Revision Library

revisionlibrary.co.uk

GCSE · Data Representation

P2.5 Data Representation: Number Systems

OCR J277 · Calculators not allowed · about 100 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Computers represent all data internally using binary digits (bits), where each digit is either 0 or 1. Which of these 4-bit binary numbers represents the denary value 9?

A) 1001
B) 1010
C) 0110
D) 1100

(Total for Question 1 is 1 mark)

Whole numbers can be converted between denary (base 10) and binary (base 2) using the binary place values 128, 64, 32, 16, 8, 4, 2 and 1.

(a)Convert the denary number 6 to a 4-bit binary number.(1)

(b)Convert the 4-bit binary number 1011 to denary.(1)

(c)Convert the denary number 250 to an 8-bit binary number. Show your working.(2)

(Total for Question 2 is 4 marks)

Hexadecimal (base 16) uses the digits 0-9 and the letters A-F, where A=10, B=11, C=12, D=13, E=14 and F=15.

(a)Convert the denary number 58 to hexadecimal. Show your working.(2)

(b)Convert the hexadecimal number 2F to denary. Show your working.(2)

(c)State the largest denary value that can be represented using a single hexadecimal digit.(1)

(Total for Question 3 is 5 marks)

Because 4 bits (a nibble) can represent exactly 16 different values (0-15), each hexadecimal digit maps directly onto one 4-bit nibble.

(a)Convert the 8-bit binary number 10110101 to hexadecimal by splitting it into nibbles. Show your working.(2)

(b)Convert the hexadecimal number E3 to an 8-bit binary number. Show your working.(2)

(Total for Question 4 is 4 marks)

One byte of data is made up of two nibbles.

(a)State how many hexadecimal digits are needed to represent one byte of data.(1)

(b)Explain why two hexadecimal digits can represent exactly one byte, with no bits left over.(2)

(Total for Question 5 is 3 marks)

Programmers often display memory addresses, error codes and colour values in hexadecimal rather than in binary or denary.

(Total for Question 6 is 3 marks)

Units of data storage are built up from the bit: 1 nibble = 4 bits, 1 byte = 8 bits, and larger units (kilobyte, megabyte, gigabyte) are each 1000 times the previous unit.

(a)Put these units in order of size, smallest first: byte, bit, kilobyte, nibble.(1)

(b)State the number of bits in one byte.(1)

(c)A downloaded document is 4 kilobytes (KB) in size. Using 1 KB = 1000 bytes, calculate the number of bits in the document. Show your working.(2)

(Total for Question 7 is 4 marks)

The table below shows the same three numbers written in denary, 8-bit binary and hexadecimal. Some values are missing.

Denary	Binary (8-bit)	Hexadecimal
100	?	?
?	01011001	?
?	?	7D

Complete the table by filling in every missing value. Show your working for at least one conversion in each row.

(Total for Question 8 is 6 marks)

The number of different values that can be represented using a fixed number of bits, n, can be calculated using a simple formula.

(a)State, in the form 2ⁿ, the formula for the number of different values that can be represented using n bits.(1)

(b)Calculate the number of different values that can be represented using 5 bits. Show your working.(2)

(c)A games console stores each player's score as an unsigned 16-bit binary number. Calculate the maximum denary score that can be stored. Show your working.(2)

(Total for Question 9 is 5 marks)

Show that the 8-bit binary number 11001001 represents the denary value 201.

(Total for Question 10 is 2 marks)

Binary addition follows the same column method as denary addition, but any column that totals 2 or more requires a carry into the next column.

(a)Add the two 8-bit binary numbers 01001101 and 00110110. Show your carries and give your answer as an 8-bit binary number.(2)

(b)State the denary value of the result to check your answer to part (a) is correct.(1)

(Total for Question 11 is 3 marks)

A colour in a graphics program is stored as the hexadecimal code #2F8C00, giving the Red, Green and Blue (RGB) intensities as three pairs of hex digits.

(a)Calculate the denary value of the Green component, given by the hex pair 8C. Show your working.(2)

(b)Give one everyday example, other than colour codes, where hexadecimal is used to represent data.(1)

(Total for Question 12 is 3 marks)

A binary shift moves every bit in a binary number a fixed number of places to the left or right, filling any empty places with zeros.

