Logic gates, truth tables, K-maps and logic diagrams – IB Computer Science exam questions with mark scheme

1Standard5 marks

The logic circuit shown has inputs A, B and C and output Q.

(a) State the Boolean expression for Q. [2]

(b) Construct a truth table for the circuit, showing an intermediate column for (A AND NOT B) and the output Q for all input combinations. [3]

A AND (NOT B) correctly formed (allow $A\cdot\overline{B}$, A ∧ ¬B or equivalent notation)(1)
Q = (A AND NOT B) OR C (allow $A\overline{B} + C$, or C OR (A AND NOT B); reject A AND (NOT B OR C))(1)
Truth table has all 8 input combinations (000 to 111), in any consistent order(1)
Intermediate column A AND NOT B correct: 0,0,0,0,1,1,0,0 (for ABC = 000 to 111)(1)
Q column correct: 0,1,0,1,1,1,0,1 (follow through from the student's intermediate column; allow T/F for 1/0)(1)
Answer: Q = (A AND NOT B) OR C; Q = 0,1,0,1,1,1,0,1 for ABC = 000 to 111
Examiner tip: Add an intermediate column for each gate output; it makes errors easy to spot and the Q column can earn follow-through credit.

2Core3 marks

(a) Identify the logic gate whose output is 1 only when both of its inputs are 0. [1]

(b) Copy and complete the truth table below for an XOR gate and an XNOR gate, each with inputs A and B. [2]

A	B	A XOR B	A XNOR B
0	0
0	1
1	0
1	1

(a) NOR (accept NOT OR / OR followed by NOT; reject NAND)(1)
(b) XOR column correct: 0, 1, 1, 0 (for AB = 00, 01, 10, 11)(1)
(b) XNOR column correct: 1, 0, 0, 1 (allow follow through as the exact inverse of the candidate’s XOR column; allow T/F for 1/0)(1)
Answer: (a) NOR. (b) XOR: 0, 1, 1, 0; XNOR: 1, 0, 0, 1
Examiner tip: XOR is 1 when the inputs are different; XNOR is its inverse, 1 when the inputs are the same. Students often swap the two.

3Core4 marks

A lift control system uses three inputs:

D = 1 when the lift door is fully closed
I = 1 when a floor button inside the lift is pressed
O = 1 when a call button outside the lift is pressed

The motor output M = 1 only when the door is closed and at least one of the buttons is pressed.

(a) Construct a Boolean expression for M. [2]

(b) Copy and complete the output column of the truth table for M. [2]

D	I	O
0	0	0
0	0	1
0	1	0
0	1	1
1	0	0
1	0	1
1	1	0
1	1	1

(a) I OR O (as a bracketed term) / $I + O$(1)
(a) M = D AND (I OR O) / $M = D\cdot(I + O)$ (accept $M = D\cdot I + D\cdot O$; reject D AND I OR O without brackets unless precedence made clear)(1)
(b) M = 0 for all four rows where D = 0(1)
(b) M = 0, 1, 1, 1 for DIO = 100, 101, 110, 111 (allow T/F for 1/0)(1)
Answer: (a) M = D AND (I OR O). (b) M = 0,0,0,0,0,1,1,1 for DIO = 000 to 111
Examiner tip: Brackets matter: D AND I OR O is read as (D AND I) OR O, which would let the motor run with the door open.

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4Standard4 marks

A logic circuit is defined by the Boolean expression

$$Q = \overline{(A + B)} \oplus \left(B\cdot\overline{C}\right)$$

(a) Construct the logic diagram for this expression, using standard gate symbols. [3]

(b) State the value of Q when A = 0, B = 1 and C = 0. [1]

(a) A NOR gate with inputs A and B (accept an OR gate followed by a NOT gate)(1)
(a) A NOT gate on C, whose output feeds a 2-input AND gate with B(1)
(a) Outputs of the NOR and AND gates feed an XOR gate whose output is Q (XOR symbol must show the extra curved line; ignore wire neatness, allow labels on gates if symbols are recognisable)(1)
(b) Q = 1(1)
Answer: (a) NOR(A, B) and AND(B, NOT C) feeding an XOR gate. (b) Q = 1
Examiner tip: The XOR symbol is an OR shape with an extra curved line at the input side; without it the gate is read as OR and the final mark is lost.

