Question

If \(|A| = 25, |B| = 30, |C| = 45, |A \cap B| = 5, |B \cap C| = 10, |A \cap C| = 6, |A \cap B \cap C| = 2\) then arrange the following in non-decreasing order:
A. \(|A \cup B|\)
B. \(|B \cup C|\)
C. \(|A \cup B \cup C|\)
D. \(|C \cup A|\)
E. \(|A - B|\)
Choose the correct answer from the options given below:

• A. C, B, A, E, D
• B. E, A, D, C, B
• C. E, A, D, B, C
• D. E, D, A, B, C

Hint:

We know that \(|A-B| = 20\) must be the smallest value as it is only a subset of \(A\) (size 25).
Also, the union of all three sets, \(|A \cup B \cup C| = 81\), must logically be the largest value.
Thus, the sequence must begin with E and end with C.
Looking at the options, only (C) and (D) satisfy this, and verifying the order of D (64) < B (65) points uniquely to (C).

Answer 1

The Correct Option is C