Question

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Answer 1

$\it \dfrac{\overline{ab}}{\overline{cd}}=\overline{c,15}>c\ \Rightarrow \overline{ab}>c\cdot\overline{cd} \Rightarrow c\leq3\ \ \ \ \ \ (1)\\ \\ \\ \dfrac{\overline{ab}}{\overline{cd}}=\overline{c,15} \Rightarrow \dfrac{\overline{ab}}{\overline{cd}}=c+\dfrac{\ 15^{(5}}{100} \Rightarrow \dfrac{\overline{ab}}{\overline{cd}}= c+\dfrac{3}{20} \Rightarrow \overline{ab}=(c+\dfrac{3}{20})\overline{cd} \Rightarrow\\ \\ \\ \Rightarrow \overline{ab} =c\cdot\overline{cd}+\dfrac{3}{20}\cdot\overline{cd}\ \ \ \ \ \ (2)$

$\it \overline{ab} \in\mathbb{N}\ \stackrel{(2)}{\Longrightarrow}\ \dfrac{3}{20}\cdot\overline{cd}\in\mathbb{N}\ \ \ \ \ (3)\\ \\ \\ (1),\ (3) \Rightarrow \overline{cd}=20\ \ \ \ \ \ \ \ \ \ (4)$

$\it (2),\ (4) \Rightarrow \overline{ab}=2\cdot20+3\Rightarrow \overline{ab}=43\\ \\ Deci,\ \ \dfrac{\overline{ab}}{\overline{cd}}=\dfrac{43}{20}$

Answer 2

$\it \overline{c,15}=c+\dfrac{\ 15^{(5}}{100}=c+\dfrac{3}{20}=\dfrac{20c+3}{20}\ \ ireductibil\breve a\\ \\ \\ \dfrac{\overline{ab}}{\overline{cd}}=\dfrac{20c+3}{20} \Rightarrow \overline{cd}=20 \Rightarrow c=2 \Rightarrow \overline{ab}=20\cdot2+3=43\\ \\ \\ Prin\ \ urmare,\ \ \dfrac{\overline{ab}}{\overline{cd}}=\dfrac{43}{20}$