Find the HCF of $\frac{2}{3}, \frac{4}{6}, \frac{8}{27}$

Firstly, we will simplify $\frac{4}{6}$ which will become $\frac{2}{3}$. Hence, we are dealing with 2 unique fractions:

$\Rightarrow \frac{2}{3}, \frac{8}{27}$

HCF of Fractions $=\frac{\text{HCF of the Numerators}}{\text{LCM of the Denominators}}$

HCF of Numerators $=$ HCF of $2, 8 = 2$

LCM of Denominators $=$ LCM of $3, 27 = 27$

Hence, HCF of the fractions $=\frac{2}{27}$

Formula to remember:

$\text{HCF}\left( \dfrac{x}{y}, \dfrac{a}{b} \right) = \dfrac{\text{HCF}(x, a)}{\text{LCM}(y, b)}$

$\text{LCM}\left( \dfrac{x}{y}, \dfrac{a}{b} \right) = \dfrac{\text{LCM}(x, a)}{\text{HCF}(y, b)}$