(a)State the effect on the denary value of a binary number when a single left binary shift is performed.(1)

(b)The 8-bit register value 00001101 (denary 13) is shifted left by 2 places. State the new 8-bit binary value and its denary equivalent.(2)

(c)The 8-bit register value 11000000 (denary 192) is shifted left by 2 places. State the resulting 8-bit binary value, and explain why this shift no longer represents multiplying the denary value by 4.(2)

(Total for Question 13 is 5 marks)

A processor uses 8-bit registers to store unsigned binary numbers, so the largest value it can hold is 255.

(a)Add the binary numbers 11110101 (denary 245) and 00010000 (denary 16). Show your carries and give your answer as a 9-bit binary number.(2)

(b)State the 8-bit value that would actually be stored in the register, and explain why this is described as an overflow error.(2)

(Total for Question 14 is 4 marks)

A network technician reads the hexadecimal pair B4 from part of a device's MAC address and needs to check it against a filter rule written in binary.

(a)Convert the hexadecimal value B4 to an 8-bit binary number. Show your working.(2)

(b)Convert your answer to part (a) to denary. Show your working.(2)

(Total for Question 15 is 4 marks)

A digital camera stores the colour of each pixel using 24 bits, made up of 8 bits each for the red, green and blue components.

(a)Calculate the total number of different colours that can be represented using 24 bits. Give your answer both in the form 2ⁿ and as a denary value. Show your working.(2)

(b)Explain why increasing the number of bits used to store each pixel's colour increases the overall file size of an image.(2)

(Total for Question 16 is 4 marks)

A programmer writes the following algorithm, using the OCR Exam Reference Language, to convert a positive denary number entered by the user into its binary representation, using the divide-by-2 remainder method:

number = input("Enter a positive whole number: ")
number = int(number)
binary = ""
WHILE number > 0
    remainder = number MOD 2
    binary = str(remainder) + binary
    number = number DIV 2
ENDWHILE
print(binary)

(a)Complete a trace table for this algorithm when the user enters the value 22, showing the value of number at the start of each pass of the loop, the remainder calculated, the value of binary after it is updated, and the new value of number after the DIV 2 step.(5)

(b)State the value that would be output by print(binary) once the algorithm has finished running.(1)

(Total for Question 17 is 6 marks)

A programmer wants to write an algorithm, using the OCR Exam Reference Language, that inputs two positive integers, num1 and num2, calculates their total, and outputs whether adding these two numbers together would cause an overflow if the result were stored in an 8-bit register (i.e. whether the total is greater than 255). Write this algorithm. Any correct logic will be credited; minor syntax errors will be ignored.

(Total for Question 18 is 4 marks)

Discuss the benefits of using hexadecimal, rather than binary or denary, when programmers need to work with large binary values such as memory addresses, error codes or colour values. In your answer, you should refer to the readability of the numbers, the length of the numbers, and how easily hexadecimal can be converted to and from binary.

(Total for Question 19 is 6 marks)

A microcontroller stores unsigned integers using 8-bit registers.

(a)A register holds the binary value 10101100. Convert this value to hexadecimal. Show your working.(2)

(b)The values 10101100 and 01100101 are added together in the register. Perform the binary addition, showing your carries, and state whether an overflow occurs, giving a reason.(3)

(c)State the 8-bit binary value that would actually be stored in the register once the overflow has occurred, and give this value in denary.(2)

(Total for Question 20 is 7 marks)

Mark scheme · P2.5 Data Representation: Number Systems

Question 1

B1 A cao
Answer: A) 1001

Question 2

(a) B1 0110 cao
(a) Answer: 0110
(b) B1 11 cao
(b) Answer: 11
(c) M1 correct place values identified, e.g. 128+64+32+16+8+2 = 250
(c) A1 11111010 cao
(c) Answer: 11111010

Question 3

(a) M1 58 divided by 16 to give 3 remainder 10 (i.e. 3 and A) shown
(a) A1 3A cao
(a) Answer: 3A
(b) M1 2x16 and F(=15)x1 identified
(b) A1 47 cao
(b) Answer: 47
(c) B1 15 (F) cao
(c) Answer: 15

Question 4

(a) M1 binary correctly split into nibbles 1011 and 0101
(a) A1 B5 cao
(a) Answer: B5
(b) M1 E = 1110 and 3 = 0011 identified
(b) A1 11100011 cao
(b) Answer: 11100011

Question 5

(a) B1 2 cao
(a) Answer: 2
(b) B1 one byte is made up of 8 bits, and each hexadecimal digit represents exactly 4 bits (one nibble) oe
(b) B1 2 hex digits x 4 bits = 8 bits, which exactly matches the size of a byte, so no bits are left over oe
(b) Answer: One byte is 8 bits, and each hexadecimal digit represents exactly 4 bits (a nibble). Since 2 x 4 = 8, two hexadecimal digits map exactly onto one byte with no bits left over.

