Question 22826
 suppose five bales of hay are weighed two at a time in all possible ways. the weights in pounds are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121. how much does each bale weigh

 Let them be A,B,C,D & E. (in increasing order)
 Since each of them appeared in the list of partial sums 4 times(why ?)
 we have 4(A+B+C+D+E)={{{SIGMA}}} 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121 = 1156.

 So, A+B+C+D+E = 289.

 We see that A+B = 110 & D+E = 121, so C = 289-(110+121) = 58.
 Also, since A+C = 112, we get A = 112-58= 54,B = 110 - 54 = 56.
 Similarly, C+E = 120, E = 120 - 58 = 62, D =  121 - 62 = 59.

 Answer: 54, 56,  58, 59, 62. 
 
 Kenny