A natural number is chosen at random from the first 100 natural numbers. The probability that `x+(100)/x > 50` is `1//10` b. `11//50` c. `11//20` d. none of these

Options

A. `1//10`
B. `11/50`
C. `11/20`
D. none of these

Explanation

To solve the problem, we need to find the probability that a randomly chosen natural number \( x \) from the first 100 natural numbers satisfies the inequality \( x + \frac{100}{x} > 50 \).

### Step-by-Step Solution:

1. **Start with the Inequality**:
\[ x + \frac{100}{x} > 50 \]

2. **Multiply through by \( x \)** (assuming \( x > 0 \)):
\[ x^2 + 100 > 50x \]

3. **Rearrange the Inequality**:
\[ x^2 - 50x + 100 > 0 \]

4. **Use the Quadratic Formula**:
To find the roots of the equation \( x^2 - 50x + 100 = 0 \), we can use the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1, b = -50, c = 100 \).

5. **Calculate the Discriminant**:
\[ b^2 - 4ac = (-50)^2 - 4 \cdot 1 \cdot 100 = 2500 - 400 = 2100 \]

6. **Find the Roots**:
\[ x = \frac{50 \pm \sqrt{2100}}{2} \]
Simplifying \( \sqrt{2100} \):
\[ \sqrt{2100} = \sqrt{100 \cdot 21} = 10\sqrt{21} \]
Thus,
\[ x = \frac{50 \pm 10\sqrt{21}}{2} = 25 \pm 5\sqrt{21} \]

7. **Approximate \( \sqrt{21} \)**:
\( \sqrt{21} \approx 4.58 \), so:
\[ 5\sqrt{21} \approx 22.9 \]
Therefore, the roots are approximately:
\[ x \approx 25 + 22.9 \quad \text{and} \quad x \approx 25 - 22.9 \]
This gives us:
\[ x \approx 47.9 \quad \text{and} \quad x \approx 2.1 \]

8. **Determine the Intervals**:
The inequality \( x^2 - 50x + 100 > 0 \) holds for:
\[ x < 2.1 \quad \text{or} \quad x > 47.9 \]
Since \( x \) must be a natural number, we have:
\[ x \leq 2 \quad \text{or} \quad x \geq 48 \]

9. **Count the Favorable Cases**:
- For \( x \leq 2 \): The values are \( 1, 2 \) (2 cases).
- For \( x \geq 48 \): The values are \( 48, 49, 50, \ldots, 100 \) (which gives \( 100 - 48 + 1 = 53 \) cases).

10. **Total Favorable Cases**:
\[ 2 + 53 = 55 \]

11. **Total Possible Cases**:
There are 100 natural numbers from 1 to 100.

12. **Calculate the Probability**:
\[ P = \frac{\text{Favorable Cases}}{\text{Total Cases}} = \frac{55}{100} = \frac{11}{20} \]

### Final Answer:
The probability that \( x + \frac{100}{x} > 50 \) is \( \frac{11}{20} \).