H446 Revision

1.4.1

Data Types

Exam questions: 11 questions, 111 marks

Revision notes: Data types

Question 1

13 marks

1(a)

Two's complement can be used to represent negative binary numbers.

1(a)(i)

Convert the denary number -124 into an 8-bit two's complement binary number.

Mark scheme 1(a)(i)

1000 0100

Guidance

CAO

Examiner’s Comments

This was generally well answered with the majority of candidates gaining full marks. Those that did not had either got it completely wrong or had missed or added an extra zero.

1(a)(ii)

State one other way to represent negative binary numbers.

Mark scheme 1(a)(ii)

Sign and magnitude

Guidance

Accept: One’s complement

Examiner’s Comments

This was generally well answered and the majority of candidates gained one mark.

1(b)

Convert the denary number 298 into hexadecimal.

Mark scheme 1(b)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

12A

  • 1 mark for 12

  • 1 mark for A

Guidance

Must be in correct order.

Examiner’s Comments

This was generally well answered and the majority of candidates gained two marks.

out of 2

1(c)

The binary number 10011101 0110 is stored in normalised floating point form with an 8-bit mantissa and a 4-bit exponent both written in two's complement.

Convert this binary number into denary.

You must show your working.

Mark scheme 1(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark per bullet to max 3:

  • Converting exponent to 6

  • Converting mantissa to 1001110.1 / moving binary point 6 to the right

  • -49.5

Guidance

MP2 - if flipped = 110001.1

Correct answer, with any binary working gets full marks

Examiner’s Comments

The majority of candidates were able to gain a mark for converting the exponent to 6 and many were able to gain the second mark for moving the point 6 places to the right. Many candidates gained full marks although some were unable to convert the binary number back to denary to give -49.5

out of 3

1(d)

The binary number 00001011 0111 is represented in unnormalised floating point form with an 8-bit mantissa and a 4-bit exponent both written in two's complement.

Convert this binary number into a normalised floating point binary number, represented using an 8-bit mantissa and 4-bit exponent format.

You must show your working.

Mark scheme 1(d)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark per bullet to max 4:

  • Reducing exponent to 4 / reducing exponent by 3

  • New mantissa starts 01011...

  • ... and is written out to 8 bits making 01011000

  • New exponent of 0100

Guidance

Correct mantissa and exponent, with any binary working gets full marks

Examiner’s Comments

There were a range of responses with more successful candidates gaining full marks. Some candidates lost marks for incorrectly writing the final mantissa to 8 bits.

out of 4

1(e)

Calculate the binary addition of these two 8-bit (unsigned) binary numbers.

Show your working.

[Diagram redrawn from the original paper by an AI model (Gemma 4 12B). It may contain mistakes.]

0011101111
100110011 +
-----------
Mark scheme 1(e)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark per bullet to max 2:

  • 1101 0000

  • Correct working out/ 6 carries shown

Guidance

0 mark for denary calculations with no carries shown.

Examiner’s Comments

The majority of candidates were able to gain full marks on this question showing accurate working out and a correct response.

out of 2

Question 2

12 marks

2(a)

Using the binary value 1001 1101, convert this into:

2(a)(i)

A positive denary number.

Mark scheme 2(a)(i)

157

Guidance

CAO

Examiner’s Comments

Most candidates answered this question correctly.

2(a)(ii)

A negative denary number using two's complement.

Mark scheme 2(a)(ii)

  • 99

Guidance

CAO

Examiner’s Comments

Most candidates answered this question correctly.

2(a)(iii)

A hexadecimal value.

Mark scheme 2(a)(iii)

9D

Guidance

CAO

Examiner’s Comments

Most candidates answered this question correctly.

2(b)

Binary values stored by a computer can represent different types of data.

Discuss the different types of data that can be stored in binary and why computer systems store data in binary format.

You should refer to the following in your answer:

  • what types of data can be represented in binary
  • examples of how binary is used to represent this data
  • why computers store data in binary format.
Mark scheme 2(b)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

Mark Band 3 - High Level (7-9 marks)
The candidate demonstrates a thorough knowledge and
understanding of why data is stored in binary with points covering a range of examples of data types. All detail is generally accurate and relevant.
The candidate is able to apply their knowledge and understanding directly and consistently to the context provided.
Evidence/examples will be explicitly relevant to the explanation. The candidate will come to a clear conclusion that must be justified by their comments.

There is a well-developed line of reasoning which is clear and logically structured. The information presented is relevant and

substantiated.

