In the given number $7549823184$, the position of the first and tenth digits are interchanged, similarly, the position of the second and ninth digits are interchanged, and so on till the end. Then, what is the sum of the digits which are placed at odd numbered positions from right to left?

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**Solution:**

1. The original number is: $7 5 4 9 8 2 3 1 8 4$
   - Positions from left to right: 1 to 10

2. According to the problem, the digits at positions are swapped as follows:
   - 1st and 10th digits swapped
   - 2nd and 9th digits swapped
   - 3rd and 8th digits swapped
   - 4th and 7th digits swapped
   - 5th and 6th digits swapped

3. Perform the swaps:
   - Swap 1st (7) and 10th (4): positions 1 and 10 become 4 and 7
   - Swap 2nd (5) and 9th (8): positions 2 and 9 become 8 and 5
   - Swap 3rd (4) and 8th (1): positions 3 and 8 become 1 and 4
   - Swap 4th (9) and 7th (3): positions 4 and 7 become 3 and 9
   - Swap 5th (8) and 6th (2): positions 5 and 6 become 2 and 8

4. After swapping, the new number is:
   - Position 1: 4
   - Position 2: 8
   - Position 3: 1
   - Position 4: 3
   - Position 5: 2
   - Position 6: 8
   - Position 7: 9
   - Position 8: 4
   - Position 9: 5
   - Position 10: 7

So, the new number is $4 8 1 3 2 8 9 4 5 7$

5. Now, find the digits at odd numbered positions from right to left.
   - Positions from right to left:
     - 1st digit from right: position 10 from left → digit = 7
     - 3rd digit from right: position 8 from left → digit = 4
     - 5th digit from right: position 6 from left → digit = 8
     - 7th digit from right: position 4 from left → digit = 3
     - 9th digit from right: position 2 from left → digit = 8

6. Sum these digits:
   $$7 + 4 + 8 + 3 + 8 = 30$$

**Answer:** $30$

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