A natural number 'n' is selected at random from the set of first 100 numbers. The probability that `n+(100)/(n)le 50` equal to

Options

A. `(9)/(10)`
B. `(39)/(50)`
C. `(19)/(20)`
D. `(9)/(20)`

Answer

To solve the problem, we need to find the probability that a randomly selected natural number \( n \) from the set of the first 100 natural numbers satisfies the condition \( n + \frac{100}{n} \leq 50 \).

### Step-by-step Solution:

1. **Set Up the Inequality**:
We start with the inequality:
\[ n + \frac{100}{n} \leq 50 \]

2. **Multiply Through by \( n \)**:
To eliminate the fraction, multiply both sides by \( n \) (note that \( n \) is a natural number and thus positive):
\[ n^2 + 100 \leq 50n \]

3. **Rearrange the Inequality**:
Rearranging gives us:
\[ n^2 - 50n + 100 \leq 0 \]

4. **Solve the Quadratic Inequality**:
We can find the roots of the quadratic equation \( n^2 - 50n + 100 = 0 \) using the quadratic formula:
\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = -50, c = 100 \):
\[ n = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]
\[ n = \frac{50 \pm \sqrt{2500 - 400}}{2} \]
\[ n = \frac{50 \pm \sqrt{2100}}{2} \]
\[ n = \frac{50 \pm 14.49}{2} \]
This gives us two roots:
\[ n_1 = \frac{64.49}{2} \approx 32.245, \quad n_2 = \frac{35.51}{2} \approx 17.755 \]

5. **Determine the Interval**:
The quadratic opens upwards (since the coefficient of \( n^2 \) is positive), so the inequality \( n^2 - 50n + 100 \leq 0 \) holds between the roots:
\[ 17.755 \leq n \leq 32.245 \]

6. **Identify Integer Solutions**:
The integer values of \( n \) that satisfy this inequality are:
\[ n = 18, 19, 20, \ldots, 32 \]
This gives us the integers from 18 to 32, inclusive.

7. **Count the Favorable Outcomes**:
The count of integers from 18 to 32 is:
\[ 32 - 18 + 1 = 15 \]

8. **Total Outcomes**:
The total number of natural numbers from 1 to 100 is 100.

9. **Calculate the Probability**:
The probability \( P \) that a randomly selected natural number \( n \) satisfies the condition is:
\[ P = \frac{\text{Number of favorable outcomes}}{\text{Total outcomes}} = \frac{15}{100} = \frac{3}{20} \]

### Final Answer:
The probability that \( n + \frac{100}{n} \leq 50 \) is:
\[ \frac{3}{20} \]