# Shifting Average

What happens to the average if you subtract 8 from all of the numbers?

The average of four numbers is 48.

If 8 is subtracted from each number, what is the average of the four new numbers?

*This problem is taken from the World Mathematics Championships*

**Working numerically**

To find the average, the 4 numbers were added together and then divided by 4. That gave 48.

So before the total was divided by 4, it must have been 48$\times$4 = 192.

Subtracting 8 from each of the numbers will reduce the total by 4$\times$8 = 32. So the new total will be 192$-$32 = 160.

So the new average will be 160$\div$4 = 40.

**Average as balancing**

The average of the 4 numbers can be thought of as the point balanced between them, so that the distances above and below the average are balanced. This is shown in the diagram below - the sum of the red and blue lengths is equal to the sum of the yellow and green lengths.

So when the numbers are all reduced by 8, all of the points and the distances between them move 8 units down, as shown in the diagram. So the average will also go down by 8.

Image

So the new average is 40.

**Average as sharing**

Imagine that 4 people each have some money, and the average amount of money is £48. That means that if the 4 people pooled their money and then shared it out equally, they would each get £48.

If they all spend £8, that would decrease each of their shares by £8. So each person's share would be £40.

So the new average is 40.

**Working algebraically**

The four numbers can be written as $a,b,c$ and $d.$

Then $(a+b+c+d)\div4=48$, and we need to find $((a-8)+(b-8)+(c-8)+(d-8))\div4$.

$$\begin{split}(a-8)+(b-8)+(c-8)+(d-8)&=a+b+c+d-8-8-8-8\\

&=a+b+c+d-32\end{split}$$So $$\begin{split}((a-8)+(b-8)+(c-8)+(d-8))\div4&=(a+b+c+d-32)\div4\\

&=((a+b+c+d)\div4)-(32\div4)\end{split}$$But $(a+b+c+d)\div4=48$. So $((a+b+c+d)\div4)-(32\div4)=48-(32\div4)=48-8=40$

So the new average is $40$.