Okay, let's slightly generalize this
Average of [tex]n[/tex] numbers is [tex]a[/tex]
and then [tex]r[/tex] numbers are removed, and you're asked to find the sum of these [tex]r[/tex] numbers.
Solution:
If average of [tex]n[/tex] numbers is [tex]a[/tex] then the sum of all these numbers is [tex]n\cdot a[/tex]
Now we remove [tex]r[/tex] numbers, so we're left with [tex](n-r)[/tex] numbers. and their. average will be [tex]{\text{sum of these } (n-r) \text{ numbers} \over (n-r)}[/tex] let's call this new average [tex] a^{\prime}[/tex]
For simplicity, say, sum of these [tex]r[/tex] numbers, which are removed is denoted by [tex]x[/tex] .
so the new average is [tex]\frac{\text{Sum of } n \text{ numbers} - x}{n-r}=a^{\prime}[/tex]
or, [tex] \frac{n\cdot a -x}{n-r}=a^{\prime}[/tex]
Simplify the equation, and solve for [tex]x[/tex] to get,
[tex] x= n\cdot a -a^{\prime}(n-r)=n(a-a^{\prime})+ra^{\prime}[/tex]
Hope you understand it :)
