Mark schemes
Q1.
Marks are for AO2 (apply)
1 mark for having A and B connected to (different) NOT gates;
1 mark for an AND gate connected to C and to the output of a NOT gate;
1 mark for an OR gate connected to NOT A and NOT B AND C and outputting to Q;
 
Max 2 if circuit not fully correct.
[3]
Q2.
 
(a)     All marks AO2 (apply)
 
1 mark: inputs D and M connected to an OR gate;
1 mark: inputs K and output of OR gate connected to AND gate plus output connected to A;
A a logically equivalent circuit
2
(b)     All marks AO2 (apply)
A = (D + M) · K
1 mark: D + M somewhere in expression, even if full expression incorrect
1 mark: fully correct expression
A A logically equivalent expression
2
(c)     1 mark for AO1 (understanding), 1 mark for AO2 (application) and 1 mark for AO1 (knowledge)
AO1 (understanding):1 mark: Flip-flop will store the state of its input / / Flip-flop acts as memory;
AO2 (application):1 mark: Insert into circuit between the output of the OR gate and the AND gate / / after the AND gate;
AO1 (knowledge):1 mark: Clock signal / / trigger / / signal to indicate when the value (of the input) should be stored / read;
3
[7]
Q3.
Mark is for AO1 (understanding)
OR;
A. A+B // +
[1]
Q4.
Marks are for AO2 (apply)
Marking guidance for examiners
•   Award marks for working out until an incorrect step has been made.
•   If, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. Example, if the expression P .P .(P + Q) + P .P .1 was changed to P .(P + Q) + P .0, the candidate would get one mark for simplifying the first part to P .(P + Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P .0 (ie to 0)
Mark as follows:
1 mark for final answer X
Max 3 marks for working:
•   1 mark for each application of an identity other than cancelling NOTs that produces a simpler expression.
•   1 mark for expanding brackets
•   1 mark for putting an expression into brackets that would lead to a simpler expression.
Note: a simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.
Max 3 if answer is correct but any incorrect working or significant steps of working is missing.
Example working (1)
X. X + X. Y̅ + Y. X + Y. Y̅     [expansion of brackets]
X + X. Y̅ + Y. X + 0       [use of X • X = X and Y. Y̅ = 0]
X(1 + Y̅ + Y) or X + X(Y + Y̅)  [taking X outside of brackets]
Alternative example working (2)
X + (Y. Y̅)          [Use of distributive law]
X + 0             [Y. Y̅ = 0]
X              [Recognising X + 0 = X]
[4]
Q5.
Marks are for AO2 (apply)
Marking guidance for examiners
•        Award marks for working out until an incorrect step has been made.
•        If, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. Example, if the expression P.P.(P + Q) + P.P.1 was changed to P.(P+Q) + P.0 the candidate would get one mark for simplifying the first part to P.(P+Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P.0 (i.e. to 0).
Mark as follows
1 mark for final answer:
Max 3 for working; Max 3 if answer is correct but any incorrect working or significant steps of working is missing:
1 mark for a successful application of De Morgan’s Law that would lead to a simpler expression. Max 2 for applications of De Morgan’s Law.
1 mark for applying an identity other than cancelling nots that produces a simpler expression. Max 2 for applying identities.
1 mark for expanding brackets or putting an expression into brackets that would lead to a simpler expression. Max 2 for expanding brackets or putting an expression into brackets.
Note: a simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.
Example working (1)
 
	[application of De Morgan’s law]	
	[expansion of brackets]	
	[use of identities  = 0 and X+0 = X]	
	[expansion of brackets]	
	[use of identity X.X = X twice]	
	[use of identity X + X = X]	
Alternative example working (2)
 
	[application of De Morgan’s Law]	
	[application of De Morgan’s Law twice]	
	[expansion of brackets]	
	[use of identities  and X + 0 = X]	
	[application of De Morgan’s Law]	
	[application of De Morgan’s Law]	
	[expansion of brackets]	
	[use of identity X.X = X twice]	
	[use of identity X + X = X]	
Alternative example working (3)
 
