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User: 24530 Boolean algebra INTRODUCTIONBoolean algebra is a form of algebra named after George Boole who originally developed it in the mid-1800s. The study of Boolean algebra is closely linked to logic gates, so this chapter should be read in conjunction with Chapter 31. The basic principle is that logical expressions can be evaluated that will result in one of two results/outcomes – either TRUE or FALSE. For example, the following are examples of Boolean expressions: The button has been pressed5 < 10 Age > 17 and hold a driving licence SPECIFICATION COVERAGE 3.6.4 Logic gates3.6.5 Boolean algebra LEARNING OBJECTIVES In this chapter you will learn: •that Boolean algebra produces a result that either equals TRUE or FALSE•how truth tables are used to represent Boolean expressions •how to use the AND, OR and NOT operators on their own or groupedtogether •how to use NAND, NOR and XOR operators•how to simplify Boolean expressions •how De Morgan’s Law allows Boolean expressions to be created usingonly NAND or NOR operators.
Although Boolean logic predates computers, it has become an important aspect of computing, as the result is one of two states, which equates to binary and the way the electronic circuitry of the processor works. For example, for the first expression: The button has been pressed , we could represent the input as A as follows: ●A = 1 where 1 means that the statement is TRUE ●A = 0 where 0 means the statement is FALSE.KEYWORD Boolean expression: an equation made up of Boolean operations.246 30 Boolean algebraAs there is one input there are two possible results, 0 and 1. Boolean logic can be used to evaluate statements with any number of inputs to return a TRUE or FALSE value. The statement about the driving licence above has two inputs. To evaluate it: At least 17 = 1Under 17 = 0andHold a licence = 1No licence = 0In this case there are four possible inputs: 00, 01, 10, 11.
●Truth tablesWe can use a truth table to combine the permutations of 0s and 1s and workout which shows whether the answer is TRUE or FALSE. Table 30.1 shows all the possible inputs and the output of each combination as follows.In this example: A = at least 17 and B = Holds a licence.
Q shows the possible results.
Inputs OutputAB Q 00 01 0 10 011 1T able 30.1 A truth table This particular truth table is an example of an AND gate as both inputs need to be 1 in o rder to generate an output of 1. In our example above you have to be at least 17 and hold a driving licence to legally drive a car. The two inputs need to be ANDed together to generate the final result.In this case there is only one combination of A and B that will lead to a TRUE statement being returned, which is where A and B are both 1. When creating Boolean statements it is possible to use the relational operators in Table 30.2. Operator Name of operator‹l ess than‹= les s than or equal to== equal to!= not equal to ›= greater than or equal to›g reater thanTable 30.2 Relational operatorsKEYWORDS Truth table: a method of representing/calculating the result of every possible combination of inputs in a Boolean expression.
AND gate: result is true if both inputs are true.
KEYWORDBoolean operation: a single Boolean function.Statements can be combined to form more complex expressions and this is done using six main Boolean operations: AND, OR, NOT, NAND, NOR and XOR. We will look at each of these in turn.OR OPERATION ●AND operation As we saw in the first example, in an AND statement, all conditions (inputs) must be TRUE to generate a TRUE output. For example in an embeddedsystem to control a lift, you might evaluate this statement: Button has been pressed AND Door is closedwhere: A = Button pressedB = Doors closed.A = 1 means the button has been pressed, A = 0 means the button has notbeen pressed.B = 1 means the door is closed, B = 0 means the door is not closed.Q is whether or not the lift should move.A AND B = Q, which can also be notated as A.B = Q KEYWORD OR: Boolean operation that outputs true if either of its inputs are true.Inputs OutputAB Q 00 01 010 011 1T able 30.3 Truth table for AND The expression therefore is only TRUE when A and B are both 1. You could also say that the expression is TRUE when Q = 1. ●OR operationAn OR expression can return a TRUE result when any of the inputs are true. Consider the following expression that could be used to validate an employee’s ID: Proof of ID is Passport OR Driving Licencewhere: A = PassportB = Driving Licence.A = 1 means the employee has a passport, A = 0 means the employee doesnot have a passport.B = 1 means the employee has a driving licence, B = 0 means the employee does not have a driving licence.Q means that the employee has either a passport or a driving licence and therefore has at least one valid form of ID. This is written as A+B = Q with the + representing the OR expression.24724830 Boolean algebraInputs OutputAB Q 00 01 110 111 1T able 30.4 Truth table for ORAQ 0110Table 30.5 Truth table for NOTKEYWORD NOT: Boolean operation that inverts the result so true becomes false and false becomes true.
Inputs Intermediate step Output (from A AND NOT B)AB N OT B Q00 1 0 01 0 010 1 1 11 0 0T able 30.6 Extended truth table for A AND NOT BIn this example there are three possible TRUE results. As long as the employee has one of the types of ID, then this results in a TRUE result. ●NOT operationThe NOT statement inverts the input so that TRUE becomes FALSE and FALSE becomes TRUE.
This is written as Q = NOT A or Q = A. Notice the overbar above the A,which is standard notation for NOT. Notice that the results are inverted so FALSE becomes TRUE and TRUE becomes FALSE. The NOT statement can be used in combination with other Boolean expressions to create more complex selections. For example, when searching the web, it is possible to use the minus key (–) to exclude results that are nothing to do with your topic. This effectively uses a NOT operation. Consider the following expression for a web search about Python programming: Python – snake is the same as Python NOT snakeA = 1 if Python is found; A = 0 if Python is not found.B = 1 if snake is found; B = 0 if snake is not found.When we NOT the B input we get a TRUE if snake is not found (as we have inverted it).
This is then ANDed with the A input to generate the final outcome of TRUE or FALSE. This is because in a search engine there is an assumed AND between the Python and the NOT which would make the full statement read as Python and NOT snake . As you can see this is the logic we require for the search.
The overall search result can be represented as in Table 30.6.NOR OPERATION249●Combining AND and OR expressions AND expressions can be combined with OR expressions to create morecomplex statements. For example, to input data from a barcode: Barcode scanner on and barcode scannedorBarcode number input manuallyA = Barcode scanner on: A = 1 means yes, A = 0 means no.B = Barcode scanned: B = 1 means yes, B = 0 means no.C = Barcode number input manually: C = 1 means yes, C = 0 means no.Q indicates whether the data from the barcode has been read or not, eitherautomatically or manually. For data to be input from the barcode, the scanner must be on and barcodescanned, or the barcode must be input manually.In this case,Q = A AND B OR Cwhich can also be written as: Q = A.B+CIn Boolean notation, + means OR and . means AND. ●NAND operation NAND is a combination of NOT and AND and produces a TRUE resultif any of the inputs are false. It is commonly used to create NAND gates on inte grated circuits, which can be used for example, to create solid statedrives (see Chapter 35).The truth table would be as shown in Table 30.7. KEYWORDSNOR: Boolean operation that outputs true if all of its inputs are false. NOR gate: result is true if both inputsare false.Inputs OutputAB Q 00 101 110 111 0T able 30.7 Truth table for NANDKEYWORDS NAND: Boolean operation that outputs true if any of the inputs are false. NAND gate: result is true if any of the inputs are false. This is written as A.B, which means NOT A.B. This could be described as the inverted form of A.B.
●NOR operationThe NOR or NOT OR expression results in a TRUE value only if all inputs are FALSE. It means that the answer is TRUE if it is neither A nor B. It is used to create NOR gates on integrated circuits, which can be used forexam ple to make CMOS devices (see Chapter 35).25030 Boolean algebraInputs OutputAB Q 00 101 010 011 0T able 30.8 Truth table for NOR KEYWORDXOR: Boolean operation that is true if either input is true but not if both inputs are true.The truth table would be as shown in Table 30.8.
