The smallest package weighed only 1/3 more than 1/2 the weight of the heaviest package. if the two packages combined weighed 100 kilograms, how much did the smaller package weigh?

First of all, let us assign variables to represent the given. Let us make 

s = the smallest package 
h = heaviest package

Next, we translate the given into mathematical terms

"the smallest package weighed only 1/3 more than 1/2 the weight of the heaviest package" can be represented as

s = 1/2*h + 1/3

Since the given also stated that the sum of the two packages is 100 kg

s + h = 100 which can be written as

h = 100-s

We have two unknowns and two equations, so this problem can be solved.

Substituting the second equation to the first equation yields

s = 1/2(100-s) + 1/3
s = 50-(1/2)s+(1/3)
s+(1/2)s = 50+(1/3)
(3/2)s = (151/3)
s = 302/9 = 33.56 kg

The smaller package weighed 33.56 kg.

Let us assume the weight of the heaviest package = a kg
Then
Weight of the smallest package in question = (a/2) + [ (a/2) * (1/3) ] kg
                                                                      = (a/2) + (a/6) kg
Combined weight of the two packages = 100 kg
So
a + (a/2) + (a/6) = 100
(6a + 3a + a)/6 = 100
10a = 100 * 6
10a = 600
a = 600/10
   = 60 kg
So weight of the smaller package = 100 - 60 kg
                                                      = 40 kg.

The answer is the first one