Why is "Reading binary bits from the wrong end, so the place values are applied in reverse." a common mistake in Converting between number bases?

Why it happens: Place values are usually recited smallest-first as '1, 2, 4, 8…', so students write them left to right in that order and end up with the largest value on the wrong side. How to avoid it: The **leftmost** bit always carries the **largest** place value — 128 in an 8-bit number. Write the row as **128, 64, 32, 16, 8, 4, 2, 1** left to right before touching the bits. Check by converting back: reading `1011` in reverse gives 13 instead of the correct 11. Source: Reversed-bit errors are one of the three faults examiner reports identify most often in conversion questions.

Why is "Giving a binary answer with fewer than 8 bits when the question asked for an 8-bit number." a common mistake in Converting between number bases?

Why it happens: The value is genuinely correct, so the answer feels complete — the requirement to fill a fixed number of bit positions reads as a formatting detail rather than part of the answer. How to avoid it: Pad with **leading zeros** on the left: 12 becomes `00001100`, not `1100`. Leading zeros never change the value but are required when a bit length is specified, and they are essential for aligning nibbles when converting to hexadecimal. Source: AQA conversion questions routinely specify an 8-bit answer, matching the specification's maximum of binary 1111 1111.

Why is "Swapping the quotient and remainder when converting decimal to hexadecimal by division." a common mistake in Converting between number bases?

Why it happens: Both numbers are produced by the same division and neither is obviously 'first', so the order is easy to reverse — and the resulting answer still looks like a valid hex value. How to avoid it: The **quotient** is the **first** digit and the **remainder** is the **second**: 203 ÷ 16 = 12 r 11 gives `CB`, not `BC`. Always **convert back** — (12 × 16) + 11 = 203 ✓ confirms it in seconds, whereas `BC` would give 188. Source: This is the standard error in the division method and is why converting back is worth building as a habit.

Why is "Splitting binary into nibbles from the left, leaving an incomplete group at the right-hand end." a common mistake in Converting between number bases?

Why it happens: Reading left to right is the natural habit for text, so students group in the same direction without considering where the place values are anchored. How to avoid it: Always group **from the right**, so any incomplete group ends up on the **left** where it can be padded with leading zeros. `101101` becomes `0010 1101` = `2D`; grouping from the left would give `1011 01`, which is not a valid pair of nibbles. Source: Nibble alignment errors produce plausible-looking but wrong hexadecimal answers and appear in error-identification questions.

Why is "Using a hexadecimal letter directly in arithmetic without converting it to its decimal value first." a common mistake in Converting between number bases?

Why it happens: Digits 0–9 can be used straight away, so the letters feel like they should behave the same way — the conversion step is easy to skip when working quickly. How to avoid it: Convert every letter **before** calculating: A=10, B=11, C=12, D=13, E=14, F=15. For `2F`, that means (2 × 16) + **15** = 47. Writing the decimal value above each letter takes two seconds and removes the risk entirely. Source: Hex-letter errors are common in decimal conversions and are caught immediately by a range check, since results should not exceed 255.

Why is "Writing only the final answer, with no place-value row, nibble split or division shown." a common mistake in Converting between number bases?

Why it happens: The conversions feel simple enough to do mentally, and showing working seems like wasted time in an exam. How to avoid it: Conversion questions carry **method marks**, so a bare wrong answer scores **nothing** whereas correct working with a small slip still scores. Writing the place-value row takes about five seconds and also prevents the miscounted-column errors that cause most losses in the first place. Source: Mark schemes for 3.3.2 award marks for correct method as well as the final value.

AQA GCSE Computer Science (8525)

Converting between number bases

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Detailed Study Notes

Detailed notes on Fundamentals of Data Representation for AQA GCSE Computer Science, covering key concepts, explanations, examples, and exam-focused revision points.

Converting Between Number Bases — AQA GCSE Computer Science 8525 (3.3.2)

Six conversions, all within a fixed ceiling: decimal 0–255, binary up to 1111 1111, hexadecimal up to FF. These are pure method marks — every one is a mechanical procedure you can drill to automatic recall, and the specification caps the difficulty so there are no surprises. The single highest-value habit is writing the place-value row before you start, and the single best shortcut is that binary ↔ hex needs no arithmetic at all.

