0
$\begingroup$

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:

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

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

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

  • 2
    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.2012-10-06

2 Answers 2

0

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 $(n=9)$

1

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?

  • 0
    @guru: To help the reader you should define the variables: "Let x1,x2,...xn be the original scores. Let X' be" what? In your first line, clearly 161*n does not equal 160*n. It would be easier to let $N$ be the total of the original scores, instead of x1+x2+...xn. The numerator should be $161(n+1)$ as there are now $n+1$ scores, but after that fix your second line is correct.2012-10-06