| Exam Board | OCR MEI |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Solving quadratics and applications |
| Type | Geometric area leads to quadratic |
| Difficulty | Standard +0.3 This is a straightforward application of the trapezium area formula leading to a simple quadratic equation. The 'show that' part guides students to the equation, and solving x² + 2x - 35 = 0 factors easily to (x+7)(x-5). Slightly easier than average due to the structured guidance and standard factorisation. |
| Spec | 1.02f Solve quadratic equations: including in a function of unknown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2} \times x \times (x + 2 + 3x + 6)\) oe | M1 | correct statement of area of trap; may be rectangle \(\pm\) triangle, or two triangles; eg \(2x(x+2) + \frac{1}{2} \times 2x \times (2x+4)\); condone missing brackets for M1; condone also for A1 if expansion is treated as if they were there |
| \(x(4x+8) = 140\) oe and given ans \(x^2 + 2x - 35 = 0\) obtained correctly with at least one further interim step | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([AB =]\ 1\) www | 3 | |
| or B2 for \(x = [-7\) or\(]\ 5\) cao www or for \(AB = 21\) or \(-15\) | B2 | 5 on its own or \(AB = 5\) with no working scores 0; we need to see \(x = 5\) |
| or M1 for \((x+7)(x-5)\ [=0]\) or formula or completing square used eg \((x+1)^2 - 36\ [=0]\); condone one error eg factors with sign wrong or which give two terms correct when expanded | M1 | may be done in (i) if not here |
| or M1 for showing \(f(5) = 0\) without stating \(x = 5\) | M1 |
## Question 5(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2} \times x \times (x + 2 + 3x + 6)$ oe | M1 | correct statement of area of trap; may be rectangle $\pm$ triangle, or two triangles; eg $2x(x+2) + \frac{1}{2} \times 2x \times (2x+4)$; condone missing brackets for M1; condone also for A1 if expansion is treated as if they were there |
| $x(4x+8) = 140$ oe and given ans $x^2 + 2x - 35 = 0$ obtained correctly with at least one further interim step | A1 | |
## Question 5(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[AB =]\ 1$ www | 3 | |
| or **B2** for $x = [-7$ or$]\ 5$ cao www or for $AB = 21$ or $-15$ | B2 | 5 on its own or $AB = 5$ with no working scores 0; we need to see $x = 5$ |
| or **M1** for $(x+7)(x-5)\ [=0]$ or formula or completing square used eg $(x+1)^2 - 36\ [=0]$; condone one error eg factors with sign wrong or which give two terms correct when expanded | M1 | may be done in (i) if not here |
| or **M1** for showing $f(5) = 0$ without stating $x = 5$ | M1 | |
---5 Fig. 9 shows a trapezium ABCD , with the lengths in centimetres of three of its sides.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d22f53f5-ba80-4065-a94b-2a9c92c20dfb-1_462_877_1796_684}
\captionsetup{labelformat=empty}
\caption{Fig. 9}
\end{center}
\end{figure}
This trapezium has area $140 \mathrm {~cm} ^ { 2 }$.\\
(i) Show that $x ^ { 2 } + 2 x - 35 = 0$.\\
(ii) Hence find the length of side AB of the trapezium.
\hfill \mbox{\textit{OCR MEI C1 Q5 [5]}}