At a glance

You must convert both ways between: binary↔decimal, binary↔hexadecimal, decimal↔hexadecimal.
The ceiling is fixed: decimal 255 = binary 1111 1111 = hexadecimal FF.
Binary → decimal: write the place values above, add up wherever there is a 1.
Decimal → binary: work left to right through 128, 64, 32, 16, 8, 4, 2, 1 — does it fit? subtract if yes.
Binary ↔ hex is the easy one: split into groups of 4 bits, convert each group. No arithmetic.
Decimal ↔ hex: the safest route is to go via binary, since both halves are then mechanical.
Direct decimal → hex: divide by 16 — the quotient is the first digit, the remainder is the second.
Always pad to 8 bits with leading zeros when a question asks for an 8-bit answer.
Check every answer by converting back, and confirm it lies within 0–255.

What you’ll learn

Mapped to the AQA GCSE 8525 syllabus (2027-2029).

Understand how binary can be used to represent whole numbers, and be able to represent decimal values between 0 and 255 in binary.
Understand how hexadecimal can be used to represent whole numbers, and be able to represent decimal values between 0 and 255 in hexadecimal.
Be able to convert in both directions between binary and decimal, binary and hexadecimal, and decimal and hexadecimal, using maximum equivalent values of decimal 255, binary 1111 1111 and hexadecimal FF.

Binary ↔ decimal

One direction is addition, the other is subtraction. Both start with the place-value row.

Binary → decimal (the easy direction)

Method: write the place values above the bits, then add the values wherever there is a 1.

Convert 10110101:

Place value	128	64	32	16	8	4	2	1
Bit	1	0	1	1	0	1	0	1

$128 + 32 + 16 + 4 + 1 = 181$

That is the whole method. Write the row, circle the 1s, add.

Decimal → binary (the reverse)

Method: work left to right through the place values, asking "does it fit?" Subtract when it does.

Convert 181:

Place value	Does it fit into what's left?	Bit	Remaining
128	128 ≤ 181 ✓	1	181 − 128 = 53
64	64 > 53 ✗	0	53
32	32 ≤ 53 ✓	1	53 − 32 = 21
16	16 ≤ 21 ✓	1	21 − 16 = 5
8	8 > 5 ✗	0	5
4	4 ≤ 5 ✓	1	5 − 4 = 1
2	2 > 1 ✗	0	1
1	1 ≤ 1 ✓	1	0

Reading the bit column downwards: 10110101

Two checks that catch almost every error:

The remaining value must reach exactly 0 by the end. If it does not, you have made a mistake.
Convert your answer back to decimal — it takes ten seconds and confirms the result.

Padding to 8 bits

Convert 11: only the 8, 2 and 1 columns are used, giving 1011. But if the question asks for an 8-bit answer, pad with leading zeros:

$11 = \texttt{00001011}$

Leading zeros do not change the value — 1011 and 00001011 are the same number — but an 8-bit answer must show all eight bits.

AQA tip. Never attempt these mentally. Writing the place-value row and the subtraction column takes twenty seconds and turns a question where you might slip into one you cannot get wrong. Mark schemes award method marks for the working.

Binary → decimal: write place values, add wherever there is a 1.
Decimal → binary: work left to right, 'does it fit?', subtract when yes.
The remainder must reach exactly 0 — if not, there is an error.
Pad to 8 bits with leading zeros when the question asks for 8 bits.
Always check by converting back.

See the full worked example for converting between number bases →

Binary ↔ hexadecimal — the easy pair

Split into groups of four. No arithmetic required in either direction.

This is the conversion the whole of hexadecimal exists for, and it needs no calculation at all.

The 4-bit lookup table

Worth memorising — it makes both directions instant:

Binary	Hex	Binary	Hex
0000	0	1000	8
0001	1	1001	9
0010	2	1010	A
0011	3	1011	B
0100	4	1100	C
0101	5	1101	D
0110	6	1110	E
0111	7	1111	F

If you cannot recall a row, work it out with the place values 8, 4, 2, 1: 1011 = 8 + 2 + 1 = 11 = B.

Binary → hexadecimal

Method: split into groups of 4 from the RIGHT, then convert each group.

Convert 11010110:

1101  0110
 ↓     ↓
 D     6      →  D6

Each nibble is converted independently. This is exactly why hexadecimal is used as shorthand for binary.

Hexadecimal → binary

Method: convert each hex digit to 4 bits, then join them.

