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
Create a FREE account and get:
CAT Averages, Ratios & Proportions
CAT Logarithms, Surds & Indices
CAT Functions, Graphs & Statistics
CAT DI Data Change Over Period
CAT Tables With Missing Values
CAT LR Selections With Conditions
2 weeks, 2 days ago
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.
2 months ago
You added 119 in your addition
2 years, 10 months ago
Where did that 4 come from?
1 year, 9 months ago
Where is it written boxes are counted 4 times
2 years, 9 months ago
each box has been weighted 4 time