# Problem of the Week

Problem D and Solution

Another Average

## Problem

The numbers \(2124\), \(1984\), \(1742\), \(2344\), \(2074\), and \(1632\) are each written on a card. Daniyal takes four of the cards and calculates the mean (average) of their numbers to be \(2021\). Determine the mean of the numbers on the remaining two cards.

Extra Problem: Can you interpret the following picture puzzle?

## Solution

At the outset, it should be noted that we could “play” with the numbers to determine which of the four numbers have an average of \(2021\). We could then easily determine the average of the remaining two numbers. This method works decently well on a problem with a small number of numbers. However, if we were to increase the size of the list by just a few more numbers, then the task would not be easily solved using this approach. It turns out, we can solve this problem without actually figuring out which four numbers Daniyal used.

The sum of all six numbers is \[2124+1984+1742+2344+2074+1632=11\,900\] Since the average of four of the numbers is 2021, then the sum of those four numbers is \(4\times 2021=8084\).

The sum of the two remaining numbers is \(11\,900-8084=3816\). Since there are two numbers in the sum, the average of the two numbers is calculated by dividing the sum by 2. The average of the remaining two numbers is then \(3816 \div 2 = 1908\).

Although not required, the two numbers that sum to \(3816\) are \(1742\) and \(2074\). It is then easily verified that the average of the four other numbers, \(2124\), \(1984\), \(2344\), and \(1632\), is \(2021\).

**Extra Problem Answer:** Ten percent above average.