What is the HCF of \(\frac{7}{16}\), \(\frac{21}{32}\), \(\frac{49}{8}\) and \(\frac{63}{32}\)?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\frac{7}{32}\)
For fractions in lowest terms, HCF \(=\) HCF of the numerators \(\div\) LCM of the denominators.
Numerators 7, 21, 49, 63: each is a multiple of 7, and \(1, 3, 7, 9\) have no common factor, so HCF \(=7\)
Denominators 16, 32, 8, 32: 32 is a multiple of the others, so LCM \(=32\)
HCF of the fractions \(\displaystyle=\frac{7}{32}\)
Check: \(\displaystyle\frac{7}{16}\div\frac{7}{32}=2\), \(\displaystyle\frac{21}{32}\div\frac{7}{32}=3\), \(\displaystyle\frac{49}{8}\div\frac{7}{32}=28\), \(\displaystyle\frac{63}{32}\div\frac{7}{32}=9\), all whole numbers
The HCF is \(\frac{7}{32}\).