Question 935166
<pre>
I think this is more complete, as it starts from scratch:

Let 100h+10t+u be the number that was reversed,

Let S = the sum of the 9 three-digit numbers that weren't reversed.

Then 

average BEFORE reversing the number = {{{(S+100h+10t+u)/10}}}

average AFTER reversing the number = {{{(S+100u+10t+h)/10}}}

We are told that

{{{(S+100u+10t+h)/10}}}{{{""=""}}}{{{(S+100u+10t+h)/10}}}{{{""+""}}}{{{29.7}}}

Multiply through by 10

{{{S+100u+10t+h}}}{{{""=""}}}{{{S+100u+10t+h}}}{{{""+""}}}{{{297}}}

That simplifies to

{{{99u=99h+297}}}

Divide through by 99

{{{u=h+3}}}

{{{u-h=3}}}

They differ by 3.

Edwin</pre>