Equal Equations

$\begin{aligned} 3A & = 6B ........(i) \\ 3B & = 14C ........(ii) \\ 7C & = 15D ........(iii) \\ \end{aligned}$

Dividinf $(i)$ by $(ii)$

$\begin{aligned} \frac{3A}{3B} & = \frac{6B}{14C} \\ \Rightarrow \frac AB & = \frac{3B}{7C} \\ \Rightarrow A & = \frac{3B^2}{15D} ~~~~~~~~[7C = 15D ~ \text{from equation} ~ (iii)] \\ \Rightarrow A & = \dfrac{ \left(\frac A2 \right)^2}{5D} ~~~~[B = \frac A2 ~ \text{from equation} ~ (i)] \\ \Rightarrow 5AD & = \frac{A^2}4 \\ \Rightarrow 20D & = \dfrac{A^2}A \\ \implies A & = 20D\\ \end{aligned}$

Hence, the answer is 20.

Dividing $3A = 6B$ and $14C = 3B$ we get, $\dfrac{3A}{14C} = \dfrac{6B}{3B} \\ 3A = 28C$

On dividing $3A = 28C$ and $15D = 7C$ we get, $\dfrac{3A}{15D} = \dfrac{28C}{7C} \\ \dfrac{A}{5D} = 4 \\ \boxed{A = 20D}$

Double 15D=7C to get 30D=14C=3B, then double again: 60D=28C=6B=3A, so 60D=3A, or 20D=A. And so, the answer is 20.

7C = 15D which means 30D = 14C also 14C = 3B which means 3B = 30D so 6B = 60D and 3A = 6B which means 3A = 60D and that means 1A = 20D