Explanation
Solution
Step 1: Define the individual weights
Let the weights of the five boxes in ascending order be p,q,r,s and t.
p < q < r < s < t
Step 2: Understand the pairwise sums
When 5 items are weighed in pairs, there are 5C2=10 possible combinations.
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The smallest sum must be: p+q=110
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The second smallest sum must be: p+r=112
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The largest sum must be: s+t=121
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The second largest sum must be: r+t=120
Step 3: Calculate the total sum of all boxes
Each box is weighed exactly 4 times across the 10 pairs.
4(p+q+r+s+t)=110+112+113+114+115+116+117+118+120+121
p+q+r+s+t=289—(Equation 1)
Step 4: Find the weight of the heaviest box (t)
Substitute the smallest pair (p+q=110) and the second largest pair (r+t=120) into Equation 1 to solve for s:
Now, use the largest pair sum (s+t=121) to find t:
Conclusion:
The weight of the heaviest box is 62 kg.
Correct Option: 2 (62)
