Sum of the numbers is 128 and their HCF and LCM are 8 and 504 respectively. The sum of the reciprocal of those numbers will be

Options

A. `2//63`
B. `1//28`
C. `8//27`
D. `16//63`

Answer

To solve the problem, we need to find two numbers whose sum is 128, with a Highest Common Factor (HCF) of 8 and a Least Common Multiple (LCM) of 504. We will also find the sum of the reciprocals of these numbers.

### Step-by-Step Solution:

1. **Let the two numbers be represented as:**
\[ a = 8m \quad \text{and} \quad b = 8n \]
where \(m\) and \(n\) are co-prime integers (since the HCF is 8).

2. **Using the sum of the numbers:**
\[ a + b = 128 \]
Substituting the values of \(a\) and \(b\):
\[ 8m + 8n = 128 \]
Dividing the entire equation by 8:
\[ m + n = 16 \]

3. **Using the relationship between HCF, LCM, and the product of the numbers:**
The formula states that:
\[ \text{HCF} \times \text{LCM} = a \times b \]
Substituting the known values:
\[ 8 \times 504 = (8m) \times (8n) \]
Simplifying:
\[ 4032 = 64mn \]
Dividing both sides by 64:
\[ mn = \frac{4032}{64} = 63 \]

4. **Now we have two equations:**
\[ m + n = 16 \quad \text{(1)} \]
\[ mn = 63 \quad \text{(2)} \]

5. **We can find \(m\) and \(n\) by solving these equations.**
From equation (1), we can express \(n\) in terms of \(m\):
\[ n = 16 - m \]
Substituting into equation (2):
\[ m(16 - m) = 63 \]
Expanding and rearranging:
\[ 16m - m^2 = 63 \implies m^2 - 16m + 63 = 0 \]

6. **Now we can solve this quadratic equation using the quadratic formula:**
\[ m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{16 \pm \sqrt{(-16)^2 - 4 \cdot 1 \cdot 63}}{2 \cdot 1} \]
\[ = \frac{16 \pm \sqrt{256 - 252}}{2} = \frac{16 \pm \sqrt{4}}{2} = \frac{16 \pm 2}{2} \]
This gives us:
\[ m = \frac{18}{2} = 9 \quad \text{or} \quad m = \frac{14}{2} = 7 \]

7. **Thus, the values of \(m\) and \(n\) are:**
\[ m = 9, \quad n = 7 \quad \text{(or vice versa)} \]

8. **Now we can find the numbers \(a\) and \(b\):**
\[ a = 8m = 8 \times 9 = 72 \]
\[ b = 8n = 8 \times 7 = 56 \]

9. **Finally, we calculate the sum of the reciprocals:**
\[ \text{Sum of reciprocals} = \frac{1}{a} + \frac{1}{b} = \frac{1}{72} + \frac{1}{56} \]
To add these fractions, we need a common denominator:
\[ \text{LCM of 72 and 56} = 504 \]
Converting the fractions:
\[ \frac{1}{72} = \frac{7}{504}, \quad \frac{1}{56} = \frac{9}{504} \]
Adding them together:
\[ \frac{7 + 9}{504} = \frac{16}{504} = \frac{2}{63} \]

### Final Answer:
The sum of the reciprocals of the two numbers is:
\[ \frac{2}{63} \]