SECTION 05 / 5

How to Solve Week 3 Exam Questions

On this page 12 headings

PAST-PAPER PRACTICE

Choose a paper

15 of 15 questions shown

Start here “What is the minimum number of gates?” comes in every quiz and end-term. This page teaches the method from zero, then solves every Week 3 question from the 11 past papers step by step, each with the exact paper it came from.

The 4-step gate-count recipe — Most Week 3 questions follow these four steps: simplify, check for a shortcut, build, then count.

How to read this page: each grey “Asked in” box is a real past-paper question. Below it, How to solve shows every step.

Type 1: Minimum number of 2-input NAND / NOR / XNOR gates

Method

Learn the method

Step 1: simplify the function first. Many answers turn out to be just 1 gate.

Step 2: learn this table by heart (every number was checked by a computer search over all possible circuits):

Function	NAND gates	NOR gates
NOT A	1	1
$A \cdot B \left(\operatorname{AND}\right)$	2	3
$A + B \left(\operatorname{OR}\right)$	3	2
NAND	1	4
NOR	4	1
$A \oplus B \left(\operatorname{XOR}\right)$	4	5
$A \odot B$ (XNOR, “ $A = B$ ”)	5	4

Why these numbers (Level 1 → 3):

NOT from NAND: tie both inputs together. $\operatorname{NAND} \left(A , A\right) = \operatorname{NOT} \left(A \cdot A\right) = \operatorname{NOT} A \to$ 1 gate.
AND from NAND: NAND gives NOT(AB). Flip it with one more NAND → 2 gates.
OR from NAND: De Morgan says $A + B = \operatorname{NOT} \left(\overline{A} \cdot \overline{B}\right) = \operatorname{NAND} \left(\overline{A} , \overline{B}\right)$ . Making $\overline{A}$ and $\overline{B}$ costs 2 → total 3.
NOR is the mirror image: $\operatorname{OR} = 2 , \operatorname{AND} = 3$ .

Step 3: shortcuts

$\operatorname{NOR} \left(X , Y\right) = \overline{X} \cdot \overline{Y}$ : one NOR gives a “both are 0” term for free.
$\operatorname{NAND} \left(X , Y\right) = \overline{X} + \overline{Y}$ : one NAND gives an “either is 0” term for free.

Step 4: count every gate, including NOTs made from NAND/NOR.

Past paper questions

Asked in

2025 T2 Quiz 1Q1054 marks

Question: Minimum number of 2-input NAND gates to realize $F = \left(\overline{X} + \overline{Y}\right) \left(Z + W\right)$ ?

Show solution

Spot the shortcut: $\overline{X} + \overline{Y} = \operatorname{NAND} \left(X , Y\right) \to$ G1, 1 gate, no NOTs needed.
Multiply out: $F =$ G1· $Z +$ G1·W
SOP → NAND–NAND: G2 = NAND(G1, Z), G3 = NAND(G1, W), F = NAND(G2, G3)
Count: G1, G2, G3, $F =$ 4 ✔ (a computer search confirmed 3 is impossible)

Asked in

2025 T2 End-term (AN)Q155, 2025 T2 Quiz 2Q138, 2026 T2 End-termQ83 marks each

Question: Minimum number of 2-input NOR gates for $F = \Sigma \left(0 , 2 , 8 , 1 0\right)$ (4 inputs)?

Show solution

Write the rows in binary $\left(A B C D\right) : 0 = 0 0 0 0 , 2 = 0 0 1 0 , 8 = 1 0 0 0 , 1 0 = 1 0 1 0$
Which letters never change? $B = 0$ and $D = 0$ in all four rows. A and C take both values.
So $F = \overline{B} \cdot \overline{D}$
$\overline{B} \cdot \overline{D} = \operatorname{NOT} \left(B + D\right) =$ NOR(B, D) → 1 gate ✔

Asked in

2025 T2 End-term (FN)Q1233 marks

Question: Minimum number of 2-input NOR gates for $F = \Sigma \left(5 , 7 , 1 3 , 1 5\right)$ ?

Show solution

Binary: 0101, 0111, 1101, 1111 → $B = 1$ and $D = 1$ always
$F = B \cdot D$ (an AND)
AND from NOR costs 3: $\operatorname{NOR} \left(B , B\right) = \overline{B} , \operatorname{NOR} \left(D , D\right) = \overline{D} , \operatorname{NOR} \left(\overline{B} , \overline{D}\right) = B D$ ✔

Asked in

2025 T3 End-term (AN)Q103 marks

Question: Minimum number of 2-input NAND gates for the minimized SOP of $f \left(x , y , z\right) = \Sigma m \left(0 , 2 , 5 , 7\right)$ ?

