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The mean of a list of $n$ numbers is $6$. When the number $17$ is added to the list, the mean becomes $7$. What is the value of $n$?

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You need to say what parts you do and don't understand. "Mean"? "Added to the list"? – Steve Jessop Jul 12 '14 at 15:34


  1. $(x_1+x_2+\cdots+x_n)/n=6$

  2. $(x_1+x_2+\cdots+x_n+17)/(n+1)=7$

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Let $x_1,x_2,\ldots,x_n$ be such numbers, then $$\frac{x_1+\cdots+x_n}{n}=6\Longrightarrow x_1+\cdots+x_n=6n$$ and $$\frac{x_1+\cdots+x_n+17}{n+1}=7.$$ Hence $$\frac{6n+17}{n+1}=7\Rightarrow 6n+17=7n+7.$$ Thus $$n=10.$$

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I don't understand how you get 6n and n+1 – Sam Jul 12 '14 at 2:01
First you multiply by $n$ the first equality. The second equality is because you have $n+1$ numbers $(x_1,x_2,\ldots,x_n,17)$. – DiegoMath Jul 12 '14 at 2:03
We aren't here to give people free answers to homework questions. Please try to limit your responses to hints, when responding to a [homework] question. – Justin Jul 12 '14 at 6:23
@Quincunx Perhaps you aren't, but the general consensus on meta is that people are free to answer and vote on questions however they want. If you don't like the question, downvote and/or vote to close. If you don't like the answer, downvote it. I personally think that in general, hints are better, but given that the OP didn't understand this worked solution, I doubt the hints were at all helpful. Sometimes one clear, fully explained worked solution is far more helpful than any number of hints could be, especially if you haven't seen a solution before. – Tom Oldfield Jul 12 '14 at 15:20

The mean of a list of n numbers is 6.

This problem involves a list of numbers. $n$ of them. So lets give them a name: let's call the numbers $x_i$, where $i$ ranges over the integers from $1$ through $n$.

Then, this is saying

$$ \mathrm{mean}(x_1, x_2, \ldots, x_n) = 6 $$

You know a formula for the mean, so you could plug that in.

When the number 17 is added to the list, the mean becomes 7.

This is saying

$$ \mathrm{mean}(x_1, x_2, \ldots, x_n, 17) = 7 $$

What is the value of n?

You will now attempt to solve these equations for $n$. Despite there being many free variables, this will turn out to be possible. This will involve recognizing a common expression that appears in both equations. It may help to create a new variable $s$ to refer to that common expression.

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