Answer: Option A
$$\eqalign{
& \frac{a}{b}{\text{ = }}\frac{2}{3}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{4}{5}\,\,\left( {{\text{Given}}} \right) \cr
& or\,\frac{c}{b} = \frac{5}{4} \cr
& \frac{{a + b}}{{b + c}} \cr
& = \frac{{b\left( {\frac{a}{b} + 1} \right)}}{{b\left( {\frac{c}{b} + 1} \right)}} \cr
& = \frac{{\frac{a}{b} + 1}}{{\frac{c}{b} + 1}} \cr
& = \frac{{\left( {\frac{2}{3} + 1} \right)}}{{\left( {\frac{5}{4} + 1} \right)}} \cr
& = \frac{{\frac{2 + 3}{3}}}{{\frac{{5 + 4}}{4}}} \cr
& = \frac{{5 \times 4}}{{3 \times 9}} \cr
& = \frac{{20}}{{27}} \cr
& \therefore \frac{{a + b}}{{b + c}} = \frac{{20}}{{27}} \cr
& {\bf{Alternate:}} \cr
& a{\text{ }}\,\,\,{\text{ }}\,{\text{ }}\,\,\,\,{\text{ }}b{\text{ }}\,\,\,\,{\text{ }}\,\,\,\,{\text{ }}\,\,\,{\text{ }}c \cr
& {2_{ \times \left( 4 \right)}}\,\,\,\,\,{\text{ }}{3_{ \times \left( 4 \right)}} \cr
& \,{\text{ }}\,{\text{ }}\,\,\,\,\,\,\,\,\,\,\,\,\,{4_{ \times \left( 3 \right)}}{\text{ }}\,\,\,\,\,\,{5_{ \times \left( 3 \right)}} \cr
& \overline {\underline {8{\text{ }}\,\,\,\,\,\,\,\,\,\,\,{\text{ }}12{\text{ }}\,\,\,\,\,\,\,\,\,\,{\text{ }}15\,\,\,\,} } \cr
& \therefore \frac{{a + b}}{{b + c}} = \frac{{8 + 12}}{{12 + 15}} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{20}}{{27}} \cr} $$