I have 5 bales of hay and they are weighed 2 @ a time.
The bales weigh 80,82,83,84,85,86,87,88,90,and 91 lbs.
they are weighed without marking which bales weighed what.
They are wighed in a sequence as bale 1 and 2, 1&3, 1&4,
1&5, 2&3, 2&4, 2&5 and so on.
In table form, we have the weights as:
A B C D
---------------------
B A+B
C A+C B+C
D A+D B+D C+D
E A+E B+E C+E D+E
Plugging in the values, we have:
A B C D
-------------------
B 80
C 82 85
D 83 86 88
E 84 87 90 91
Now, we can begin solving multiple systems of equations
to calculate individual weights.
Take two equations with a common term and subtract one
from the other to eliminate the common term.
B + C = 85
-(A + C = 82)
------------
B - A = 3
Add another equation from the table to isolate B.
B + A = 80
B - A = 3
---------
2B = 83
B = 41.5
Solve for A.
B + A = 80
41.5 + A = 80
A = 80 - 41.5
A = 38.5
Let's use the A column in the table to come up with
values we can use for the other bales.
(C + A = 82) - (B + A = 80) = (C - B = 2) C = B + 2
(D + A = 83) - (B + A = 80) = (D - B = 3) D = B + 3
(E + A = 84) - (B + A = 80) = (E - B = 4) E = B + 4
Scanning down col A, we can see from the table that
C is 2 lbs heavier than B, D is 3 lbs heavier than B,
and E is 4 lbs heavier than B, so our calculations
prove out.
C = B + 2 C = 41.5 + 2 C = 43.5
D = B + 3 D = 41.5 + 3 D = 44.5
E = B + 4 E = 41.5 + 4 E = 45.5
Try verifying all those values.
Row C, col B: B+C = 85 41.5 + 43.5 = 85 checks
Row D, col B: B+D = 86 41.5 + 44.5 = 86 checks
Row E, col B: B+E = 87 41.5 + 45.5 = 87 checks
Row D, col C: C+D = 88 43.5 + 44.5 = 88 checks
Row E, col C: C+E = 90 43.5 + 45.5 = 89 XXXXX
Row E, col D: D+E = 91 44.5 + 45.5 = 90 XXXXX
There is something wrong with the last two entries
in the table. Let's examine the values in the table.
In row D, going from col col B to col C, we find an
increase of 2. This matches the difference between
weight B and weight C, 2 lbs. However, in row E,
going from col B to col C, we find an increase of 3!
That does not agree with the difference between
weight B and weight C. Now, look at row E, going from
col B to col D. It shows an increase of 4. We have
calculated the difference between weight B and weight D
to be 3, not 4. In fact, the only way to get 91 (the
last value in the table) is to have two bales that
weight 45.5, which clearly we don't have.
Without access to the book that this problem came from,
there is no way to decide what caused the discrepancy.