Convert 2F:

⚠️ Each digit must become exactly 4 bits, including leading zeros. Writing 2 as 10 instead of 0010 breaks the alignment and gives the wrong answer.

AQA tip. If the binary has fewer than 8 bits, pad on the left before splitting. 101101 becomes 0010 1101 = 2D. Splitting from the left instead would give 1011 01, which is wrong — always group from the right.

Binary → hex: split into groups of 4 FROM THE RIGHT, convert each group.
Hex → binary: convert each digit to exactly 4 bits, including leading zeros.
Nibble place values are 8, 4, 2, 1 — use them if you forget the lookup.
Pad the binary on the left before splitting if it is not a multiple of 4.
No arithmetic on the whole number is needed in either direction.

See the full worked example for converting between number bases →

Decimal ↔ hexadecimal

Two routes: go via binary (safest), or divide by 16 (faster once fluent).

Route 1 — go via binary (recommended)

Because both halves are mechanical, this is the most reliable method under exam pressure:

$\text{decimal} \;\longleftrightarrow\; \text{binary} \;\longleftrightarrow\; \text{hexadecimal}$

Convert 181 to hexadecimal:

Decimal → binary: 181 = 10110101 (worked earlier)
Split into nibbles: 1011 0101
Convert each: 1011 = 11 = B, 0101 = 5 = 5
Answer: B5

Convert B5 to decimal:

Hex → binary: B = 1011, 5 = 0101 → 10110101
Binary → decimal: 128 + 32 + 16 + 4 + 1 = 181 ✓

Route 2 — divide by 16 (direct)

Decimal → hexadecimal: divide by 16. The quotient is the first hex digit, the remainder is the second.

Convert 181:

$181 \div 16 = 11 \text{ remainder } 5$

Quotient 11 → B
Remainder 5 → 5
Answer: B5 ✓

This is exactly the DIV and MOD pair from 3.2.3: 181 DIV 16 = 11 and 181 MOD 16 = 5.

Hexadecimal → decimal: multiply the first digit by 16 and add the second.

Convert B5: (11 × 16) + 5 = 176 + 5 = 181 ✓

Which route to use

	Via binary	Divide by 16
Speed	Slower	Faster
Reliability	Very high — both steps mechanical	Depends on confident division
Shows working?	Yes, clearly	Yes

Use whichever you can perform without hesitating. Both earn full marks. If you are unsure of your division, the binary route is safer because each step is a lookup rather than a calculation.

AQA tip. Whichever route you take, check by converting back. It takes seconds and catches the single most common error — mixing up which of the quotient and remainder is the first digit.

Route 1: decimal ↔ binary ↔ hex — both steps mechanical, most reliable.
Route 2: decimal → hex by dividing by 16; quotient = first digit, remainder = second.
Hex → decimal: (first digit × 16) + second digit.
Dividing by 16 is exactly DIV and MOD from 3.2.3.
Both routes earn full marks — always check by converting back.

See the full worked example for converting between number bases →

The exam method: working, padding and checking

Method marks are available. Show the row, pad correctly, convert back.

Always show the working

Conversion questions carry method marks. A bare answer that is wrong scores nothing; correct working with a slip in the final arithmetic still scores.

Minimum working to show:

Binary → decimal: the place-value row and the addition.
Decimal → binary: the place values with the subtraction, or the "does it fit" column.
Binary ↔ hex: the nibble split, clearly grouped.
Decimal ↔ hex: the division with quotient and remainder, or the intermediate binary.

Padding rules

Situation	What to do
Question asks for 8 bits	Pad with leading zeros: 11 → `00001011`
Binary has fewer than 4 bits per group	Pad on the left before splitting: `101101` → `0010 1101`
Hex digit converts to fewer than 4 bits	Pad to exactly 4: `2` → `0010`, not `10`

Leading zeros never change the value — but they are required when the question specifies a bit length, and they are essential for correct nibble alignment.

The three checks

Range check. Every answer must sit within the specification's ceiling: decimal 0–255, binary at most 1111 1111, hex at most FF. Anything larger means an error.
Convert back. The most powerful check available. Ten seconds, and it verifies the whole conversion.
Zero check. In decimal → binary, the remaining value must reach exactly 0.

