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Question 39 : Find the HCF of \\frac{2}{3}, \frac{4}{6}, \frac{8}{27}\\)

- \\frac{2}{27}\\)
- \\frac{8}{3}\\)
- \\frac{2}{3}\\)
- \\frac{8}{27}\\)

\\frac{2}{27}\\)

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HCF of fractions = \\frac{HCF of numerators}{LCM of denominators}\\)

But we'll have to write the fractions in the simplest form first. So, \\frac{4}{6} = \frac{2}{3}\\)

HCF of (\\frac{2}{3}, \frac{4}{6}, \frac{8}{27}\\) ) = HCF of (\\frac{2}{3}, \frac{2}{3}, \frac{8}{27}\\) ) = \\frac{HCF(2,2,8))}{LCM(3,3,27))}\\) = \\frac{2}{27}\\).

Therefore, HCF of (\\frac{2}{3}, \frac{4}{6}, \frac{8}{27}\\)) = \\frac{2}{27}\\).

The question is **"Find the HCF of \\frac{2}{3}, \frac{4}{6}, \frac{8}{27}\\)" **

Choice A is the correct answer

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