<PRE><B>Question 1132302</B>
F-70
S-50
I=30
two complaints=44=F+S, S+I, F+I
150 complaints altogether.
88 of them are with 44 patients
That leaves 56 patients.
x have 3 complaints
56-x have 1 complaint
The total complaints for that group is 3x+56-x, and that equals 62 complaints
2x+56=62
2x=6
x=3 patients with all 3 complaints  ANSWER

3 have 3 for total of 9
44 have 2 for total of 88
53 have 1 for total of 53</PRE>