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Find the H.C.F. of \(\dfrac{2}{3}, \dfrac{4}{6}, \dfrac{8}{27}\)

  • A.\(\dfrac{2}{27}\)
  • B.\(\dfrac{8}{3}\)
  • C.\(\dfrac{2}{3}\)
  • D.\(\dfrac{8}{27}\)

Answer: A

Formula,
\(H.C.F. = \dfrac{H.C.F. of Numerators}{L.C.M. of Denominators}\)
So,
\(\dfrac{H.C.F. of (2,4,8) }{L.C.M. of (3,6,27)} \\~\\=\dfrac{2}{27}\)

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