# Problem of the Week

Problem E and Solution

Just an Average Sum

## Problem

Faisal chooses four numbers. When each number is added to the mean (average) of the other three, the following sums are obtained: \(25\), \(37\), \(43\), and \(51\).

Determine the mean of the four numbers Faisal chose.

Extra Problem: Can you interpret the following picture puzzle? You may need to research the meanings of some mathematical symbols used in the puzzle.

## Solution

Let \(a,b,c,\) and \(d\) represent the four numbers. It is possible to precisely determine the four numbers, but the problem asks for only their average, which is \(\frac{a+b+c+d}{4}\).

When the first number is added to the average of the other three numbers, the result is \(25\). Thus,

\[a+\frac{b+c+d}{3}=25\]

which can be rewritten as

\[3a+b+c+d=75 \tag{1}\]

When the second number is added to the average of the other three numbers, the result is \(37\). Thus,

\[b+\frac{a+c+d}{3}=37\]

which can be rewritten as

\[a+3b+c+d=111 \tag{2}\]

When the third number is added to the average of the other three numbers, the result is \(43\). Thus,

\[c+\frac{a+b+d}{3}=43\]

which can be rewritten as

\[a+b+3c+d=129 \tag{3}\]

When the fourth number is added to the average of the other three numbers, the result is \(51\). Thus,

\[d+\frac{a+b+c}{3}=51\]

which can be rewritten as

\[a+b+c+3d=153 \tag{4}\]

Adding equations \((1)\), \((2)\), \((3)\), and \((4)\), we obtain \(6a+6b+6c+6d=468\). Dividing this equation by \(6\) gives \(a+b+c+d=78\). It follows that \(\dfrac{a+b+c+d}{4}=19.5\).

Therefore, the average of the four numbers is \(19.5\).

Although it is not required, we could solve the system of equations to determine that the numbers are: \(-1.5\), \(16.5\), \(25.5\), and \(37.5\).

**Extra Problem Solution:**

The notation \(\dfrac{1}{n} \displaystyle \sum_{i=1}^{n} x_i\) is a mathematical short form which represents the average of the \(n\) numbers \(x_1,x_2,\ldots,x_n\). So the picture puzzle can be interpreted as “Be greater than average”.