Six numbers a, b, c, d, e, f are such that ab = 1, bc = $\dfrac{1}{2}$, cd = 6, de = 2 and ef = $\dfrac{1}{2}$. What is the value of (ad : be : cf)?

1. 4 : 3 : 27

2. 6 : 1 : 9

3. 8 : 9 : 9

4. 72 : 1 : 9

Given,

ab = 1, bc = $\dfrac{1}{2}$, cd = 6, de = 2, ef = $\dfrac{1}{2}$.

$\Rightarrow ef = \dfrac{1}{2} \\[1em] \Rightarrow e = \dfrac{1}{2f} \\[1em] \Rightarrow de = 2 \\[1em] \Rightarrow d = \dfrac{2}{e} = \dfrac{2}{\dfrac{1}{2f}} = 4f \\[1em] \Rightarrow cd = 6 \\[1em] \Rightarrow c = \dfrac{6}{d} = \dfrac{6}{4f} = \dfrac{3}{2f} \\[1em] \Rightarrow bc = \dfrac{1}{2} \\[1em] \Rightarrow b = \dfrac{1}{2c} = \dfrac{1}{2 \times \dfrac{3}{2f}} = \dfrac{f}{3} \\[1em] \Rightarrow ab = 1 \\[1em] \Rightarrow a = \dfrac{1}{b} = \dfrac{1}{\dfrac{f}{3}} = \dfrac{3}{f}$

ad : be : cf = $\dfrac{3}{f} \times 4f : \dfrac{f}{3} \times \dfrac{1}{2f}: \dfrac{3}{2f} \times f$

= 12 : $\dfrac{1}{6} : \dfrac{3}{2}$

Since, L.C.M. of 2 and 6 is 6.

= 12 × 6 : $\dfrac{1}{6} \times 6 : \dfrac{3}{2} \times 6$

⇒ 72 : 1 : 9.

Hence, option 4 is the correct option.