[\dfrac{3}{1400}] The H.C.F of two 9/10,12/25,18/35 and 21/40 is:

For Competitive Exams

The H.C.F of two 9/10,12/25,18/35 and 21/40 is:

$ \dfrac{3}{5} $

$ \dfrac{252}{5} $

$ \dfrac{3}{1400} $

$ \dfrac{63}{700} $

Explanation:

Required H.C.F. =$ \dfrac{H.C.F. of 9, 12, 18, 21}{L.C.M. of 10, 25, 35, 40} $=$ \dfrac{3}{1400} $

Additional Questions

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The H.C.F of two 9/10,12/25,18/35 and 21/40 is:	Answer

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