SECTION 06 / 6

How to Solve Week 2 Exam Questions

On this page 25 headings

PAST-PAPER PRACTICE

Choose a paper

20 of 20 questions shown

Start here Every Week 2 question from the 11 past papers is on this page, with the exact paper it came from. For each type you first learn the method from zero, then you see every real question solved step by step, writing out every line. Three official answer keys turned out to be wrong. Those are marked so you don’t get confused.

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

Type 1: Find the output of a chain of gates

Method

Learn the method

Work one gate at a time, from left to right. Write each gate’s output in terms of the input. Use these rules:

Rule	Why
$X \oplus X = 0 , X \odot X = 1$	same inputs → XOR says “not different” (0), XNOR says “same” (1)
$X \oplus 0 = X , X \oplus 1 = \overline{X}$	XOR with 1 flips
$X \odot 1 = X , X \odot 0 = \overline{X}$	XNOR with 0 flips
$\operatorname{NAND} \left(1 , X\right) = \overline{X}$	AND with 1 gives X, then flip

Level 1: one gate. XOR with both inputs $X \to X \oplus X =$ 0. Level 2: two gates. G1 = $X \odot X = 1$ , then G2 = $1 \oplus X =$ $\overline{X}$ .

Check trick: put $X = 0$ , then $X = 1$ , and push the numbers through. If you get 1, 0 → the answer is $\overline{X}$ . If you get 0, 1 → X.

Past paper questions

Asked in

2025 T2 Quiz 1Q964 marks

Question: X feeds a chain: XNOR → XNOR → XNOR → XOR. Each gate’s second input is X. What is the output Y? Options: $X / \overline{X} / 1 / 0$

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

Show solution

G1 = $X \odot X =$ 1
G2 = G1 ⊙ $X = 1 \odot X =$ X
G3 = G2 ⊙ $X = X \odot X =$ 1
$Y =$ G3 ⊕ $X = 1 \oplus X =$ $\overline{X}$
Check with $X = 0 : 1 \to 0 \to 1 \to Y = 1$ ✔ $\left(\overline{X} = 1\right)$

Asked in

2025 T3 Quiz 1Q1664 marks

Question: X feeds a chain: XOR → XNOR → XNOR → XOR. Each gate’s second input is X. What is the output? Options: $X / \overline{X} / 1 / 0$

Show solution

G1 = $X \oplus X =$ 0
G2 = $0 \odot X =$ $\overline{X}$ (XNOR with 0 flips)
G3 = $\overline{X} \odot X =$ 0 (they are always different)
Out = $0 \oplus X =$ X

Asked in

2026 T2 Quiz 1Q124 marks

Question: X feeds a chain: XNOR → XOR → XOR → NAND. Each gate’s second input is X. What is the output? Options: $\overline{X} / 1 / 0 / X$

Show solution

G1 = $X \odot X =$ 1
G2 = $1 \oplus X =$ $\overline{X}$
G3 = $\overline{X} \oplus X =$ 1
Out = $\operatorname{NAND} \left(1 , X\right) = \operatorname{NOT} \left(1 \cdot X\right) =$ $\overline{X}$

Type 2: Probability that the output is 1 (or 0)

Method

Learn the method

“Inputs are uniformly random” means every row of the truth table is equally likely.

P(output = 1) = (number of rows where output = 1) ÷ (total rows)

Total rows = 2ⁿ for n inputs: 3 inputs → 8 rows, 4 inputs → 16 rows.

Probability from a truth table — $F = \left(A \oplus B\right) \cdot C$ is 1 in 2 of the 8 rows, so $P = 2 / 8 = 0$ .25.

Level 1: $F = A \cdot B$ . Only row 11 gives 1. $P = 1 / 4 =$ 0.25. Level 2: $F = A + B$ . Rows 01, 10, 11 give 1. $P = 3 / 4 =$ 0.75. Level 3: big circuits. First simplify the circuit to a formula, then count. A big circuit often simplifies to just one letter, and then $P = 0$ .5.

