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}
Any 10 elements of the Set A are transferred to the Set B, the the average of Set B :

Options

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

Answer

To solve the problem step by step, we will first identify the elements of each set and then calculate the average of Set B before and after transferring 10 elements from Set A.

### Step 1: Identify the Elements of Each Set
- **Set A** (Prime numbers up to 100):
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}

- **Set B** (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}

- **Set C** (Non-prime odd numbers up to 100):
C = {1, 9, 15, 21, 25, 27, 33, 35, 39, 45, 49, 51, 55, 57, 63, 65, 69, 77, 81, 83, 85, 87, 91, 93, 95, 99}

### Step 2: Calculate the Average of Set B
To find the average of Set B, we need to calculate the sum of its elements and divide by the number of elements.

- **Number of elements in Set B**:
There are 49 elements in Set B.

- **Sum of elements in Set B**:
We can use the formula for the sum of an arithmetic series:
\[ S_n = \frac{n}{2} \times (2a + (n-1)d) \]
where:
- \( n = 49 \) (number of terms)
- \( a = 4 \) (first term)
- \( d = 2 \) (common difference)

Plugging in the values:
\[ S_{49} = \frac{49}{2} \times (2 \times 4 + (49 - 1) \times 2) = \frac{49}{2} \times (8 + 96) = \frac{49}{2} \times 104 = 49 \times 52 = 2548 \]

- **Average of Set B**:
\[ \text{Average} = \frac{\text{Sum of elements}}{\text{Number of elements}} = \frac{2548}{49} \approx 52 \]

### Step 3: Transfer 10 Elements from Set A to Set B
Assuming we transfer the smallest 10 prime numbers from Set A to Set B:
- Transferred Elements: {2, 3, 5, 7, 11, 13, 17, 19, 23, 29}

### Step 4: Calculate the New Average of Set B
- **New Number of Elements in Set B**:
\( 49 + 10 = 59 \)

- **New Sum of Elements in Set B**:
Original sum of Set B is 2548. The sum of transferred elements is:
\[ 2 + 3 + 5 + 7 + 11 + 13 + 17 + 19 + 23 + 29 = 129 \]
New sum:
\[ 2548 + 129 = 2677 \]

- **New Average of Set B**:
\[ \text{New Average} = \frac{2677}{59} \approx 45.37 \]

### Conclusion
The average of Set B decreases from approximately 52 to approximately 45.37 after transferring 10 elements from Set A.