Question 885833
WITH CALCULATOR:
For the six numbers:
{{{average=SUM/6=28}}}--->{{{SUM=28*6=168}}}
 
For the two numbers averaging 18:
{{{average=sum/2=18}}}--->{{{sum=18*2=36}}}
 
The sum of the other four numbers is
{{{168-36=132}}} , and the average of those four numbers is
{{{152/4=highlight(33)}}}
 
IN MY HEAD:
Those two numbers averaging 18 are (on the average)
{{{28-18=10}}} lower than average.
The {{{2*10=20}}} units missing must have been made up by the other {{{4}}} numbers
That means that the average of those other {{{4}}} numbers must be
{{{20/4=5}}} units higher than the average of all six numbers.
So the average of the other four numbers is
{{{28+5=33}}}