The HCF of \(\frac{2}{3}\), \(\frac{2}{9}\), \(\frac{64}{81}\) and…
2024
The HCF of \(\frac{2}{3}\), \(\frac{2}{9}\), \(\frac{64}{81}\) and \(\frac{10}{27}\) is:
Answer: C. \(\frac{2}{81}\) — ConceptFor fractions written in lowest terms, HCF of the fractions = HCF of their numerators ÷ LCM of their denominators. Reducing first is essential because…
- A.
\(\frac{18}{81}\)
- B.
\(\frac{2}{27}\)
- C.
\(\frac{2}{81}\)
- D.
\(\frac{2}{3}\)
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Correct answer: C
Concept
For fractions written in lowest terms, HCF of the fractions = HCF of their numerators ÷ LCM of their denominators.
Reducing first is essential because a common factor hidden in an unreduced numerator and denominator can distort either calculation.
Application
The fractions are already in lowest terms: \(\frac{2}{3}\), \(\frac{2}{9}\), \(\frac{64}{81}\), and \(\frac{10}{27}\).
Find the numerator HCF: HCF(2, 2, 64, 10) = 2.
Find the denominator LCM: LCM(3, 9, 81, 27) = 81, because each denominator divides 81.
Apply the fraction rule: \(\frac{\operatorname{HCF}(2,2,64,10)}{\operatorname{LCM}(3,9,81,27)}=\frac{2}{81}\).
Cross-check
Dividing the four given fractions by \(\frac{2}{81}\) gives 27, 9, 32, and 15, all integers. No larger common fraction can work because the numerator HCF is already 2 while the denominator must accommodate 81.
Therefore, the HCF is \(\frac{2}{81}\).