Question 6

B1 hexadecimal numbers are much shorter/more compact than the equivalent binary number oe
B1 long strings of binary digits are difficult for people to read/write accurately, so hexadecimal reduces the chance of transcription errors oe
B1 each hexadecimal digit maps directly onto a 4-bit nibble, so converting between binary and hexadecimal is quick and does not need a calculator oe
Answer: Hexadecimal is much shorter than binary for the same value, so it is quicker and less error-prone for people to read and write. It also converts to and from binary very quickly, because each hex digit represents exactly one 4-bit nibble.

Question 7

(a) B1 bit, nibble, byte, kilobyte cao (all four correct, in this order)
(a) Answer: bit, nibble, byte, kilobyte
(b) B1 8 cao
(b) Answer: 8
(c) M1 4 x 1000 x 8 shown, or equivalent staged working
(c) A1 32000 cao
(c) Answer: 32000 bits

Question 8

B1 row 1 binary: 01100100 cao
B1 row 1 hexadecimal: 64 cao (ft from their binary if consistent)
B1 row 2 denary: 89 cao
B1 row 2 hexadecimal: 59 cao (ft from the given binary)
B1 row 3 denary: 125 cao
B1 row 3 binary: 01111101 cao (ft from their denary if consistent)
Answer: Row 1: 100 = 01100100 = 64 (hex); Row 2: 01011001 = 89 = 59 (hex); Row 3: 125 = 01111101 = 7D (hex)

Question 9

(a) B1 2ⁿ cao
(a) Answer: 2ⁿ
(b) M1 2⁵ shown, or doubling shown (2,4,8,16,32)
(b) A1 32 cao
(b) Answer: 32
(c) M1 2¹⁶ identified or calculated as 65536
(c) A1 65535 cao
(c) Answer: 65535

Question 10

M1 correct place values identified: 128, 64, 8 and 1
A1 128+64+8+1 = 201 shown cso
Answer: 201 (given, cso)

Question 11

(a) M1 correct column addition attempted with carries shown/propagated correctly
(a) A1 10000011 cao
(a) Answer: 10000011
(b) B1 131 cao (ft from their answer to (a))
(b) Answer: 131

Question 12

(a) M1 8x16 and C(=12)x1 identified
(a) A1 140 cao
(a) Answer: 140
(b) B1 any valid example, e.g. MAC addresses, error/exception codes, memory addresses, or web page/URL character encoding oe
(b) Answer: MAC addresses (or memory addresses, or error codes)

Question 13

(a) B1 it multiplies the denary value by 2 oe
(a) Answer: It multiplies the value by 2
(b) M1 00110100 stated
(b) A1 52 stated (ft from their binary answer)
(b) Answer: 00110100 = 52
(c) B1 00000000 stated as the resulting 8-bit value
(c) B1 the two leftmost 1 bits are shifted off the end of the 8-bit register and lost (overflow), because the true result (192x4=768) needs more than 8 bits to store, so the register cannot hold it correctly oe
(c) Answer: 00000000; the two leading 1 bits are shifted off the left-hand end of the register and lost, because the true result of 192x4=768 needs more than 8 bits, so an overflow occurs and the stored value is incorrect.

Question 14

(a) M1 correct column addition attempted with carries shown/propagated correctly
(a) A1 100000101 cao
(a) Answer: 100000101
(b) B1 00000101 (5) stated as the stored 8-bit value, ft from their 9-bit answer to (a)
(b) B1 the true result (261) is greater than the maximum value (255) that an 8-bit register can hold, so the extra (9th) bit is lost/discarded and the stored value is incorrect oe
(b) Answer: 00000101 (denary 5) would be stored; this is an overflow error because the true result, 261, is too large to fit in 8 bits (max 255), so the leading bit is lost and the stored value is wrong.

