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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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Ricky went to the store and bought 8 boxes of apples The lightest box weighed 33 pounds The heaviest box weighed 49 pounds Which could be the total weight of the apple boxes?
It is possible that only 1 box weighed 33 lbs and 7 boxes weighed 49 lbs giving 376lbs. It is also possible that 7 boxes weigh 33lbs and only 1 box weighs 49 lbs giving 280lbs.
