Question: A shipping clerk has five boxes of different but unknown weights, each weighing less than 100 kg. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kg. What is the weight of the second heaviest box?
Explanation:
Let the weights in increasing order be a, b, c, d and e. (a is the lightest and e the heaviest)
Since all possible pairs are formed, each box will form a pair with each of the other 4 boxes. Hence, each box will have 4 pairs.
a + b = 110
Box 'a' will make a total of 4 pairs (one each with b, c, d and e)
∴ When we add all such pairs, each box will get added 4 times.
∴ 4(a + b + c + d + e) = 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121.
⇒ a + b + c + d + e = 289 …(1)
Since a and b are the two lightest boxes
Also, since d and e are the two heaviest boxes
From (1), (2) and (3) we get,
The second lightest pair will be a and c.
From (2) and (4) we get,
The second heaviest pair will be c and e.
From (3) and (5) we get,
∴ The second heaviest box is 59 kgs.
Hence, option (d).