This is written as A+B, which means it is NOT A+B. This could be described as the inverted form of A+B. ●XOR operationThe exclusive OR expression produces a TRUE result only when one of theinputs is TRUE and the other is FALSE. If they are both TRUE it returns aFALSE result. It can be used to carry out bitwise operations and to createan adder in logic circuits. There is an example of this in the next chapter. The truth table would be as shown in Table 30.9.
Inputs OutputAB Q00 0 01 110 111 0T able 30.9 Truth table for XORThis is written as Q = A ⊕B, which means Q is true when either A or B are true, but not when both are true.
●Simplifying Boolean expressionsWhen using Boolean expressions it is good practice to reduce the expression into its simplest form. As Boolean algebra is used to create logicgates, simplifying the expressions also simplifies the actual circuit that willbe built, reducing the number of components needed, which in turn willmake the circuit cheaper to make, more efficient in operation and morereliable as fewer gates are being used.To help visualise the process for ensuring that Boolean expressions are in their simplest form you can run it through a truth table. For example, takethe expression A.B+A. The values of A can be 0 or 1 and the values of Bcan be 0 or 1 leading to four possible inputs: 00, 01, 10, 11.
The first part of the statement is A.B so the result is true when A and B = 1.
The truth table would look like Table 30.10.SIMPLIFYING BOOLEAN EXPRESSIONS251Next we look at the +A part of the expression, which means OR A in Boolean expressions. This means that (A.B)+A will be true when A and B is 1 or A is 1.AB A .B000 010100111Table 30.10 Truth table for A AND BA B A.B (A.B)+A00 0 0 10 0 101 0 011 1 1Table 30.11 Truth table for (A.B)+A Looking at the final column of Table 3.11 you can see that (A.B)+A is only true when A is true. Therefore the expression can be reduced to A: (A.B)+A = AAn expression may be made up of many variables, usually referenced as letters (A, B, C etc) each of which can produce a result of 0 or 1. This can lead to the creation of complex Boolean expressions. Therefore rules have been developed as a method of simplifying expressions. Table 30.12 shows the common rules associated with what are known as Boolean identities. In this section we will look at how the rules can be used to simplify an expression. Note that De Morgan’s Law is covered separately in the next section. Table 30.12 Common rules associated with Boolean identitiesIdentity name AND form OR form Identity A.1 = A A+0 = ANull (or Dominance) Law A.0 = 0 A+1 = 1 Idempotence Law A.A = A A+A = AInverse Law A.A = 0 A+A = 1 Commutative Law A.B = B.A A+B = B+Associative Law (A.B).C = A.(B.C) (A+B)+C = A+(B+C) Distributive Law A+B.C = (A+B).(A+C) A.(B+C) = A.B+A.CAbsorption Law A.(A+B) = A A+A.B = ADe Morgan’s Law (A.B) = A+B (A+B) = A.B Double Complement Law A = A252 30 Boolean algebraTable 30.13 Explanations of the main identities and rulesA.B = B.A The order in which two variables are ANDed makes no differenceA+B = B+A The order in which two variables are ORed makes no differenceA.0 = 0 A variable ANDed with 0 equals 0 A+1 = 1 A variable ORed with 1 equals 1A+0 = A A variable ORed with 0 equals the variable A.1 = A A variable ANDed with 1 equals the variableA.A = A A variable ANDed with itself equals the variable A+A = A A variable ORed with itself equals the variableA.A = 0 A variable ANDed with its inverse equals 0 A+A = 1 A variable ORed with its inverse equals 1 A = A A variable that is double inversed equals the variable (A.B).C = A.(B.C) It makes no difference how the variables are grouped together when ANDed (A+B)+C = A+(B+C) It makes no difference how the variables are grouped together when ORed A.(B+C) = A.B+A.C The expression can be distributed or factored out, meaning that variables can be moved in and out of brackets either side of the expression. In English this expression would be A AND (B OR C) = (A AND B) OR (A AND C). The rules can be used to simplify expressions. A+A.B = A This means A OR (A AND B) = A and can be proved by looking at the truth table (Table 30.11) above.To use an example with an inverse: A+ A.B = A+B This means that A OR (NOT A AND B) = A OR B. It could be proved that this is true by drawing out a truth table for the two expressions. Alternatively this can be deduced by logical reasoning: ●Suppose that A is 1, then A+ A.B will be 1.
●On the other hand, suppose that A is 0. Then A+ A.B will only evaluate to1 if A.B is 1, which will only be true if B is 1.
●So, A+ A.B = 1 when A = 1 or when B = 1, hence it is equivalent to A+B.To use an example that uses distribution: (A+B).(A+C) = A+B.C This means (A OR B) AND (A OR C) is the same as A OR (B AND C). This can be achieved by factoring out the A.
The more complex example below shows the stages that you might go through to simplify an expression: (A+B).C. C+(A+ A).BStarting with the first part of the expression (A+B).C. C: (A+B).0 Any value ANDed with its inverse = 00 Any value ANDed with 0 = 0DE MORGAN’S LAW253Now taking the second part of the expression (A+ A).B: (A+A) Any value ORed with its inverse = 1.B Any value ANDed with 1 is the variable = BSo putting both parts of the expression together we get 0+B. When you OR a variable with 0 you get the variable so the answer is B. Therefore we can say that the simplified expression of (A+B).C.
C+(A+ A).B is B. ●De Morgan’s LawDe Morgan’s Law is another way of simplifying Boolean statementsby invert ing all the variables, changing ANDs to OR and ORs to ANDs and then inverting the whole expression. One application is to simplifystatements so that only NAND or NOR gates are used. This makes it muchsimpler to create logic gates and circuits, which in turn makes it easierto design and build microprocessors. For example, solid state drives aremade up of NAND gates.In simple terms this means that ANDs can replace ORs and ORs can replace ANDs. This works as long as the rest of the expression is changed,or negated to take account of this.
The basic principles are: ●Rule 1: NOT (A AND B) is the same as (NOT A) OR (NOT B)●Rule 2: NOT (A OR B) is the same as (NOT A) AND (NOT B)In a lgebraic notation: ●Rule 1: A.B is the same as A+B●Rule 2: A+B is the same as A.B The Venn diagram in Figure 30.1 shows the concept. The area outside the Venn diagram is X. We can define X as being: ●NOT in A+B and●NOT in A and also NOT in B.AAB B + Figure 30.1 Venn diagram representing NOT (A and B)KEYWORDDe Morgan’s Law: a process for simplifying Boolean expressions This could be written as follows: X = A+BX = A.B A.B = A+B254 30 Boolean algebraWhen using De Morgan’s Law to write Boolean expressions, the following steps must be taken: ●You can only apply De Morgan’s Law to one operator at a time.
●If the operator is an OR change it to an AND, and vice versa.
●Invert the terms on either side of the operator.
●Invert the entire expression.
For example, Figure 30.2 shows how to simplify the expression: A+ A.B.A + A B ApplyDe Morgan's Law to this operatorApplyDe Morgan's Law to this operator= ======A + A + B A + A + B A A + B A (A + B)A A + A B O + A BA B Figure 30.2 Applying De Morgan’s Law This can also be shown by a truth table: AB A B A.B A+A.B A+A.B 0011 1 1 0110 0 0 1 10010 1 0 1100 0 1 0The final column of the truth table only has a 1 in the row where A = 0 and B = 1, therefore the result obtained using De Morgan’s Law is confirmed by the truth table, i.e. that the expression is equivalent to A.B. Practice questions can be found at the end of the section on page 264.
