If `A={x|x` is prime number and `x<30},` find the number of different rational numbers whose numerator and denominator belong to `Adot`

Options

A. 70
B. 80
C. 90
D. 100

Answer

To solve the problem, we need to find the number of different rational numbers whose numerator and denominator belong to the set A, where A consists of prime numbers less than 30.

### Step 1: Identify the Set A
First, we need to list all the prime numbers less than 30. The prime numbers are:
- 2
- 3
- 5
- 7
- 11
- 13
- 17
- 19
- 23
- 29

Thus, the set A can be defined as:
\[ A = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29\} \]

### Step 2: Count the Elements in Set A
Next, we count the number of elements in set A. There are 10 prime numbers in total:
\[ |A| = 10 \]

### Step 3: Form Rational Numbers
A rational number can be formed by taking any two elements from the set A as the numerator (p) and the denominator (q). The rational number can be represented as \( \frac{p}{q} \).

### Step 4: Calculate the Number of Ways to Choose p and q
To find the number of different rational numbers, we need to select 2 elements from the set A. The number of ways to choose 2 elements from a set of n elements is given by the combination formula:
\[ nCr = \frac{n!}{r!(n-r)!} \]
where \( n \) is the total number of elements in the set, and \( r \) is the number of elements to choose.

Here, \( n = 10 \) and \( r = 2 \):
\[ 10C2 = \frac{10!}{2!(10-2)!} = \frac{10!}{2! \cdot 8!} \]

### Step 5: Simplify the Combination
Now we simplify \( 10C2 \):
\[ 10C2 = \frac{10 \times 9}{2 \times 1} = \frac{90}{2} = 45 \]

### Step 6: Consider the Order of p and q
Since \( \frac{p}{q} \) and \( \frac{q}{p} \) are considered different rational numbers (unless \( p = q \), which is not possible here as all elements are distinct primes), we multiply the number of combinations by 2:
\[ \text{Total rational numbers} = 45 \times 2 = 90 \]

### Final Answer
Thus, the total number of different rational numbers whose numerator and denominator belong to set A is:
\[ \boxed{90} \]