A shipping clerk has five boxes of different but unknown weights each weighing less than 100 kgs. The clerk weighs the boxes in pairs. The weights obtained are 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121 kgs. What is the weight, in kgs, of the heaviest box?

Let the individual weights be a,b,c,d,e in increasing order such that e is max and a is min. Adding all the addition of weight together we get 4*(a+b+c+d+e) = 1156 so a+b+c+d+e = 289 . Out of these a+b will be lowest sum and d+e will be the max . so a+b=110 and d+e=121 so we get value of c as 58 . now c have the 3rd highest weight so addition of c and e must give the second largest total i.e 120 . hence e = 120-58 = 62