Data representation (binary, hex, storing data) – IB Computer Science exam questions with mark scheme

1Core4 marks

(a) Convert the decimal number 45 into an 8-bit binary number. [1]

(b) Convert your answer to part (a) into hexadecimal. [1]

(a) 00101101 (must be 8 bits as the question asks; 101101 alone does not score)(1)
(b) 2D (follow through from an incorrect (a) if grouped into nibbles correctly)(1)
(c) 9 × 16 + 15 (or 9F = 1001 1111 with place values shown, 128+16+8+4+2+1)(1)
(c) = 159(1)
Answer: (a) 00101101 (b) 2D (c) 159
Examiner tip: For binary to hex, split into groups of four bits from the right (0010 1101 = 2D); in hex, F = 15, not 16.

2Standard3 marks

A bitmap image is 1920 pixels wide and 1080 pixels high. Each pixel is stored using 24 bits (a colour depth of 24 bits). Calculate the uncompressed file size, in megabytes (MB), ignoring metadata. (Use 1 MB = 1 000 000 bytes.) [3]

Number of pixels = 1920 × 1080 = 2 073 600(1)
Size = 2 073 600 × 24 = 49 766 400 bits ÷ 8 = 6 220 800 bytes (allow 2 073 600 × 3 bytes)(1)
= 6.22 MB (accept 6.2 MB or 6.2208 MB; unit required)(1)
Answer: 6.22 MB (6 220 800 bytes)
Examiner tip: Convert bits to bytes (÷8) before converting to MB; the final unit must be stated.

3Standard4 marks

A 3-minute stereo (2-channel) recording is made using a sample rate of 44.1 kHz (44 100 samples per second for each channel) and a sample resolution of 16 bits per sample. Calculate the uncompressed file size in MB (use 1 MB = 1 000 000 bytes). Show your working. [4]

Time = 3 × 60 = 180 s(1)
Bits = 44 100 × 16 × 2 (channels) × 180 (factors in any order)(1)
= 254 016 000 bits = 31 752 000 bytes (÷ 8)(1)
= 31.75 MB (accept 31.8 MB or 31.752 MB; unit required)(1)
Answer: 31.75 MB (31 752 000 bytes)
Examiner tip: Remember to multiply by the number of channels (2 for stereo) and divide by 8.

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4Core3 marks

A web designer sets the colour of a button using the hexadecimal colour code #2E8B57. The first pair of hex digits gives the red value, the second pair the green value and the third pair the blue value.

(a) Convert the green value, 8B, into 8-bit binary. [1]

(b) Convert the green value, 8B, into denary. [1]

(a) 10001011 (8 = 1000, B = 1011) (allow a space between nibbles; must be 8 bits)(1)
(b) 139 (8 × 16 + 11) (allow follow-through from a correct conversion of their binary in (a))(1)
(c) Hex is shorter/more compact (one hex digit represents 4 bits), so it is easier for humans to read, remember or type / fewer errors when copying (reject 'easier to convert to binary' on its own – that compares hex with denary, not with binary; reject 'uses less memory/storage' – the computer still stores it in binary)(1)
Answer: (a) 10001011 (b) 139 (c) e.g. shorter and easier for people to read/write without errors; each hex digit maps to exactly 4 bits.
Examiner tip: In (c) 'hex takes up less memory' scores zero: the computer stores the same bits either way; the benefit is for the human reading it.

5Core3 marks

A school stores each student's locker number as an unsigned binary integer.

(a) State the largest denary number that can be stored using 6 bits. [1]

(b) The school has 300 lockers, each needing a different code. Calculate the minimum number of bits needed for each locker code. Justify your answer. [2]

(a) 63 (allow 2⁶ − 1; reject 64)(1)
(b) 8 bits give only 2⁸ = 256 different codes, which is fewer than 300 / idea that 256 < 300(1)
9 bits give 2⁹ = 512 codes, which is enough, so 9 bits (answer 9 with no justification scores 1 mark only)(1)
Answer: (a) 63 (b) 9 bits, because 2⁸ = 256 < 300 ≤ 2⁹ = 512.
Examiner tip: n bits give 2ⁿ different patterns but the largest value is 2ⁿ − 1, because one pattern is used for zero.

6Standard4 marks

A car park display stores the text Gate B12 as a string. The string contains 8 characters, including the space.

(a) In ASCII the character A has the code 65 and the letters are coded in alphabetical order. Write the 8-bit binary code for the character D. [1]

(b) Calculate the number of bits needed to store the string Gate B12 using 8 bits per character. [1]

(c) The display is upgraded so that it can show place names in Arabic, Greek and Chinese. Explain why the system would need to use Unicode rather than ASCII. [2]

(a) 01000100 (= 68) (must be 8 bits; accept 1000100 only if the leading 0 is shown elsewhere in working – otherwise 0)(1)
(b) 64 bits (8 × 8) (accept 8 bytes)(1)
(c) ASCII uses 7/8 bits per character so can represent at most 128/256 characters, which is mainly the English/Latin alphabet, digits and punctuation(1)
Unicode uses more bits per character (e.g. up to 16/32 bits; variable-length encodings such as UTF-8) so it can represent many more (over a million possible) characters, covering the scripts of most world languages, so Arabic, Greek and Chinese characters each get a unique code(1)
Answer: (a) 01000100 (b) 64 bits (c) ASCII has too few codes (128/256) for non-Latin scripts; Unicode uses more bits per character so it has unique codes for characters from all these languages.
Examiner tip: For (c) you must link the number of bits to the number of characters it can represent; 'Unicode has more characters' alone gets only one mark.

