| Exam Board | AQA |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2013 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Solve equations with surds |
| Difficulty | Moderate -0.8 This is a routine C1 surds question requiring standard techniques: simplifying surds using prime factorization, solving a linear equation with surds, and rationalizing a denominator. All steps are algorithmic with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure and algebraic manipulation required. |
| Spec | 1.02b Surds: manipulation and rationalising denominators |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \((\sqrt{48} =) 4\sqrt{3}\) | B1 | condone \(n = 4\). No ISW. |
| (ii) \(\sqrt{12} = 2\sqrt{3}\) and \(\sqrt{48} = 4\sqrt{3}\) | M1 | (FT 'their'\(n\)) \(2x\sqrt{3} = 7\sqrt{3} - 4\sqrt{3}\) |
| \(x = \frac{7\sqrt{3} - 4\sqrt{3}}{2\sqrt{3}}\) | A1 | correct quotient unsimplified or correct equation in integers e.g. \(6x = 21 - 12\) |
| \(= \frac{3}{2}\) | A1cso | 3 marks total; accept 1.5 but not \(\frac{9}{6}\) etc |
| Alternative 1: | ||
| \(x = \frac{7\sqrt{3} - \sqrt{48}}{\sqrt{12}} \times \frac{\sqrt{12}}{\sqrt{12}}\) | M1 | |
| integer terms \(= \frac{42 - 24}{12}\) | A1 | |
| \(= \frac{3}{2}\) | A1 | |
| (b) \(\frac{11\sqrt{3} + 2\sqrt{5}}{2\sqrt{3} + \sqrt{5}} \times \frac{2\sqrt{3} - \sqrt{5}}{2\sqrt{3} - \sqrt{5}}\) | M1 | |
| (numerator \(=\) ) \(22 \times 3 + 4\sqrt{15} - 11\sqrt{15} - 2 \times 5\) | A1 | correct unsimplified but must simplify \((\sqrt{3})^2\), \((\sqrt{5})^2\) and \(\sqrt{3} \times \sqrt{5}\) correctly |
| (denominator \(= 12 - 5 =\) ) 7 | B1 | must be seen or identified as denominator giving \(\frac{56 - 7\sqrt{15}}{7}\) |
| (Answer \(=\) ) \(8 - \sqrt{15}\) | A1cso | 4 marks total; \(m = 8\) |
**(a)(i)** $(\sqrt{48} =) 4\sqrt{3}$ | B1 | condone $n = 4$. No ISW.
**(ii)** $\sqrt{12} = 2\sqrt{3}$ and $\sqrt{48} = 4\sqrt{3}$ | M1 | (FT 'their'$n$) $2x\sqrt{3} = 7\sqrt{3} - 4\sqrt{3}$
$x = \frac{7\sqrt{3} - 4\sqrt{3}}{2\sqrt{3}}$ | A1 | correct quotient unsimplified or correct equation in integers e.g. $6x = 21 - 12$
$= \frac{3}{2}$ | A1cso | 3 marks total; accept 1.5 but not $\frac{9}{6}$ etc
| | | **Alternative 1:**
| | | $x = \frac{7\sqrt{3} - \sqrt{48}}{\sqrt{12}} \times \frac{\sqrt{12}}{\sqrt{12}}$ | M1 |
| | | integer terms $= \frac{42 - 24}{12}$ | A1 |
| | | $= \frac{3}{2}$ | A1 |
**(b)** $\frac{11\sqrt{3} + 2\sqrt{5}}{2\sqrt{3} + \sqrt{5}} \times \frac{2\sqrt{3} - \sqrt{5}}{2\sqrt{3} - \sqrt{5}}$ | M1 |
(numerator $=$ ) $22 \times 3 + 4\sqrt{15} - 11\sqrt{15} - 2 \times 5$ | A1 | correct unsimplified but must simplify $(\sqrt{3})^2$, $(\sqrt{5})^2$ and $\sqrt{3} \times \sqrt{5}$ correctly
(denominator $= 12 - 5 =$ ) 7 | B1 | must be seen or identified as denominator giving $\frac{56 - 7\sqrt{15}}{7}$
(Answer $=$ ) $8 - \sqrt{15}$ | A1cso | 4 marks total; $m = 8$
**Total: 8 marks**
---2
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Express $\sqrt { 48 }$ in the form $n \sqrt { 3 }$, where $n$ is an integer.
\item Solve the equation
$$x \sqrt { 12 } = 7 \sqrt { 3 } - \sqrt { 48 }$$
giving your answer in its simplest form.
\end{enumerate}\item Express $\frac { 11 \sqrt { 3 } + 2 \sqrt { 5 } } { 2 \sqrt { 3 } + \sqrt { 5 } }$ in the form $m - \sqrt { 15 }$, where $m$ is an integer.
\end{enumerate}
\hfill \mbox{\textit{AQA C1 2013 Q2 [8]}}