A) 43
B) 76
C) 49
D) None of these
Correct Answer: A
Solution :
Let B, H, F denote the sets of members who are on the basketball team, hockey team and football team respectively. Then we are given \[n\,(B)=21,\,n\,(H)=26,n\,(F)=29\] \[n\,(H\cap B)=14\], \[n\,(H\cap F)=15\], \[n\,(F\cap B)=12\] and \[n\,(B\cap H\cap F)=8\]. We have to find \[n\,(B\cup H\cup F)\]. To find this, we use the formula \[n\,(B\cup H\cup F)=n\,(B)+n\,(H)+n\,(F)\] \[-n\,(B\cap H)-n\,(H\cap F)-n\,(F\cap B)+n\,(B\cap H\cap F)\] Hence,\[n\,(B\cup H\cup F)=(21+26+29)-(14+15+12)+8=43\] Thus these are 43 members in all.You need to login to perform this action.
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