Past paper questions

Asked in

2025 T2 Quiz 1Q1005 marks

Question: X and Y go into an XOR. Separately, X and Y go into an AND. The XOR output and Z go into a second AND. Both AND outputs go into an OR → F. Inputs are uniformly random. $P \left(F = 1\right)$ ? Options: 0.75 / 0.33 / 0.25 / 0.50 / 1 / None of these

Show solution

Write the formula: $F = \left(X \oplus Y\right) \cdot Z + X \cdot Y$
Make the table (8 rows):

X Y Z	$X \oplus Y$	$\left(X \oplus Y\right) \cdot Z$	$X \cdot Y$	F
000	0	0	0	0
001	0	0	0	0
010	1	0	0	0
011	1	1	0	1
100	1	0	0	0
101	1	1	0	1
110	0	0	1	1
111	0	0	1	1

$F = 1$ in 4 rows → $P = 4 / 8 =$ 0.50 (F is 1 when at least two inputs are 1. This is the “majority” function.)

Asked in

2025 T3 Quiz 1Q1705 marks (and the same circuit in 2026 T2 Quiz 1Q94 marks)

Question: Three AND gates feed an OR → Y. AND1 = $\overline{X} \cdot Y$ . AND2 = $X \cdot Y \cdot Z$ . AND3 = $X \cdot Y \cdot \overline{Z}$ . Inputs are uniformly random. $P \left(Y = 0\right)$ ? (The 2026 paper asks $P \left(f = 1\right)$ .)

Show solution

$Y = \overline{X} Y + X Y Z + X Y \overline{Z}$
Join the last two: $X Y Z + X Y \overline{Z} = X Y \left(Z + \overline{Z}\right) = X Y \cdot 1 =$ XY
$Y = \overline{X} Y + X Y = Y \left(\overline{X} + X\right) = Y \cdot 1 =$ Y
The whole circuit is just the input Y! Y is 0 in half the rows → $P \left(0\right) =$ 0.5, and $P \left(1\right) =$ 0.5 too.

Asked in

2025 T2 End-term (AN)Q156 and 2025 T2 End-term (FN)Q1243 marks each

Question: A 4-input circuit (x, y, w, z) with five AND gates, two OR gates and three NOT gates. $P \left(f = 1\right)$ ?

Show solution

Read every gate from the diagram: A1 = $\overline{x} \cdot y$ , A2 = $\overline{w} \cdot z$ , O1 = A2 + $\overline{z}$ , A3 = A1·O1 A4 = $\overline{x} \cdot w \cdot z$ , O2 = A4 + x, A5 = O2· $y f =$ A3 + A5
Simplify: O1 = $\overline{w} z + \overline{z} =$ $\overline{w} + \overline{z}$ . O2 = $\overline{x} w z + x =$ $x + w z$ .
$f = \overline{x} y \left(\overline{w} + \overline{z}\right) + y \left(x + w z\right)$
$I f y = 0 , f = 0$ . $I f y = 1 : f = \overline{x} \left(\overline{w} + \overline{z}\right) + x + w z$ . $I f x = 1 \to 1$ . $I f x = 0 \to \left(\overline{w} + \overline{z}\right) + w z = 1$ (one of them is always true).
So $f = y$ , and $P = 8 / 1 6 =$ 0.5. (We checked all 16 rows by computer.)

Type 3: Σm → “for which inputs is f always 1?”

Method

Learn the method

Write out every row number in binary.
Mark the rows where $f = 1$ .
Look for the pattern: count the 1s in each row.
Test every option. An option is correct only if $f = 1$ in every row that the option describes.

Count of 1s	Rows (3 variables)
0 ones	0 (000)
1 one	1 (001), 2 (010), 4 (100)
2 ones	3 (011), 5 (101), 6 (110)
3 ones	7 (111)

Quick facts: $\Sigma m \left(1 , 2 , 4 , 7\right) = A \oplus B \oplus C$ (odd number of 1s). $\Sigma m \left(0 , 3 , 5 , 6\right) =$ its opposite (even number of 1s).

Past paper questions

Asked in

2025 T3 Quiz 1Q1674 marks

Question: $f \left(x , y , z\right) = \Sigma m \left(1 , 2 , 4 , 7\right)$ . For which condition is f always 1? Options: 2 or more inputs at 0 / 2 or more inputs at 1 / any odd number of inputs at 0 / any odd number of inputs at 1

Show solution

$f = 1$ rows: 001, 010, 100, 111 → counts of 1s: 1, 1, 1, 3. All odd.
Test the other options:
- “2 or more at 0” includes $0 0 0 \to f = 0$ ✗
- “2 or more at 1” includes $0 1 1 \to f = 0$ ✗
- “Odd number of 0s” includes 000 (three 0s) → $f = 0$ ✗

Asked in

2025 T2 Quiz 1Q974 marks. The official key is incomplete.Key issue

Question: $f \left(x , y , z\right) = \Sigma m \left(1 , 2 , 4 , 7\right)$ . For which condition is f always 1? Options: exactly one input at 1 / all inputs at 1 / any two inputs at 0 / any even number of inputs at 1

Show solution

“Exactly one input at 1” → rows $0 0 1 , 0 1 0 , 1 0 0 \to f = 1$ in all ✔
But also “all inputs at 1” → row $1 1 1 \to f = 1$ ✔
And “any two inputs at 0” means exactly one at 1 → the same rows → $f = 1$ ✔
“Even number of 1s” includes $0 0 0 \to f = 0$ ✗
So three options are true. The key accepted “exactly one input at 1”. In the exam, pick that one, the most direct description.