Mark Band 2- Mid Level
(4-6 marks)
The candidate demonstrates reasonable knowledge and understanding of why data is stored in binary with points covering a range of examples of data types stored in binary. The material is generally accurate but at times underdeveloped.
The candidate is able to apply their knowledge and understanding directly to the context provided although one or two opportunities are missed.
Evidence/examples are for the most part implicitly relevant to the explanation.
The candidate will attempt to come to a conclusion, although it may not be fully justified by their answer.

There is a line of reasoning presented with some structure. The information presented is in the most part relevant and supported by some evidence.

Mark Band 1- Low Level (1-3 marks)
The candidate demonstrates a basic knowledge of why data is stored in binary with points covering some examples of data types stored in binary. The material is basic and contains some inaccuracies. The candidate makes a limited attempt to apply acquired knowledge and understanding to the context provided.
The candidate provides nothing more than an unsupported assertion.

The information is basic and communicated in an unstructured way. The information is supported by limited evidence and the relationship to the evidence may not be clear.

0 mark
No attempt to answer the question or response is not worthy of credit.

Guidance

AO1
Types of data stored in binary:
E.g.

  • Numeric
  • Alphanumeric/text/characters/symbols
  • Sound
  • Images
  • Video

AO2:
Examples:

  • Text - Use of character sets such as ASCII and Extended ASCII. Each character is given a separate binary code. Increasing the number of bits for each character increase the number of possible characters that can be stored
  • Numbers - Numbers can be stored using binary. Sign and Magnitude and Two’s Compliment can be used to

store negative binary numbers. Floating Point binary can be used to store data with a decimal point.

Other answers outside of the specification should still be accepted. E.g.

  • Images are split into pixels. Each pixel is given a binary value which represents a colour.
  • Analogue sound is sampled. A sample is a reading that is taken at certain points and is stored as a binary number.

AO3
Reasons computers use binary:

  • Computers made up of switches/logic gates
  • They will have on/off/two states
  • This easily translates to binary values

Examiner’s Comments

Candidates were assessed on the quality of their extended response in this question. Most candidates were able to explain why computers store data in a binary format and many discussed character sets being used to store test using ASCII and UNICODE. Many candidates also further discussed storing numbers using sign and magnitude and floating point, etc.

Few candidates discussed other points such as sound, video and images being stored in binary, and few mentioned that computers are made up of logic gates/switches.

out of 9

Question 3

14 marks

3(a)(i)

Convert the denary number 189 to hexadecimal.

Mark scheme 3(a)(i)

  • BD

Guidance

Correct answer only

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

3(a)(ii)

Convert the unsigned binary number 1010101111 to hexadecimal.

Mark scheme 3(a)(ii)

  • 2AF

Guidance

Correct answer only

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

3(b)

Negative binary values can be represented using either sign and magnitude or two’s complement.

3(b)(i)

Convert the denary number -107 to an 8-bit binary number using sign and magnitude.

Mark scheme 3(b)(i)

  • 1110 1011

Guidance

Correct answer only

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

3(b)(ii)

Convert the denary number -107 to an 8-bit binary number using two’s complement.

Mark scheme 3(b)(ii)

  • 1001 0101

Guidance

Correct answer only

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

3(b)(iii)

Give one advantage of storing values using two’s complement instead of sign and magnitude.

Mark scheme 3(b)(iii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Calculations are more easily performed on two’s complement

  • Two’s complement allows for a (negligible) larger range of numbers to be stored / by example

  • No additional hardware is required in two’s complement / Addition and subtraction are carried out using only an adder

  • Two’s complement has only one representation for 0

Guidance

Accept the reverse of the MP

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

out of 1

3(c)

Show how the denary value -15.75 can be represented as a normalised floating point binary number using 8 bits for the mantissa and 4 bits for the exponent.

You must show your working.

Mark scheme 3(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • -15.75 is 10000.01 (in fixed point two's complement)

  • Binary point moved 4 places left

  • Mantissa 1000 0010

  • Exponent 0100

Guidance

Mantissa must be 8 bits, exponent must be 4 bits

Accept alternative working for MP2 if appropriate.

If mantissa AND exponent are correct with any working, 4 marks

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

out of 4

3(d)

The normalised floating point binary number 0100 1110 is stored using 4 bits for the mantissa and 4 bits for the exponent, both in two’s complement.

Convert this number to denary.

You must show your working.

