A) 26
B) 27
C) 28
D) Any of the three values 26, 27, 28 is possible
Correct Answer: D
Solution :
[d]We have |
\[n(A\cup B\cup C)=n(A)+n(B)+n(C)-\] |
\[n(A\cap B)-n(B\cap C)-n(C\cap A)+n(A\cap B\cap C)\] |
\[=10+15+20-8-9-n(C\cap A)+n(A\cap B\cap C)\] |
\[=28-\{n(C\cap A)-n(A\cap B\cap C)\}\] (i) |
Since \[n(C\cap A)\ge n(A\cap B\cap C)\] |
We have \[n(C\cap A)\ge n(A\cap B\cap C)\ge 0\] (ii) |
From (i) and (ii): \[n(A\cup B\cup C)\le 28\] (iii) |
Now, \[n(A\cup B)=n(A)+n(B)-n(A\cap B)=10+15-8=17\]and \[n(B\cup C)=n(B)+n(C)-n(B\cap C)=15+20-9=26\]since, \[n(A\cup B\cup C)\ge n(A\cup C)\] and \[n(A\cup B\cup C)\ge n(B\cup C)\] we have |
\[n(A\cup B\cup C)\ge 17\] and \[n(A\cup B\cup C)\ge 26\] |
Hence \[n(A\cup B\cup C)\ge 26\] (iv) |
From (iii) and (iv) we obtain |
\[26\le n(A\cup B\cup C)\le 28\] |
Also \[n(A\cup B\cup C)\] is a positive integer \[\therefore \,\,\,n(A\cup B\cup C)=26\,\,or\,\,27\,\,or\,\,28\] |
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