# Embarrassingly-basic fraction question

I'm trying to do a calculation at work to figure out what the average # of pages a visitor is viewing.

I am given:

47,000 visits

12% of visits do not bounce (that is, 12% navigate the site at least one link after stumbling upon it)

1.23 pages/visit is the average amount of pages a visitor sees.

My current attempt:

( 47,000 visits ) * (.12) = (1.23 pages)/(1 visit )

Solving for: the number of pages the average non-bouncing visitor is viewing, in order to bump the average to 1.23

Notice the glaring lack of variable... I have been out of hard math for maybe 4-5 years now, and I'm sure this is far below the average question you guys get, but I don't necessarily need you to tell me the solution so much as remind me of the problem-solving logic I should apply to solve this on my own... thank you very much for your time!

• So if someone bounces, they only viewed one page, but if they did not bounce, they viewed at least 2 pages (Including the initial one)? – Vladhagen Oct 29 '13 at 19:09
• Yes, that is essentially what it means (and for this example, exactly). It is a measure of a page's ability to "capture the interest" of a viewer -- there are some caveats (e.g. a News Site may have a high bounce rate if people are coming there only for 1 specific article). – HC_ Oct 29 '13 at 19:12

The number of non-bouncers is $(.12)(47000)=5640$, which also means that the number who do bounce is $47000-5640 = 41360$. Let $A$ represent the average number of pages seen by a non-bouncer (what you want to solve for). Then you know that the total number of pages seen by the 47000 users is $(A)(5640)+(1)(41360) = 5640A +41360$. Therefore the average number of pages seen by all users is $$\frac{5640A + 41360}{47000}$$ but you also know that this number is $1.23$. So, solve for $A$ in $$\frac{5640A + 41360}{47000} = 1.23$$