TIFR CSE 2022 | Part A | Question: 9

Question

You are given the following properties of sets $A, B, X$, and $Y$. For notation, $|A|$ denotes the cardinality of set $A$ (i.e., the number of elements in $A$ ), and $A \backslash B$ denotes the set of elements that are in $A$ but not in $B$.

1. $A \cup B=X \cup Y$
2. $A \cap B=X \cap Y=\emptyset$
3. $|Y \backslash A|=2$
4. $|A \backslash X|=4$

Which of the following statements $\text{MUST}$ then be $\text{FALSE?}$

• A. $|X|=5$
• B. $|Y|=5$
• C. $|A \cup X|=|B \cup Y|$
• D. $|A \cap X|=|B \cap Y|$
• E. $|A|=|B|$

Answer 1

We’re only concerned with sets $A,B,X,Y$.

And it is given that $A \cup B = X \cup Y$.

So, we define a Universal Set $U = A \cup B = X \cup Y$ which contains all sets that we’re conderned with and nothing else.

Given – $|Y \setminus A| = |Y \cap A^c| = 2$ and $|A \setminus X| = |A \cap X^c| = |A \cap Y| = 4$.

Since, $(X \cap Y) = \phi$ and $(X \cup Y) = U$.

$Y = (Y \cap A) \cup (Y \cap A^c)$.

$\therefore |Y| = |(Y \cap A)| \cup |(Y \cap A^c)| = 4 + 2 = 6$.

Answer :- B.

Comments

answer not making sense

why you put Y intersection A = 4

also please give the anser of option A

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$|A \cap Y| = 4$ follows from $|A \setminus X| = 4, X \cap Y = \phi, X \cup Y = U$.