Mark scheme 3(d)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Exponent is -2

  • Binary point moved 2 places left (0.001) / 0.5 x 2-2

  • 0.125 / 1/8 (one eighth)

Guidance

MP2 is for correct working of whichever method is used.

If answer is correct and working is shown, 3 marks

Examiner’s Comments

Binary and hexadecimal questions were generally well answered with many candidates being able to gain full marks across the question parts.

out of 3

3(e)

Complete each of the following sentences relating to the storage of floating point binary numbers with an appropriate word.

Increasing the number of bits used for the mantissa increases the …… of the number that can be stored.

Increasing the number of bits used for the exponent increases the …… of the number that can be stored.

Mark scheme 3(e)

  • precision / accuracy

  • range / size / magnitude

Guidance

Examiner’s Comments

Many candidates attempted this question with many showing a good understanding of floating-point binary. The most common error was to get the answers the wrong way round giving size or range for the first gap and precision or accuracy for the second.

Question 4

10 marks

4(a)

State the benefit of using a normalised form when representing data as a floating point number.

Mark scheme 4(a)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Allows for more accuracy/precision from the given number of bits

  • The representation of each binary value is unique

Guidance

Examiner’s Comments

This question was not answered well as many candidates seemed to misread the question. The majority of candidates that answered this question correctly stated that normalised form allowed for greater accuracy/precision from the number of bits.

out of 1

4(b)

Describe what is meant by the term ‘character set’.

Mark scheme 4(b)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • List/mapping of characters (that can be understood by the hardware/software/computer)

  • Each character is given a unique binary/numeric code

that is stored instead of the character.

Guidance

Allow a list of letters /symbols

Examiner’s Comments

Most candidates were given at least 1 mark for this question. Some did not gain the

second mark as they missed stating that each character has a unique binary value.

out of 2

4(c)(i)

Convert the hexadecimal number 66 into a denary number. Show your working.

Mark scheme 4(c)(i)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • 102 (correct answer)

  • Suitable working out

Guidance

Examiner’s Comments

This question was generally answered well. Candidates tended to be awarded either full or no marks. If the working was correct for calculating the denary value, the response was usually correct as this only required simple addition.

out of 2

4(c)(ii)

State two reasons why a programmer would prefer to use hexadecimal numbers rather than binary numbers.

1

2

Mark scheme 4(c)(ii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Hexadecimal values are shorter than binary as 4 bits/nibble can be represented by one hex character.

  • Hexadecimal values are faster / more reliable to communicate / enter / write down / read.

Guidance

Do not accept Hexadecimal is “easy” communicate/enter etc…

Examiner’s Comments

Many candidates achieved 1 mark for this question. Many candidates correctly stated that hexadecimal numbers are easier/faster to read/write, but were not given the second mark as they didn’t state that hexadecimal numbers are shorter than binary numbers.

out of 2

4(d)

Show the denary value 6.25 as a floating point binary number representing the mantissa and exponent. Both of these should be stored in two’s complement representation.

You should use as few bits as possible.

Show your working.

Mark scheme 4(d)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Fixed point is 0110.01

  • Mantissa becomes 0.11001

  • Exponent of three needed (represented at 011)

Guidance

Final Answer 011001 011

Correct answer with any working gets full marks

Examiner’s Comments

Many candidates were given the first mark by working out the fixed-point equivalent of 6.25 (0110.01) but were not given further marks because they did not read the question correctly. The question states the response must use as few bits as possible. Although some candidates correctly calculated the mantissa, many had extra digits in the exponent which could have been removed. See Exemplar 2 which was given full marks.

Exemplar 2

[Image redrawn from the original mark scheme by an AI model (Gemma 4 12B). It may contain mistakes.]

A candidate's handwritten answer:

Show your working. 3

8 4 2 1 1/2 1/4 1/8 4 2 1

0 1 1 0 0 1 0 0 1 1

The candidate clearly shows the fixed-point value of 6.25 to gain the first mark. The second mark is given for showing the

mantissa with six bits and the final mark for showing the exponent with three bits. All of these values are displayed clearly, and the candidate uses arrows to demonstrate the movement of the decimal point as additional working.

out of 3

Question 5

1 mark

5

A bubble sort will sort an array of 50 integer values called numberArray.

State why the integer values are stored in an array instead of separate variables.

