\(p : q = \frac{1}{3} : \frac{1}{2}\)

\(\frac{p}{q} = \frac{2}{3}\)

\(2q = 3p ... (1)\)

\(q : r = \frac{2}{5} : \frac{4}{7}\)

\(\frac{q}{r} = \frac{2}{5} \times \frac{7}{4} = \frac{7}{10}\)

\(10q = 7r ... (2)\)

Eliminating q, we have

(1) : \(2q = 3p \)

\(10q = 15p\)

\(\implies 15p = 7r\)

\(\therefore \frac{p}{r} = \frac{7}{15}\)

\(p : r = 7 : 15\)