If $$\frac{a}{b} = \frac{4}{5}$$   and $$\frac{b}{c} = \frac{{15}}{{16}}{\text{,}}$$   then $$\frac{{18{c^2} - 7{a^2}}}{{45{c^2} + 20{a^2}}}$$   is equal to?

Solution (By Examveda Team)

$$\eqalign{ & \frac{a}{b}{\text{ = }}\frac{4}{5}{\text{ and }}\frac{b}{c}{\text{ = }}\frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{b} \times \frac{b}{c}{\text{ = }}\frac{4}{5} \times \frac{{15}}{{16}} \cr & \Rightarrow \frac{a}{c} = \frac{3}{4} \cr & \therefore \frac{{18{c^2} - 7{a^2}}}{{45{c^2} + 20{a^2}}} \cr & = \frac{{{c^2}\left( {18 - 7\frac{{{a^2}}}{{{c^2}}}} \right)}}{{{c^2}\left( {45 + 20\frac{{{a^2}}}{{{c^2}}}} \right)}} \cr & = \frac{{18 - 7{{\left( {\frac{a}{c}} \right)}^2}}}{{45 + 20{{\left( {\frac{a}{c}} \right)}^2}}} \cr & = \frac{{18 - 7 \times \frac{9}{{16}}}}{{45 + 20 \times \frac{9}{{16}}}} \cr & = \frac{{18 - \frac{{63}}{{16}}}}{{45 + \frac{{45}}{4}}} \cr & = \frac{{225 \times 4}}{{16 \times 225}} \cr & = \frac{1}{4} \cr} $$