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There are 10 possible combinations of pairs in five bales: A+B, A+C, A+D, A+E, B+C, B+D, B+E, C+D, C+E, and D+E. Each bale shows four times in the above possible pairs. So, the sum of all possible pairs equals the sum of all the weights; or

4A+4B+4C+4D+4E = 1156 or

4(A+B+C+D+E) =1156; therefore

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

The two lightest bales equal the lightest pair weight of 110; the two heaviest produce the heaviest pair weight of 121. Combining these two produces the following:

A+B+D+E = 231

Subtract the second conclusion from the first and you conclude

C = 58.

If A+B is the lightest pair, A+C is second-lightest at 112 lb; therefore A=54. Solving for the remaining values is now a matter of moments.

A =54

B = 56

C = 58

D = 59

E = 62.

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Q: 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?
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