A) 16
B) 6
C) 8
D) 20
Correct Answer: A
Solution :
[a] Given, \[n\,(C)=12,\]\[n\,(P)=16,\]\[n\,(H)=18,\] \[n\,(C\cup P\cup H)=30\] From \[n\,(C\cup P\cup H\cup )=n\,(C)+n\,(P)+(H)-n\,(C\cap P)\] \[-\,n\,(P\cap H)-n\,(C\cap H)+n\,(C\cap P\cap H)\] \[\therefore \]\[n\,(C\cap P)+n\,(P\cap H)+n\,(C\cap H)=16\] Now, number of pupils taking two subjects \[=n\,(C\cap P)+n\,(P\cap H)+n\,(C\cap H)-\,3n\,(C\cap P\cap H)\] \[=16-0=16\] |
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