TASKS1 Write an example of a Boolean expression and draw the corresponding truth table for each of the following expressions.
a)ANDb)OR c)NOTd)NOR e)NANDf)XOR2 Write an example of a Boolean expression for a real-life situation where you could use any combination of: a)ANDb)OR c)NOT3 Give an example of where you could use the following Boolean expressions.
a)NANDb)NOR c)XORDE MORGAN’S LAW2554 Simplify the following expressions. a)(A+¯A).Bb)(A+B)+B c )A.(B+B) d)B.(A+B) e)A.B.C+ A.B5 What are the principles of De Morgan’s Law? STUDY / RESEARCH TASKS1 Use Venn diagrams to represent each of the six main expressions.
2 Research the relationship between the NAND expression and solid state drives.3 Research the relationship between the NOR expression and CMOS.
4 How is Boolean logic applied to web searching?5 Set yourself some Boolean expressions and then go through the process of simplifying them, using De Morgan’s Law where necessary.KEY POINTS•Boolean algebra returns avalues that is either TRUE orFALSE. •Truth tables are a visualmethod of showing the resultsof a Boolean expression. •You need to know how toconstr uct AND, OR, NOT,NAND, NOR and XORstatements and combinethem to create more complexexpressions. •You should always try tocreate Boolean expressions in their simplest form.
•De Morgan’s Law is a methodthat can be used to simplify Boolean algebra expressions.
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3.6.4 Logic gates
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Construct truth tables for the following logic gates:
NOT
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Students should know and be able to use ANSI/IEEE standard 91-1984 Distinctive shape logic gate symbols for these logic gates.
Be familiar with drawing and interpreting logic gate circuit diagrams involving one or more of the above gates.
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3.6.5 Boolean algebra
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Be familiar with the use of Boolean identities and De Morgan’s laws to manipulate and simplify Boolean expressions.
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De Morgan', 'index7': '', 'marks': '4 marks', 'time': '4', 'ttext': 'Internal components', 'fld': 'AA_COM', 'fil': 'QS18S29P04', 'year': 2018, 'res': '', 'flag1': 1}, {'index1': '|C.E$|', 'index2': '|2$|', 'index3': '2$', 'index4': 'S6$.5', 'index5': '|', 'index6': 'Q23S207p3', 'qid': 'Q23S207p3', 'des': 'Simplify Boolean expression; De Morgan', 'index7': '', 'marks': '4 marks', 'time': '5', 'ttext': 'Logic and logic gates', 'fld': 'AA_COM', 'fil': 'Q23S207P3', 'year': 2023, 'res': '', 'flag1': 0}, {'index1': '|C.E$|', 'index2': '|2$|', 'index3': '2$', 'index4': 'S6$.5', 'index5': '|', 'index6': 'QS17S25C', 'qid': 'QS17S25C', 'des': 'Simplification; expression', 'index7': '', 'marks': '4 marks', 'time': '5', 'ttext': 'Logic and logic gates', 'fld': 'AA_COM', 'fil': 'QS17S25C', 'year': 2017, 'res': '', 'flag1': 1}, {'index1': '|C.E$|', 'index2': '|2$|', 'index3': '2$', 'index4': 'S6$.5', 'index5': '|', 'index6': 'QS19S26p04', 'qid': 'QS19S26p04', 'des': "Boolean algebra simplification; De Morgan's laws", 'index7': '', 'marks': '4 marks', 'time': '5', 'ttext': 'Logic and logic gates', 'fld': 'AA_COM', 'fil': 'QS19S26P04', 'year': 2019, 'res': '', 'flag1': 1}]}
User: These are the content of said exam questions, these are the ONLY exam questions, you will use these later, delivered in form 'fil', JSON: {'Q23S207P1': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="questiona">(a)  The figure below shows a circuit diagram.</p>\r\n<p class="graph"><img src="AA_COM/Q23S207P1_files_Q/img01.jpg"/> </p>\r\n<p class="indent2new">Complete the truth table below for the circuit shown in the figure above.</p>\r\n<p class="bottom"> </p>\r\n<table class="data-table left2">\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded"><b>A</b></p></td>\r\n\t\t<td class="col-2"><p class="padded"><b>B</b></p></td>\r\n\t\t<td class="col-2"><p class="padded"><b>C</b></p></td>\r\n\t\t<td class="col-3"><p class="padded"><b>L</b></p></td>\r\n\t\t<td class="col-2"><p class="padded"><b>M</b></p></td>\r\n\t\t<td class="col-3"><p class="padded"><b>N</b></p></td>\r\n\t\t<td class="col-2"><p class="padded"><b>X</b></p></td>\r\n\t\t<td class="col-3"><p class="padded"><b>Y</b></p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">0</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t\t<td class="col-2"><p class="padded">1</p></td>\r\n\t\t<td class="col-2"><p class="padded"> </p></td>\r\n\t</tr>\r\n</table>\r\n<p class="mark">(3)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">(b)  Using the figure above, write a Boolean expression for output <b>Y</b> in terms of inputs <b>A</b>, <b>B</b> and <b>C</b>.</p>\r\n<p class="indent2new answer-toggle"><b>Y</b> = _______________________________________________________________</p>\r\n<p class="mark">(2)</p>\r\n<p class="mark">(Total 5 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="questiona">(a)  <b>Marks are for AO2 (application)</b></p>\r\n<p class="indent2new"><b>1 mark</b> for each highlighted column L, N and Y completed correctly.</p>\r\n<p class="indent2new"><img src="AA_COM/M23S207P1_files_M/img01.jpg"/> </p>\r\n<p class="indent2new"><b>A</b>. Follow through for Y if column N is completed incorrectly.</p>\r\n<p class="levelms">3</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1 indentright">(b)  <b>Marks are for AO2 (application)</b></p>\r\n<p class="indent2new"><b>2 marks:</b> <img class="vert-middle" src="AA_COM/M23S207P1_files_M/img02.jpg"/></p>\r\n<p class="indent2new">//</p>\r\n<p class="indent2new"><b>1 mark</b> for one of the following somewhere in the expression:</p>\r\n<p class="indent2 m-t-0">• <img src="AA_COM/M23S207P1_files_M/img03.jpg"/> <b>I.</b> presence / absence of brackets around <img class="vert-middle" src="AA_COM/M23S207P1_files_M/img04.jpg"/></p>\r\n<p class="indent2 m-t-0">• <img class="vert-middle" src="AA_COM/M23S207P1_files_M/img05.jpg"/></p>\r\n<p class="indent2 m-t-0">• <img class="vert-middle" src="AA_COM/M23S207P1_files_M/img06.jpg"/></p>\r\n<p class="indent2new"><b>Note:</b> If using a different algebraic notation refer to team leader.</p>\r\n<p class="levelms">2</p>\r\n<p class="mark">[5]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="questiona">(a)  This truth table question was well answered with over three-quarters of students gaining full marks. Some students incorrectly completed the L column (A XOR B) with the most common incorrect response being for A NOR B. However, these students often went on to complete the N and Y columns correctly, as they were not dependent on the output from the XOR gate (column L).</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">(b)  Students were asked to write a Boolean expression for part of the logic circuit diagram (output Y). This was well answered with 60% of students scored both marks, although over 7% did not attempt the question.</p>\r\n<p class="indent2new">It is worth noting that this year we accepted responses written using AND, OR and XOR, but in future it is expected that students will write Boolean expressions with the syntax used in AQA examination papers: ∙ (AND), + (OR), ⨁ (XOR) and overbar (NOT).