Higher November 2022 Paper 3 Q16
16 \(H\) is inversely proportional to the cube root of \(L\).
\(H = 7 \quad\) when \(\quad L = 64\)
(a) Work out an equation connecting \(H\) and \(L\). [3 marks]
(b) Work out the value of \(H\) when \(\quad L = 2744\) [2 marks]
| Answer | Mark | Comments |
|---|---|---|
| Alternative method 1 | ||
| \(H \propto \dfrac{1}{\sqrt[3]{L}}\) or \(H = \dfrac{k}{\sqrt[3]{L}}\) | M1 | oe equation any letter implied by \(7 = \dfrac{k}{\sqrt[3]{64}}\) |
| \((k =)\ 7 \times \sqrt[3]{64}\) or \((k =)\) 28 | M1dep | oe |
| \(H = \dfrac{28}{\sqrt[3]{L}}\) | A1 | oe equation SC1 \(H = \dfrac{7}{4}\sqrt[3]{L}\) or \(\dfrac{4}{7}H = \sqrt[3]{L}\) |
| Alternative method 2 | ||
| \(H \propto \dfrac{1}{\sqrt[3]{L}}\) or \(cH = \dfrac{1}{\sqrt[3]{L}}\) | M1 | oe equation any letter implied by \(7c = \dfrac{1}{\sqrt[3]{64}}\) |
| \((c =)\ \dfrac{1}{7 \times \sqrt[3]{64}}\) or \((c =)\ \dfrac{1}{28}\) | M1dep | oe |
| \(\dfrac{H}{28} = \dfrac{1}{\sqrt[3]{L}}\) | A1 | oe equation SC1 \(H = \dfrac{7}{4}\sqrt[3]{L}\) or \(\dfrac{4}{7}H = \sqrt[3]{L}\) |
Additional guidance
| Up to M2 may be awarded for correct work, with no or incorrect answer, even if this is seen amongst multiple attempts | |
| \((k =)\) 28 or \((k \propto)\) 28 or \((c =)\ \dfrac{1}{28}\) or \((c \propto)\ \dfrac{1}{28}\) | M1M1 |
| Condone use of \(\propto\) for up to M1M1A0 eg \(H \propto \dfrac{k}{\sqrt[3]{L}}\) \(k \propto 28\) \(H \propto \dfrac{28}{\sqrt[3]{L}}\) | M1 M1dep A0 |
| Answer | Mark | Comments |
|---|---|---|
| \(\dfrac{\text{their } 28}{\sqrt[3]{2744}}\) or \(\dfrac{\text{their } 28}{14}\) | M1 | oe |
| 2 | A1ft | ft their equation of the form \(H = \dfrac{k}{\sqrt[3]{L}}\) or \(cH = \dfrac{1}{\sqrt[3]{L}}\) SC1 24.5 |
Additional guidance
| \(k = 56\) in part (a) then \(H = \dfrac{56}{\sqrt[3]{2744}}\) and \(H = 4\) | M1A1ft |