OCR MEI Paper 2 2023 June — Question 3 3 marks

Exam BoardOCR MEI
ModulePaper 2 (Paper 2)
Year2023
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicIndices and Surds
TypeEvaluate numerical powers
DifficultyModerate -0.8 This question requires applying negative and fractional index laws to evaluate a numerical expression, then simplifying to lowest terms. While it involves multiple steps (dealing with the negative power, fractional power, and simplification), these are all standard procedural techniques from the indices topic with no problem-solving insight required. It's slightly easier than average because it's purely mechanical application of well-drilled rules, though the negative exponent and need to find 'smallest possible' integers adds minor complexity beyond the most basic index questions.
Spec1.02a Indices: laws of indices for rational exponents4.01a Mathematical induction: construct proofs

3 In this question you must show detailed reasoning.
Find the smallest possible positive integers \(m\) and \(n\) such that \(\left( \frac { 64 } { 49 } \right) ^ { - \frac { 3 } { 2 } } = \frac { m } { n }\).

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
take reciprocal; calculate cube; calculate square root to obtain \(m = 343, n = 512\) isw or \(\frac{343}{512}\) iswB1, B1, B1 Operations may be in any order, but 3 distinct numerical steps required for 3 marks. If taking reciprocal and one other step combined, allow B1B1B0. If B0B0 for cubing and square rooting, allow SC1 for \(\left(\sqrt{\frac{49}{64}}\right)^3 = \frac{343}{512}\) or \(\left(\sqrt{\frac{64}{49}}\right)^3 = \frac{512}{343}\) seen
[3]
*eg* \(\left(\frac{49}{64}\right)^{\frac{3}{2}}\)B1 1.1 — taking reciprocal; may be awarded after simplification
\(\left(\frac{7}{8}\right)^3\)B1 1.1 — square roots found; may be seen before taking reciprocal
\(\frac{343}{512}\) iswB1 1.1 — dependent on award of both preceding marks
[3]
*Alternatively:* \(\left(\frac{49}{64}\right)^3 = \left(\frac{m}{n}\right)^2\)
\(117649 = m^2\) and \(262144 = n^2\)B1
\(m = 343\) and \(n = 512\)B1 dependent on award of both preceding marks
[3]
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| take reciprocal; calculate cube; calculate square root to obtain $m = 343, n = 512$ isw or $\frac{343}{512}$ isw | B1, B1, B1 | Operations may be in any order, but 3 **distinct numerical** steps required for 3 marks. If taking **reciprocal** and **one** other step combined, allow **B1B1B0**. If **B0B0** for cubing and square rooting, allow **SC1** for $\left(\sqrt{\frac{49}{64}}\right)^3 = \frac{343}{512}$ or $\left(\sqrt{\frac{64}{49}}\right)^3 = \frac{512}{343}$ seen |
| **[3]** | | |
| *eg* $\left(\frac{49}{64}\right)^{\frac{3}{2}}$ | B1 | 1.1 — taking reciprocal; may be awarded after simplification |
| $\left(\frac{7}{8}\right)^3$ | B1 | 1.1 — square roots found; may be seen before taking reciprocal |
| $\frac{343}{512}$ isw | B1 | 1.1 — dependent on award of both preceding marks |
| **[3]** | | |
| *Alternatively:* $\left(\frac{49}{64}\right)^3 = \left(\frac{m}{n}\right)^2$ | | |
| $117649 = m^2$ and $262144 = n^2$ | B1 | |
| $m = 343$ and $n = 512$ | B1 | dependent on award of both preceding marks |
| **[3]** | | |
3 In this question you must show detailed reasoning.\\
Find the smallest possible positive integers $m$ and $n$ such that $\left( \frac { 64 } { 49 } \right) ^ { - \frac { 3 } { 2 } } = \frac { m } { n }$.

\hfill \mbox{\textit{OCR MEI Paper 2 2023 Q3 [3]}}