The sum and difference of a number with it sreciprocal are `113/56` and `15/56` respectively, the number is :

Options

A. `11/4`
B. `13/6`
C. `14/8`
D. `7/8`

Answer

To solve the problem, we need to find a number \( x \) such that the sum and difference of the number and its reciprocal satisfy the given conditions.

1. **Set up the equations**:
We know that:
\[ x + \frac{1}{x} = \frac{113}{56} \quad \text{(1)} \]
\[ x - \frac{1}{x} = \frac{15}{56} \quad \text{(2)} \]

2. **Add the two equations**:
By adding equations (1) and (2), we can eliminate the reciprocal:
\[ (x + \frac{1}{x}) + (x - \frac{1}{x}) = \frac{113}{56} + \frac{15}{56} \]
This simplifies to:
\[ 2x = \frac{113 + 15}{56} = \frac{128}{56} \]
Simplifying \( \frac{128}{56} \):
\[ 2x = \frac{64}{28} = \frac{16}{7} \]
Therefore:
\[ x = \frac{16}{14} = \frac{8}{7} \quad \text{(3)} \]

3. **Verify with the reciprocal**:
Now we need to find the reciprocal of \( x \):
\[ \frac{1}{x} = \frac{7}{8} \]

4. **Check the sum**:
Now, let's check the sum:
\[ x + \frac{1}{x} = \frac{8}{7} + \frac{7}{8} \]
Finding a common denominator (which is 56):
\[ = \frac{8 \times 8}{56} + \frac{7 \times 7}{56} = \frac{64}{56} + \frac{49}{56} = \frac{113}{56} \]
This matches equation (1).

5. **Check the difference**:
Now, let's check the difference:
\[ x - \frac{1}{x} = \frac{8}{7} - \frac{7}{8} \]
Finding a common denominator (which is 56):
\[ = \frac{8 \times 8}{56} - \frac{7 \times 7}{56} = \frac{64}{56} - \frac{49}{56} = \frac{15}{56} \]
This matches equation (2).

Thus, the number is \( \frac{8}{7} \).