Question : H.C.F of $\frac{2}{3},\frac{4}{5}$ and $\frac{6}{7}$ is:
Option 1: $\frac{48}{105}$
Option 2: $\frac{2}{105}$
Option 3: $\frac{1}{105}$
Option 4: $\frac{24}{105}$
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Correct Answer: $\frac{2}{105}$
Solution : Given: $\frac{2}{3},\frac{4}{5},\frac{6}{7}$ Apply formula, HCF of the fraction = $\frac{\text{HCF of numerator}}{\text{LCM of denominator}}$ HCF of 2, 4, and 6 is 2. LCM of 3, 5, and 7 is 105. Hence, the correct answer is $\frac{2}{105}$.
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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: $\frac{1}{48}$
Question : If $6088x=7610$, then value of $x$ is:
Option 1: $\frac{5}{4}$
Option 2: $\frac{4}{5}$
Option 3: $\frac{6}{7}$
Option 4: $\frac{7}{6}$
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:
Option 1: 4
Option 2: 5
Option 3: 14
Option 4: 20
Question : The value of $1 \frac{3}{4}+1 \frac{5}{7} \div 2 \frac{3}{7} \times 2 \frac{3}{7}=$?
Option 1: $1 \frac{11}{28}$
Option 2: $2 \frac{13}{28}$
Option 3: $3 \frac{13}{28}$
Option 4: $4 \frac{23}{28}$
Question : Simplify the following expression. $\frac{7}{10} \div \frac{3}{7}$ of $\left(2 \frac{3}{10}+2 \frac{3}{5}\right)+\frac{1}{5} \div 1 \frac{2}{5}-\frac{2}{7}$
Option 1: $-\frac{4}{21}$
Option 2: $\frac{5}{21}$
Option 3: $\frac{4}{21}$
Option 4: $-\frac{5}{21}$
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