	[application of De Morgan’s Law]	
	[De Morgan’s Law twice]	
	[Expansion]	
	[Identity  and A + 0 = A]	
	[application of De Morgan’s Law]	
	[application of De Morgan’s Law]	
	[Expansion]	
	[Identity  and B.B = B]	
	[Final answer via identity A + A = A]	
[4]
Q6.
Marks are for AO2 (apply)
Marking guidance for examiners
•   Award marks for working out until an incorrect step has been made.
•   If, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. Example, if the expression P.P.(P+Q) + P.P.1 was changed to P.(P+Q)+P.0, the candidate would get one mark for simplifying the first part to P.(P+Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P.0 (i.e. to 0)
Mark as follows
1 mark for final answer A
3 marks for working
Max 3 for working. Award up to two marks for applying each one of the three techniques (one mark per application):
•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression
•   applying an identity other than cancelling NOTs that produces a simpler expression
•   successfully expanding brackets
Note: A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.
Note: Any application of De Morgan’s Law or expanding brackets which result in an expression which should be bracketed must be shown with brackets to be awarded a mark.

[4]
Q7.
Marks are for AO2 (apply)
Marking guidance for examiners
•   Award marks for working out until an incorrect step has been made.
•   If, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. Example, if the expression P.P.(P+Q) + P.P.1 was changed to P.(P+Q)+P.0, the candidate would get one mark for simplifying the first part to P.(P+Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P.0 (ie to 0)
Mark as follows
MAX 3 marks for working
Award one mark each for applying the techniques below:
•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression
•   successfully expanding brackets
Award one mark for each application of a Boolean identity MAX 2.
Note: A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.
Example working (1)
 
Example working (2)
 
Example working (3)
 
1 mark for final answer A XOR B // A Exclusive OR B // A EOR B // A EXOR B // A ⨁ B
[4]
Q8.
All marks AO2 (apply)
Award up to four marks for the working shown, but Max 3 if the response does not show that (A + B) · (B + C · (D + D̅)) = A · C + B
1 mark for each application of an identity or theorem that produces an expression that is logically equivalent to the original expression but uses fewer logical operators.
1 mark for a successful application of the distribution law – only one mark, regardless of how many times this has been applied
Continue marking until an incorrect step is encountered. If a student misses out some steps but does not make an error then continue marking.
Example Solution 1
 
(A + B) ⋅ (B + C ⋅ (D + D̅))
= (A + B) ⋅ (B + C ⋅ 1)
= (A + B) ⋅ (B + C)
= A ⋅ B + A ⋅ C + B ⋅ B + B ⋅ C
= A ⋅ B + A ⋅ C + B + B ⋅ C
= A ⋅ B + A ⋅ C + B
= A ⋅ C + B	 
By identity X + X̅ = 1
By identity X ⋅ 1 = X
Using distribution law
By identity X ⋅ X = X
By redundancy theorem X + X ⋅ Y = X
By redundancy theorem X + X ⋅ Y = X	
Example Solution 2
 
(A + B) ⋅ (B + C ⋅ (D + D̅))
= (A + B) ⋅ (B + C ⋅ 1)
= (A + B) ⋅ (B + C)
= A ⋅ C + B	 
By identity X + X̅ = 1
By identity X ⋅ 1 = X
Using distribution law (this jump is worth 2 marks)	
[4]
Q9.
All marks AO2 (apply)
Marking guidance for examiners
•   Award marks for working out until an incorrect step has been made.
•   If, in any one step, a candidate is simplifying different parts of an expression simultaneously award all relevant marks for this multiple stage but don’t award any further marks for working in any parts simplified incorrectly. For example, if the expression P .P .(P + Q) + P .P .1 was changed to P .(P + Q) + P .0, the candidate would get one mark for simplifying the first part to P .(P + Q) and could get further marks for correctly simplifying this part of the expression further but should not be awarded marks for simplifying the incorrectly changed part P .0 (ie to 0)
1 mark: for final answer: B + C
MAX 3 for working. Award up to two marks for applying each one of the three techniques (one mark per application):
•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression.
•   applying an identity other than cancelling NOTs that produces a simpler expression.
•   successfully expanding brackets.
Note: A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.
Example Working (1)
 