Asked in

2026 T2 Quiz 1Q114 marks

Question: $f \left(x , y , z\right) = \Sigma m \left(0 , 3 , 5 , 6\right)$ . For which condition is f always 1? Options: 2 or more inputs at 0 / 2 or more inputs at 1 / any odd number of inputs at 0 / any odd number of inputs at 1

Show solution

$f = 1$ rows: 000, 011, 101, 110. Count the 0s: 3, 1, 1, 1 → all odd.
Test the others: “2 or more at 0” includes $0 0 1 \to f = 0$ ✗; “2 or more at 1” includes $1 1 1 \to f = 0$ ✗; “odd number of 1s” includes $0 0 1 \to f = 0$ ✗.

Asked in

2026 T2 Quiz 1Q84 marks. Two options are true here.Key issue

Question: Given $f \left(A , B , C\right) = A \oplus B \oplus C = 1$ , which statement is always true? Options: $A = B \oplus C / A + B + C = 1 / A = B \odot C / A B C = 1$ / None of these

Show solution

XOR both sides with $\left(B \oplus C\right) : A = 1 \oplus \left(B \oplus C\right) = \operatorname{NOT} \left(B \oplus C\right) =$ $B \odot C$ ✔
Check “ $A + B + C = 1$ ”: $A \oplus B \oplus C = 1$ means an odd number of 1s, so at least one is 1. So $A + B + C = 1$ is also always true.
“ $A = B \oplus C$ ” is the opposite ✗. “ $A B C = 1$ ” fails for 100 ✗.
The key accepts only $A = B \odot C$ , so choose that in the exam.

Type 4: Count the minterms

Method

Learn the method

Each row where $F = 1$ is one minterm. Count the rows. For “exactly k of n inputs”, use ⁿCₖ = n! ÷ (k! × (n − k)!).

Level 1: exactly one 1 among 3 inputs → ³C₁ = 3 (001, 010, 100). Level 2: exactly two 1s among 4 inputs → ⁴C₂ = (4 × 3) ÷ (2 × 1) = 6.

Past paper question

Asked in

2025 T2 End-term (AN)Q159, 2025 T2 End-term (FN)Q127, 2026 T2 End-termQ123 marks each

Question: A circuit has inputs A, B, C, D. $F = 1$ if and only if exactly two of the four inputs are 0 (the AN paper says “are 1”). Minimum number of minterms in the SOP for F?

Show solution

“Exactly two 0s” among four inputs = “exactly two 1s”. Both count the same rows.
Count: ⁴C₂ = (4 × 3) ÷ (2 × 1) = 6
List to check: 0011, 0101, 0110, 1001, 1010, 1100 → 6 rows ✔
Trap: “minimum number of minterms” is still 6. Minterms can’t be simplified away; each 1-row is one minterm.

Type 5: Expression → list of minterms (Σ)

Method

Learn the method

Take one product term at a time.
The letters in the term are fixed. Missing letters can be 0 or 1 (both).
Write all the rows, convert to numbers, and join all the lists.

Level 1: $F \left(A , B\right) = A$ . B is missing → rows $1 0 , 1 1 \to \Sigma m \left(2 , 3\right)$ . Level 2: $F \left(A , B , C\right) = A \overline{B}$ . C is missing → $1 0 0 , 1 0 1 \to \Sigma m \left(4 , 5\right)$ .

Past paper questions

Asked in

2025 T2 Quiz 2Q1324 marks

Question: Find the minterms of $F \left(A , B , C , D\right) = \overline{A} \overline{B} + A C \overline{D} + B \overline{C} \overline{D}$ .

