\(p : q = \frac{1}{3} : \frac{1}{2}\) \(\frac{p}{q} = \frac{2}{3}\) \(2q = 3p ... (1)\) \(q : r = \frac{2}{5} : \frac{4}{7}\) \(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\) \(10q = 7r ... (2)\) Eliminating q, we have (1) : \(2q = 3p \) \(10q = 15p\) \(\implies 15p = 7r\) \(\therefore \frac{p}{r} = \frac{7}{15}\) \(p : r = 7 : 15\)