05 \times 30 + 4.8\) leading to \(V =\) | M1 | Substituting \(t = 30\) into \(V = pq^t\) using their values |
| \(= \) awrt £2,000,000 | A1 | Accept awrt £1.99 million or £2.00 million; condone omission of £ sign |
## Question 13(a):
For a correct equation in $p$ or $q$: $p = 10^{4.8}$ or $q = 10^{0.05}$ | M1 | Usually $p = 10^{4.8}$ or $q = 10^{0.05}$; may be $\log q = 0.05$ or $\log p = 4.8$
For $p =$ awrt 63100 or $q =$ awrt 1.122 | A1 |
For correct equations in both $p$ and $q$: $p = 10^{4.8}$ and $q = 10^{0.05}$ | dM1 | Links two equations; both values imply M1 M1
For $p =$ awrt 63100 and $q =$ awrt 1.122 | A1 |
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## Question 13(b):
(i) The value of the painting on 1st January 1980 (is £63,100) | B1 | Accept original value/cost of the painting
(ii) The proportional increase in value each year | B1 | Value rises 12.2% a year; "1.122 is the decimal multiplier representing year on year increase"; do **not** accept "the amount by which it is rising"
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## Question 13(c):
Uses $V = 63100 \times 1.122^{30}$ or $\log V = 0.05 \times 30 + 4.8$ leading to $V =$ | M1 | Substituting $t = 30$ into $V = pq^t$ using their values
$= $ awrt £2,000,000 | A1 | Accept awrt £1.99 million or £2.00 million; condone omission of £ sign
---13.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f7935caa-6626-4ba8-87ef-e9bb59e1ac3e-36_563_1019_244_523}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
The value of a rare painting, $\pounds V$, is modelled by the equation $V = p q ^ { t }$, where $p$ and $q$ are constants and $t$ is the number of years since the value of the painting was first recorded on 1st January 1980.
The line $l$ shown in Figure 3 illustrates the linear relationship between $t$ and $\log _ { 10 } V$ since 1st January 1980.
The equation of line $l$ is $\log _ { 10 } V = 0.05 t + 4.8$
\begin{enumerate}[label=(\alph*)]
\item Find, to 4 significant figures, the value of $p$ and the value of $q$.
\item With reference to the model interpret
\begin{enumerate}[label=(\roman*)]
\item the value of the constant $p$,
\item the value of the constant $q$.
\end{enumerate}\item Find the value of the painting, as predicted by the model, on 1st January 2010, giving your answer to the nearest hundred thousand pounds.
\end{enumerate}
\hfill \mbox{\textit{Edexcel AS Paper 1 2018 Q13 [8]}}