A) 11
B) 12
C) 13
D) 14
Correct Answer: D
Solution :
Let M,P and C be the sets of students taking examinations in Mathematics, Physics and Chemistry, respectively. We have, \[n(M\cup P\cup C)=50,n(M)=37,n(P)=24,\] \[n(C)=43\] \[n(M\cap P)<19,n(M\cap C)\le 29,n(P\cap C)\le 20\] Now, \[n(M\cup P\cup C)=n(M)+n(P)+n(C)\] \[-n(M\cap P)\]\[-n(M\cap C)-n(P\cap C)+n\] \[(M\cap P\cap C)\] \[\Rightarrow \]\[50=37+24+43-\{n(M\cap P)+n(M\cap C)\] \[+n(P\cap C)\}\]\[+n(M\cap P\cap C)\] \[\Rightarrow \]\[n(M\cap P\cap C)=n(M\cap P)+n(M\cap C)\] \[+n(P\cap C)-54\] \[\Rightarrow \]\[n(M\cap P)+n(M\cap C)+n(P\cap C)\] \[=n(M\cap P\cap C)+54\] ? (i) Now, \[n(M\cap P)\le 19,n(M\cap C)\le 29,\] \[n(P\cap C)\le 20\] \[\Rightarrow \] \[n(M\cap P)+n(M\cap C)+n(P\cap C)\] \[\le 19+29+20\] [using Eq. (i)] \[\Rightarrow \] \[n(M\cap P\cap C)+54\le 68\] \[\Rightarrow \] \[n(M\cap P\cap C)\le 14\]You need to login to perform this action.
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