Mark scheme 5

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark for:

  • Can refer to all 50 only using one identifier / all values can be indexed in one array

  • The numbers can be passed as a single parameter

  • Does not need 50 variables to be declared/passed

Guidance

Examiner’s Comments

The concept of having a single indexable identifier seemed to be poorly understood with few candidates gaining the mark. While many candidates understood that an array can be iterated through, they did not link this to the fact that elements in an array are indexable. Similarly, unqualified points such as ‘an array is easier to sort’ were not given marks for the same reason.

There was a common misconception that an array would use far less memory space than 50 separate variables. There would be minimal difference since each separate value would take the same amount of storage space as the corresponding value in an array. The only overhead for separate variables would be the extra pointers to the memory locations for each, compared to one pointer for an array.

out of 1

Question 6

6 marks

6

A programmer creates another function to count and return how many capital letters are in a string that is passed into the function as a parameter.

The asc() function takes in a character and returns its ASCII value. For example asc("A") returns 65. Capital letters have ASCII values between 65 and 90 inclusive.

6(i)

Complete the function below.

function countCapitals(text)
  / initialise counter to 0
  capCount = 0
  / loop through each character in the string passed in
  for x = 0 to text.length-1
     c = text.subString(x, 1)
     / check if character is a capital
     if asc(c) >= 65 ……
       / if so, increment counter
       ……
     endif
  next x
  ……
endfunction
Mark scheme 6(i)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

function countCapitals(text)
  / initialise counter to 0
  capCount = 0
  / loop through each character in
the string passed in
  for x = 0 to text.length-1
    c = text.subString(x, 1)
    / check if character is a
capital
    if asc(c) >= 65 and asc(c) <= 90
/ if so, increment counter
      capCount = capCount + 1
    endif
  next x
  return capCount
endfunction

Guidance

Accept alternative answers in high-level languages (e.g. capCount ++ or capCount +=1 / && for and for BP1)

Accept countCapitals = capCount for BP3 (Returning via assigning to function identifier is used in VB / Pascal)

Accept and asc(c) < 91 instead of <=90

Allow FT for returning the value they increment as a counter if it isn’t capCount

out of 3

6(ii)

Give one similarity between ASCII and Unicode.

Mark scheme 6(ii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Both (use binary) to represent characters / are character sets

  • The first 7/8 bits of Unicode is the same as ASCII (overlaps)

out of 1

6(iii)

Give two differences between ASCII and Unicode.

Difference 1

Difference 2

Mark scheme 6(iii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • ASCII has fewer characters (128/256) / Unicode has more characters

  • ASCII is 7/8 bits whereas Unicode can be larger 16/32 / can have variable sized characters

  • ASCII limited to Latin / English / European characters whereas Unicode can represent other symbols (e.g. Chinese/Cyrillic/Emojis)

Guidance

Only mark the 1st answer for each difference

For BP2&3, must have both sides to get the mark

Examiner’s Comments
Candidates should be encouraged to give a complete answer. Just saying ‘Unicode uses more bits than ASCII’ is not enough to be given a mark at this level.

out of 2

Question 7

13 marks

7(a)(i)

Convert the hexadecimal value B7E to a binary number.

Mark scheme 7(a)(i)

  • 1011 0111 1110

Guidance

CAO

7(a)(ii)

110010101 is a binary number that is represented using sign and magnitude.

Convert this binary number to a denary number.

Mark scheme 7(a)(ii)

  • -149

Guidance

CAO

7(a)(iii)

Complete this binary subtraction. Both numbers are 8-bit integer values represented using two’s complement.

Show the result in the same format and show your working.

[Diagram redrawn from the original paper by an AI model (Gemma 4 12B). It may contain mistakes.]

0110 1101  —
0011 0100
Mark scheme 7(a)(iii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

0011 1001

  • One mark for correct left hand nibble (CAO)

  • One mark for correct right hand nibble (CAO)

  • One mark for working clearly shown

Guidance

Working could include showing “borrowing” values from other columns or making the second number negative and adding.

Answer must be 8 bits to achieve full marks (stated in question).

No marks if only working is denary

Examiner’s Comments
Most candidates were able to gain some marks, with many gaining full marks. A popular method was to do two’s complement addition. Candidates should be encouraged to show their working in binary and not do the subtraction in denary and then just give the answer in binary. The question asks them to complete a binary subtraction.

out of 3

7(b)

The normalised floating point number 1010 1110 is stored using 4 bits for the mantissa and 4 bits for the exponent, both in two’s complement.

Give the denary version of this number, showing your working.

