Constructing an average equation - Standardized Test

Question

A question in the standardized test is:

The average of $n$ numbers is $a$. If $x$ is subtracted from each number the average will be

a) $(ax)/n\quad$ b) $(an)/x\quad$ c) $an-x\quad$ d) $n-x\quad$ e) $a-x$

The answer to this is e) $a-x$.

Could anyone please help me figure how the author got this answer ??

Answer 1

Let $n_1, \dots, n_m$ be the $m$ numbers. That the average of the numbers is $a$ means that $$\frac{n_1 + \dots + n_m}{m} = a. $$ Now subtract $x$ from each number and compute the average and you get:

$$\begin{align} \frac{(n_1- x) + \dots +(n_m - x) }{m} &= \frac{n_1 + \dots + n_m - mx}{m} \\ &= \frac{n_1 + \dots n_m}{m} - \frac{mx}{m} \\ &= a - x \end{align} $$

Answer 2

Suppose the original numbers were $x_1$, $x_2$, ..., $x_n$, so that the average is $$a=\frac{x_1+x_2+\cdots+x_n}{n}.$$

Now, if we subtract $x$ from each of the numbers, we're going to be averaging $x_1-x$, $x_2-x$, ..., $x_n-x$, which gives $$\begin{align} \frac{(x_1-x)+(x_2-x)+\cdots+(x_n-x)}{n} &=\frac{x_1+x_2+\cdots+x_n-n\cdot x}{n}\\ &=\frac{x_1+x_2+\cdots+x_n}{n}-\frac{nx}{n}\\ &=a-x. \end{align}$$