- A$\frac{1}{6}$ and $\frac{2}{6}$
- B$\frac{1}{2}$ and $\frac{2}{1}$
- ✓$\frac{5}{6}$ and $\frac{7}{6}$
- D$\frac{2}{3}$ and $\frac{4}{3}$
We have,
$\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{5}{3}=\frac{5\times2}{3\times2}=\frac{10}{6}$
And, $\frac{1}{2}=\frac{1\times3}{2\times3}=\frac{3}{6}$ and $\frac{2}{1}=\frac{2\times6}{1\times6}=\frac{12}{6}$
Also, $\frac{2}{3}=\frac{2\times2}{3\times2}=\frac{4}{6}$ and $\frac{4}{3}=\frac{4\times2}{3\times2}=\frac{8}{6}$
Since, $\frac{1}{6}<\frac{2}{6}<\frac{3}{6}\Big(\frac{1}{2}\Big)<\frac{4}{6}\Big(=\frac{2}{3}\Big)<\frac{5}{6}<\frac{7}{6}\\<\frac{8}{6}\Big(=\frac{4}{3}\Big)<\frac{10}{6}\Big(=\frac{5}{3}\Big)<\frac{12}{6}\Big(=\frac{2}{1}\Big)$
So, the two rational numbers between $\frac{2}{3}$ and $\frac{5}{3}$ are $\frac{5}{6}$ and $\frac{7}{6}.$
Hence, the correct opion is $(c).$
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