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The average (arithmetic mean) bowling score of $n$ bowlers is $160$. The average of these $n$ scores together with a score of $170$ is $161$. What is the number of bowlers, $n$?

I tried this:

$$X'*n = 160*n = x_1 + \dots + x_n$$

$$X'*n = 161*n = x_1 + \dots + x_n + 170$$

$$\frac{161*n}{160*n} = \frac{x_1 + \dots + x_n + 170}{x_1 + \dots + x_n}$$

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What have you tried ? –  Belgi Oct 6 '12 at 2:43
    
@Belgi, I have edited the original post. –  guru Oct 6 '12 at 2:52
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While writing down an equation is a good way to solve the problem, one can do it in an algebra-free way. Suppose that the new bowler had bowled $160$. Then the average wouldn't change. But she bowled $170$, which gave everybody, including herself, $1$ extra point on average. Thus, including the new bowler, there must be $10$ people. –  André Nicolas Oct 6 '12 at 2:52
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2 Answers 2

up vote 0 down vote accepted

Assume the total score of the $n$ bowlers is $S$, then the average is

$$A = \frac{S}{n} = 160\,,\quad (1) $$

The second statement is telling you that the average of these $n$ scores adding to them the score (170) of another player (you will have $n+1$ bowlers) is 161. That translates to

$$ 161 = \frac{ S+170 }{n+1}\,,\quad (2) \,. $$

Now, solve the two equations to get $n$. Solution $(n=9)$

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Hint: what is the total score of the original $n$ bowlers? If you add another bowler who hits $170$, what is the total? How many bowlers are there now?

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I can't understand. –  guru Oct 6 '12 at 2:59
    
The definition of the average score is the total score divided by the number of bowlers. –  Ross Millikan Oct 6 '12 at 3:01
    
I know what average is. –  guru Oct 6 '12 at 3:02
    
@guru: Then what don't you understand? The total score is $160n$. Having done that, try the next two. –  Ross Millikan Oct 6 '12 at 3:05
    
What i did is correct? See the original post again please. –  guru Oct 6 '12 at 3:32
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