Higher June 2025 Paper 2 Q24
24 A study suggests a student’s exam mark, \(m\), is directly proportional to the cube root of total revision time, \(t\) hours.
A student doubles their total revision time.
Work out the percentage increase in their exam mark. [3 marks]
| Answer | Mark | Comments |
|---|---|---|
| \(m = k\sqrt[3]{t}\) or \(km = \sqrt[3]{t}\) or \(m = k\sqrt[3]{2t}\) or \(km = \sqrt[3]{2t}\) | M1 | oe equation eg \(m = k \times \sqrt[3]{t}\) or \(\dfrac{m}{\sqrt[3]{t}} = k\) |
| Expression or calculation for \(\dfrac{m_2}{m_1}\ (\times 100)\) or \(\dfrac{m_2 - m_1}{m_1}\ (\times 100)\) | M1dep | \(m_1\) is the mark for time \(t\) \(m_2\) is the mark for time \(2t\) eg \(k\sqrt[3]{2t} \div k\sqrt[3]{t}\) or \(\dfrac{\sqrt[3]{8} - \sqrt[3]{4}}{\sqrt[3]{4}}\) |
| [25.9, 26] | A1 |
Additional guidance
| M1 may be awarded for correct work with no answer or incorrect answer, even if this is seen amongst multiple attempts | |
| A common approach used (i) Work out values of \(m_1\), \(k\) and \(t\) that satisfy \(m = k\sqrt[3]{t}\) (ii) Double their value of \(t\) and use their value of \(k\) to work out a value for \(m_2\) (iii) Write down a calculation for \(\dfrac{m_2}{m_1}\ (\times 100)\) or \(\dfrac{m_2 - m_1}{m_1}\ (\times 100)\) eg \(20 = 4 \times \sqrt[3]{125}\) (using \(m_1 = 20 \quad k = 4 \quad t = 125\)) \(m_2 = 4 \times \sqrt[3]{250}\) \(\dfrac{4 \times \sqrt[3]{250} - 20}{20}\) (award M1dep at this stage so 1st M1 is implied) 25.99 | M1M1dep A1 |
| For M1 accept use of different letters if intention is clear eg \(y = c\sqrt[3]{x}\) | M1 |
| \(m \propto \sqrt[3]{t}\) or \(m \propto k\sqrt[3]{t}\) etc with no further correct working | M0 |
| Writing square roots for cube roots must be recovered |