June 2023 Paper 2 Q3
3 In this question you must show detailed reasoning.
Find the smallest possible positive integers \(m\) and \(n\) such that \(\left(\dfrac{64}{49}\right)^{-\frac{3}{2}} = \dfrac{m}{n}\). [3]
| Scheme | Marks | AO |
|---|---|---|
| take reciprocal calculate cube calculate square root to obtain \(m = 343,\ n = 512\) isw or \(\frac{343}{512}\) isw | B1 B1 B1 | |
| [3] |
Notes
operations may be in any order, but 3 distinct numerical steps required for 3 marks
if taking reciprocal and one other step are combined into one step, 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
eg
| Scheme | Marks |
|---|---|
| \(\left(\frac{49}{64}\right)^{\frac{3}{2}}\) | B1 |
| \(\left(\frac{7}{8}\right)^3\) | B1 |
| \(\frac{343}{512}\) isw | B1 |
B1: taking reciprocal; may be awarded after simplification
B1: square roots found; may be seen before taking reciprocal
B1: dependent on award of both preceding marks
eg Alternatively
| Scheme | Marks |
|---|---|
| \(\left(\frac{64}{49}\right)^{-3} = \left(\frac{m}{n}\right)^2\) \(\left(\frac{49}{64}\right)^3 = \left(\frac{m}{n}\right)^2\) | B1 |
| \(117649 = m^2\) and \(262144 = n^2\) | B1 |
| \(m = 343\) and \(n = 512\) | B1 |
B1: dependent on award of both preceding marks