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CAT 2000 | Question: 103

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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, in kg, of the heaviest box?

  1. $60$ 
  2. $62$
  3. $64$ 
  4. Cannot be determined

 

1 Answer

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To find the weights of the five boxes a, b, c, d, and e given the following conditions:


The boxes are arranged in increasing order of their weights: a < b < c < d < e.
a+b (lightest)
a+c
a+d
a+e
b+c
b+d
b+e
c+d
c+e
d+e (heaviest)

 

Sum of all combinations=4(a+b+c+d+e)=1156

a+b+c+d+e=1156/4​=289    - i

 

Use the given pairs:

  • (heaviest)    - iii

 

i-(ii+iii)

(a+b+c+d+e) - (a+b+d+e) 
c = 289 - 231
c = 58

 

Finding e:
Using c+e=120:
58+e=120
e=120−58=62


Finding d:
Using d+e=121:
d+62=121
d=121−62=59

 

Finding a:
Using a+c=112:
a+58=112
a=112−58=54


Finding b:
Using a+b=110:
54+b=110
b=110−54=56

 

Final weights of the boxes: a=54,b=56,c=58,d=59,e=62

Thus, weight, in kg, of the heaviest box is 62.

 

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