Example Working (2)
 
[4]
Q10.
(a)     Mark is for AO1 (knowledge)
XOR // EOR // Exclusive OR;
1
(b)     Mark is for AO2 (apply)

1
(c)     Mark is for AO2 (apply)
C;
1
(d)     Marks are for AO2 (apply)
 
C	B	A	T	S	R	
0	0	0	0	0	0	
0	0	1	0	0	1	
0	1	0	0	1	1	
0	1	1	0	1	0	
Mark as follows:
1 mark: column T correct;
1 mark: column S correct;
1 mark: column R correct;
3
(e)     Mark is for AO2 (analyse)
Use this circuit on the binary number to be subtracted and add the result to the other binary number;
A. Any equivalent answers
R. Number to be added IS negative
1
(f)     3 marks for AO1 (knowledge) and 3 marks for AO1 (understanding)
1 mark for AO1 (knowledge): (increase the) data bus width;
1 mark for AO1 (understanding): enables more bits (A. data) to be transferred between main memory and the processor at one time (so fewer read/write operations needed);
1 mark for AO1 (knowledge): (increase the) clock speed;
1 mark for AO1 (understanding): enables more instructions to be executed per unit of time/second (A. calculations/operations/commands instead of instructions) // each individual instruction could be executed sooner / more quickly (A. calculation/operation/command instead of instruction);
1 mark for AO1 (knowledge): (increase the) amount of cache memory;
1 mark for AO1 (understanding): cache memory is faster than main memory so the more that can be stored in cache memory the less frequently the main memory needs to be accessed;
1 mark for AO1 (knowledge): (increase the) word length;
1 mark for AO1 (understanding): larger word size means that the processor can process more bits in one go;
1 mark for AO1 (knowledge): (change the) type of cache memory;
1 mark for AO1 (understanding): some types of cache memory can be accessed faster;
A. using memory with a faster access speed
1 mark for AO1 (knowledge): (increase the) number of general purpose registers;
1 mark for AO1 (understanding): more intermediate results/variables can be kept in processor registers rather than in main memory;
1 mark for AO1 (knowledge): (increase the) address bus width;
1 mark for AO1 (understanding): enables the processor to access a larger number of main memory locations (meaning it will not need to make as much use of virtual memory this will mean that system performance is improved);
A. allows more main memory to be installed
R. How improves mark if it is not relevant for the factor stated.
NE How improves of "program will execute faster"
Note: Marks for the factor can be awarded in either the "factor" or "how improves" part of an answer
6
[13]
Q11.
(a)  All marks AO2 (apply)
 
Input	 	Outputs	
A	B	C	D	E	F	G	H	
0	0	0	0	0	0	0	0	
0	0	1	1	0	0	0	1	
0	1	1	0	1	0	1	0	
1	1	1	0	1	0	1	1	
1 mark: Column D correct or column E or column F correct
1 mark: Column G correct
1 mark: Column H correct
Max 2 if any incorrect values in table
3
(b)  All marks AO2 (apply)
B⋅C+A⋅(B⨁C)
1 mark: B⋅C or B⨁C somewhere in expression
1 mark: A is ANDed with B⨁C
1 mark: Fully correct expression
A. award second mark even if brackets around B⊕C are missing
A. use of AND, OR, XOR instead of symbols
A. (B̅⋅C)+(B⋅C̅) for (B⨁C)
If a student has written an expression but then tried to simplify it and made an error then mark the original expression that the student has written down and ignore the simplification.
3
(c)  Mark is for AO2 (analyse)
It adds together its inputs // it is a full adder circuit;
N.E. half-adder, adder
1
[7]