Question 39

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

Solution

The HCF of $$\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{27}$$

$$\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{8}{27}$$

HCF of fractions = HCF of Numerator/ LCM of Denominator

HCF of $$\dfrac{2}{3}, \dfrac{2}{3}, \dfrac{8}{27}$$ = HCF of (2,2,8) / LCM of (3,3,27) = $$\dfrac{2}{27}$$

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