Mark scheme 7(b)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Exponent is -2

  • Move decimal place 2 places left

  • Fill with 1s giving 1.1101

  • Denary answer is -0.1875 /

Guidance

Accept alternative method of -0.75 x 2-2 for BP2 and BP3. BP2 credited for -0.75, BP3 credited for multiplying by 2-2

Correct answer with valid working gets full marks

Examiner’s Comments
Most candidates correctly identified the exponent and that the point needed to be moved to the left. Any valid method of

working was given marks here, where candidates got the correct answer.

out of 4

7(c)

Table 3 here shows floating point numbers that are stored using 6 bits for the mantissa and 3 bits for the exponent, both in two’s complement.

Tick (✓) one box in each row to state whether each number is normalised or not normalised.

Binary numberNormalisedNot normalised
010101 100
110101 111
011010 010
101010 110

Table 3

Mark scheme 7(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

Binary numberNormalisedNot normalised
010101 100x
110101 111x
011010 010x

101010 110 x

Guidance

One mark per row. No mark if both/neither box is ticked.

Accept other marks that clearly indicate choice (e.g. X)
Examiner’s Comments
This question was mostly well answered. However, some candidates were able to identify 01 as being normalised but not 10 being normalised.

out of 4

Question 8

9 marks

8(a)

State why computer systems store data in binary.

Mark scheme 8(a)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark for any of the following bullet points, e.g:

  • Computers use binary logic for on/off or 1/0

  • Computer systems are based on switches/transistors

  • Binary is high tolerance

Guidance

Examiner’s Comments
This question was generally answered well.

out of 1

8(b)(i)

Convert the denary number 97 into an 8-bit binary number.

Mark scheme 8(b)(i)

  • 0110 0001

Guidance

Must get the whole answer right for the mark to be awarded.

Examiner’s Comments
To gain the mark, the candidate was required to use all 8 bits. Some used 7 bits and therefore did not achieve the mark. Candidates should be reminded to read through each question carefully.

8(b)(ii)

Convert the denary number 171 into a hexadecimal number.

Mark scheme 8(b)(ii)

  • AB

Guidance

Must get the whole answer right for the mark to be awarded.

Examiner’s Comments
This question was answered well with most errors relating to a miscalculation.

8(b)(iii)

Convert the denary number –97 into an 8-bit binary number using two’s complement.

Mark scheme 8(b)(iii)

  • 1001 1111

Guidance

Must get the whole answer right for the mark to be awarded.

Examiner’s Comments
This was generally answered well although some candidates were confused between two’s complement and sign and magnitude.

8(b)(iv)

Convert the denary number –17 into an 8-bit binary number using sign and magnitude.

Mark scheme 8(b)(iv)

  • 1001 0001

Guidance

Must get the whole answer right for the mark to be awarded.

Examiner’s Comments
As with Question 4 (b) (iii), this was

answered well although some candidates gave an answer using two’s complement.

8(c)

State one advantage of using two’s complement instead of sign and magnitude.

Mark scheme 8(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • – Can be easily used in binary arithmetic or

  • – Increased range of numbers available

Guidance

Award 0 or 2 marks

Examiner’s Comments
This question was worth either 2 or 0 marks. It was generally not answered well with most candidates being too vague.

out of 2

8(d)

Calculate the addition of these two 8-bit (unsigned) binary numbers.

Show your working.

[Diagram redrawn from the original paper by an AI model (Gemma 4 12B). It may contain mistakes.]

  11011011
+ 10001001
-----------
Mark scheme 8(d)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark per bullet up to a maximum of 2 marks, e.g:

  • (1) 0110 0100

  • suitable working out

Guidance

Award mark for bullet point one with or without the overflow stated.

Examiner’s Comments
As with the other binary calculation, this was generally answered correctly. Some candidates gave the wrong answer but gained a mark for evidence of a suitable method. The question reminds candidates to show their working and most candidates did.

out of 2

Question 9

14 marks

9(a)(i)

Convert the denary number 231 to an unsigned 8-bit binary number.

Mark scheme 9(a)(i)

11100111

9(a)(ii)

Convert the hexadecimal number 6F to an unsigned 8-bit binary number.

Mark scheme 9(a)(ii)

0110 1111

9(b)

The floating point number 001101 0100 is stored using 6 bits for the mantissa and 4 bits for the exponent, both in two’s complement. This number is not normalised.

9(b)(i)

Give the normalised version of this number, showing your working.

