Question 1046545
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If the average of *[tex \Large n] numbers is *[tex \Large x], then the sum of the *[tex \Large n] numbers must be *[tex \Large nx].


The sum of the ages of all 6 people is then 6 times 27 or 162.


The sum of the ages of the first 4 people is 4 times 23 or 92.


So the sum of the ages of the last 2 people must be 162 minus 92 or 70.


Since the sum of the ages of the last three must be 3 times 34 or 92, the age of the 4th person has to be 92 minus 70, or 22.


John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
<img src="http://c0rk.blogs.com/gr0undzer0/darwin-fish.jpg">
*[tex \Large \ \
*[tex \LARGE \ \ \ \ \ \ \ \ \ \  

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