Much like the

third Pumpkins puzzle six pumpkins, each having a different weight, are weighed two at a time in all 15 sets of two. This time exactly four pairs of duplicate values occurred. Those four values are 60, 100, 110, and 120 pounds.

How much did each individual pumpkin weigh?

(In reply to

re(3): Analytical solution (Bonus puzzle question) by Charlie)

True, but the complementary equality c+f=d+e is false for {100,110,116,120}. I suppose if I really wanted to force a+d=b+c to be false, I could just use {100,104,110,120}, but that is merely a mirror image of the former set.

Either way, assuming the largest double and the smallest double are derived from the largest four and smallest four pumpkins respectively is no longer applicable, which renders Steve's solution inapplicable to these variations.