The reference triple

$255 \;=\; \texttt{1111 1111} \;=\; \texttt{FF}$

Learn this and a few anchors around it, so you can sanity-check any answer at a glance:

Decimal	Binary	Hex
0	0000 0000	00
10	0000 1010	0A
16	0001 0000	10
100	0110 0100	64
128	1000 0000	80
255	1111 1111	FF

AQA tip. 16 in hexadecimal is 10 — this looks alarming but follows directly from the place values: (1 × 16) + 0. Recognising it prevents a moment of panic in the exam.

Conversion questions carry method marks — always show the working.
Pad to 8 bits with leading zeros when the question asks for 8 bits.
Pad binary on the left before splitting into nibbles.
Check the range (0–255), convert back, and confirm the remainder reaches 0.
Learn the anchor triples: 0, 16, 100, 128, 255.

See the full worked example for converting between number bases →

Quick recap

You must convert both ways between binary↔decimal, binary↔hexadecimal and decimal↔hexadecimal.
The ceiling is decimal 255 = binary 1111 1111 = hexadecimal FF.
Binary → decimal: write the place values, add wherever there is a 1.
Decimal → binary: work left to right through the place values, subtracting when a value fits.
Binary ↔ hex: split into groups of four from the right; no arithmetic needed.
Decimal → hex: go via binary, or divide by 16 (quotient first, remainder second).
Hex → decimal: (first digit × 16) + second digit.
Pad with leading zeros when an 8-bit answer is required or to align nibbles.
Always check by converting back and confirming the answer lies within 0–255.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

Binary place values: 128, 64, 32, 16, 8, 4, 2, 1. Nibble place values: 8, 4, 2, 1.
255 = 1111 1111 = FF — the ceiling for every conversion.
Binary ↔ hex: group into 4 bits FROM THE RIGHT.
Hex → decimal: (first digit × 16) + second digit.
Decimal → hex: divide by 16 — quotient is the FIRST digit, remainder is the SECOND.
Hex nibbles: 1010=A, 1011=B, 1100=C, 1101=D, 1110=E, 1111=F.
16 in hexadecimal is written 10.
Always show working — method marks are available.

How it’s examined

3.3.2 is assessed on Paper 2 and is the most predictable source of marks in the entire specification — every series contains conversion questions, and the specification caps them at decimal 255 / binary 1111 1111 / hexadecimal FF, so there is no scope for surprise. Expect: (1) direct conversions, 1–2 marks each, in any of the six directions (AO2); (2) conversions embedded in other questions — a colour code in hexadecimal, a character code from an ASCII table, a shifted binary value in 3.3.4 — where the conversion is a step rather than the question (AO2); (3) multi-step items, 3–4 marks, such as converting decimal to hexadecimal via binary, where method marks are awarded for the intermediate working. Because these are procedural marks, the return on practice is exceptionally high: drill each direction until it is automatic, always write the place-value row, and check every answer by converting back. Examiner reports consistently identify the same three losses — reading bits right-to-left, forgetting to pad to 8 bits, and swapping the quotient and remainder when dividing by 16.

Sources: AQA GCSE Computer Science 8525 specification (v1.3, 16 June 2025) — 3.3.2 Converting between number bases; AQA GCSE Computer Science Paper 2 past papers, mark schemes and examiner reports. Last reviewed 2026-07-29.

Take this whole topic with you

Step-by-step worked examples — Converting between number bases

Step-by-step solutions to past-paper-style questions on converting between number bases, written exactly the way a tutor would explain them at the board.

Question type:

1Convert binary to decimal and back (4 marks)
Getting startedDirect calculation• Paper 2 style — 4 marks (AO2)• binary, decimal, place-value
Question
(a) Convert the binary number 01101110 to decimal. (b) Convert the decimal number 45 to an 8-bit binary number. (4 marks)
Step-by-step solution
1. Step 1
  (a) Write the place-value row above the bits, then add the values wherever the bit is 1: 64 + 32 + 8 + 4 + 2.
  $64 + 32 + 8 + 4 + 2 = 110$
2. Step 2
  (b) Work left to right through the place values, asking whether each fits into what remains. 128 > 45 ✗, 64 > 45 ✗, 32 ≤ 45 ✓ → subtract, leaving 13.
3. Step 3
  Continue: 16 > 13 ✗, 8 ≤ 13 ✓ → 5 left, 4 ≤ 5 ✓ → 1 left, 2 > 1 ✗, 1 ≤ 1 ✓ → 0 left.
4. Step 4
  The remainder reached exactly 0, confirming the conversion. Read the bit column downwards and pad to 8 bits.
Answer
(a)

Place value 128 64 32 16 8 4 2 1
Bit 0 1 1 0 1 1 1 0

64 + 32 + 8 + 4 + 2 = 110 (2 marks — 1 for the place values, 1 for the total.)

