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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 ??

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you should show what you have attempted. –  picakhu May 23 '12 at 0:53
    
I cant figure out how to approach this problem. As I already mentioned its a standardized test question.It is very unlikely that i would post questions here without attempting and re-attempting them first. –  Rajeshwar May 23 '12 at 1:01
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Probably common sense would take care of it. The average amount of money in the bank accounts of $47$ people is $2000$. If $15$ is taken from each account, what is the average left? –  André Nicolas May 23 '12 at 1:17
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@Rajeshwar, I have looked at all your questions. I think if you spend a little more time (and effort), you should be able to solve them. –  picakhu May 23 '12 at 2:07
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@picakhu, at first I was not inclined to agree with you. However, after looking at his questions, I also do not feel he is posting in the spirit of Math.SE. –  000 May 23 '12 at 6:31
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2 Answers 2

up vote 1 down vote accepted

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:

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

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Thanks for the explanation. –  Rajeshwar May 23 '12 at 1:10
    
@Rajeshwar: Glad to help. –  Thomas May 23 '12 at 1:12
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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 $$\begin{align} \frac{(x_1-x)+(x_2-x)+\cdots+(x_n-x)}{n} &=\frac{x_1+x_2+\cdots+x_n-n\cdot x}{n}\\ &=\frac{x_1+x_2+\cdots+x_n}{n}-\frac{nx}{n}\\ &=a-x. \end{align}$$

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thanks for the answer I think i am almost there.How did you get a-x from the second last step ? Could you please elaborate ? –  Rajeshwar May 23 '12 at 1:09
    
Thanks that helped. –  Rajeshwar May 23 '12 at 1:10
    
@Rajeshwar: I added a step in between what had been the last two steps—does that make it clearer? –  Isaac May 23 '12 at 1:10
    
Yes it does. I really appreciate the help –  Rajeshwar May 23 '12 at 1:14
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