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