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Problems on H.C.F and L.C.M
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The H.C.F. of $$\frac{{9}}{{10}},$$ $$\frac{{12}}{{25}},$$ $$\frac{{18}}{{35}}$$ and $$\frac{{21}}{{40}}$$ is:

Answer & Solution
Correct Answer: Option C
$$\eqalign{ & {\text{Required}}\,{\text{H}}{\text{.C}}{\text{.F}}{\text{.}} \cr & = \frac{{{\text{H}}{\text{.C}}{\text{.F}}{\text{.}}\,{\text{of}}\,9,\,12,\,18,\,21}}{{{\text{L}}{\text{.C}}{\text{.M}}{\text{.}}\,{\text{of}}\,10,\,25,\,35,\,40}} \cr & = \frac{3}{{1400}} \cr} $$
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