Question 15

(a) M1 B = 1011 and 4 = 0100 identified
(a) A1 10110100 cao
(a) Answer: 10110100
(b) M1 correct place values identified, e.g. 128+32+16+4
(b) A1 180 cao (ft from their answer to (a))
(b) Answer: 180

Question 16

(a) M1 2²⁴ identified
(a) A1 16777216 cao
(a) Answer: 2²⁴ = 16777216 colours
(b) B1 more bits are needed to store the colour of each individual pixel oe
(b) B1 since the total file size is (bits per pixel x number of pixels), increasing the bits per pixel increases the total number of bits needed for the whole image oe
(b) Answer: Each pixel needs more bits to store its colour, and since total file size = bits per pixel x number of pixels, increasing the bits per pixel increases the total size of the file.

Question 17

(a) B1 pass 1: number=22, remainder=0, binary="0", new number=11, all correct
(a) B1 pass 2: number=11, remainder=1, binary="10", new number=5, all correct
(a) B1 pass 3: number=5, remainder=1, binary="110", new number=2, all correct
(a) B1 pass 4: number=2, remainder=0, binary="0110", new number=1, all correct
(a) B1 pass 5: number=1, remainder=1, binary="10110", new number=0, all correct (loop then ends as number is no longer > 0)
(a) Answer: Pass 1: 22,0,"0",11 | Pass 2: 11,1,"10",5 | Pass 3: 5,1,"110",2 | Pass 4: 2,0,"0110",1 | Pass 5: 1,1,"10110",0
(b) B1 10110 cao, ft from their trace table
(b) Answer: 10110

Question 18

B1 inputs num1 and num2 and converts them to integers, e.g. using input() and int() oe
B1 calculates the total, e.g. total = num1 + num2 oe
B1 uses selection to compare total against 255, e.g. IF total > 255 THEN ... ELSE ... ENDIF oe
B1 outputs an appropriate message for both the overflow branch and the no-overflow branch oe

Answer:

num1 = input("Enter the first number: ")
num2 = input("Enter the second number: ")
num1 = int(num1)
num2 = int(num2)
total = num1 + num2
IF total > 255 THEN
    print("Overflow: the result cannot be stored in an 8-bit register")
ELSE
    print("No overflow. Result = " + str(total))
ENDIF

Question 19

Level 1 (1-2): Basic, isolated point(s) with little or no development, e.g. a simple unsupported statement that hexadecimal is shorter than binary. Answer may be a simple, undeveloped list.
Level 2 (3-4): A reasonable discussion covering at least two relevant points (e.g. length, readability, conversion), with some development but limited linkage between ideas or minor inaccuracies.
Level 3 (5-6): A detailed, coherent discussion covering multiple relevant points, clearly linked to why they matter for programmers (e.g. reduced transcription errors, quick conversion via nibbles), with logically structured reasoning throughout.
Indicative content:
- Hexadecimal numbers are much shorter/more compact than the equivalent binary number (e.g. an 8-bit binary number needs only 2 hex digits), making them quicker to read and write.
- Long strings of binary digits are difficult for humans to read accurately and are more likely to lead to transcription/typing errors; hexadecimal reduces this risk.
- Each hexadecimal digit corresponds exactly to a 4-bit nibble, so conversion between binary and hexadecimal is quick and does not require a calculator or long division.
- Hexadecimal is still closely related to the underlying binary representation (unlike denary), so it remains meaningful to programmers working with bit patterns, memory addresses and colour codes.
- Converting between denary and binary requires more complex division/remainder calculations and is less directly linked to the bit pattern actually stored by the computer.
- A balanced answer might note that hexadecimal is still less immediately meaningful to non-programmers than denary, so it is a trade-off between compactness/accuracy for programmers and general readability.

Question 20

(a) M1 binary split into nibbles 1010 and 1100
(a) A1 AC cao
(a) Answer: AC
(b) M1 correct column addition attempted with carries shown/propagated correctly
(b) A1 100010001 cao (9-bit result)
(b) B1 overflow occurs, because a carry is generated out of the most significant (8th) bit / because the true total, 273, is greater than the maximum 8-bit value of 255 oe
(b) Answer: 100010001; overflow occurs because the true sum, 273, is greater than 255, so a carry is generated out of the register's most significant bit
(c) B1 00010001 stated (ft from their 9-bit answer to (b), dropping the leading bit)
(c) B1 17 cao (ft from their binary answer)
(c) Answer: 00010001 = 17

Data Representation: Number Systems - Worksheets, Questions and Revision