A) 6
B) 5
C) 8
D) None of the above
Correct Answer: D
Solution :
[d] \[n\,(M)=23,\]\[n\,(P)=24,\]\[n\,(C)=19\] \[n\,(M\cap P)=12,\]\[n\,(M\cap C)=9,\] \[n\,(P\cap C)=7\]\[n\,(M\cap P\cap C)=4\] We have to find \[n\,(M\cap P\cap C'),\]\[n\,(P\cap M'\cap C'),\] \[n\,(C\cap M'\cap P')\] Now,\[n\,(M\cap P\cap C')=n\,[M\cap (P\cup C)']\] \[=n\,(M)-n\,(M\cap (P\cup C)\] \[=n\,(M)-n\,[(M\cap P)\cup (M\cap C)]\] \[=n\,(M)-n\,(M\cap P)-n\,(M\cap C)+n\,(M\cap P\cap C)\] \[=23-12-9+4=27-21=6\] \[n\,(P\cap M'\cap C')=n\,[(P\cap (M\cup C)']\] \[=n\,(P)-n\,[P\cap (M\cup C)]\] \[=n\,(P)-n\,[(P\cap M)\cup (P\cap C)]\] \[=n\,(P)-n\,(P\cap M)-n\,(P\cap C)+n\,(P\cap M\cap C)\] \[=24-12-7+4=9\] \[n\,(C\cap M'\cap P')=n\,(C)-n\,(C\cap P)\] \[-\,n\,(C\cap M)+n\,(C\cap P\cap M)\] \[=19-7-9+4=32-16=7\] \[\therefore \]n (exactly one subject) = 6 + 9 + 7 = 22 |
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