If `f (x) + f (1 - x)` is equal to 10 for all real numbers x then `f ((1)/(100)) + f ((2)/(100)) + f ((3)/( 100))+...+f ((99)/(100))` equals

Answer

To solve the problem, we need to find the value of the sum \( f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + f\left(\frac{3}{100}\right) + \ldots + f\left(\frac{99}{100}\right) \) given that \( f(x) + f(1 - x) = 10 \) for all real numbers \( x \).

### Step-by-step Solution:

1. **Understanding the Given Condition**:
We know that for any real number \( x \), the function satisfies:
\[ f(x) + f(1 - x) = 10 \]

2. **Substituting Specific Values**:
Let's substitute \( x = \frac{1}{100} \):
\[ f\left(\frac{1}{100}\right) + f\left(1 - \frac{1}{100}\right) = f\left(\frac{1}{100}\right) + f\left(\frac{99}{100}\right) = 10 \]

3. **Continuing the Pattern**:
Now, substitute \( x = \frac{2}{100} \):
\[ f\left(\frac{2}{100}\right) + f\left(1 - \frac{2}{100}\right) = f\left(\frac{2}{100}\right) + f\left(\frac{98}{100}\right) = 10 \]

4. **Continuing Until \( x = \frac{49}{100} \)**:
We can continue this process up to \( x = \frac{49}{100} \):
\[ f\left(\frac{49}{100}\right) + f\left(\frac{51}{100}\right) = 10 \]

5. **Pairing the Terms**:
Notice that we can pair the terms in the sum:
\[ \begin{align*} & \left( f\left(\frac{1}{100}\right) + f\left(\frac{99}{100}\right) \right) + \left( f\left(\frac{2}{100}\right) + f\left(\frac{98}{100}\right) \right) + \ldots + \left( f\left(\frac{49}{100}\right) + f\left(\frac{51}{100}\right) \right) + f\left(\frac{50}{100}\right) \end{align*} \]

6. **Calculating the Number of Pairs**:
There are 49 pairs, each summing to 10:
\[ 49 \times 10 = 490 \]

7. **Adding the Middle Term**:
The middle term is \( f\left(\frac{50}{100}\right) \):
\[ f\left(\frac{50}{100}\right) + f\left(\frac{50}{100}\right) = 10 \implies 2f\left(\frac{50}{100}\right) = 10 \implies f\left(\frac{50}{100}\right) = 5 \]

8. **Final Calculation**:
Now, we add the contributions from the pairs and the middle term:
\[ 490 + 5 = 495 \]

Thus, the value of \( f\left(\frac{1}{100}\right) + f\left(\frac{2}{100}\right) + \ldots + f\left(\frac{99}{100}\right) \) is \( \boxed{495} \).