H.C.F. of frac{2}{3}, frac{4}{5} and frac{6}{ | vedclass

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(B) To find the $H.C.F.$ of fractions,we use the formula:$H.C.F. = \frac{H.C.F. 	ext{ of numerators}}{L.C.M. 	ext{ of denominators}}$Given fractions

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$\frac{2}{105}$

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(B) To find the $H.C.F.$ of fractions,we use the formula:$H.C.F. = \frac{H.C.F. 	ext{ of numerators}}{L.C.M. 	ext{ of denominators}}$Given fractions: $\frac{2}{3}, \frac{4}{5}, \frac{6}{7}$Step $1$: Find the $H.C.F.$ of the numerators $(2, 4, 6)$.The factors of $2$ are $1, 2$.The factors of $4$ are $1, 2, 4$.The factors of $6$ are $1, 2, 3, 6$.The greatest common factor is $2$.Step $2$: Find the $L.C.M.$ of the denominators $(3, 5, 7)$.Since $3, 5,$ and $7$ are all prime numbers,their $L.C.M.$ is their product:$3 	imes 5 	imes 7 = 105$.Step $3$: Combine the results.$H.C.F. = \frac{2}{105}$.

$H.C.F.$ of $\frac{2}{3}, \frac{4}{5}$ and $\frac{6}{7}$ is

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