</p></h5>', 'notes': '', 'resources': ''}, 'QS17S25B': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Complete the truth table below to prove that <b>A +</b> <img class="vert-middle vert-up-1" src="AA_COM/QS17S25B_files_Q/overlineb.jpg"/> is equivalent to <img src="AA_COM/QS17S25B_files_Q/img01.jpg"/></p></p>\r\n<p class="bottom"> </p>\r\n<table class="data-table left1">\r\n\t<tr>\r\n\t\t<td class="col-3"><p class="padded"><b>A</b></p></td>\r\n\t\t<td class="col-3"><p class="padded"><b>B</b></p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3"><p class="padded">0</p></td>\r\n\t\t<td class="col-3"><p class="padded">0</p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3"><p class="padded">0</p></td>\r\n\t\t<td class="col-3"><p class="padded">1</p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3"><p class="padded">1</p></td>\r\n\t\t<td class="col-3"><p class="padded">0</p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3"><p class="padded">1</p></td>\r\n\t\t<td class="col-3"><p class="padded">1</p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t\t<td class="col-3"><p class="padded"> </p></td>\r\n\t</tr>\r\n</table>\r\n<p class="mark">(Total 3 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"><b>Marks are for AO2 (apply)</b></p>\r\n<p class="bottom"> </p>\r\n<table class="data-table left1">\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded"> </p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"> </p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded comp-top font12"><b>1</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded comp-top font12"><b>2</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded comp-top font12"><b>3</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded comp-top font12"><b>4</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded comp-top font12"><b>5</b></p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><b>A</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><b>B</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><img class="vert-middle vert-up-1" src="AA_COM/MS17S25B_files_M/overlineb.jpg"/></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><b>A + </b><img class="vert-middle vert-up-1" src="AA_COM/MS17S25B_files_M/overlineb.jpg"/></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><img class="vert-middle vert-up-1" src="AA_COM/MS17S25B_files_M/overlinea.jpg"/></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><img class="vert-middle vert-up-1" src="AA_COM/MS17S25B_files_M/overlinea.jpg"/> • <b>B</b></p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded"><img class="vert-middle vert-up-1" src="AA_COM/MS17S25B_files_M/img01.jpg"/></p></td>\r\n\t</tr>\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t</tr>\r\n\t\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t</tr>\r\n\t\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t</tr>\r\n\t\r\n\t<tr>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">0</p></td>\r\n\t\t<td class="col-3 vert-top"><p class="padded">1</p></td>\r\n\t</tr>\r\n</table></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Mark as follows:</b></p>\r\n<p class="indent1new indentright m-t-0"><b>1 mark</b> for column 1 and 3 correct</p>\r\n<p class="indent1new indentright m-t-0"><b>1 mark</b> for column 4 correct</p>\r\n<p class="indent1new indentright m-t-0"><b>1 mark</b> for columns 2 and 5 correct and identical</p>\r\n<p class="indent1new indentright"><b>I</b>. order of columns</p>\r\n<p class="mark">[3]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Most students were able to secure some marks on the truth table with over half securing full marks. Where marks were dropped students tended to not follow a logical sequence of steps to calculate NOT B, then NOT A, leading to A OR (NOT B), (NOT A) AND B and then finally NOT ((NOT A) AND B).</p></h5>', 'notes': '', 'resources': ''}, 'QS1726B': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Represent the Boolean equation <img class="vert-down-1" src="AA_COM/QS1726B_files_Q/img01.jpg"/> in the form of a logic circuit by drawing a diagram.</p>\r\n<p class="graph"><img src="AA_COM/QS1726B_files_Q/img02.jpg"/> </p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"> </p>\r\n<p class="indent1new indentright m-t-0"><img src="AA_COM/MS1726B_files_M/img01.jpg"/> </p>\r\n<p class="indent1new indentright">1 mark - inputs A and B connected to AND gate</p>\r\n<p class="indent1new indentright m-t-0">1 mark - inputs A and NOT B connected to AND gate</p>\r\n<p class="indent1new indentright m-t-0">1 mark - NOT gates after AND gates</p>\r\n<p class="indent1new indentright m-t-0">1 mark - output of NOT gates into OR gate and connected to Q (allow mark if NOT gate(s) missing)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Alternative using NAND gates:</b></p>\r\n<p class="indent1new indentright"><img src="AA_COM/MS1726B_files_M/img02.jpg"/> </p>\r\n<p class="indent1new indentright">1 mark - inputs A and B connected to NAND gate</p>\r\n<p class="indent1new indentright m-t-0">1 mark - input B connected to NOT gate</p>\r\n<p class="indent1new indentright m-t-0">1 mark - inputs A and NOT B connected to NAND gate</p>\r\n<p class="indent1new indentright m-t-0">1 mark - output of NAND gates into OR gate and connected to Q (allow mark if NAND gates are drawn as AND gates)</p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '', 'notes': '', 'resources': ''}, 'QS17L24A': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">A computer process, X, can only start executing once processes A and B have finished executing and either communication channel C or communication channel D or both are available to use.</p>\r\n<p class="indent1new">The states of processes and communication channels can be read using the following Boolean variables:</p>\r\n<p class="indent1">•   <b>A</b> is set to TRUE if process A has completed and FALSE if process A is still running.</p>\r\n<p class="indent1 m-t-0">•   <b>B</b> is set to TRUE if process B has completed and FALSE if process B is still running.</p>\r\n<p class="indent1 m-t-0">•   <b>C</b> is set to TRUE if communication channel C is available and FALSE if it is not available.</p>\r\n<p class="indent1 m-t-0">•   <b>D</b> is set to TRUE if communication channel D is available and FALSE if it is not available.</p>\r\n<p class="indent1new">The Boolean variable <b>X</b> should be set to TRUE if the values of the variables <b>A</b>, <b>B</b>, <b>C</b> and <b>D</b> indicate that process X can start and to FALSE if they indicate that process X cannot start yet.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">(a) Draw a logic circuit that will represent the logic of the system described above for the inputs <b>A</b>, <b>B</b>, <b>C</b> and <b>D</b> and the output X.</p>\r\n<p class="graph"><img src="AA_COM/QS17L24A_files_Q/img01.jpg"/> </p>\r\n<p class="mark">(3)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">(b) Write a Boolean expression to represent the logic used to start process X.</p>\r\n<p class="indent2new"><b>X =</b><span class="answer-toggle">________________________________________________________________</span></p>\r\n<p class="mark">(2)</p>\r\n<p class="mark">(Total 5 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="questiona indentright">(a)  <b>All marks AO2 (apply)</b></p>\r\n<p class="indent2new m-t-s"><img src="AA_COM/MS17L24A_files_M/img01.jpg"/> </p>\r\n<p class="indent2new indentright"><b>1 mark:</b> inputs A and B connected to an AND gate;</p>\r\n<p class="indent2new indentright m-t-0"><b>1 mark:</b> inputs C and D connected to an OR gate;</p>\r\n<p class="indent2new indentright m-t-0"><b>1 mark:</b> output of an AND gate (but not the same one as connected to inputs A and B) connected to X;</p>\r\n<p class="indent2new indentright"><b>MAX 2</b> if circuit does not fully represent the logic of the system OR the circuit diagram contains any errors</p>\r\n<p class="levelms">3</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1 indentright">(b)  <b>All marks AO2 (apply)</b></p>\r\n<p class="indent2new indentright">X = A • B • (C + D)</p>\r\n<p class="indent2new indentright m-t-0"><b>1 mark:</b> either A • B or C + D somewhere in an incorrect expression</p>\r\n<p class="indent2new indentright m-t-0"><b>2 marks:</b> fully correct expression</p>\r\n<p class="indent2new indentright m-t-0"><b>A</b>. A logically equivalent expression for <b>2 marks</b></p>\r\n<p class="levelms">2</p>\r\n<p class="mark">[5]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="questiona">(a) This question part was well tackled with over three quarters of students drawing a fully correct logic circuit. If students make a mistake whilst drawing a logic circuit and they cannot correct it in a clear way then they are advised to redraw it on a new sheet of paper as it can be difficult to discern what type of gate the student had drawn if, for example, an AND gate is drawn on top of an OR gate.</p>\r\n<p class="indent1">(b) This question part was well tackled. Students need to be aware of the significance of brackets as the expression (A.B).(C+D) is not logically equivalent to A.B.C+D as the AND (.) operation has a higher order of precedence than OR (+). Three quarters of students achieved both available marks.