Show solution

$\overline{A} \overline{B} \to A = 0 , B = 0 , C$ and D free → 0000, 0001, 0010, 0011 → 0, 1, 2, 3
$A C \overline{D} \to A = 1 , C = 1 , D = 0 , B$ free → 1010, 1110 → 10, 14
$B \overline{C} \overline{D} \to B = 1 , C = 0 , D = 0 , A$ free → 0100, 1100 → 4, 12
Join: Σ(0, 1, 2, 3, 4, 10, 12, 14) ✔

Asked in

2026 T2 Quiz 2Q114 marks

Question: Find the minterms of $A B + \overline{A} C \overline{D} + B \overline{C} D$ .

Show solution

AB → 11XX → 1100, 1101, 1110, 1111 → 12, 13, 14, 15
$\overline{A} C \overline{D} \to$ 0X10 → 0010, 0110 → 2, 6
$B \overline{C} D \to$ X101 → 0101, 1101 → 5, 13 (13 is already there)
Join: Σ(2, 5, 6, 12, 13, 14, 15) ✔

Type 6: Expression → truth table

Method

Learn the method

For each row, put the values into each product term. If any term is $1 , F = 1$ .

Asked in

2025 T3 End-term (FN)Q143 marks

Question: Choose the correct truth table for $F \left(x , y , z\right) = \overline{x} \overline{y} \overline{z} + \overline{x} y + x \overline{z}$ .

Show solution

$\overline{x} \overline{y} \overline{z} \to$ only 000 → row 0
$\overline{x} y \to 0 1 X \to$ rows 2, 3
$x \overline{z} \to$ 1X0 → rows 4, 6
$F = 1$ in rows 0, 2, 3, 4, 6 → F column: 1, 0, 1, 1, 1, 0, 1, 0 ✔
Pick the option whose F column matches exactly. One wrong row means a wrong option.

Type 7: Which Boolean identity is correct?

Method

Learn the method

Test each identity with numbers. One row that gives different sides proves it wrong. If it survives all rows, it is right.

True identity	Name
$\left(A + B\right) \left(A + C\right) = A + B C$	distributive (OR over AND)
$\left(A + B\right)$ ’ = A’·B’	De Morgan
$\left(A \oplus B\right)$ ’ = $A B + A$ ’B’	XNOR
$\left(A \oplus B\right) \oplus \left(B \oplus C\right) = A \oplus C$	$B \oplus B$ cancels

Past paper questions

Asked in

2025 T2 Quiz 2Q1335 marks (multiple select)

Question: Choose the correct identities: $\left(A \oplus B\right)$ ’ = $A B + A$ ’B’ / $\left(A + B\right)$ ’ = A’ + B’ / $A \left(B + C\right) = A B + C / \left(A + B\right) \left(A + C\right) = A + B C$

Show solution

$\left(A \oplus B\right)$ ’ is 1 when $A = B$ , which is exactly $A B + A$ ’B’ ✔
$\left(A + B\right)$ ’ = A’ + B’? Try $A = 1 , B = 0$ : left = (1)’ = 0, right = 0 + 1 = 1 ✗ (correct De Morgan: A’·B’)
$A \left(B + C\right) = A B + C$ ? Try $A = 0 , C = 1$ : left = 0, right = 1 ✗ (correct: $A B + A C$ )
$\left(A + B\right) \left(A + C\right) = A A + A C + A B + B C = A \left(1 + C + B\right) + B C =$ $A + B C$ ✔

Asked in

2026 T2 Quiz 2Q145 marks

Question: Choose the correct option: $\left(A \oplus B\right)$ ’ = $A \oplus B / \left(A + B\right) \oplus C = A + \left(B \oplus C\right) / \left(A \oplus B\right) \oplus \left(B \oplus C\right) = A \oplus C / \left(A \cdot B\right) \oplus \left(A \cdot C\right) = A$

Show solution

$\left(A \oplus B\right)$ ’ = $A \oplus B$ ? A thing can’t equal its own opposite ✗
$\left(A + B\right) \oplus C = A + \left(B \oplus C\right)$ ? Try $A = 1 , B = 0 , C = 1$ : left = 1 ⊕ 1 = 0, right = 1 + 1 = 1 ✗
$\left(A \oplus B\right) \oplus \left(B \oplus C\right) = A \oplus B \oplus B \oplus C = A \oplus$ 0 ⊕ $C = A \oplus C$ ✔ $\left(B \oplus B = 0\right)$
$\left(A B\right) \oplus \left(A C\right) = A$ ? Try $A = 1 , B = 1 , C = 1$ : left = 1 ⊕ 1 = 0, right = 1 ✗

Type 8: Fill in the missing gates (GT1, GT2)

Method

Learn the method

Write what the circuit computes, with GT1 and GT2 as unknowns.
Try each option. Put it in and simplify.
If unsure, test with numbers: pick a row, compute the circuit and the required output. They must match in every row.

