SOLUTION: A farmer brought 5 boxes of melons to the market. Instead of being weighed individually, they were weighed in all possible combinations of two: boxes 1 and 2, 1 and 3, 1 and 4, 1

Question 770173: A farmer brought 5 boxes of melons to the market. Instead of being weighed individually, they were
weighed in all possible combinations of two: boxes 1 and 2, 1 and 3, 1 and 4, 1 and 5, boxes 2 and 3, boxes 2
and 4, etc. The weights of each of these combinations were written down and arranged in numeric order,
without keeping track of which weight matched which pair of boxes. The weights in kilograms for each pair of
boxes are 110, 112, 113, 114, 115, 116, 117, 118, 120, and 121.
How much does each box of melons weight? Explain in writing how you calculated your answers and show
that your answers are correct.
Answer by DrBeeee(684) (Show Source): You can put this solution on YOUR website!
Pretty tricky isn't it?
I name the 5 boxes a,b,c,d,e and assume they are in order from least to most heavy. We can write at least four equations for sure. The two lightest boxes, a and b, must be
(1) a + b = 110 the smallest sum. The next heavier pair must be
(2) a + c = 112
The two heaviest boxes, d and e, must be
(3) d + e = 121 and the second heaviest pair is
(4) c + e = 120
If we subtract (4) from (3) we get
(5) d = c + 1 which when we apply to (2) we get
(6) a + c + 1 = 112 + 1 or
(7) a + d = 113
This equation (7) is not independent of the others because we used (3) and (4) to generate it. So we still only have 4 independent equations, but 5 unknowns. We need another independent equation to solve this problem.
The sum of the next pair
(8) a + e = {114,115,116,117,118} and likewise
(9) c + d = {114,115,116,117,118}
To be honest I don't have a good mathematical reason for choosing either pair. I sort of argue that (8) should be the median of the set and (9) could be {117,118}. The only method I can prove is trial and error!
I used
(10) c + d = 117
This gives us the following 5 independent equations:
(11) a + b = 110
(12) a + c = 112
(13) c + d = 117
(14) c + e = 120
(15) d + e = 121
To solve add (13) and (14) to get
(16) 2c + (d + e) = 117 + 120 or
(17) 2c + (d+e) = 237
Use (15) in (17) to get
(18) 2c + 121 = 237 or
(19) 2c = 116 or
(20) c = 58
Now we can use (12) to get
(21) a = 54 and (13) to get
(22) d = 59 and (14) to get
(23) e = 62 and (11) to find
(24) b = 56
Let's check the answer.
Is (a+b=110)?
Is (54+56=110)?
Is (110=110)? Yes
Is (a+c=112)?
Is (54+58=112)?
Is (112=112)? Yes
Is (a+d=113)?
Is (54+59=113)?
Is (113=113)? Yes
Is (a+e=116)?
Is (54+62=116)?
Is (116=116)? Yes
Is (b+c=114)?
Is (56+58=114)?
Is (114=114)? Yes
Is (b+d=115)?
Is (56+59=115)?
Is (115=115)? Yes
Is (b+e=118)?
Is (56+62=118)?
Is (118=118)? Yes
Is (c+d=117)?
Is (58+59=117)?
Is (117=117)? Yes
Is (c+e=120)?
Is (58+62=120)?
Is (120=120)? Yes
Is (d+e=121)?
Is (59+62=121)?
Is (121=121)? Yes
Are all of the weightings used?
Is {110,112,113,116,114,115,118,117,120,121} = {110,112,113,114,115,116,117,118,120,121) Yes
Are the weights in order?
Is (a,b,c,d,e) = (54,56,58,59,62) Yes
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