Question : The HCF of $\frac{1}{2}, \frac{3}{4}, \frac{5}{6},$ and $\frac{7}{8}$ is:
Option 1: $\frac{105}{2}$
Option 2: $\frac{1}{24}$
Option 3: $\frac{7}{24}$
Option 4: $\frac{1}{48}$
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Correct Answer: $\frac{1}{24}$
Solution : Given: $\frac{1}{2}, \frac{3}{4}, \frac{5}{6},$ and $\frac{7}{8}$ HCF of fractions = $\frac{\text{HCF of numerators}}{\text{LCM of denominators}}$ = $\frac{\text{HCF of 1,3,5,7}}{\text{LCM of 2,4,6,8}}$ = $\frac{1}{24}$ Hence, the correct answer is $\frac{1}{24}$.
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Question : H.C.F of $\frac{2}{3},\frac{4}{5}$ and $\frac{6}{7}$ is:
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Question : The value of $3 \frac{1}{5} \div 4 \frac{1}{2}$ of $5 \frac{1}{3}+\frac{1}{8} \div \frac{1}{2}$ of $\frac{1}{4}-\frac{1}{4}\left(\frac{1}{2} \div \frac{1}{8} \times \frac{1}{4}\right)$ is:
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Question : The arithmetic mean of the following numbers 1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7 is:
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