#
Problem
of the Week

Problem
D and Solution

A
Weightier Problem

## Problem

George has three objects, each of a different mass. He weighs the
objects in pairs and records the mass of each pair of objects. Later, he
realizes that he forgot to weigh the objects individually, but no longer
has access to a scale. The three recorded masses are \(2986\) g, \(3464\) g, and \(3550\) g. Determine the mass of the
heaviest object.

## Solution

Let \(a\) represent the mass, in
grams, of the lightest object

Let \(c\) represent the mass, in grams,
of the heaviest object.

Let \(b\) represent the mass, in grams,
of the third object.

The smallest recorded mass is created by adding the masses of the two
lightest objects together. Therefore, \[a+b=2986\tag{1}\]

The largest recorded mass is created by adding the masses of the two
heaviest objects together. Therefore, \[b+c=3550\tag{2}\]

Thus, it must be the case that \[a+c=3464\tag{3}\]

At this point we could solve a system of equations involving three
equations and three unknowns. Instead, we will add equations \((1)\), \((2)\), and \((3)\) together. \[\begin{align*}
(a+b)+(b+c)+(a+c) &= 2986+3550+3464\\
2a+2b+2c &= 10\,000\\
2(a+b+c) &= 10\,000\\
a+b+c &= 5000\tag{4}
\end{align*}\]

From equation \((4)\), we know that
the total mass of the three objects is \(5000\) g. But from equation \((1)\), the mass of the two lighter objects
is \(2986\) g. We can subtract equation
\((1)\) from equation \((4)\) to obtain the mass of the heaviest
object. \[\begin{aligned}
(a+b+c)-(a+b) & = 5000 - 2986\\
a+b+c-a-b & = 2014\\
c & = 2014
\end{aligned}\]

Therefore, the heaviest object has a mass of \(2014\) g. Although we are not asked to,
from here, we could determine that the other objects have mass of \(1450\) g and mass of \(1536\) g.