Past paper questions

Asked in

2025 T3 Quiz 1Q1643 marks

Question: Y must equal $A B + B C + C A$ . A and B go into GT1. GT1’s output and C go into an AND. A and B also go into GT2. The AND output and GT2’s output go into an OR → Y. What are GT1 and GT2?

Show solution

Circuit: $Y =$ (GT1(A, B) · C) + GT2(A, B)
Try $\operatorname{XOR} , \operatorname{AND} : Y = \left(A \oplus B\right) C + A B$
Check the rows where Y should be 1 (two or more 1s):
- 011: (0⊕1)·1 + 0 = 1 ✔
- 101: (1⊕0)·1 + 0 = 1 ✔
- 110: (1⊕1)·0 + 1 = 1 ✔
- 111: (0)·1 + 1 = 1 ✔
- 001, 010, 100, 000 all give 0 ✔
All 8 rows match → XOR, AND

Asked in

2025 T3 End-term (AN)Q83 marks. Flawed question: no option fits.Key issue

Question: f must be $\left(\overline{x} + \overline{y}\right) z w$ . x goes through a NOT into GT2. y and z go into an AND, whose output also goes into GT2. z and w go into GT1. GT1 and GT2 go into a final AND → f.

Show solution

With GT1 = AND, GT2 = $\operatorname{NAND} : f = \operatorname{NAND} \left(\overline{x} , y z\right) \cdot z w = \left(x + \overline{y} + \overline{z}\right) \cdot z w =$ $\left(x + \overline{y}\right) z w$
That is $\left(x + \overline{y}\right) z w$ , not $\left(\overline{x} + \overline{y}\right) z w$ . Test $x = 0 , y = 1 , z = w = 1$ : required = (1 + 0)·1 = 1, but the circuit gives $\operatorname{NAND} \left(1 , 1\right) \cdot 1 =$ 0 ✗
We tried every pair of gates by computer, and none gives $\left(\overline{x} + \overline{y}\right) z w$ . The question has an extra NOT on x.
In the exam, write the official answer AND, NAND (it works if the NOT on x is removed).

Asked in

2025 T3 End-term (FN)Q83 marks. Flawed question.Key issue

Question: Same circuit, f must be $\left(x + \overline{y}\right) z w + \overline{z} w$ .

Show solution

With GT1 = OR, GT2 = $\operatorname{NAND} : f = \left(x + \overline{y} + \overline{z}\right) \left(z + w\right)$
Test $x = 1 , z = 1 , w = 0$ : required = (1)·1·0 + 0 = 0, circuit = (1)(1) = 1 ✗
No gate pair matches in all 16 rows (checked by computer). The key’s choice is OR, NAND, so write that.

Type 9: True or false on combinational vs sequential

Asked in

2025 T2 Quiz 2Q129 and 2026 T2 Quiz 2Q61 mark each

Question: “Sequential circuits do not utilize feedback loops, while combinational circuits do.” True or False?

Show solution

it’s the other way round. Sequential circuits use feedback (that’s how they remember). Combinational circuits have no feedback and no memory.

Practice set (new questions, same types)

X goes through XNOR(X, X), then XOR with X, then XNOR with X. Output?

1 \to 1 \oplus X = \overline{X} \to \overline{X} \odot X = 0

. Output is 0.

$F = A \cdot B + C$ , inputs random. $P \left(F = 1\right)$ ?

F = 0

only when

C = 0

and

A B = 0

: rows 000, 010, 100 → 3 zeros.

P = 5 / 8 = 0

.625.

$f = \Sigma m \left(3 , 5 , 6 , 7\right)$ . Always 1 when…?

Rows with two or more 1s → “2 or more inputs at logic 1”.

Exactly three of five inputs are 1. How many minterms?

⁵C₃ = 10.

Minterms of $F \left(A , B , C\right) = A B + \overline{C}$ ?

AB → 6, 7.

\overline{C} \to 0 , 2 , 4 , 6

. Join: Σ(0, 2, 4, 6, 7).

In one line: Push values through gate chains, simplify before counting probability, test every option row by row, and remember the papers’ Week 2 keys were wrong three times. Exam words: gate chain, uniformly random, probability, $\Sigma m$ , parity, ⁿCₖ, minterm list, Boolean identity, missing gate, feedback.

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