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 can not be equal to the total number of items possessed by all the 6 persons?

To find the possible total number of items, let us express the number of items each person has in terms of E:

• F = E + 6 (since E has 6 items less than F)
• D = F - 3 = (E + 6) - 3 = E + 3 (since F has 3 items more than D)
• B = D + 4 = (E + 3) + 4 = E + 7 (since D has 4 items less than B)
• C = E + 2 (given)
• A = C + 3 = (E + 2) + 3 = E + 5 (since A has 3 items more than C)

The total number of items is A + B + C + D + E + F = (E + 5) + (E + 7) + (E + 2) + (E + 3) + E + (E + 6) = 6E + 23.

Since the number of items E must be an integer, the total number of items minus 23 must be divisible by 6. Checking the given options:

• 41 - 23 = 18 (divisible by 6)
• 47 - 23 = 24 (divisible by 6)
• 53 - 23 = 30 (divisible by 6)
• 58 - 23 = 35 (not divisible by 6)

Therefore, 58 cannot be the total number of items.