# Section-1: Numerical Logic Solved QuestionLogical Reasoning Discussion

 Q. 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?
 ✖ A. 60 ✔ B. 62 ✖ C. 64 ✖ D. Cannot be determined

Solution:
Option(B) is correct

Let $a, b, c, d$ and $e$ be the weights, in kg ,of the five boxes with the shipping clerk, where, $a \leq b \leq c \leq d \leq e.$

$110 = a + b < a + c < ... < c + e < d + e = 121$

i.e. $a + c = 112$ and $c + e = 120$

Each box is weighed 4 times.

$\therefore 4a + 4b + 4c + 4d + 4e$ $= 110 + 112 + 113 + 114 + 115$ $+ 116 + 117 + 118 + 120 + 121$

$= 1156$

$\therefore a + b + c + d + e = 289$

Now it is clear that $a + b = 110$ and $d + e = 121$

$\therefore 110 + c + 121 = 289$

$\therefore c = 58$

Substituting this value in $c + e = 120$

$\Rightarrow e = 62$

Hence, option B is the correct choice.

## (1) Comment(s)

Mathew
()

if we take c+d=118, then we get e as 61 and c as 59, then how can we be sure that which pair corresponds to which value