(b) Working through the place values for 45:

Place value Fits? Bit Left
128 ✗ 0 45
64 ✗ 0 45
32 ✓ 1 13
16 ✗ 0 13
8 ✓ 1 5
4 ✓ 1 1
2 ✗ 0 1
1 ✓ 1 0

45 = 00101101 (2 marks — 1 for the method, 1 for the correct 8-bit answer.)
Examiner tip
Two habits protect these marks. Write the place-value row every time — it earns a method mark and prevents miscounted columns. And check that the remainder reaches exactly 0 in part (b); if it does not, there is an error. Note the answer is padded to the full 8 bits as the question requires.
2Convert binary to hexadecimal and back (4 marks)
Getting startedDirect calculation• Paper 2 style — 4 marks (AO2)• binary, hexadecimal, nibble
Question
(a) Convert the binary number 11010110 to hexadecimal. (b) Convert the hexadecimal number 3A to an 8-bit binary number. (4 marks)
Step-by-step solution
1. Step 1
  (a) Split into groups of four from the right: 1101 and 0110.
2. Step 2
  Convert each nibble using place values 8, 4, 2, 1. 1101 = 8 + 4 + 1 = 13 = D. 0110 = 4 + 2 = 6 = 6.
  $1101 = 8+4+1 = 13 = D$
3. Step 3
  (b) Convert each hex digit to exactly four bits. 3 = 0011, A = 10 = 1010.
4. Step 4
  Join them. Note the leading zeros in 0011 are essential — writing 11 would break the alignment.
Answer
(a) Split into nibbles: 1101 0110

1101 = 8 + 4 + 1 = 13 = D (1)

0110 = 4 + 2 = 6 = 6 (1)

Answer: D6

(b) Convert each digit to 4 bits:

3 = 0011 (1)

A = 10 = 1010 (1)

Answer: 00111010
Examiner tip
This is the conversion that requires no arithmetic on the whole number — each nibble is handled independently, which is precisely why hexadecimal is used as shorthand for binary. The critical detail in part (b) is padding each digit to exactly four bits: writing 3 as 11 rather than 0011 misaligns everything that follows.
3Convert hexadecimal to decimal (3 marks)
Getting startedDirect calculation• Paper 2 style — 3 marks (AO2)• hexadecimal, decimal, place-value
Question
Convert the following hexadecimal numbers to decimal, showing your working.

(a) 2F (b) B5 (c) FF (3 marks)
Step-by-step solution
1. Step 1
  Use the place values 16 and 1. The value is (first digit × 16) + second digit.
2. Step 2
  (a) 2F: F = 15, so (2 × 16) + 15 = 32 + 15 = 47.
  $(2 \times 16) + 15 = 47$
3. Step 3
  (b) B5: B = 11, so (11 × 16) + 5 = 176 + 5 = 181.
  $(11 \times 16) + 5 = 181$
4. Step 4
  (c) FF: (15 × 16) + 15 = 240 + 15 = 255 — the specification's stated maximum.
  $(15 \times 16) + 15 = 255$
Answer
(a) 2F = (2 × 16) + 15 = 32 + 15 = 47 (1) (b) B5 = (11 × 16) + 5 = 176 + 5 = 181 (1) (c) FF = (15 × 16) + 15 = 240 + 15 = 255 (1)
Examiner tip
Part (c) is the specification's stated maximum and is worth knowing by heart, since it also acts as a range check — no two-digit hexadecimal number can exceed 255. The letters must be converted to their decimal values before multiplying: treating B as 11 rather than as the letter itself is where errors creep in.
4Convert decimal to hexadecimal by two routes (5 marks)
Building confidenceMulti-step problem• Paper 2 style — 5 marks (AO2)• decimal, hexadecimal, divide-by-16
Question
Convert the decimal number 203 to hexadecimal.