</p></h5>', 'notes': '', 'resources': ''}, 'QS17S25A': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question-mid"><img src="AA_COM/QS17S25A_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new">What is the name of the logic gate represented by the truth table and symbol shown above?</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="mark">(Total 1 mark)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"><b>Mark is for AO1 (knowledge)</b></p>\r\n<p class="indent1new indentright">NAND;</p>\r\n<p class="indent1new indentright"><b>A</b>. NOT AND</p>\r\n<p class="mark">[1]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">The majority of students were able to correctly identify the NAND symbol.</p></h5>', 'notes': '', 'resources': ''}, 'Q20S205P2': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">A XOR B can be implemented as a logic circuit without using an XOR gate.</p>\r\n<p class="indent1new">Using <b>only</b> AND, OR and NOT gates draw a circuit that will produce an output <b>Q</b> which is logically equivalent to <b>A XOR B</b>.</p>\r\n<p class="graph"><img src="AA_COM/Q20S205P2_files_Q/img01.jpg"/></p>\r\n<p class="mark">(Total 3 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>3 marks are for AO2 (apply)</b></p>\r\n<p class="indent1new indentright m-b-0"><b>1 mark</b> for getting Part 1 <b>or</b> Part 2 correct on any of the three diagrams.</p>\r\n<p class="indent1new indentright m-t-0 m-b-0"><b>1 mark</b> for getting <b>corresponding</b> Part 1 <b>or</b> Part 2 correct on the same diagram.</p>\r\n<p class="indent1new indentright m-t-0"><b>1 mark</b> for getting <b>corresponding</b> Part 3 correct on the same diagram.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright m-b-0"><b>MAX 2</b> if not fully correct</p>\r\n<p class="indent1new indentright m-t-0 m-b-0">Mark response against diagram that will give the highest mark.</p>\r\n<p class="indent1new indentright m-t-0">Mark point 3 can only be awarded if at least one other mark point has been awarded.</p>\r\n<p class="indent1new indentright m-t-0">Alternative Diagram 1</p>\r\n<p class="indent1new indentright"><img class="vert-middle" src="AA_COM/M20S205P2_files_M/img01.jpg"/></p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright">Alternative Diagram 2</p>\r\n<p class="indent1new indentright"><img class="vert-middle" src="AA_COM/M20S205P2_files_M/img02.jpg"/></p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright">Alternative Diagram 3</p>\r\n<p class="indent1new indentright"><img class="vert-middle" src="AA_COM/M20S205P2_files_M/img03.jpg"/></p>\r\n<p class="mark">[3]</p></h5>', 'examinerreport': '', 'notes': '', 'resources': ''}, 'QS19S26P02': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">A factory has a machine for filling bottles on a conveyor belt.</p>\r\n<p class="indent1">•   Q represents the signal to move the conveyor belt on. When Q is set to true the belt will move on.</p>\r\n<p class="indent1 m-t-0">•   A is a sensor which outputs true if a bottle is present.</p>\r\n<p class="indent1 m-t-0">•   B is a sensor which outputs true if a bottle is full.</p>\r\n<p class="indent1 m-t-0">•   C is a sensor which outputs true if a bottle is correctly positioned.</p>\r\n<p class="indent1 m-t-0">•   D is a sensor which outputs true if the next section has a bottle in it.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new">The conveyor belt is able to move if both of these conditions are true:</p>\r\n<p class="indent1">•   a bottle is full and correctly positioned or there is no bottle present</p>\r\n<p class="indent1 m-t-0">•   there is no bottle in the next section.</p>\r\n<p class="indent1new">In the box below, draw a logic circuit for the machine.</p>\r\n<p class="graph"><img src="AA_COM/QS19S26P02_files_Q/img01.jpg"/> </p>\r\n<p class="mark">(Total 3 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>Marks are for AO2 (apply)</b></p>\r\n<p class="indent1new"><b>Mark as follows</b></p>\r\n<p class="indent1new"><b>1 mark</b> for B and C into AND gate</p>\r\n<p class="indent1new m-t-0"><b>1 mark</b> for the result of B and C (<b>I.</b> incorrect gate) as one input and a NOT gated A as a second input to an OR gate</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new m-t-0"><b>1 mark</b> D connected to NOT gate and output of this to an AND gate, the results of A, B and C (<b>I.</b> previously incorrect gates) as the other input, with the output going into Q</p>\r\n<p class="indent1new"><b>Max 2</b> if not fully correct</p>\r\n<p class="indent1new"><img src="AA_COM/MS19S26P02_files_M/img01.jpg"></p>\r\n<p class="mark">[3]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">There were a lot of successful answers to this question part. Where students did not achieve high marks, the most common mistake was missing NOT gate symbols. Some students struggle to clearly differentiate when drawing AND and OR gates symbols. More conscientious students labelled the symbols as well which, on some occasions, allowed marks to be given which otherwise may not have been.</p></h5>', 'notes': '', 'resources': ''}, 'QS18L210': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the laws of Boolean algebra, show that:</p>\r\n<p class="graph"><img src="AA_COM/QS18L210_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new">You <b>must</b> show your working.</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright">All marks AO2 (apply)</p>\r\n<p class="indent1new indentright">Award up to four marks for the working shown, but <b>Max 3</b> if the response does not show that (A + B) ∙ (B + C ∙ (D + D̅)) = A ∙ C + B</p>\r\n<p class="indent1new indentright"><b>1 mark</b> 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.</p>\r\n<p class="indent1new indentright m-t-0"><b>1 mark</b> for a successful application of the distribution law – only one mark, regardless of how many times this has been applied</p>\r\n\r\n<p class="indent1new indentright">Continue marking until an incorrect step is encountered. If a student misses out some steps but does not make an error then continue marking.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example Solution 1</b></p>\r\n<p class="bottom"> </p>\r\n<table class="data-table left1">\r\n\t<tr>\r\n\t\t<td class="col-10 vert-top no-border">\r\n\t\t\t<p class="padded text-left">(A + B) ⋅ (B + C ⋅ (D + D̅))</p>\r\n\t\t\t<p class="padded text-left m-t-0">= (A + B) ⋅ (B + C ⋅ 1)</p>\r\n\t\t\t<p class="padded text-left m-t-0">= (A + B) ⋅ (B + C)</p>\r\n\t\t\t<p class="padded text-left m-t-0">= A ⋅ B + A ⋅ C + B ⋅ B + B ⋅ C</p>\r\n\t\t\t<p class="padded text-left m-t-0">= A ⋅ B + A ⋅ C + B + B ⋅ C</p>\r\n\t\t\t<p class="padded text-left m-t-0">= A ⋅ B + A ⋅ C + B</p>\r\n\t\t\t<p class="padded text-left m-t-0">= A ⋅ C + B</p>\r\n\t\t</td>\r\n\t\t<td class="col-14 vert-top no-border">\r\n\t\t\t<p class="padded text-left"> </p>\r\n\t\t\t<p class="padded text-left m-t-0">By identity X + X̅ = 1</p>\r\n\t\t\t<p class="padded text-left m-t-0">By identity X ⋅ 1 = X</p>\r\n\t\t\t<p class="padded text-left m-t-0">Using distribution law</p>\r\n\t\t\t<p class="padded text-left m-t-0">By identity X ⋅ X = X</p>\r\n\t\t\t<p class="padded text-left m-t-0">By redundancy theorem X + X ⋅ Y = X</p>\r\n\t\t\t<p class="padded text-left m-t-0">By redundancy theorem X + X ⋅ Y = X</p>\r\n\t\t</td>\r\n\t</tr>\r\n</table></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example Solution 2</b></p>\r\n<p class="bottom"> </p>\r\n<table class="data-table left1">\r\n\t<tr>\r\n\t\t<td class="col-11 vert-top no-border">\r\n\t\t\t<p class="padded text-left">(A + B) ⋅ (B + C ⋅ (D + D̅))</p>\r\n\t\t\t<p class="padded text-left m-t-0">= (A + B) ⋅ (B + C ⋅ 1)</p>\r\n\t\t\t<p class="padded text-left m-t-0">= (A + B) ⋅ (B + C)</p>\r\n\t\t\t<p class="padded text-left m-t-0">= A ⋅ C + B</p>\r\n\t\t</td>\r\n\t\t<td class="col-15 vert-top no-border">\r\n\t\t\t<p class="padded text-left"> </p>\r\n\t\t\t<p class="padded text-left m-t-0">By identity X + X̅ = 1</p>\r\n\t\t\t<p class="padded text-left m-t-0">By identity X ⋅ 1 = X</p>\r\n\t\t\t<p class="padded text-left m-t-0">Using distribution law (<u>this jump is worth 2 marks</u>)</p>\r\n\t\t</td>\r\n\t</tr>\r\n</table>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">This is the first time that students have been asked to produce a proof that two Boolean expressions are equivalent. It was well tackled, with over 40% of students achieving full marks. A common mistake was to believe that D + D̅ = 0.