If a : (b + c) = 1 : 3 and c : (a + b) = 5 : 7, n b : (a + c) is equal to.

$$\eqalign{ & = \frac{a}{{b + c}} = \frac{1}{3} \cr & \Rightarrow a = \frac{{b + c}}{3} \cr & \Rightarrow \frac{c}{{a + b}} = \frac{5}{7} \cr & \Rightarrow 7c = 5a + 5b \cr & \Rightarrow 7c = \frac{{5\left( {b + c} \right)}}{3} + 5b \cr & \Rightarrow 7c - \frac{5}{3}c = 5b + \frac{5}{3}b \cr & \Rightarrow \frac{{16c}}{3} = \frac{{20b}}{3} \cr & \Rightarrow 16c = 20b \cr & \Rightarrow b = \frac{4}{5}c. \cr & a = \frac{{b + c}}{3} = \frac{{\frac{4}{5}c + c}}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{9c}}{5} \times \frac{1}{3} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{5}c. \cr & \therefore \frac{b}{{a + c}} = \frac{{\left( {\frac{4}{5}c} \right)}}{{\left( {\frac{3}{5}c + c} \right)}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{4c}}{5} \times \frac{5}{{8c}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{1}{2} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 1:2 \cr} $$