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(5): Analytical solution (Bonus puzzle question) by Steve Herman)

Ok, I concede. While I got a solution to the bonus puzzle, I was not making progress on a generalized method.

So I looked at the solution in the queue, and I am very impressed. Brian, this time I do dub you the Pumpkin Master. I like your solution and I was not close.

I was heading down two different paths:

a) setting up linear equations without any assumptions about a being less than b, etc.

b) Looking at the 11 pairs that could be part of a double weighing, and ordering them:

a+d < a+e < a+f

a+d < b+e < b+f etc.

Neither of these were bearing fruit.

p.s. -- did you know that a pumpkin is a fruit?

Nice bonus puzzle!