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Explanation:

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

a < b < c < d < 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.

∴ 4a + 4b + 4c + 4d + 4e = 110 + 112 + 113 + 114 + 115 + 116 + 117 + 118 + 120 + 121 = 1156

∴ a + b + c + d + e = 289

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

∴ 110 + c + 121 = 289

∴ c = 58

Substituting this value in c + e = 120

∴ e = 62

Hence, option (b).

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