Help me please

Question

You have five bales of hay. For some reason, instead of being weighed individually, they were weighed in all possible combinations of two. The weights of each of these combinations were written down and arranged in numerical order, without keeping track of which weight matched which pair of bales. The weights, in kilograms, were 80, 82, 83, 84, 85, 86, 87, 88, 90, and 91. How much does each bale weigh? Is there a solution?

Answer 1

There are five bales of hay. Label them A,B,C,D,E. Since we don’t yet know the weights of each, let’s assume the weights of A, B, C, D, and E are in order from lightest to heaviest.

When they are weighed in all combinations of two, there are ten weights [PLZ note: “all combinations of two”].

This means that each bale was weighed four times (once with each other bale). The ten combinations are AB, AC, AD, AE, BC, BD, BE, CD, CE, DE [note: for example, AC is the same as CA].

So, the sum of the given weights is four times the total weight of the five bales:
(80+82+83+84+85+86+87+88+90+91) = 856
4(A+B+C+D+E) = 856
(A+B+C+D+E) = 214 [eq1]

We assumed A-E was lightest to heaviest, so:
A+B = 80 [eq2]
A+C = 82 [eq3]
A+D = 83 [eq4]
A+E = 84 [eq5]
B+C = 85 [eq6]
B+D = 86 [eq7]
B+E = 87 [eq8]
C+D = 88 [eq9]
C+E = 90 [eq10]
D+E = 91 [eq11]

Now, substitute pairs:
(A+B+C+D+E) = 214 [eq1 again]
((A+B)+(C+D)+E) = 214
( 80 + 88 + E ) = 214 [substitute using eq2 and eq9]
168 + E = 214
E = 46

… and use appropriate eq[s] for the others to find that:
D = 44
C = 43
B = 42
A = 38

Now, check that the total of A-E is 214.

Note: there is no other solution.

Answer 2

a, b, c, d, and e represents the weight of each of the five bales.

a + b = 80

a + c = 82

a + d = 83

a + e = 84

b + c = 85

b + d = 86

b + e = 87

c + d = 80

d + e = 91

 

Using substitution or elimination solve the system of equations.