(a) Using the division method. (b) Using binary as an intermediate step. (c) State how you would check your answer. (5 marks)
Step-by-step solution
1. Step 1
  (a) Divide by 16. 203 ÷ 16 = 12 remainder 11, since 12 × 16 = 192 and 203 − 192 = 11.
  $203 \div 16 = 12 \text{ r } 11$
2. Step 2
  Convert both parts to hex digits. Quotient 12 = C (the first digit); remainder 11 = B (the second digit). Answer CB.
3. Step 3
  (b) Decimal → binary first. 203 = 128 + 64 + 8 + 2 + 1, giving 11001011.
4. Step 4
  Split into nibbles and convert. 1100 = 12 = C, 1011 = 11 = B. Same answer, CB ✓
5. Step 5
  (c) Check by converting back: (12 × 16) + 11 = 192 + 11 = 203 ✓
Answer
(a) Division method. 203 ÷ 16 = 12 remainder 11 (1)

Quotient 12 → C (first digit)

Remainder 11 → B (second digit)

Answer: CB (1)

(b) Via binary. 203 = 128 + 64 + 8 + 2 + 1 = 11001011 (1) Split into nibbles: 1100 1011

1100 = 8 + 4 = 12 = C

1011 = 8 + 2 + 1 = 11 = B

Answer: CB ✓ (1)

(c) Convert back to decimal: (12 × 16) + 11 = 192 + 11 = 203 ✓ (1)
Examiner tip
Both routes are fully creditworthy — use whichever you can perform without hesitating. The most common error in the division method is swapping the quotient and remainder, producing BC instead of CB; converting back catches this immediately, which is why part (c) matters as a habit rather than just as a question.
5Spot the errors in a set of conversions (5 marks)
Building confidenceIdentify & classify• Paper 2 style — 5 marks (AO3)• checking, debugging, conversion
Question
A student has completed these conversions. Identify which are wrong, explain the error, and give the correct answer.

(a) 1001 0110 = 150 (b) 2F = 31 (c) 45 = 00101101 (d) 1111 1111 = EE (e) 200 = C8 (5 marks)
Step-by-step solution
1. Step 1
  (a) Place values: 128 + 16 + 4 + 2 = 150. ✓ Correct.
2. Step 2
  (b) The student has treated F as its position rather than its value, or simply concatenated. Correctly: (2 × 16) + 15 = 47, not 31. ✗
3. Step 3
  (c) 32 + 8 + 4 + 1 = 45. ✓ Correct, and properly padded to 8 bits.
4. Step 4
  (d) Each nibble 1111 = 15 = F, so the answer is FF, not EE. The student has used E (14) instead of F (15). ✗
5. Step 5
  (e) Check: (12 × 16) + 8 = 192 + 8 = 200. ✓ Correct — C = 12.
Answer
(a) Correct — 128 + 16 + 4 + 2 = 150 ✓ (1)

(b) Wrong (1). The student has not used the correct value for F. Since F = 15, the conversion is (2 × 16) + 15 = 32 + 15 = 47, not 31 (1).

(c) Correct — 32 + 8 + 4 + 1 = 45, correctly padded to 8 bits ✓

(d) Wrong (1). Each nibble 1111 equals 8 + 4 + 2 + 1 = 15, which is F, not E (14). The correct answer is FF — which is also the specification's stated maximum for 8 bits (1).

(e) Correct — C = 12, so (12 × 16) + 8 = 192 + 8 = 200 ✓
Examiner tip
Error (d) is worth noting because 1111 1111 = FF = 255 is the specification's own maximum triple — recognising it instantly gives a free check on any 8-bit conversion. Error (b) is the classic hex-letter slip: the letters must be converted to their decimal values (A=10 … F=15) before any arithmetic.
6Convert through all three bases (5 marks)
Building confidenceMulti-step problem• Paper 2 style — 5 marks (AO2)• conversion, multi-step, all-bases
Question
Complete the table below, showing your working for each conversion. (5 marks)

Decimal Binary (8-bit) Hexadecimal
100
10011100
4D
Step-by-step solution
1. Step 1
  Row 1 — decimal 100 → binary. 64 ≤ 100 ✓ (36 left), 32 ≤ 36 ✓ (4 left), 4 ≤ 4 ✓ (0 left). So 01100100.
2. Step 2
  Row 1 → hex. Split 0110 0100: 4+2 = 6, and 4 = 4. So 64. (A useful coincidence to notice, not a rule.)
3. Step 3
  Row 2 — binary 10011100 → decimal. 128 + 16 + 8 + 4 = 156.
  $128 + 16 + 8 + 4 = 156$
4. Step 4
  Row 2 → hex. Split 1001 1100: 8+1 = 9, and 8+4 = 12 = C. So 9C.
5. Step 5
  Row 3 — hex 4D → binary. 4 = 0100, D = 13 = 1101, giving 01001101. → decimal: 64 + 8 + 4 + 1 = 77.
  $64 + 8 + 4 + 1 = 77$
Answer
Decimal Binary (8-bit) Hexadecimal
100 01100100 64
156 10011100 9C
77 01001101 4D