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new">Some responses were difficult to mark as the way that students laid them out made it unclear what order they had done things in. It is particularly important when presenting a proof that the examiner can see what steps the student took and in what order these were made. In other responses, students would miss out steps, making jumps in their working that were not convincing enough to be a proof.</p></h5>', 'notes': '', 'resources': ''}, 'Q22S207P5': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the rules and identities of Boolean algebra, simplify the following Boolean expression.</p>\r\n<p class="graph"><img src="AA_COM/Q22S207P5_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">Answer ________________________________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>Marks are for AO2 (application)</b></p>\r\n<p class="indent1new indentright"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1">•   Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 m-t-0">•   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).</p>\r\n<p class="indent1new indentright"><b>1 mark</b> for final answer of <img class="vert-middle vert-up-1" src="AA_COM/M22S207P5_files_M/img01.jpg"/>;</p>\r\n<p class="indent1new indentright"><b>3 marks</b> for working</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>MAX 3</b> for working. Award up to two marks for applying each of the three techniques (one mark per application) to produce a simpler expression.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">•   Applying De Morgan’s Theorem.</p>\r\n<p class="indent1">•   Multiply and/or factorise brackets.</p>\r\n<p class="indent1">•   Using a law or identity.</p>\r\n<p class="indent1new indentright"><b>Note:</b> A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example 1:</b></p>\r\n<p class="indent1new"><img src="AA_COM/M22S207P5_files_M/img02.jpg"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example 2:</b></p>\r\n<p class="indent1new"><img src="AA_COM/M22S207P5_files_M/img03.jpg"/> </p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Simplifying Boolean expressions has been assessed on previous papers and this was a relatively simple simplification. However, less than a third of students got full marks, with many of those who did not multiplying out the brackets incorrectly. In this question one half of the expression was an AND and the other was an OR. When multiplying out the brackets many students incorrectly placed an OR between all parts, eg <i>A · A</i> + <i>A</i> · <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> + <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> · <i>A</i> + <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> + <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> instead of <i>A · A</i> · <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> + <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/> · <i>A</i> · <img class="vert-middle" src="AA_COM/E22S207P5_files_E/img01.jpg"/>. Students need to have practiced on a large variety of expressions.</p></h5>', 'notes': '', 'resources': ''}, 'QW10202': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Simplify the Boolean expression:</span></p>\r<p class="graph"><img src="AA_COM/QW10202_files_Q/img01.png"> </p>\r<p class="indent1new">Show your working.</p>\r<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r<p class="mark">(Total 3 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>Algebraic Solution:</b></span></p>\r<p class="bottom"> </p>\r<table class=MsoTableGrid cellspacing=0 cellpadding=5 style=\' margin-left:20.0pt;border-collapse:collapse\'> \r\t<tr> \r\t\t<td valign=top style=\'width:100.0pt;border:solid windowtext 1.0pt;padding:0cm 5.4pt 0cm 5.4pt;\'><p class=boxl><b>Method 1</b></p></td>\r\t\t<td valign=top style=\'width:100.0pt;border:solid windowtext 1.0pt;padding:0cm 5.4pt 0cm 5.4pt;\'><p class=boxl><b>Method 2</b></p></td> \r\t</tr>\r\t<tr> \r\t\t<td valign=top style=\'width:100.0pt;border:solid windowtext 1.0pt;padding:0cm 5.4pt 0cm 5.4pt;\'><p class=boxl> <img align="middle;" src="AA_COM/MW10202_files_M/ab.png"/> + A <br/><br/> = <img align= "middle;" src="AA_COM/MW10202_files_M/a.png"> + <img align="middle;" src="AA_COM/MW10202_files_M/b.png"> + A <br/><br/> = 1 + <img align="middle;" src="AA_COM/MW10202_files_M/b.png"/> <br/><br/> = 1</p></td>\r\r\t\t\t\t<td valign=top style=\'width:100.0pt;border:solid windowtext 1.0pt;padding:0cm 5.4pt 0cm 5.4pt;\'><p class=boxl> <img align="middle;" src="AA_COM/MW10202_files_M/ab.png"/> + A <br/><br/> = <img align="middle;" src="AA_COM/MW10202_files_M/aba.png"/> <br/><br/> = <img align="middle;" src="AA_COM/MW10202_files_M/0b.png"/> <br/><br/> = <img align= "middle;" src="AA_COM/MW10202_files_M/0.png"/><br/></br/> = 1</p></td>\r\t</tr>\r</table></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><i>1 mark for an application of a DeMorgan’s law<br/>1 mark for realisation that</i> <i>A</i> + <img align="middle;" src="AA_COM/MW10202_files_M/aital.png"/> + <img align="middle;" src="AA_COM/MW10202_files_M/bital.png"/> = 1 + <img align="middle;" src="AA_COM/MW10202_files_M/bital.png"/> or <img align="middle;" src="AA_COM/MW10202_files_M/obital.png"/> = <img align="middle;" src="AA_COM/MW10202_files_M/0.png"/> (must be written in method, not just inferred that student has done this if arrives at correct answer)<br/><i>1 mark for correct answer</i></p>\r\r<p class= "indent1new indentright"><b>Truth table solution:</b></p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><img align="middle;" src="AA_COM/MW10202_files_M/img01.png"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><i>1 mark for column Y correct<br/>1 mark for column Z correct<br/>1 mark for correct answer</i></p>\r<p class="indent1new indentright"><i><b>Any other method:</b><br/>If student has used any other method to arrive <u>at correct answer</u> then award marks as follows:<br/>1 mark for correct answer, no working out<br/>2 marks for correct answer with working out, not all steps shown.<br/>3 marks for correct answer with all steps of working out shown.</i></p>\r<p class="indent1new indentright"><b>A</b> True for 1, False for 0<br/><b>A</b> alternative notations :</p>\r<p class="indent1 indentright m-t-0">• For X.Y allow X AND Y, X∧Y ,X∩Y, XY</p>\r<p class="indent1 indentright m-t-0">• For X+Y allow X OR Y, X∨Y , X∪Y</p>\r<p class="indent1 indentright m-t-0">• For <img align="middle;" src="AA_COM/MW10202_files_M/x.png"/> allow NOT X, ¬X</p>\r<p class="mark">[3]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">This question was tackled using either truth tables or the laws of Boolean algebra, with the number of candidates using each method being approximately equal. Candidates who used the truth table method appeared to make fewer errors. Many candidates showed the correct steps, using either method, but then failed to state the final answer explicitly. Candidates using the rules of Boolean algebra often made the first step by applying De Morgan’s law, but then ground to a halt. It would appear that Boolean algebra is still a topic that candidates find difficult. Often candidates would reach A OR NOT A and then make these terms completely disappear instead of equating them to ‘1’. Care was needed with the answer as moving the terms around to simplify it often resulted in transposition errors which lead to the wrong answer. There were many false starts with answers crossed out and the whole answer being rewritten.</span></p></h5>', 'notes': '', 'resources': ''}, 'QS17L24C': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the rules of Boolean algebra, simplify the following Boolean expression.</p>\r\n<p class="graph"><img src="AA_COM/QS17L24C_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new">You <b>must</b> show your working.