7Standard4 marks

A memory location holds the 16-bit pattern 1011 0110 0011 1101.

(a) Convert this bit pattern into hexadecimal. [2]

(b) Convert the hexadecimal value 7A0F into 16-bit binary. [1]

(a) Any two nibbles correctly converted, e.g. 1011 = B and 0110 = 6 / working shows groups of 4 converted separately(1)
B63D (all four digits correct and in the correct order; accept lower case b63d)(1)
(b) 0111 1010 0000 1111 (must be exactly 16 bits including the leading 0 and the 0000 nibble; spaces ignored)(1)
(c) 16 = 2⁴, so each hex digit has exactly 16 possible values, matching the 16 possible patterns of 4 bits / each group of four bits maps to exactly one hex digit (allow 'a nibble has 16 combinations, the same as one hex digit')(1)
Answer: (a) B63D (b) 0111101000001111 (c) Because 2⁴ = 16: every 4-bit group corresponds to exactly one hex digit.
Examiner tip: Students drop the leading 0 of 7 (0111) or write 0 as a single 0 instead of 0000, ending up with too few bits; always check you have 4 bits per hex digit.

8Hard5 marks

The graph shows an analogue sound wave that has been sampled 8 times (samples 0 to 7). Each sample is stored using a sample resolution of 3 bits, so the value of each sample is rounded to one of the quantisation levels 0 to 7. The dots show the stored sample values.

(a) State the 3-bit binary values stored for samples 1, 2 and 5. [2]

(b) Calculate the total number of bits needed to store all 8 samples, ignoring any file header. [1]

(c) The sample resolution is increased from 3 bits to 8 bits. Explain how this would affect the quality of the recording and the size of the file. [2]

(a) Sample 1 = 101 and sample 2 = 111 (both needed for this mark)(1)
sample 5 = 001 (must be written as 3 bits; reject '1')(1)
(b) 24 bits (8 × 3) (accept 3 bytes)(1)
(c) Quality improves: there are now 256 (2⁸) levels instead of 8, so each sample can be stored closer to the true amplitude / smaller rounding (quantisation) error, so the recording sounds more like the original(1)
File size increases: each sample needs 8 bits instead of 3, so the file is 8/3 (about 2.7) times larger (allow 'more bits per sample so a bigger file' for this mark)(1)
Answer: (a) 1: 101, 2: 111, 5: 001 (b) 24 bits (c) More levels (256), so samples are closer to the original wave (better quality), but each sample uses 8 bits so the file is about 2.7 times bigger.
Examiner tip: In (c) do not describe sample rate (how often samples are taken); resolution is about how many levels each sample can take.

9Hard6 marks

A security camera records uncompressed video. Each frame is an image 1280 pixels wide and 720 pixels high with a colour depth of 24 bits. The camera records 25 frames per second. Sound is not recorded.

(a) Calculate the size, in gigabytes (GB), of a 30-second clip. Use 1 GB = 1 000 000 000 bytes and give your answer to 2 decimal places. Show your working. [4]

(b) To save storage, the colour depth is reduced from 24 bits to 16 bits. Describe the effect of this change on the stored video. [2]

(a) Pixels per frame = 1280 × 720 = 921 600(1)
Size of one frame = 921 600 × 24 = 22 118 400 bits / × 3 = 2 764 800 bytes(1)
Number of frames = 25 × 30 = 750 and total = 750 × frame size (= 2 073 600 000 bytes / 16 588 800 000 bits)(1)
2.07 GB (correct conversion from bytes to GB; answer must be 2.07; allow follow-through for one arithmetic slip if the method is complete; a final answer of 16.59 (failing to divide by 8) scores max 3)(1)
(b) File size decreases (to two-thirds of the original, about 1.38 GB for the clip)(1)
Fewer colours can be represented (65 536 instead of about 16.7 million), so colour accuracy/quality is lower, e.g. visible banding in smooth gradients(1)
Answer: (a) 2.07 GB (b) Smaller file (about two-thirds the size) but fewer possible colours (2¹⁶ instead of 2²⁴), so lower colour quality.
Examiner tip: The most common lost mark is forgetting to divide by 8: colour depth is in bits but the answer is wanted in bytes.

10Stretch6 marks

A small embedded controller has a memory in which every location stores one byte. Memory addresses are written in hexadecimal and run from 0000 to FFFF.

(a) State the number of bits needed to store one memory address. [1]

(b) Calculate the total memory size in kibibytes (KiB). Use 1 KiB = 1024 bytes. Show your working. [2]

(c) A program is loaded into the addresses 4000 to 4FFF inclusive. Calculate how much memory, in KiB, the program uses. Show your working. [2]

(d) State, in hexadecimal, the first address after the end of the program. [1]

(a) 16 bits (4 hex digits × 4 bits each)(1)
(b) Number of locations = FFFF + 1 = 65 536 (= 2¹⁶) (accept 16⁴)(1)
65 536 ÷ 1024 = 64 KiB(1)
(c) 4FFF − 4000 + 1 = 1000₁₆ = 4096 locations/bytes (the +1 for an inclusive range must be seen or implied by the answer 4096; 4095 scores 0 for this mark)(1)
4096 ÷ 1024 = 4 KiB (allow follow-through from 4095 bytes giving 3.999 KiB for this mark only)(1)
(d) 5000 (accept 5000₁₆, 0x5000)(1)
Answer: (a) 16 (b) 65 536 bytes = 64 KiB (c) 1000₁₆ = 4096 bytes = 4 KiB (d) 5000
Examiner tip: When counting addresses in a range, include both ends: the number of addresses is last − first + 1, which is why FFFF gives 65 536, not 65 535.

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