Number systems, number bases, bits, bytes and units – A-level Computer Science exam questions with mark scheme

1Core4 marks

The diagram shows how the sets of natural numbers ($\mathbb{N}$), integers ($\mathbb{Z}$), rational numbers ($\mathbb{Q}$) and real numbers ($\mathbb{R}$) relate to each other. Four regions are labelled A, B, C and D.

For each value below, state the letter of the region in which it belongs. [4]

(i) $0$ (ii) $-7$ (iii) $0.375$ (iv) $\sqrt{5}$

(i) A (0 is a natural number in this specification: $\mathbb{N} = \{0, 1, 2, \dots\}$)(1)
(ii) B (an integer that is not natural)(1)
(iii) C (rational but not an integer; allow working shown as $\tfrac{3}{8}$)(1)
(iv) D (irrational: real but not rational)(1)
Answer: (i) A (ii) B (iii) C (iv) D
Examiner tip: Zero is a natural number in AQA's definition; many students wrongly put it in region B.

2Core3 marks

A ticket system at a concert hall stores data about each customer in a queue for returned tickets.

(a) A customer is told they are 14th in the queue. State the type of number that describes this position. [1]

(b) The hall also records the length of time, in minutes, that each customer waited. Explain why this value should be treated as a real number rather than a natural number. [2]

(a) Ordinal (number); ignore 'natural' alongside it only if ordinal is clearly given as the answer(1)
(b) Time is a measurement / a continuous quantity(1)
(b) so it can take non-whole (fractional) values, e.g. 7.5 minutes, which natural numbers cannot represent; accept natural numbers are only used for counting whole items(1)
Answer: (a) Ordinal (b) waiting time is measured, can be fractional, so real; natural numbers only count whole things.
Examiner tip: For (b) say what kind of quantity it is (measured) AND what that means for the values it can take.

3Core4 marks

A pixel's red component is stored as the denary value 203.

(a) Convert 203 into 8-bit unsigned binary. [1]

(b) Convert 203 into hexadecimal. [1]

(a) 11001011(1)
(b) CB (accept lower case cb; allow follow-through from an incorrect (a) if correctly grouped into nibbles)(1)
(c) Hex is shorter / easier for humans to read, write and remember; fewer errors when copying(1)
(c) Each hex digit maps directly to 4 bits (one nibble), so converting back to binary is easy. Do NOT accept 'hex uses less memory / is faster for the computer'(1)
Answer: (a) 11001011 (b) CB (c) shorter and less error-prone for humans; one hex digit = 4 bits so easy to convert
Examiner tip: Never claim hex takes up less storage: the computer still stores the value in binary.

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4Standard3 marks

A 12-bit status register holds the unsigned binary value

1011 0110 1101

(a) Convert this value to hexadecimal. [1]

(b) Convert this value to denary. You must show your working. [2]

(a) B6D(1)
(b) Method: correct place values used, e.g. $2048+512+256+64+32+8+4+1$ or $11\times256 + 6\times16 + 13$(1)
(b) 2925(1)
Answer: (a) B6D (b) 2925
Examiner tip: Converting via the hex value (B6D) is quicker than adding twelve binary place values and less error-prone.

5Standard3 marks

A wildlife charity attaches a tracking tag to each animal. Each tag stores a unique ID as an unsigned binary number. The charity currently needs 3000 different IDs.

(a) Calculate the minimum number of bits needed for the ID. Show how you reached your answer. [2]

(b) The charity plans to tag up to 20 000 animals. State how many extra bits would be needed compared with your answer to (a). [1]

(a) Shows that 11 bits ($2^{11} = 2048$) is too few / $2^{12} = 4096 \ge 3000$(1)
(a) 12 (bits)(1)
(b) 3 (15 bits needed since $2^{14} = 16\,384 < 20\,000 \le 2^{15} = 32\,768$); allow follow-through: 15 minus their (a)(1)
Answer: (a) 12 bits (b) 3 extra bits (15 in total)
Examiner tip: With $n$ bits you get $2^n$ different values; find the smallest power of 2 that is at least the number of items.

6Standard3 marks

A solid-state drive is advertised as having a capacity of 500 GB. When connected, the operating system reports its capacity using binary prefixes.

(a) Calculate the capacity of the drive in GiB. Give your answer to 1 decimal place. [2]

(b) Explain why the two figures are different even though no storage has been lost. [1]

(a) $500 \times 10^{9} \div 2^{30}$ (or $\div 1\,073\,741\,824$)(1)
(a) 465.7 (GiB)(1)
(b) GB uses decimal prefix giga $= 10^{9}$ bytes but GiB uses binary prefix gibi $= 2^{30}$ bytes, so 1 GiB is larger than 1 GB / fewer GiB fit in the same number of bytes(1)
Answer: (a) 465.7 GiB (b) giga = $10^9$, gibi = $2^{30}$, and a gibibyte is bigger
Examiner tip: Convert to bytes first using the decimal prefix, then divide by the binary prefix; don't mix up which way round they go.

