A natural number x is chosen at random from the first 100 natural numbers. Then the probability, for the equation `x+(100)/(x)gt50` is

Options

A. `(1)/(20)`
B. `(11)/(20)`
C. `(1)/(3)`
D. `(3)/(20)`

Answer

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:

**Step 1: Rearranging the Inequality**

Start by rearranging the inequality:

\[ x + \frac{100}{x} > 50 \]

Multiply both sides by \( x \) (since \( x \) is a natural number and thus positive):

\[ x^2 + 100 > 50x \]

**Step 2: Forming a Quadratic Equation**

Rearranging gives us:

\[ x^2 - 50x + 100 > 0 \]

**Step 3: Finding the Roots of the Quadratic Equation**

To find the roots of the quadratic equation \( x^2 - 50x + 100 = 0 \), we use the quadratic formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -50, c = 100 \):

\[ x = \frac{50 \pm \sqrt{(-50)^2 - 4 \cdot 1 \cdot 100}}{2 \cdot 1} \]

Calculating the discriminant:

\[ b^2 - 4ac = 2500 - 400 = 2100 \]

So the roots are:

\[ x = \frac{50 \pm \sqrt{2100}}{2} \]

**Step 4: Simplifying the Roots**

Calculating \( \sqrt{2100} \):

\[ \sqrt{2100} = \sqrt{100 \cdot 21} = 10\sqrt{21} \]

Thus, the roots are:

\[ x = \frac{50 \pm 10\sqrt{21}}{2} = 25 \pm 5\sqrt{21} \]

**Step 5: Approximate the Roots**

Calculating \( \sqrt{21} \approx 4.58 \):

\[ 5\sqrt{21} \approx 22.9 \]

So the roots are approximately:

\[ x_1 \approx 25 + 22.9 \approx 47.9 \quad \text{and} \quad x_2 \approx 25 - 22.9 \approx 2.1 \]

**Step 6: Analyzing the Quadratic Inequality**

The quadratic \( x^2 - 50x + 100 \) opens upwards (since the coefficient of \( x^2 \) is positive). Therefore, the inequality \( x^2 - 50x + 100 > 0 \) holds true outside the interval defined by the roots:

\[ x < 2.1 \quad \text{or} \quad x > 47.9 \]

**Step 7: Finding Valid Natural Numbers**

Since \( x \) must be a natural number, the valid values for \( x \) are:

- From 1 to 2 (which gives 1 and 2)
- From 48 to 100 (which gives 48, 49, 50, ..., 100)

Counting these:

1. From 1 to 2: 2 values (1, 2)
2. From 48 to 100: \( 100 - 48 + 1 = 53 \) values (48 through 100)

Total favorable outcomes:

\[ 2 + 53 = 55 \]

**Step 8: Calculating the Probability**

The total number of natural numbers from 1 to 100 is 100. Thus, the probability \( P \) is:

\[ P = \frac{\text{Favorable outcomes}}{\text{Total outcomes}} = \frac{55}{100} = \frac{11}{20} \]

### Final Answer:

The probability is \( \frac{11}{20} \).

---