question about an old Math JJ from 2002, Aug.-Sept.

justag

119，共150人，其中60%有A，50%有B，30%有C，非A非B非C=5，A且B且C=5，求有且只有2种的总共有多少。
==== 55（原70）

150-5=150*(0.6+0.5+0.3)-x+5==>x=70, why the answer is 55? Thanks!!!

hilisa

Total - A - B - C + AB+BC+AC + 2ABC = NON(A,B,C)
  150 -90 -75 -45 + AB+BC+AC + 2x5 = 5
             - 60 + AB+BC+AC + 10 = 5
SO:                 AB+BC+AC = 55

justag

Thanks!Lisa

But isn't this formula right?
A U B U C = A+ B+ C -(AB+BC+CA) +ABC ?

If it's right, then it seems to me:

total=150, A=150*0.6, B=75, C=45, and A U B U C = 150 - 5 = 145, ABC=5. Then there must be something misunderstanding about the 非A非B非C=5, right? Can you please help me out with that? Thanks a lot!