Hints:

1. Start by arranging all ten given numbers (pairwise sums of the packages) in increasing order—from the lightest combination to the heaviest.

2. Add up all ten numbers. The total sum is 276.

3. Realize that each of the five original package weights was used in exactly four of the pairwise sums, so:

   Sum of all pairwise sums=4×Sum of the five package weights

   Therefore:

   Sum of package weights=476/4​=69

4. From the ten pairwise sums, identify:

   • The lightest pair (smallest number) — this is the sum of the two lightest packages.

   • The heaviest pair (largest number) — this is the sum of the two heaviest packages.

   Suppose the lightest pair is 22, and the heaviest is 33.

5. Since the total of all five weights is 69, and the sum of the lightest two is 22, and the heaviest two is 33:

   Middle weight=69−22−33=14

6. That means:

   • The lightest + middle = 22 + 14 = 36

   • The heaviest + middle = 14 + 33 = 47

This middle value helps you piece together the original weights one step at a time.

Solution and Exploration:

When we weigh every pair of five boxes, each individual box appears in exactly four pairwise combinations. So, if we add up all ten pairwise weights, we get four times the total weight of all five boxes.

Let’s begin:

Sum of all pairwise weights:

22+24+25+26+27+28+29+30+32+33=276

Since this is four times the total weight of the boxes:

Total weight of boxes=276/4=69

Let’s label the box weights from lightest to heaviest as:

A<B<C<D<E

Then:

A+B+C+D+E=69

From the smallest pairwise sum (22), we know:

A+B=22

From the largest pairwise sum (33), we know:

D+E=33

Substitute into the total:

A+B+C+D+E=(A+B)+C+(D+E)=22+C+33=55+C

So:

69=55+C⇒C=14

Now go back and find the remaining values:

• From A + C = 24, substitute C=14:
  A=24−14=10

• From E + C = 32, substitute C=14:
  E=32−14=18

• From A+B=22, substitute A=10:
  B=12

• From D+E=33, substitute E=18:
  D=15

Final Weights of the Boxes (in order):

10, 12, 14, 15, 18​

These steps and reasoning work for any similar problem involving the pairwise sums of five weights.