Higher June 2025 Paper 1 Q11
11 \(T\) is inversely proportional to \(w\).
\(w\) is directly proportional to the cube root of \(d\).
When \(w = 6\), \(T = 20\)
When \(w = 1\), \(d = 8\)
Find the value of \(d\) when \(T = 48\) (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 125 | P1 | for setting up an equation with a constant term, eg \(T = \dfrac{k}{w}\) or \(w = K\sqrt[3]{d}\) |
| P1 | for a process to substitute values in one equation, eg \(20 = \dfrac{k}{6}\) or \(k = 120\) or \(1 = K\sqrt[3]{8}\) or \(K = \dfrac{1}{2}\) oe | |
| P1 | (dep P2) for combining the two equations ft their values of \(k\) and \(K\), eg \(T = \dfrac{\text{``}120\text{''}}{\text{``}\frac{1}{2}\text{''}\sqrt[3]{d}}\) oe OR for a correct process to find the value of \(w\) when \(T = 48\), eg \(w = \dfrac{\text{``}120\text{''}}{48}\) (= 2.5 oe) | |
| P1 | for substitution into their formula, eg \(48 = \dfrac{\text{``}120\text{''}}{\text{``}\frac{1}{2}\text{''}\sqrt[3]{d}}\) OR for substitution of found value of \(w\) in the equation for \(d\), eg \(\text{``}2.5\text{''} = \text{``}\dfrac{1}{2}\text{''}\sqrt[3]{d}\) | |
| A1 | cao |
Additional guidance
Condone the use of ‘\(\alpha\)’ instead of ‘=’ for the first two P marks
Equation can be implied by correct substitution
\(T = \dfrac{120}{w} \qquad w = \dfrac{1}{2}\sqrt[3]{d}\)