There are 3 sets of natural numbers 1 ot 100. Set A contains all the natural numbers which are prime, upto 100. Set B contains all the non-prime even natural numbers upto 100. Set C contains all the non-prime odd natural numbers upto 100 i.e.,
A = {2,3,5,7,11,...89, 97}
B = {4,6,8,10,12,...98,100}
C = {1,9,15,21,25,27,33,95,99}
If an element less than 50 belongs to Set A is transferred to Set B, then the average of Set B :

Options

A. a. increases
B. b. decreases
C. c. remains
D. d. can't say

Answer

To solve the problem, we will follow these steps:

### Step 1: Identify the elements of Set A (Prime numbers less than 100)
Set A contains all prime numbers up to 100. The prime numbers are:
\[ A = \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97\} \]
The prime numbers less than 50 in Set A are:
\[ \{2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47\} \]

### Step 2: Count the number of elements in Set A
The number of prime numbers less than 50 is 15.

### Step 3: Calculate the sum of elements in Set A (less than 50)
The sum of the prime numbers less than 50 is:
\[ 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 + 31 + 37 + 41 + 43 + 47 = 328 \]

### Step 4: Identify the elements of Set B (Non-prime even numbers up to 100)
Set B contains all non-prime even numbers up to 100:
\[ B = \{4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100\} \]
The total number of elements in Set B is 49.

### Step 5: Calculate the sum of elements in Set B
The sum of the even numbers from 4 to 100 can be calculated using the formula for the sum of an arithmetic series:
\[ \text{Sum} = \frac{n}{2} \times (a + l) \]
Where:
- \( n \) = number of terms = 49
- \( a \) = first term = 4
- \( l \) = last term = 100

Calculating:
\[ \text{Sum} = \frac{49}{2} \times (4 + 100) = \frac{49}{2} \times 104 = 49 \times 52 = 2548 \]

### Step 6: Calculate the average of Set B
The average of Set B before transferring any elements is:
\[ \text{Average of B} = \frac{\text{Sum of B}}{\text{Number of elements in B}} = \frac{2548}{49} \approx 52 \]

### Step 7: Transfer elements from Set A to Set B
We transfer the 15 prime numbers less than 50 (sum = 328) to Set B. The new sum of Set B becomes:
\[ \text{New Sum of B} = 2548 + 328 = 2876 \]
The new number of elements in Set B becomes:
\[ \text{New Number of elements in B} = 49 + 15 = 64 \]

### Step 8: Calculate the new average of Set B
The new average of Set B is:
\[ \text{New Average of B} = \frac{\text{New Sum of B}}{\text{New Number of elements in B}} = \frac{2876}{64} \approx 44.89 \]

### Conclusion
The average of Set B after transferring the elements from Set A is approximately 44.89.

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