CAT 2000 Question Paper Question 153

Question 153

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

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2 weeks, 2 days ago


Aman Kumar

7 months ago

The above solution is wrong as upon adding the paired weights from 110, 112,….till 121 we get 1275 not 1154. Please look into this.

Rishabh Sahu

2 months ago

You added 119 in your addition


2 years, 10 months ago

Where did that 4 come from?

Amit Kumar

1 year, 9 months ago

Where is it written boxes are counted 4 times

Kumar Aakash

2 years, 9 months ago

each box has been weighted 4 time


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