7Hard4 marks

A data logger in a greenhouse takes a 16-bit temperature reading every 5 seconds for exactly 24 hours. No other data is stored.

(a) Calculate the amount of storage needed for the readings, in KiB. Show your working. [3]

(b) The logger has 32 KiB of memory for readings. State, with a reason, whether all of the readings can be stored. [1]

(a) Number of readings: $24 \times 60 \times 60 \div 5 = 17\,280$; allow 17 281 (counting a reading at both the start and the end)(1)
(a) Converts to bytes: $17\,280 \times 2 = 34\,560$ bytes (or $17\,280 \times 16 = 276\,480$ bits then $\div 8$)(1)
(a) $34\,560 \div 1024 = 33.75$ KiB; allow 33.75 (2 d.p.) from 17 281 readings ($34\,562 \div 1024 = 33.752$); do not accept 34.56, which uses $\div 1000$(1)
(b) No, because 33.75 KiB is greater than 32 KiB (allow follow-through from their (a))(1)
Answer: (a) 33.75 KiB (b) No: 33.75 KiB > 32 KiB
Examiner tip: A KiB is 1024 bytes, not 1000; dividing by 1000 gives 34.56 and loses the final mark.

8Hard4 marks

In a byte-addressable memory, every address refers to one byte. Addresses are written in hexadecimal.

(a) A small embedded system uses addresses 0000 to FFFF. State how many bytes of memory can be addressed. [1]

(b) Express your answer to (a) using the most appropriate binary prefix. [1]

(c) Another system has 1 MiB of byte-addressable memory starting at address 0. State the highest address in hexadecimal. [1]

(d) State the minimum number of hexadecimal digits needed to write every address in a 4 GiB byte-addressable memory. [1]

(a) 65 536 (allow $2^{16}$)(1)
(b) 64 KiB (do not accept 64 KB / 65.5 kB unless KiB also given; allow follow-through)(1)
(c) FFFFF (allow 0FFFFF or leading zeros; do not accept 100000)(1)
(d) 8 (since 4 GiB $= 2^{32}$ bytes and each hex digit represents 4 bits)(1)
Answer: (a) 65 536 (b) 64 KiB (c) FFFFF (d) 8
Examiner tip: Addresses start at 0, so the highest address is one less than the number of locations.

9Stretch4 marks

The subroutine below converts a positive denary integer into hexadecimal. String indexing starts at 0, so digits[10] is "A".

FUNCTION ToHex(n)
  digits ← "0123456789ABCDEF"
  result ← ""
  WHILE n > 0
    r ← n MOD 16
    result ← digits[r] + result
    n ← n DIV 16
  ENDWHILE
  RETURN result
ENDFUNCTION

(a) Complete a trace table with columns n, r and result for the call ToHex(1000). Show the value of n at the start of each iteration. [3]

(b) State the value returned. [1]

(a) First iteration: n = 1000, r = 8, result = "8"(1)
(a) Second iteration: n = 62, r = 14, result = "E8" (r and result both correct; accept r shown as 14, not 'E')(1)
(a) Third iteration: n = 3, r = 3, result = "3E8", and loop ends with n = 0 (no further rows)(1)
(b) 3E8 (allow follow-through from their final result)(1)
Answer: (a) (1000, 8, "8"), (62, 14, "E8"), (3, 3, "3E8"), then n = 0 (b) 3E8
Examiner tip: The new digit is put in front of result, so digits come out in reverse order of the remainders; check by converting back: $3\times256+14\times16+8=1000$.

10Stretch4 marks

A student says: “1 TB and 1 TiB are basically the same amount of storage.”

(a) Calculate exactly how many more bytes there are in 1 TiB than in 1 TB. [1]

(b) Express this difference as a percentage of 1 TB, to 3 significant figures. [1]

(c) The percentage difference between kibi and kilo is only 2.4%. Explain why the percentage difference becomes larger for each larger prefix. [2]

(a) 99 511 627 776 (bytes) (from $2^{40} - 10^{12}$)(1)
(b) 9.95(%)(1)
(c) Each step up in binary prefix multiplies by 1024, but each step up in decimal prefix multiplies by only 1000(1)
(c) so the ratio is $1.024^{k}$ for the $k$th prefix; the 2.4% difference compounds / is multiplied in at each step (kilo 2.4%, mega 4.86%, giga 7.37%, tera 9.95%)(1)
Answer: (a) 99 511 627 776 bytes (b) 9.95% (c) ratio is $1.024^k$, so the gap compounds with each prefix
Examiner tip: In (c), say that the 1024 vs 1000 factor is applied once per prefix step, not just that the binary prefix is 'bigger'.

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