If $$\frac{a}{b} = \frac{1}{3}\,,$$ $$\,\,\frac{b}{c} = 2\,,$$ $$\,\,\frac{c}{d} = \frac{1}{2}\,,$$ $$\,\,\frac{d}{e} = 3$$   and $$\,\frac{e}{f} = \frac{1}{4}\,,$$   then what is the value of $$\frac{{abc}}{{def}} = ?$$

Solution(By Examveda Team)

$$\eqalign{ & \frac{a}{b} = \frac{1}{3}\,\, \Rightarrow b = 3a; \cr & \frac{b}{c} = 2\,\,\, \Rightarrow c = \frac{b}{2} = \frac{{3a}}{2}; \cr & \frac{c}{d} = \frac{1}{2}\,\,\, \Rightarrow d = 2c = 2\left( {\frac{{3a}}{2}} \right) = 3a; \cr & \frac{d}{e} = 3\,\,\, \Rightarrow e = \frac{d}{3} = \left( {\frac{{3a}}{3}} \right) = a; \cr & \frac{e}{f} = \frac{1}{4}\,\,\, \Rightarrow f = 4e = 4a \cr & \therefore \frac{{abc}}{{def}} = \frac{{\left( a \right)\left( {3a} \right)\left( {\frac{{3a}}{2}} \right)}}{{\left( {3a} \right)\left( a \right)\left( {4a} \right)}} \cr & \frac{{abc}}{{def}} = \frac{9}{2}{a^3} \times \frac{1}{{12{a^3}}} \cr & \frac{{abc}}{{def}} = \frac{3}{8} \cr} $$