H.C.F of `(2)/(3), (4)/(5) and (6)/(7)` is

Options

A. `(48)/(105)`
B. `(2)/(105)`
C. `(1)/(105)`
D. `(24)/( 105)`

Answer

To find the H.C.F (Highest Common Factor) of the fractions \(\frac{2}{3}\), \(\frac{4}{5}\), and \(\frac{6}{7}\), we can follow these steps:

### Step 1: Identify the Numerators and Denominators
The numerators of the fractions are:
- \(2\) (from \(\frac{2}{3}\))
- \(4\) (from \(\frac{4}{5}\))
- \(6\) (from \(\frac{6}{7}\))

The denominators of the fractions are:
- \(3\) (from \(\frac{2}{3}\))
- \(5\) (from \(\frac{4}{5}\))
- \(7\) (from \(\frac{6}{7}\))

### Step 2: Calculate the H.C.F of the Numerators
Now, we need to find the H.C.F of the numerators \(2\), \(4\), and \(6\).

- The factors of \(2\) are: \(1, 2\)
- The factors of \(4\) are: \(1, 2, 4\)
- The factors of \(6\) are: \(1, 2, 3, 6\)

The common factors are \(1\) and \(2\). Therefore, the H.C.F of \(2\), \(4\), and \(6\) is \(2\).

### Step 3: Calculate the L.C.M of the Denominators
Next, we calculate the L.C.M of the denominators \(3\), \(5\), and \(7\).

- The multiples of \(3\) are: \(3, 6, 9, 12, 15, 18, 21, 24, 27, 30, ...\)
- The multiples of \(5\) are: \(5, 10, 15, 20, 25, 30, ...\)
- The multiples of \(7\) are: \(7, 14, 21, 28, 35, 42, 49, 56, 63, 70, ...\)

The least common multiple of \(3\), \(5\), and \(7\) is \(105\) (the smallest number that is a multiple of all three).

### Step 4: Form the H.C.F of the Fractions
The H.C.F of the fractions is given by the formula:
\[ \text{H.C.F} = \frac{\text{H.C.F of Numerators}}{\text{L.C.M of Denominators}} \]
Substituting the values we found:
\[ \text{H.C.F} = \frac{2}{105} \]

### Final Answer
Thus, the H.C.F of \(\frac{2}{3}\), \(\frac{4}{5}\), and \(\frac{6}{7}\) is \(\frac{2}{105}\).

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