</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="rightnew answer-toggle">Answer _________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"><b>All marks AO2 (apply)</b></p>\r\n<p class="indent1new indentright"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1 indentright m-t-0">•   Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 indentright m-t-0">•   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)</p>\r\n<p class="indent1new indentright"><b>1 mark:</b> for final answer: B + C</p>\r\n<p class="indent1new indentright"><b>MAX 3</b> for working. Award up to two marks for applying each one of the three techniques (one mark per application):</p>\r\n<p class="indent2 indentright">•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression.</p>\r\n<p class="indent2 indentright m-t-0">•   applying an identity other than cancelling NOTs that produces a simpler expression.</p>\r\n<p class="indent2 indentright m-t-0">•   successfully expanding brackets.</p>\r\n<p class="indent1new indentright"><b>Note:</b> A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example Working (1)</b></p>\r\n<p class="indent1new"><img class="vert-middle" src="AA_COM/MS17L24C_files_M/img02.jpg"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example Working (2)</b></p>\r\n<p class="indent1new"><img class="vert-middle" src="AA_COM/MS17L24C_files_M/img03.jpg"/> </p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Most students achieved some marks for this question part but only a fifth achieved full marks. The most commonly made mistake was to incorrectly apply the identity A+A̅=1 to the subexpression A̅+A∙(A+B) which failed to recognise that this could not be done because of the order of precedence of AND (.) and OR (+). Another mistake made by some students was to cancel NOTs when they could not be cancelled, for example believing that the NOTs in (B̅∙C̅) could be cancelled with the longer NOT that related to the entire expression.</p></h5>', 'notes': '', 'resources': ''}, 'QS18S29P04': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the rules of Boolean algebra, simplify the following expression.</p>\r\n<p class="graph"><img src="AA_COM/QS18S29P04_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new">You <b>must</b> show your working.</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="rightnew answer-toggle">Answer _________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"><b>Marks are for AO2 (apply)</b></p>\r\n<p class="indent1new indentright"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1 indentright m-t-0">•   Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 indentright m-t-0">•   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)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Mark as follows</b></p>\r\n<p class="indent1new indentright"><b>MAX 3 marks for working</b></p>\r\n<p class="indent1new indentright">Award one mark each for applying the techniques below:</p>\r\n<p class="indent1 indentright m-t-0">•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression</p>\r\n<p class="indent1 indentright m-t-0">•   successfully expanding brackets</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright">Award one mark for each application of a Boolean identity <b>MAX 2</b>.</p>\r\n<p class="indent1new indentright"><b>Note:</b> A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p>\r\n<p class="indent1new indentright"><b>Example working (1)</b></p>\r\n<p class="indent1new"><img src="AA_COM/MS18S29P04_files_M/img01.jpg"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example working (2)</b></p>\r\n<p class="indent1new"><img src="AA_COM/MS18S29P04_files_M/img02.jpg"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example working (3)</b></p>\r\n<p class="indent1new"><img src="AA_COM/MS18S29P04_files_M/img03.jpg"/> </p>\r\n<p class="indent1new indentright"><b>1 mark</b> for final answer A XOR B // A Exclusive OR B // A EOR B // A EXOR B // A ⨁ B</p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Most commonly students did not know the identity between A̅ ⋅ B + B̅ ⋅ A and A ⨁ B. Prior to this stage of the calculation, the most common error was students misapplying De Morgan’s Law. This led to an incorrect algebraic statement which led to students being unable to access later mark points.</p></h5>', 'notes': '', 'resources': ''}, 'Q23S207P3': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the rules of Boolean algebra, simplify the following expression.</p>\r\n<p class="graph"><img src="AA_COM/Q23S207P3_files_Q/img01.jpg"/></p>\r\n<p class="indent1new">You <b>must</b> show your working.</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>Marks are for AO2 (application)</b></p>\r\n<p class="indent1new"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1 m-t-0">• Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 m-t-0">• 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).</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new"><b>Mark as follows:</b></p>\r\n<p class="indent1new"><b>MAX 3 marks for</b> working</p>\r\n<p class="indent1new">Award one mark each for applying the techniques below:</p>\r\n<p class="indent1 m-t-0">• A successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression.</p>\r\n<p class="indent1 m-t-0">• Successfully expanding brackets.</p>\r\n<p class="indent1 m-t-0">• Extracting common factors from terms.</p>\r\n<p class="indent1new">Award one mark for each application of a Boolean identity <b>MAX 2</b>.</p>\r\n<p class="indent1new"><b>Note:</b> A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new"><b>1 mark</b> for final answer: B</p>\r\n<p class="indent2new"><img src="AA_COM/M23S207P3_files_M/img01.jpg"/> </p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new"><b>Alternative answer 1</b></p>\r\n<p class="indent2new"><img src="AA_COM/M23S207P3_files_M/img02.jpg"/> </p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Simplifying Boolean expressions has been assessed on previous papers and almost half the students scored full marks. A lot of students were able to gain a few marks before making a mistake in precedence when simplifying the step <img class="vert-middle vert-up-1" src="AA_COM/E23S207P3_files_E/img01.jpg"/> incorrectly taking the B+B to B.</p></h5>', 'notes': '', 'resources': ''}, 'QS17S25C': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the laws of Boolean algebra, simplify the following Boolean expression.</p>\r\n<p class="graph"><img src="AA_COM/QS17S25C_files_Q/img04.jpg"/> </p>\r\n<p class="indent1new">You <b>must</b> show your working.</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="rightnew answer-toggle">Answer _________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question indentright"><b>Marks are for AO2 (apply)</b></p>\r\n<p class="indent1new indentright"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1 indentright m-t-0">•   Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 indentright m-t-0">•   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)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Mark as follows:</b></p>\r\n<p class="indent1new indentright"><b>1 mark</b> for final answer X</p>\r\n<p class="indent1new indentright"><b>Max 3 marks for working:</b></p>\r\n<p class="indent1 indentright m-t-0">•   <b>1 mark</b> for each application of an identity other than cancelling NOTs that produces a simpler expression.</p>\r\n<p class="indent1 indentright m-t-0">•   <b>1 mark</b> for expanding brackets</p>\r\n<p class="indent1 indentright m-t-0">•   <b>1 mark</b> for putting an expression into brackets that would lead to a simpler expression.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Note:</b> a simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Max 3</b> if answer is correct but any incorrect working or significant steps of working is missing.</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new indentright"><b>Example working (1)</b></p>\r\n<p class="indent1new indentright">X. X + X. Y̅ + Y. X + Y. Y̅     [expansion of brackets]</p>\r\n<p class="indent1new indentright m-t-0">X + X. Y̅ + Y. X + 0       [use of X • X = X and Y. Y̅ = 0]</p>\r\n<p class="indent1new indentright m-t-0">X(1 + Y̅ + Y) or X + X(Y + Y̅)  [taking X outside of brackets]</p>\r\n<p class="indent1new indentright"><b>Alternative example working (2)</b></p>\r\n<p class="indent1new indentright">X + (Y. Y̅)          [Use of distributive law]</p>\r\n<p class="indent1new indentright m-t-0">X + 0             [Y. Y̅ = 0]</p>\r\n<p class="indent1new indentright m-t-0">X              [Recognising X + 0 = X]</p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">It was pleasing to see that the majority of students tackled this question and a large number did so successfully using a number of approaches. The most common mistakes included not identifying Y AND NOT Y as being equal to 0, or not fully simplifying to a final answer. Some students dropped marks by making leaps which, whilst arithmetically correct, were not Boolean identities and could not be followed directly. In this case it is impossible to distinguish between correct understanding and a lucky guess.