They weigh 54, 56, 58, 59, 62 pounds.
Let's assume that the weight of five bales are B1, B2, B3, B4 and B5 pounds respectively. Also, B1 <= B2 <= B3 <= B4 <= B5
It is given that five bales of hay are weighed two at a time in all possible ways. It means that each of the bale is weighted four times.
Thus,
4*(B1 + B2 + B3 + B4 + B5) = (110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121)
4*(B1 + B2 + B3 + B4 + B5) = 1156
(B1 + B2 + B3 + B4 + B5) = 289 pounds
Now, B1 and B2 must add to 110 as they are the lightest one.
B1 + B2 = 110
Similarly, B4 and B5 must add to 121 as they are the heaviest one.
B4 + B5 = 121
From above three equation, we get B3 = 58 pounds
Also, it is obvious that B1 and B3 will add to 112 - the next possible higher value. Similarly, B3 and B5 will add to 120 - the next possible lower value.
B1 + B3 = 112
B3 + B5 = 120
Substituting B3 = 58, we get B1 = 54 and B5 = 62
From 2 & 3 equations, we get B2 = 56 and B4 = 59
Hence, the weight of five bales are 54, 56, 58, 59 and 62 pounds.