5Standard5 marks

The logic circuit shown has inputs A, B and C. P is the output of the NAND gate, R is the output of the XOR gate and Q is the output of the circuit.

(a) Copy and complete the truth table for the circuit. [3]

A	B	C
0	0	0
0	0	1
0	1	0
0	1	1
1	0	0
1	0	1
1	1	0
1	1	1

(b) The whole circuit could be replaced by a single gate. Identify this gate and justify your answer using the truth table. [2]

(a) P column correct: 1, 1, 1, 1, 1, 1, 0, 0(1)
(a) R column correct: 0, 1, 1, 0, 0, 1, 1, 0(1)
(a) Q column correct: 0, 0, 0, 0, 0, 0, 0, 1 (follow through: Q = NOT(P OR R) from the candidate’s P and R columns)(1)
(b) A (3-input) AND gate (with inputs A, B and C) (allow follow through from candidate’s Q column if it gives a recognisable single gate)(1)
(b) because Q = 1 only when A, B and C are all 1 / only in the row 111(1)
Answer: (a) P = 1,1,1,1,1,1,0,0; R = 0,1,1,0,0,1,1,0; Q = 0,0,0,0,0,0,0,1. (b) A 3-input AND gate: Q is 1 only when A = B = C = 1.
Examiner tip: Work out P and R row by row before Q; the NOR gate outputs 1 only when both of its inputs are 0.

6Standard3 marks

Using the laws of Boolean algebra, simplify the expression

$$X = \overline{\overline{A}\cdot\overline{B}}\cdot\left(A + \overline{B}\right)$$

Show your working. [3]

Applies De Morgan’s law (and double negation): $\overline{\overline{A}\cdot\overline{B}} = A + B$(1)
Correct further step, e.g. $(A + B)(A + \overline{B}) = A + B\cdot\overline{B}$ (distributive law) or expansion to $A + A\overline{B} + AB + B\overline{B}$(1)
$X = A$ (with $B\cdot\overline{B} = 0$ / absorption shown; a correct final answer with no working scores this mark only)(1)
Answer: $X = A$
Examiner tip: Students often apply De Morgan to only one bar; break the long bar, change AND to OR, and then cancel the double bars.

7Hard5 marks

The truth table below defines the output Q of a circuit with inputs A, B and C.

A	B	C	Q
0	0	0	1
0	0	1	0
0	1	0	1
0	1	1	0
1	0	0	1
1	0	1	1
1	1	0	1
1	1	1	0

(a) Construct a Karnaugh map for Q and show the groups you would use to simplify it. [3]

(b) Hence state the simplified Boolean expression for Q. [2]

(a) K-map drawn with Gray-code ordering (e.g. AB = 00, 01, 11, 10 against C = 0, 1) and the five 1s placed correctly (cells ABC = 000, 010, 100, 101, 110)(1)
(a) A group of four 1s covering every cell where C = 0 (accept the group shown wrapping round the edges if the map is laid out the other way)(1)
(a) A group of two covering ABC = 100 and 101 (reject a group that includes 111 or 001)(1)
(b) $\overline{C}$ term (from the group of four)(1)
(b) $Q = \overline{C} + A\overline{B}$ (accept NOT C OR (A AND NOT B); follow through from candidate’s groups if valid)(1)
Answer: $Q = \overline{C} + A\cdot\overline{B}$
Examiner tip: Groups must be powers of 2 and as large as possible; a group of four gives a single-variable term, so look for it first.

