| Exam Board | Edexcel |
|---|---|
| Module | C1 (Core Mathematics 1) |
| Year | 2009 |
| Session | January |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Indices and Surds |
| Type | Evaluate numerical powers |
| Difficulty | Easy -1.8 This is a straightforward recall question testing basic index laws with a simple cube root (125 = 5³). Part (a) requires only recognizing 125^(1/3) = 5, and part (b) applies the negative power rule to get 1/25. No problem-solving or multi-step reasoning required—pure mechanical application of index laws. |
| Spec | 1.02a Indices: laws of indices for rational exponents |
| Answer | Marks | Guidance |
|---|---|---|
| \(5\) | B1 | \(\pm 5\) is B0. (1 mark total) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{(\text{their } 5)^2}\) or \(\left(\frac{1}{\text{their } 5}\right)^2\) | M1 | Follow through their value of 5. Must have reciprocal and square. \(5^{-2}\) is not sufficient unless \(\frac{1}{5^2}\) follows. |
| \(= \frac{1}{25}\) or \(0.04\) | A1 | \(\pm\frac{1}{25}\) is A0. A negative introduced at any stage scores M1 but not A1. Correct answer with no working scores both marks. (2 marks total, [3] overall) |
## Question 1:
**(a)**
$5$ | B1 | $\pm 5$ is B0. (1 mark total)
**(b)**
$\frac{1}{(\text{their } 5)^2}$ or $\left(\frac{1}{\text{their } 5}\right)^2$ | M1 | Follow through their value of 5. Must have reciprocal and square. $5^{-2}$ is not sufficient unless $\frac{1}{5^2}$ follows.
$= \frac{1}{25}$ or $0.04$ | A1 | $\pm\frac{1}{25}$ is A0. A negative introduced at any stage scores M1 but not A1. Correct answer with no working scores both marks. (2 marks total, **[3]** overall)
---\begin{enumerate}[label=(\alph*)]
\item Write down the value of $125 ^ { \frac { 1 } { 3 } }$.
\item Find the value of $125 ^ { - \frac { 2 } { 3 } }$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel C1 2009 Q1 [3]}}