Show solution

Binary: 000, 010, 101, 111
Rows 000 and $0 1 0 : x = 0 , z = 0 \to \overline{x} \overline{z}$ . Rows 101 and $1 1 1 : x = 1 , z = 1 \to x z$ .
$f = \overline{x} \overline{z} + x z = 1$ when $x = z \to$ XNOR(x, z)
XNOR from $\operatorname{NAND} =$ 5 (from the table) ✔

Asked in

2025 T3 End-term (FN)Q93 marks

Question: Minimum number of 2-input XNOR gates for the same $f = \Sigma m \left(0 , 2 , 5 , 7\right)$ ?

Show solution

$f = x \odot z$ (from above). That is one XNOR gate → 1 ✔. Always simplify first!

Asked in

2025 T2 End-term (AN)Q157 and 2026 T2 End-termQ103 marks each

Question: A 1-bit comparator outputs 1 if $A = B$ , else 0. Minimum 2-input NOR gates?

Show solution

“1 when $A = B$ ” is XNOR.
From the table: XNOR from $\operatorname{NOR} =$ 4.
The circuit: G1 = NOR(A, B), G2 = NOR(A, G1), G3 = NOR(B, G1), $Y = \operatorname{NOR}$ (G2, G3).

Building XNOR from 4 NOR gates A 1-bit equality checker built from exactly 4 NOR gates.

Asked in

2025 T2 End-term (FN)Q1253 marks

Question: Same comparator (1 if $A = B$ ), but using only 2-input NAND gates. Minimum?

Show solution

XNOR from $\operatorname{NAND} = \operatorname{XOR}$ (4 NANDs) + one $\operatorname{NOT} \left(1 \operatorname{NAND}\right) =$ 5.

Asked in

2026 T2 Quiz 2Q123 marks

Question: $F = \Sigma \left(0 , 1 , 2 , 3 , 4 , 5 , 7\right)$ using 3-input NAND gates. Minimum?

Show solution

F is 1 in every row except 6 (110). So $F = 0$ only when $A = 1 , B = 1 , C = 0$ .
$F = \operatorname{NOT} \left(A \cdot B \cdot \overline{C}\right) =$ $\operatorname{NAND} \left(A , B , \overline{C}\right)$ → 1 gate, but it needs $\overline{C}$ .
$\overline{C} = \operatorname{NAND} \left(C , C , C\right) \to 1$ more gate.
Total 2 ✔

Type 2: Timing diagram → minimum gates

Method

Learn the method

Draw a line wherever any input changes. This cuts time into slices.
In each slice, read every input and Y. Each slice = one truth-table row.
Find the function, then count gates (Type 1).

Usual pattern in these papers: A changes slowest, C fastest, so the slices go 000, 001, 010, …, 111 in order.

Timing diagram to truth table — Cut the waves into slices; each slice is one row.

Past paper questions

Asked in

2026 T2 Quiz 1Q35 marks

Question: Timing diagram for A, B, C and Y. Minimum 2-input XOR gates to make Y?

Question figure (redrawn exactly from the paper) — Question figure

Show solution

Read the 8 slices (ABC → Y): 000→0, 001→1, 010→1, 011→0, 100→1, 101→0, 110→0, 111→1
$Y = 1$ when the number of 1s is odd (1, 1, 1, 3) → $Y = A \oplus B \oplus C$
With 2-input XORs: $\left(A \oplus B\right) \oplus C \to$ 2 gates ✔

Asked in

2025 T2 Quiz 1Q1025 marks

Question: Timing diagram for A, B, C and Y. Minimum 2-input NAND gates to make Y?

Show solution

Read the slices: $Y = 1 , 1 , 1 , 1 , 1 , 1 , 1$ , 0. Y is 0 only in slice 111.
So $Y = \operatorname{NOT} \left(A \cdot B \cdot C\right) , a 3$ -input NAND.
With 2-input NANDs: G1 = NAND(A, B), G2 = NAND(G1, G1) = $A B , Y = \operatorname{NAND}$ (G2, C)
Total 3 ✔

Asked in

2025 T3 Quiz 1Q1755 marks (the answer key is missing in the PDF; our answer is checked)

Question: Timing diagram for A, B, C and Output Y. Minimum 2-input NOR gates to make Y?

Show solution

Read the slices: $Y = 0 , 1 , 0 , 1 , 1 , 1 , 1 , 1$
$Y = 0$ only in rows 000 and 010 → there $A = 0$ and $C = 0$ . So $Y =$ $A + C$ .
OR from $\operatorname{NOR} =$ 2: G1 = $\operatorname{NOR} \left(A , C\right) , Y = \operatorname{NOR}$ (G1, G1) ✔

Type 3: SOP ↔︎ POS multiple choice

Method

Learn the method

$\Sigma m$ lists the rows where $F = 1$ . $\Pi M$ lists the rows where $F = 0$ . Change Σ ↔︎ Π by taking the other numbers.
A maxterm for a 0-row: digit 0 → plain letter, digit 1 → barred letter, joined by +.
To check an option quickly: find where the option is 0 and compare with the 0-rows.