8Stretch5 marks

A circuit with inputs A, B, C and D has output

$$Q = \overline{A}\,\overline{B}\,\overline{C}\,\overline{D} + \overline{A}\,\overline{B}\,C\,\overline{D} + A\,\overline{B}\,\overline{C}\,\overline{D} + A\,\overline{B}\,C\,\overline{D} + \overline{A}\,B\,\overline{C}\,D + \overline{A}\,B\,C\,D + A\,B\,\overline{C}\,D + A\,B\,C\,D$$

(a) Use a Karnaugh map to simplify the expression for Q. Show your groups. [4]

(b) An engineer claims the circuit could be built with a single 2-input gate. Determine whether this claim is correct. [1]

(a) 4 × 4 K-map with Gray-code order on both axes (00, 01, 11, 10) and all eight 1s placed correctly(1)
(a) Group of the four corner cells (wrapping round both edges) giving $\overline{B}\,\overline{D}$(1)
(a) Group of the four central cells (B = 1, D = 1) giving $B\,D$(1)
(a) $Q = \overline{B}\,\overline{D} + B\,D$ (accept equivalent, e.g. (NOT B AND NOT D) OR (B AND D))(1)
(b) Claim is correct: Q = B XNOR D / $\overline{B \oplus D}$, a single XNOR gate, since A and C do not affect the output (allow follow through from candidate’s (a))(1)
Answer: $Q = \overline{B}\,\overline{D} + BD = B$ XNOR $D$; the claim is valid (one XNOR gate, A and C are redundant).
Examiner tip: The four corners of a 4-variable K-map are adjacent; missing this wrap-around group leaves two pairs and a longer expression.

9Standard4 marks

A technician builds a circuit that is meant to behave as a single XOR gate with inputs A and B. She tests it and records:

A	B	Output
0	0	1
0	1	0
1	0	0
1	1	1

(a) Identify the gate that the circuit actually behaves as. [1]

(b) NAND is described as a universal gate. Explain how a NOT gate can be built using only a NAND gate. [2]

(a) XNOR (accept NOT XOR / equivalence gate)(1)
(b) Connect the single input to both inputs of the NAND gate / tie the two inputs together(1)
(b) so the inputs are always equal: input 0 gives NAND(0,0) = 1 and input 1 gives NAND(1,1) = 0, so the output is the inverse of the input(1)
(c) Any one from: any logic function / every other gate can be made from NAND gates alone; only one type of gate needs to be manufactured, which lowers cost/simplifies production; NAND is fast/needs few transistors to make(1)
Answer: (a) XNOR. (b) Join both NAND inputs to the one input; NAND(0,0)=1, NAND(1,1)=0, so it inverts. (c) e.g. every gate can be built from NAND alone, so only one gate type has to be mass-produced.
Examiner tip: For (b) state the connection and show it inverts; just writing ‘use a NAND gate’ earns nothing.

10Hard6 marks

A bank safe has three key switches, one held by each of three managers A, B and C. Each switch gives 1 when its key is turned. The safe opens (S = 1) when at least two of the three keys are turned.

(a) Construct the truth table for S. [2]

(b) Use a Karnaugh map to derive a simplified Boolean expression for S. [3]

(a) All 8 input combinations listed (000 to 111)(1)
(a) S = 0, 0, 0, 1, 0, 1, 1, 1 for ABC = 000 to 111 (allow T/F for 1/0)(1)
(b) K-map with the four 1s placed correctly (ABC = 011, 101, 110, 111)(1)
(b) Three overlapping groups of two, each including the cell ABC = 111 (allow groups shown in any valid layout)(1)
(b) $S = AB + AC + BC$ (accept (A AND B) OR (A AND C) OR (B AND C); a redundant extra term such as $ABC$ is not simplest form and does not score)(1)
(c) Three 2-input AND gates (AB, AC, BC) whose outputs feed a 3-input OR gate / two 2-input OR gates, output S (follow through from (b))(1)
Answer: (a) S = 0,0,0,1,0,1,1,1. (b) $S = AB + AC + BC$. (c) Three AND gates feeding an OR gate.
Examiner tip: Cells may belong to more than one group; reusing the 111 cell in each pair gives the simplest expression.

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