Question

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?

Answer 1

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.

Answer 2

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.