Let A, B, C be finite sets. Suppose that n(A)=10, n(B)=15, n(C)=20, n(A B)=8 and n(B C)=9. Then the possible value of n(A B C) is

Correct Answer is : Any of the three values 26, 27, 28 is possible

Any of the three values 26, 27, 28 is possible

Correct Answer is : Any of the three values 26, 27, 28 is possible

Test Index

BITSAT 2017 Paper

Section: Mathematics

Question : 109 of 150

Marks: +1, -0

n(A)=10, n(B)=15, n(C)=20, n(A \cap B)=8

n(B \cap C)=9

n(A \cup B \cup C)

Solution:

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)-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\;\text{or}\;27\;\text{or}\;28

Go to Question:

Prev Question

Next Question