| Exam Board | Edexcel |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2022 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Straight Lines & Coordinate Geometry |
| Type | Equation of line through two points |
| Difficulty | Easy -1.3 This is a straightforward substitution problem requiring students to form two simultaneous equations from given coordinates (3, 1.05) and (5, 1.65), solve for p and q, then substitute V=2.50 to find T. It's purely mechanical with no conceptual challenge—simpler than average A-level questions as it requires only basic algebraic manipulation of linear equations. |
| Spec | 1.02a Indices: laws of indices for rational exponents1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1.05 = 3p + q\) and \(1.65 = 5p + q\), e.g. \(2p = 0.6\) | M1 | Forms two simultaneous equations and proceeds to find \(p\) or \(q\). May be implied by correct \(p\) or \(q\) (allow sight of 30 or 15). Also scores for attempt to calculate \(\frac{1.65-1.05}{5-3}\). |
| \(p = 0.3\), \(q = 0.15\) | A1 | \(p=0.3\) or \(q=0.15\) |
| A1 | \(p=0.3\) and \(q=0.15\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.5 = {``}0.3{"}T + {``}0.15{"} \Rightarrow T =\) | M1 | Uses their \(p\) and \(q\) with \(V=2.5\) and rearranges to find \(T\). Must come from \(\frac{2.5 \pm q}{p}\). |
| \(T = 7.8\) | A1 | cao, isw after sight of correct answer. |
# Question 3:
## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1.05 = 3p + q$ and $1.65 = 5p + q$, e.g. $2p = 0.6$ | M1 | Forms two simultaneous equations and proceeds to find $p$ or $q$. May be implied by correct $p$ or $q$ (allow sight of 30 or 15). Also scores for attempt to calculate $\frac{1.65-1.05}{5-3}$. |
| $p = 0.3$, $q = 0.15$ | A1 | $p=0.3$ or $q=0.15$ |
| | A1 | $p=0.3$ and $q=0.15$ |
## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.5 = {``}0.3{"}T + {``}0.15{"} \Rightarrow T =$ | M1 | Uses their $p$ and $q$ with $V=2.5$ and rearranges to find $T$. Must come from $\frac{2.5 \pm q}{p}$. |
| $T = 7.8$ | A1 | cao, isw after sight of correct answer. |
---\begin{enumerate}
\item The share price of a company is monitored.
\end{enumerate}
Exactly 3 years after monitoring began, the share price was $\pounds 1.05$\\
Exactly 5 years after monitoring began, the share price was $\pounds 1.65$\\
The share price, $\pounds V$, of the company is modelled by the equation
$$V = p t + q$$
where $t$ is the number of years after monitoring began and $p$ and $q$ are constants.\\
(a) Find the value of $p$ and the value of $q$.
Exactly $T$ years after monitoring began, the share price was $\pounds 2.50$\\
(b) Find the value of $T$, according to the model, giving your answer to one decimal place.
\hfill \mbox{\textit{Edexcel P1 2022 Q3 [5]}}