Mark scheme 9(b)(i)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Move decimal place 1 place right = 0.1101 to normalise

  • Pad with 0s, mantissa is 011010

  • Exponent is 4 / Subtract 1 from exponent…

  • New exponent is 3 = 0011

out of 4

9(b)(ii)

Convert your answer to part (i) to denary, showing your working.

Mark scheme 9(b)(ii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • move dp 3 places right

  • 0110.1

  • 6.5

Guidance

Accept conversion of un-normalised number (will result in same output)

Accept method of using 0.1101 x 2^3

FT required from part (i) if this is done incorrectly.

out of 3

9(c)

Add together the two numbers below. Both numbers are stored in normalised floating point format, using 6 bits for their mantissa and 4 bits for their exponent which are both in two’s complement. Show the result in the same format and show your working.

0110000110 +
0101000100
Mark scheme 9(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • Calculating exponent of 6 for first number/Calculating exponent of 4 for second number

  • Moving binary point correct number of places in both mantissa to give 0110000.0 and 01010.0

  • Addition of 0110000.0 and 01010.0 to give 0111010.0

  • Movement of binary point six places to left to match original mantissa format and give value of 011101

  • Calculate exponent in original format to give value of 0110

Guidance

Alternative:

  • Calculation of exponents (6 and 4) / difference between exponents of 2

  • Move first mantissa 2 places left / second mantissa 2 places right to match exponents

  • …giving 011.000 + 0.10100 (exp 4) / 0.11000 + 0.00101 (exp 6)

  • Correct addition (answer of 011.1010 / 0.111010)

  • Final answer normalised (011101 exp 6)

If exponents initially matched to 4, final bullet point should only be given if answer is normalised. If exponents matched to 6, answer will already be normalised.Ensure that answer is in format described (6 bits for mantissa 4 bits for exponent) to gain final bullet point.

out of 5

Question 10

14 marks

10(a)(i)

Convert the denary number −119 to an 8-bit binary number with two’s complement representation.

Mark scheme 10(a)(i)

1000 1001

Guidance

Correct answer only

10(a)(ii)

Convert the unsigned binary number 1101101 to hexadecimal.

Mark scheme 10(a)(ii)

6D

Guidance

Correct answer only

10(a)(iii)

Convert the denary number 171 to hexadecimal.

Mark scheme 10(a)(iii)

AB

Guidance

Correct answer only

10(a)(iv)

Convert the hexadecimal number A6 to binary.

Mark scheme 10(a)(iv)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1010 0110

1 mark per nibble

Guidance

Correct answer only

out of 2

10(b)

Show how the denary value −9.125 can be represented in normalised floating point format, using 8 bits for the mantissa and 4 bits for the exponent, both in two’s complement.

Mark scheme 10(b)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • 10110.111 in fixed point

  • Move binary point four places to the left (1.0110111)

  • Gives mantissa of 10110111

  • therefore exponent is 4 = 0100

  • Forming final answer of 10110111 0100

out of 5

10(c)(i)

Show the result of applying an XOR mask of 1100 0111 to the byte 0101 1101.

[Diagram redrawn from the original paper by an AI model (Gemma 4 12B). It may contain mistakes.]

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Mark scheme 10(c)(i)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • 1001 1010

Guidance

1 mark per nibble, mark left to right

out of 2

10(c)(ii)

Describe a mask that could be applied to an 8-bit number to ensure that:

  • the most significant bit is always set to 1
  • all other bits remain unchanged.
Mark scheme 10(c)(ii)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

  • 1000 0000

  • OR

out of 2

Question 11

5 marks

11(a)

Convert the binary number 10110011 into a hexadecimal number.

Mark scheme 11(a)

B3

11(b)

Convert the two’s complement binary number 10011011 into a denary number.

Mark scheme 11(b)

  • 101

11(c)

Calculate the subtraction of the following two 8-bit binary numbers.

You must show your working.

[Diagram redrawn from the original paper by an AI model (Gemma 4 12B, corrected by Claude Sonnet). It may contain mistakes.]

11010011
01111001  -
__________
Mark scheme 11(c)

Tick each point your answer makes. Your mark is worked out from your ticks, and you can change it.

1 mark per bullet up to a maximum of 2 marks, e.g:

  • 0101 1010

  • Suitable working out

Guidance

Allow any suitable method of working out, e.g. the carry method or using two’s complement.

out of 2

11(d)

Convert the binary number 11101100 into a denary number.

END OF QUESTION PAPER

Mark scheme 11(d)

236