GATE 2020 IN Q.11: solution and answer

GATE 2020 · INQ.11 · 1 mark · MCQ

The Boolean expression for the shaded regions as shown in the figure is ____

GATE 2020 IN question 11: Venn diagram: a square (the universal set) containing two overlapping circles, labelled A (left) and B (right). — Figure from the GATE 2020 IN paper. It is part of the question.

(A)
$(A + B) \cdot (\overline{A} + \overline{B})$
(B)
$(A + \overline{B}) \cdot (\overline{A} + B)$
(C)
$(\overline{A} + B) \cdot (\overline{A} + \overline{B})$
(D)
$(\overline{A} + \overline{B}) \cdot (A + \overline{B})$

Answer and solution

ANSWER

The shaded regions are $A \overline{B} + \overline{A} B = A \oplus B$ . In product-of-sums form this is $(A + B) \cdot (\overline{A} + \overline{B})$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2020 IN (IIT Delhi)

Read the Venn diagram. The shaded area is "in $A$ but not in $B$ " together with "in $B$ but not in $A$ ". The overlap $A B$ and the area outside both circles are not shaded. Complement is written with $\overline{A}$ (an overbar), and the printed dot between brackets means AND.
Write it in sum-of-products form: $F = A \overline{B} + \overline{A} B = A \oplus B$ , the exclusive-OR.
Turn it into product-of-sums form. $F$ is 0 only in the two unshaded regions: $\overline{A} \overline{B}$ (outside both circles, $A = 0, B = 0$ ) and $A B$ (the overlap, $A = 1, B = 1$ ). The maxterms for these are $M_{0} = (A + B)$ and $M_{3} = (\overline{A} + \overline{B})$ , so $F = (A + B) (\overline{A} + \overline{B})$ .
Check by multiplying out, using $A \overline{A} = 0$ and $B \overline{B} = 0$ : $(A + B) (\overline{A} + \overline{B}) = A \overline{A} + A \overline{B} + \overline{A} B + B \overline{B} = A \overline{B} + \overline{A} B$ This is exactly the shaded area, so option (A) is correct.
Rule out the other options. (B) $(A + \overline{B}) (\overline{A} + B) = A \overline{A} + A B + \overline{A} \overline{B} + \overline{B} B = A B + \overline{A} \overline{B}$ , the XNOR, which is exactly the unshaded area. (C) $(\overline{A} + B) (\overline{A} + \overline{B}) = \overline{A} + B \overline{B} = \overline{A}$ , by the distributive law $(X + Y) (X + Z) = X + Y Z$ ; that is everything outside circle $A$ . (D) $(\overline{A} + \overline{B}) (A + \overline{B}) = \overline{B} + \overline{A} A = \overline{B}$ , by the same law; that is everything outside circle $B$ .

KEY IDEA

Shading "A only" plus "B only" is the exclusive-OR, $A \overline{B} + \overline{A} B$ . The options are in product-of-sums form, so build the POS from the regions that are NOT shaded: outside both circles gives the maxterm $(A + B)$ and the overlap gives $(\overline{A} + \overline{B})$ . Their product is the answer.

WATCH OUT FOR

Picking option (B), $(A + \overline{B}) (\overline{A} + B)$ , because it looks like a "mixed" XOR pattern. Multiplied out it is $A B + \overline{A} \overline{B}$ , the XNOR, which is exactly the unshaded area. The other trap is putting the maxterms for the shaded regions into the POS instead of the maxterms for the unshaded ones.

From the GATE 2020 IN paper. Every question of this paper · Download the full paper · All 22 Boolean logic questions, with notes

GATE 2020 IN Q.11: solution and answer

Answer and solution

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