Question: $F \left(a , b , c\right) = \Pi \left(0 , 1 , 2 , 4\right)$ . Find the minimized SOP. Options: $\overline{a} c + a b / a \overline{b} + a \overline{c} + \overline{b} c / a b + b c + c a / a \overline{b} + a \overline{c} + \overline{b} \overline{c}$ / None

Show solution

$F = 0$ in rows $0 , 1 , 2 , 4 \to F = 1$ in the other rows: 3, 5, 6, 7
Binary: 011, 101, 110, 111 → every row has two or more 1s
That is $a b + b c + c a$ (any two are 1) ✔

Asked in

2025 T3 Quiz 1Q1684 marks

Question: $F \left(a , b , c\right) = \Pi \left(1 , 2 , 3 , 7\right)$ . Find the minimized SOP.

Show solution

$F = 1$ in rows 0, 4, 5, 6 = 000, 100, 101, 110
Pair them: $0 0 0 + 1 0 0 \to \overline{b} \overline{c}$ . $1 0 0 + 1 0 1 \to a \overline{b}$ . $1 0 0 + 1 1 0 \to a \overline{c}$ .
$F = a \overline{b} + a \overline{c} + \overline{b} \overline{c}$ ✔ (each pair shares 100, and every row is covered)

Asked in

2025 T2 Quiz 1Q984 marks. The official key is wrong.Key issue

Question: $F \left(A , B , C\right) = \Sigma \left(0 , 2 , 3 , 6\right)$ . Which is the correct minimized POS? Options: $\left(A + B\right) \left(\overline{A} + \overline{C}\right) / \left(\overline{A} + B + C\right) \left(A + \overline{B}\right) / \left(A + \overline{B}\right) \left(\overline{A} + \overline{C}\right)$ / None of these

Show solution

$F = 0$ in rows 1, 4, 5, 7 = 001, 100, 101, 111
Check the keyed option $\left(A + \overline{B}\right) \left(\overline{A} + \overline{C}\right)$ . It is 0 when $A = 0$ and $B = 1$ (rows 2, 3), or when $A = 1$ and $C = 1$ (rows 5, 7). So its 0-rows are 2, 3, 5, 7, not 1, 4, 5, 7 ✗
Row 2 makes it obvious: F should be 1 at row 2 (it’s in the Σ list), but the option gives 0.
The true minimum POS needs three terms: $\left(B + \overline{C}\right) \left(\overline{A} + B\right) \left(\overline{A} + \overline{C}\right)$ . We checked by computer that no 2-term POS works.
So the correct choice is None of these. (In the exam the key gave marks to $\left(A + \overline{B}\right) \left(\overline{A} + \overline{C}\right)$ .)

Type 4: Minimum number of literals

Method

Learn the method

A literal is one letter, with or without a bar, each time it appears. Simplify fully, then count letters.

Useful steps: take out a common part, then use $X + \overline{X} = 1$ .

Asked in

2026 T2 Quiz 1Q55 marks

Question: Minimum number of literals in the SOP form: $a b + \overline{a} \overline{b} \overline{c} + \overline{a} b \overline{c}$

Show solution

The last two terms share $\overline{a} \overline{c} : \overline{a} \overline{b} \overline{c} + \overline{a} b \overline{c} = \overline{a} \overline{c} \left(\overline{b} + b\right)$
$\overline{b} + b = 1$ , so they become $\overline{a} \overline{c}$
$F = a b + \overline{a} \overline{c} \to$ letters: $a , b , \overline{a} , \overline{c} \to$ 4 ✔

Practice set

Minimum 2-input NAND gates for $A + B C$ ?

G1 = $\operatorname{NAND} \left(A , A\right) = \overline{A}$ , G2 = $\operatorname{NAND} \left(B , C\right) , F = \operatorname{NAND}$ (G1, G2).

Minimum 2-input NOR gates for $F = \Sigma \left(1 , 3 , 9 , 1 1\right)$ (4 inputs)?

Rows

0 0 0 1 , 0 0 1 1 , 1 0 0 1 , 1 0 1 1 \to B = 0 , D = 1 \to F = \overline{B} D

\operatorname{NOR} \left(B , \overline{D}\right)

with

\overline{D} = \operatorname{NOR} \left(D , D\right) \to 2

gates.

Timing slices give $Y = 1 , 1 , 0 , 0 , 1 , 1 , 0 , 0$ (ABC from 000 to 111). Function and minimum 2-input NAND?

Y = 1

exactly when

B = 0

, so

Y = \overline{B}

. NAND(B, B) → 1 gate.

$F = \Pi \left(0 , 3 , 5 , 6\right)$ . Write F as $\Sigma m$ .

\Sigma m \left(1 , 2 , 4 , 7\right) \left(= A \oplus B \oplus C\right)

In one line: Simplify first, use the NAND/NOR table and free shortcuts, read timing diagrams slice by slice, and test SOP/POS options by finding their 0-rows. Exam words: minimum gate count, universal gates, NAND, NOR, XNOR, comparator, timing diagram, $\Sigma m , \Pi M$ , maxterm, literal.

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