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
\(\frac{T_1}{T_2}\) = \(\frac{9 \times 3}{5 x 5}\) = \(\frac{27}{5}\)
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