</p></h5>', 'notes': '', 'resources': ''}, 'QS19S26P04': {'question': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Using the rules and identities of Boolean Algebra, simplify the following Boolean expression.</p>\r\n<p class="graph"><img src="AA_COM/QS19S26P04_files_Q/img01.jpg"/> </p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="indent1new answer-toggle">_______________________________________________________________________</p>\r\n<p class="mark">(Total 4 marks)</p></h5>', 'markscheme': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question"><b>Marks are for AO2 (apply)</b></p>\r\n<p class="indent1new"><b>Marking guidance for examiners</b></p>\r\n<p class="indent1 m-t-0">•   Award marks for working out until an incorrect step has been made.</p>\r\n<p class="indent1 m-t-0">•   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)</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new"><b>Mark as follows</b></p>\r\n<p class="indent1new"><b>1 mark</b> for final answer A</p>\r\n<p class="indent1new"><b>3 marks for working</b></p>\r\n<p class="indent1new"><b>Max 3</b> for working. Award up to two marks for applying each one of the three techniques (one mark per application):</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1">•   a successful application of De Morgan’s Law (and any associated cancellation of NOTs) that produces a simpler expression</p>\r\n<p class="indent1 m-t-0">•   applying an identity other than cancelling NOTs that produces a simpler expression</p>\r\n<p class="indent1 m-t-0">•   successfully expanding brackets</p></h5><h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="indent1new"><b>Note:</b> A simpler expression is one that is logically equivalent to the original expression but uses fewer logical operators.</p>\r\n<p class="indent1new m-t-0"><b>Note:</b> 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.</p>\r\n<p class="indent1new"><img src="AA_COM/MS19S26P04_files_M/img01.jpg"></p>\r\n<p class="mark">[4]</p></h5>', 'examinerreport': '<h5 class=\'contentSeparator\' style=\'page-break-inside: avoid;\'><div class=\'qnumber-hidden\'></div><p class="question">Some centres appear to have been teaching De Morgan’s laws by using an overbarred AND/OR in place of an OR/AND, e.g. <img class="vert-middle vert-up-1" src="AA_COM/ES19S26P04_files_E/img01.jpg"/>. While this may help some students learn the procedure to follow when applying De Morgan’s laws, they must remember to convert any overbarred AND/OR to OR/AND in their final answer to gain the mark(s). Many students failed to gain marks due to missing brackets when parts of equations have been simplified but required the brackets for correct order of further simplification. On occasion this has inadvertently led to a correct final answer as the student has assumed the brackets are there, but they have failed to gain working marks as full understanding is not evident.</p></h5>', 'notes': '', 'resources': ''}}
User: Now, the user will ask questions after this message. Use everything I gave you and the knowledge you have acquired, and from the notes you made, to answer their question in a very concise way that WILL score ALL marks. Just like the notes. For this, your answer will not have the same structure as the revision notes. Think of it as answering an exam question: short, consice, useful, scores ALL marks. Just like the exam questions provided to you, you will not leave your response in a 'notes' manner. You are just answering an exam question: which means it has to be very short unless explicitly specified by the user that it must be long and detailed. So, keep it short, concise, and only a couple of sentences that score ALL marks. Remember, it is an A-Level exam question and you will style it as such: incredibly concise sentences (essentially like bullet points but without the bullet points) that score marks. Most sentences won't even have commas, this does not mean you can't use them. Also do not repeat the question in your answer, just do it. Score all marks, assume how many marks there are based on the question if it is not provided to you. Imagine that for every sentence you make, you score a mark (for simpler questions), for high (>5 mark) questions that may differ.
In summary, you must:
* Answer very concisely. Kind of like bullet points but as a short para. Most of the time, a sentence, or point, will be a mark.
* Do not repeat the question at the start.
* Don't over-do your answer. Please. 1 mark (e.g., usually, 'What' questions) is like one very short sentence usually. Remember the summary from the notes? Your answer will essentially be like that for most questions but a bit (a bit!) more developed. Remember, this is just A-Levels. Use some of the exam questions that I provided as an example of how you should model your answers.
* When you're listing examples, don't list every example in the universe.
* Score all marks. Be helpful.
User: Write the notes you made
Assistant: **Boolean Algebra and Logic Gates**
**Introduction to Boolean Algebra**
Boolean algebra, developed by George Boole, is an algebraic system used in logic gates studies. It evaluates logical expressions to produce TRUE or FALSE results. E.g., "The button has been pressed," 5 < 10, Age > 17 and holds a driving license.
**Basic Operators and Truth Tables**
- **AND**: TRUE if all conditions are TRUE. (A ∙ B = Q)
- **OR**: TRUE if at least one condition is TRUE. (A + B = Q)
- **NOT**: Inverts the input's value. (Q = ¬A)
- **NAND**: Outputs TRUE if any input is FALSE.
- **NOR**: Outputs TRUE only if all inputs are FALSE.
- **XOR (Exclusive OR)**: TRUE if only one input is TRUE, not both.
**Spec Points**
**Construct Truth Tables**
Truth tables show results for all input combinations in Boolean expressions. Examples of spec points: NOT, AND, OR, XOR, NAND, and NOR gates.
**Logic Gates and Circuits**
Students should understand ANSI/IEEE standard logic gate symbols and create logic gate circuit diagrams involving these gates.
**Boolean Algebra Simplification and Optimization**
Using Boolean identities simplifies expressions to create efficient circuits with fewer components. Common rules include AND, OR, Inverse, and Commutative Laws.
**De Morgan's Laws**
- **Rule 1**: ¬(A ∙ B) = ¬A + ¬B
- **Rule 2**: ¬(A + B) = ¬A ∙ ¬B
These rules help simplify Boolean expressions, allowing easier circuit designs with NAND or NOR gates.
**Exam Practice Questions**
- Complete truth tables and logic circuits. (5 marks; Q23S207P1)
- Simplify Boolean expressions using laws and identities. (4 marks; QS19S26P04)
- Draw logic circuits for Boolean expressions. (4 marks; QS1726B)
**Definitions**
- **Boolean Expression**: An equation of Boolean operations producing TRUE or FALSE.
- **Truth Table**: A table calculating the result of inputs in a Boolean expression.
- **AND Gate**: Outputs TRUE if both inputs are TRUE.
- **OR Gate**: Outputs TRUE if any input is TRUE.
- **NOT Gate**: Inverts the input signal.
- **NAND/NOR/XOR Gate**: Specialized logic gates with unique TRUE/FALSE conditions.
- **De Morgan’s Law**: Methods for simplifying Boolean expressions using complement operations.
**Summary**
Understand and apply logic gate operations (AND, OR, NOT). Construct and interpret logic gate circuits using standard symbols. Simplify expressions with Boolean identities and optimize circuits with De Morgan's Laws. Practice applying these concepts with truth tables and exam questions.
{
"chapter": 30,
"page": 245,
"chapter_name": "Boolean Algebra",
"topics": ["Boolean expressions", "Truth tables", "Logic gates", "Boolean algebra simplification", "De Morgan’s Laws"]
}
User: exam questions must include the content and answer that scores full marks as well
Assistant: