The HCF of \(\frac{2}{3}\) , \(\frac{4}{5}\)&\(\frac{7}{8}\) is :
Answer & explanation
Correct answer: option 2
HCF of \(\frac{2}{3}\) , \(\frac{4}{5}\)&\(\frac{7}{8}\) = \(\frac{HCF \; of \; numerator}{LCM \; of \; denominator}\)
HCF of numerator= 1
LCM of denominator ( 3 × 5 × 8 ) = 120
HCF of \(\frac{2}{3}\) , \(\frac{4}{5}\)&\(\frac{7}{8}\) = \(\frac{1}{120}\)