Count "votes against" in pair-ties?

[DEMOREP1 at aol.com]

Mr. Eppley wrote:
Isn't it possible for the votes against an unbeaten candidate in a pair-tie
to be worse than the Condorcet score of a beaten candidate?

       A    B    C
  A        50=   2L
  B   50=       35
  C    1   45L
     ---- ---- ----
       ?    ?    2      <--- Maximum votes against.  Count the ties?

A (and only A) is unbeaten and wins by our definition.  But A's 50 in the tie
is larger than C's 2 in C's only loss.  Is A really less beaten than C? 

I hope I'm not muddying the waters.  The odds against a pair-tie in a large
election are negligible.
-----
Demorep1--I thought that a candidate must get more votes than each other
candidate to be the Condorcet winner (i.e. a tie is not a win). Thus, is A
the winner ? 
Does any candidate in the example have majority approval on a yes/no vote ?
Are there overlapping votes in the 6 pairings-
1  CA
2  AC
35 BC
45 CB
50 AB
50 BA
(such as an ACB vote) ?
With a low number of voters, as in many clubs, ties are a distinct
possibility.