Question

Let $A$ and $B$ be finite sets such that $n(A-B)=18$, $n(A \cap B)=25$ and $n(A \cup B)=70$. Then $n(B)$ is equal to

• A. 52
• B. 25
• C. 27
• D. 43
• E. 45

Hint:

Set Theory Tip: Split sets into three disjoint regions: $A-B$, $A\cap B$, and $B-A$. Then counting becomes very easy.

Answer 1

The Correct Option is A

Concept:
For two finite sets: $$n(A \cup B)=n(A-B)+n(A \cap B)+n(B-A)$$ Also, $$n(B)=n(B-A)+n(A \cap B)$$

Step 1: Use the union formula.
Given: $$n(A-B)=18,\quad n(A \cap B)=25,\quad n(A \cup B)=70$$ So: $$70=18+25+n(B-A)$$

Step 2: Add known values.
$$18+25=43$$ Hence: $$70=43+n(B-A)$$

Step 3: Find $n(B-A)$.
Subtract 43 from both sides: $$n(B-A)=70-43=27$$

Step 4: Find $n(B)$.
Now: $$n(B)=n(B-A)+n(A \cap B)$$ So: $$n(B)=27+25=52$$

Step 5: Choose the correct option.
Thus, $$n(B)=52$$ Hence the correct answer is (A) 52.