There are 6 persons: A, B, C, D, E and F A has 3 items more than C D has 4 items less than B E has 6 items less than F C has 2 items more than E F has 3 items more than D Which one of the following figures cannot be equal to the total number of items possessed by all the 6 persons?

To solve this, we must express all items in terms of a single variable. Let us use E as the base variable. Based on the given conditions: C has 2 items more than E. So, C equals E plus 2. A has 3 items more than C. So, A equals E plus 2 plus 3, which is E plus 5. F has 6 items more than E. So, F equals E plus 6. F has 3 items more than D. This means D equals F minus 3. So, D equals E plus 6 minus 3, which is E plus 3. D has 4 items less than B. This means B equals D plus 4. So, B equals E plus 3 plus 4, which is E plus 7. Now, we sum the items for all 6 persons: Total equals A plus B plus C plus D plus E plus F. Total equals (E plus 5) plus (E plus 7) plus (E plus 2) plus (E plus 3) plus E plus (E plus 6). Total equals 6E plus 23. Since the number of items must be a whole number, E must be an integer (0, 1, 2, 3...). If E is 3, Total is 18 plus 23 which equals 41 (Option A). If E is 4, Total is 24 plus 23 which equals 47 (Option B). If E is 5, Total is 30 plus 23 which equals 53 (Option C). The total must follow the format 6E plus 23. Subtracting 23 from the options must yield a number divisible by 6. 41 minus 23 equals 18 (divisible by 6). 47 minus 23 equals 24 (divisible by 6). 53 minus 23 equals 30 (divisible by 6). 58 minus 23 equals 35 (not divisible by 6). Therefore, 58 cannot be the total number of items. Option D is correct.