MCQ
The fraction $\frac{84}{98}$ in its lowest terms is:
- A$\frac{42}{49}$
- B$\frac{12}{14}$
- ✓$\frac{6}{7}$
- D$\frac{3}{7}$
Factors of $84: 1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84$
Factors of $98: 1, 2, 7, 14, 49, 98$
Common factors of $84$ and $98: 1, 2, 14$
$\therefore HCF$ of $84$ and $98 = 14$
Now,
$\frac{84}{98}=\frac{84\div14}{98\div14}=\frac{6}{7} ($Dividing numerator and senominator by the $HCF$ of $84$ and $98$ i.e., $14)$
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