If p : q = \(\frac{2}{3}\) : \(\frac{5}{6}\), then the sum S1 of ratio = \(\frac{2}{3}\) + \(\frac{5}{6}\) = \(\frac{9}{6}\) If q : r = \(\frac{3}{4}\) : \(\frac{1}{2}\), then the sum S2 of ratio = \(\frac{3}{4}\) + \(\frac{1}{2}\) = \(\frac{5}{4}\) Let p + q = T1, then q = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))T1 = \(\frac{5}{9}\)T1 Again, let q + r = T2, then q = (\(\frac{3}{4} \div \frac{5}{4}\))T2 = (\(\frac{3}{4} \times \frac{4}{5}\))T2 = \(\frac{3}{5}\)T2 Using q = q \(\frac{5}{9}\)T1 = \(\frac{3}{5}\)T2 5 x 5T1 = 9 x 3T2 \(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{25}\) Giving that, T1 = 27 and T2 = 25 P = (\(\frac{2}{3} \div S_1\))T1 = (\(\frac{2}{3} \div \frac{9}{6}\))T1 = (\(\frac{2}{3} \times \frac{6}{9}\))27 = 12 q = (\(\frac{5}{6} \div S_1\))T1 = (\(\frac{5}{6} \div \frac{9}{6}\))T1 = (\(\frac{5}{6} \times \frac{6}{9}\))27 = 15 and r = (\(\frac{1}{2} \div S_2\))T2 = (\(\frac{1}{2} \div \frac{5}{4}\))T2 = (\(\frac{1}{2} \times \frac{4}{5}\))25 = 10 Hence p : q : r = 12: 15 : 10