Let A, B, C, D, E be the weights of the five pumpkins in ascending order. Similar to the previous problem:
A+B = 24
A+C = 28
C+E = 44
D+E = 50
The omitted duplicate weight accounts for the sum of four of the five pumpkins. Therefore the sum of the eight weights given equals 3 times the sum of the weight of all five pumpkins plus the one pumpkin which is not in the duplicate set. If W represents that pumpkin, then:
3*(A+B+C+D+E) + W = 284
The system of equations formed can be reduced to:
B = C-4
A = 28-C
D = C+6
E = 44-C
W = 62-3C
There are five cases to consider - W being each of A-E. Substituting A-E in turn and finishing solving the system yields:
If W=A then A=11, B=13, C=17, D=23, E=27. The duplicate weight is B+E = C+D = 40. This is a solution.
If W=B then C=16.5 and B=12.5. This is enough to discard this answer since B+C = 29, which is not in the list of given weights and cannot be the omitted weight (the omitted weight must be A+E = C+D in this case).
If W=C then C=15.5 and B=11.5. With B+C = 27 and the omitted weight being A+E = B+D, this case can be discarded by the same reasoning as the prior case.
If W=D then C=14 and D=20. Then A=14, B=10 and E=30. The weight 36 does not appear as any combination of A-E, therefore this is not a solution.
If W=E then C=9 and E=35. Then A=19, B=5, and D=15. The weight 30 does not appear as any combination of A-E, therefore this is not a solution.
The pumpkins weigh 11, 13, 17, 23, and 27 kilograms. The omitted weight is 13+27 = 17+23 = 40 kilograms. |
