Paper 2 Data Representation Answers

These answers correspond to Paper 2 Data Representation Drills.

Verification note: every Python code block in this answer file has been executed locally.

Answer 1: Denary to Binary Function

Model answer:

def denary_to_binary(n):
    if n == 0:
        return "0"
 
    digits = ""
    while n > 0:
        remainder = n % 2
        digits = str(remainder) + digits
        n = n // 2
    return digits
 
 
print(denary_to_binary(0))
print(denary_to_binary(13))
print(denary_to_binary(45))

Expected output:

0
1101
101101

Mark points:

• handles zero;
• loops while the number is greater than zero;
• calculates remainder modulo 2;
• stores each remainder as a binary digit;
• builds digits in reverse or reverses at the end;
• updates the quotient using integer division by 2;
• stops when quotient becomes zero;
• returns the expected strings.

Common weak answer:

• returning the remainders in the order they are generated, which reverses the binary value.

Answer 2: Binary to Denary Function

Model answer:

def binary_to_denary(bits):
    total = 0
    for bit in bits:
        total = total * 2 + int(bit)
    return total
 
 
print(binary_to_denary("0"))
print(binary_to_denary("1101"))
print(binary_to_denary("10101100"))

Expected output:

0
13
172

Mark points:

• initializes total to zero;
• loops through bits from left to right;
• multiplies the running total by 2;
• converts each bit to an integer;
• adds the bit value;
• handles "0";
• produces all expected outputs.

Common weak answer:

• adding the digit values only, so 1011 would incorrectly become 3.

Answer 3: Denary to Hex Function

Model answer:

def denary_to_hex(n):
    digits = "0123456789ABCDEF"
    if n == 0:
        return "0"
 
    result = ""
    while n > 0:
        remainder = n % 16
        result = digits[remainder] + result
        n = n // 16
    return result
 
 
print(denary_to_hex(0))
print(denary_to_hex(58))
print(denary_to_hex(214))

Expected output:

0
3A
D6

Mark points:

• handles zero;
• uses digit symbols 0 to F;
• divides by 16 repeatedly;
• uses remainder modulo 16;
• maps remainders 10 to 15 to A to F;
• builds digits in the correct order;
• updates the quotient;
• stops correctly;
• produces all expected outputs.

Common weak answer:

• returning decimal 10 as two characters instead of hexadecimal digit A.

Answer 4: Hex to Denary Function

Model answer:

def hex_to_denary(hex_text):
    digits = "0123456789ABCDEF"
    total = 0
    for ch in hex_text:
        value = digits.index(ch)
        total = total * 16 + value
    return total
 
 
print(hex_to_denary("0"))
print(hex_to_denary("3A"))
print(hex_to_denary("D6"))

Expected output:

0
58
214

Mark points:

• maps each hex digit to a value from 0 to 15;
• scans digits from left to right;
• multiplies running total by 16;
• adds current digit value;
• handles A correctly as 10;
• handles D correctly as 13;
• returns integer results;
• produces all expected outputs.

Common weak answer:

• treating 3A as 3 + 10 = 13 instead of 3 * 16 + 10 = 58.

Answer 5: Validate Binary

Model answer:

def valid_binary(bits):
    if bits == "":
        return False
    for bit in bits:
        if bit != "0" and bit != "1":
            return False
    return True
 
 
print(valid_binary("1010"))
print(valid_binary(""))
print(valid_binary("1021"))

Expected output:

True
False
False

Mark points:

• rejects empty string;
• loops through each character;
• accepts 0;
• accepts 1;
• rejects any other character.

Common weak answer:

• checking whether the string is numeric, which would wrongly accept 1021.

Answer 6: Validate Hex

Model answer:

def valid_hex(hex_text):
    if hex_text == "":
        return False
 
    allowed = "0123456789ABCDEF"
    for ch in hex_text.upper():
        if ch not in allowed:
            return False
    return True
 
 
print(valid_hex("3A"))
print(valid_hex("ff"))
print(valid_hex(""))
print(valid_hex("G1"))

Expected output:

True
True
False
False

Mark points:

• rejects empty string;
• defines the allowed hexadecimal characters;
• converts lowercase to uppercase or otherwise accepts lowercase;
• loops through every character;
• rejects characters outside 0 to 9 and A to F;
• produces all expected outputs.

Common weak answer:

• rejecting lowercase ff when the question says lowercase should be accepted.

Answer 7: ASCII Codes

Model answer:

def char_codes(text):
    pairs = []
    for ch in text:
        pairs.append((ch, ord(ch)))
    return pairs
 
 
print(char_codes("Az0"))

Expected output:

[('A', 65), ('z', 122), ('0', 48)]

Mark points:

• loops through each character;
• uses ord;
• stores character and code as a pair;
• preserves order;
• returns the expected list.

Common weak answer:

• using int(ch), which only works for digit characters and does not give the character code.

Answer 8: Unicode Characters

Model answer:

def unicode_summary(text):
    return (len(text), len(text.encode("utf-8")))
 
 
print(unicode_summary("A€你"))

Expected output:

(3, 7)

Mark points:

• counts characters using len(text);
• encodes as UTF-8 bytes;
• counts encoded bytes;
• returns (3, 7) for the given string.

Common weak answer:

• assuming every character uses one byte. In UTF-8, A uses 1 byte, € uses 3 bytes, and 你 uses 3 bytes.

Answer 9: File Conversion

Model answer:

def denary_to_binary(n):
    if n == 0:
        return "0"
 
    digits = ""
    while n > 0:
        remainder = n % 2
        digits = str(remainder) + digits
        n = n // 2
    return digits
 
 
def convert_file(filename):
    results = []
    with open(filename, "r", encoding="utf-8") as f:
        for line in f:
            number = int(line.strip())
            results.append(str(number) + " -> " + denary_to_binary(number))
    return results
 
 
with open("numbers.txt", "w", encoding="utf-8") as f:
    f.write("5\n13\n45\n")
 
print(convert_file("numbers.txt"))

Expected output:

['5 -> 101', '13 -> 1101', '45 -> 101101']

Mark points:

• opens the file;
• loops through all lines;
• strips whitespace;
• converts each line to an integer;
• calls the conversion function;
• formats each result as requested;
• appends each result to a list;
• returns the expected list.

Common weak answer:

• reading only the first line of the file.

Answer 10: Menu Converter

Model answer:

def denary_to_binary(n):
    n = int(n)
    if n == 0:
        return "0"
 
    digits = ""
    while n > 0:
        remainder = n % 2
        digits = str(remainder) + digits
        n = n // 2
    return digits
 
 
def binary_to_denary(bits):
    total = 0
    for bit in bits:
        total = total * 2 + int(bit)
    return str(total)
 
 
def convert_choice(choice, value):
    if choice == "1":
        return binary_to_denary(value)
    if choice == "2":
        return denary_to_binary(value)
    return "INVALID CHOICE"
 
 
print(convert_choice("1", "1101"))
print(convert_choice("2", "13"))
print(convert_choice("9", "13"))

Expected output:

13
1101
INVALID CHOICE

Mark points:

• defines or calls a binary-to-denary conversion;
• defines or calls a denary-to-binary conversion;
• checks choice "1";
• checks choice "2";
• returns or prints the selected conversion result;
• handles invalid choice;
• converts value types where needed;
• does not use Python base-conversion shortcuts;
• produces the three expected outputs;
• keeps the menu choice logic clear.

Common weak answer:

• converting both choices in the same direction.