We have two responses for you

Hi Mark.

Clearly, the two heaviest add to 121 units and the two lightest add to
110 units. Since all sums are whole numbers, the individual weights
must be either all whole numbers or all whole numbers + 0.5 units.
Also, since there is no duplicate sum, all bales have different
weights. So I will name them bale A, B, C, D, and E in increasing
weight.

A + B = 110, therefore B is more than 55 units and A is less than 55 units.
D + E = 121, therefore D is less than 60.5 units and E is more than 60.5 units.

This means (A, B, C, D, E) could be:
(54.5, 55.5, 56.5, 57.5, 63.5)
(54.5, 55.5, 56.5, 58.5, 62.5)
(54.5, 55.5, 56.5, 59.5, 61.5)
(54.5, 55.5, 57.5, 58.5, 62.5)
(54.5, 55.5, 57.5, 59.5, 61.5)
(54.5, 55.5, 58.5, 59.5, 61.5)
(54, 56, 57, 58, 63)
(54, 56, 57, 59, 62)
(54, 56, 57, 60, 61)
(54, 56, 58, 59, 62)
(54, 56, 58, 60, 61)
(54, 56, 59, 60, 61)
(53.5, 56.5, 57.5, 58.5, 62.5)
(53.5, 56.5, 57.5, 59.5, 61.5)
(53.5, 56.5, 58.5, 59.5, 61.5)
(53, 57, 58, 59, 62)
(53, 57, 58, 60, 61)
(53, 57, 59, 60, 61)
(52.5, 57.5, 58.5, 59.5, 61.5)
(52, 58, 59, 60, 61)

These are the only 20 possibilities. Now rule out any that can
produce 119 or 111.

When I do this, I get 4 possibilities. If A + D = B + C, then I would
have a duplicate sum (and I have no duplicates) so for the final step,
I rule those out and I am left with just one set left: the answer.

Good luck!
Stephen La Rocque.

Hi Mark,

I want to add one observation to Stephen's response. You said that the 5 bales of hay are weighed 2 at a time in all possible ways. Thus A is weighed with B, C, D, and E and hence A got weighed 4 times. Likewise each bale got weighed 4 times. Thus if you add the results of all 10 weights (I got 1156) you have weighed each bale 4 times so (A + B + C + D + E) = 1156/4 = 289.

Penny