Question 1
8 marks
1(a) 4 marks
Ben installs burglar alarms. The alarm is made up of a door sensor and a motion sensor. When the alarm is set, the siren will sound if either the door sensor or the motion sensor detect movement.
The alarm also has a test mode setting. When the test mode setting is enabled, an engineer can check the sensors are working without the siren going off.
The inputs to the alarm are as follows.
| Input | System |
|---|---|
| A | Door sensor |
| B | Motion sensor |
| C | Alarm has been set |
| D | Test mode enabled |
Draw a logic circuit to show the logic that is used in the burglar alarm to determine if the siren goes off.
Mark scheme 1(a)
Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.
1 mark for each correct area including labelled inputs
[Image redrawn from the original mark scheme by an AI model (Claude Sonnet). It may contain mistakes.]
flowchart LR
A["A"] --> OR1["OR"]
B["B"] --> OR1
OR1 --> AND1["AND"]
C["C"] --> AND1
AND1 --> AND2["AND"]
D["D"] --> NOT1["NOT"]
NOT1 --> AND2
AND2 --> Out["Output"]
Alternatives (1 mark per area):
[Image redrawn from the original mark scheme by an AI model (Claude Sonnet). It may contain mistakes.]
flowchart LR
A["A"] --> OR1["OR"]
B["B"] --> OR1
C["C"] --> AND1["AND"]
D["D"] --> NOT1["NOT"]
NOT1 --> AND1
OR1 --> AND2["AND"]
AND1 --> AND2
AND2 --> O1["Output"]
A2["A"] --> OR2["OR"]
B2["B"] --> OR2
D2["D"] --> NOT2["NOT"]
OR2 --> AND3["AND"]
NOT2 --> AND3
AND3 --> AND4["AND"]
C2["C"] --> AND4
AND4 --> O2["Output"]
Guidance
Max 3 if any additional gates.
NOT gate must have only one input, AND/OR gates must have two inputs NOT gate must have circle
Ignore any names on gates
Examiner’s Comments
There were a range of marks on this question. More successful candidates gave all four gates accurately, but most candidates were able to gain at least one or two marks. Some candidates were unaware of what a logic circuit was and instead produced a range of different diagrams including flow charts.
1(b)(i) 1 mark
Simplify the Boolean expression ¬A v ¬B using De Morgan's First Law.
Mark scheme 1(b)(i)
¬ (A ∧ B)
Guidance
Accept NOT (A AND B) / other correct notations e.g.
¬ must be the correct way round
Examiner’s Comments
Many candidates gained full marks with the most common error being ¬(A v B)
1(b)(ii) 1 mark
Simplify the Boolean expression ¬(¬B) using double negation.
Mark scheme 1(b)(ii)
B
Guidance
Examiner’s Comments
Almost all candidates were able to gain a mark on this question.
1(b)(iii) 2 marks
Simplify the Boolean expression (A v B) ∧ (A v C) using distribution.
Mark scheme 1(b)(iii)
-
1 mark for A
-
1 mark for v
(B ∧ C)
A v (B ∧ C)
Guidance
Doesn’t need brackets
Allow other correct notation e.g. A+B.C
Allow either order e.g. (B∧C) v A
Allow Av(BC) / AvBC
Examiner’s Comments
The majority of candidates recognised that distribution meant that there was only one A following simplification but the most common error for the second mark was again to use OR instead of AND for B AND C