Question

HCF of $\frac{2}{3}$, $\frac{4}{5}$ and $\frac{6}{7}$ is

• A. $\frac{24}{105}$
• B. $\frac{2}{105}$
• C. $\frac{48}{105}$
• D. $\frac{1}{105}$

Answer 1

The correct option is B

$\frac{2}{105}$

Explanation for the correct option

H.C.F. of fraction$\mathbf{=}\frac{\mathbf{H}\mathbf{.}\mathbf{C}\mathbf{.}\mathbf{F}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{numerator}}{\mathbf{L}\mathbf{.}\mathbf{C}\mathbf{.}\mathbf{M}\mathbf{.}\mathbf{}\mathbf{of}\mathbf{}\mathbf{denominator}}$

HCF $\left(\frac{2}{3},\frac{4}{5},\frac{6}{7}\right)$$=\frac{H.C.Fof2,4,6}{L.C.M.of3,5,7}$

$$\begin{gathered} =\frac{2}{3\times 5\times 7} \\ =\frac{2}{105} \end{gathered}$$

Therefore, the H.C.F. $\left(\frac{2}{3},\frac{4}{5},\frac{6}{7}\right)$ is $\frac{2}{105}$

Hence, the correct option is B.