A) \[\frac{1}{\sqrt{3}}\]
B) \[\sqrt{3}\]
C) 1
D) \[\frac{1}{2}\]
Correct Answer: C
Solution :
\[L.C.M.\] \[\frac{6}{14}and\frac{2}{7}\] \[\Rightarrow \] \[\frac{L.C.M.(6,2)}{H.C.F.(14,7)}=\frac{6}{7}\] \[\text{7}\times \text{13}+\text{13}=\text{1}0\text{4}=\text{23}\times \text{13}\] \[\therefore \] \[\text{7}\times \text{13}+\text{13}\] \[\therefore \] \[\text{224}=\text{12}0\times \text{1}+\text{1}0\text{4}\] \[\text{12}0=\text{1}0\text{4}\times \text{1}+\text{16}\]
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