A) 16
B) 6
C) 8
D) 20
Correct Answer: A
Solution :
Given, \[n(N)=12,\,\,n(P)=16,\,\,n(H)=18\] and \[n(N\cup P\cup H)=30\] \[\because \] \[n(N\cup P\cup H)=n(N)+n(P)+n(H)\] \[-n(N\cap P)\] s\[-n(N\cap H)-n(N\cap H)+n(N\cap P\cap H)\] \[\Rightarrow \] \[30=12+16+18-n(N\cap P)-n(P\cap H)\] \[-n(N\cap H)+0\] \[\Rightarrow \] \[n(N\cap P)+n(P\cap H)+n(N\cap H)=16\] \[\therefore \] Number of pupils taking two subjects \[=\,n(N\cap P)+n(P\cap H)+n(N\cap H)\] \[-3n(N\cap P\cap H)\] \[=16-0=16\]You need to login to perform this action.
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