Working:

Row 1: 100 = 64 + 32 + 4 → 01100100; nibbles 0110=6, 0100=4 → 64 (2 marks)

Row 2: 10011100 = 128 + 16 + 8 + 4 = 156; nibbles 1001=9, 1100=12=C → 9C (2 marks)

Row 3: 4D → 4 = 0100, D = 1101 → 01001101 = 64 + 8 + 4 + 1 = 77 (1 mark)
Examiner tip
Notice the efficient route in every row: whenever hexadecimal is involved, go via binary rather than dividing, because the nibble split needs no arithmetic. The apparent coincidence that decimal 100 is hex 64 is exactly that — a coincidence — and should not be mistaken for a shortcut.
7Handle padding and nibble alignment correctly (6 marks)
StretchMulti-step problem• Paper 2 style — 6 marks (AO3)• padding, nibble, alignment, debugging
Question
A student converts the binary number 101101 to hexadecimal. They split it as 1011 and 01, giving B1.

(a) Explain why this is wrong. (b) Give the correct conversion, showing your method. (c) Explain why a binary number should be padded on the left rather than the right. (6 marks)
Step-by-step solution
1. Step 1
  (a) The split is from the wrong end. Nibbles must be grouped from the right, so the incomplete group ends up on the left, not the right.
2. Step 2
  The student's second group 01 has only two bits, which cannot be a valid nibble — every hex digit needs exactly four bits.
3. Step 3
  (b) Pad on the left to reach 8 bits: 101101 becomes 00101101.
4. Step 4
  Split and convert: 0010 = 2, 1101 = 8+4+1 = 13 = D. Answer 2D.
5. Step 5
  (c) Leading zeros do not change the value — 101101 and 00101101 are both 45 — whereas adding zeros on the right multiplies the value.
6. Step 6
  Demonstrate it: padding right would give 10110100 = 180, a completely different number.
Answer
(a) The student has grouped the bits from the left instead of from the right, leaving an incomplete group of only two bits (01) (1). Every hexadecimal digit must represent exactly four bits, so a two-bit group is not valid — the incomplete group must be at the left-hand end, where it can be padded (1).

(b) Pad on the left to 8 bits: 101101 → 00101101 Split into nibbles: 0010 1101

0010 = 2

1101 = 8 + 4 + 1 = 13 = D

Answer: 2D (2 marks)

(c) Adding zeros on the left does not change the value, because those columns have place values that contribute nothing — 101101 and 00101101 both equal 45 (1). Adding zeros on the right would shift every bit into a higher place value and so multiply the number: 10110100 would be 180, a completely different value (1).
Examiner tip
The reason nibbles are grouped from the right is the same reason we pad on the left: the least significant end must stay fixed, because that is where the place values are anchored. Part (c) can be checked instantly by converting both padded versions back to decimal, which is the general habit worth building.
8Apply conversions in a real context (6 marks)
StretchWord problem• Paper 2 style — 6 marks (AO2/AO3)• application, colour-code, conversion
Question
A graphics program stores each colour component as an 8-bit value. A designer wants a colour with red = 204, green = 51 and blue = 153.

(a) Convert each value to an 8-bit binary number. (b) Give the hexadecimal colour code. (c) Explain why the code is normally written in hexadecimal rather than decimal. (6 marks)
Step-by-step solution
1. Step 1
  (a) Red 204: 128 ✓ (76 left), 64 ✓ (12 left), 8 ✓ (4 left), 4 ✓ (0 left) → 11001100.
2. Step 2
  Green 51: 32 ✓ (19 left), 16 ✓ (3 left), 2 ✓ (1 left), 1 ✓ (0 left) → 00110011.
3. Step 3
  Blue 153: 128 ✓ (25 left), 16 ✓ (9 left), 8 ✓ (1 left), 1 ✓ (0 left) → 10011001.
4. Step 4
  (b) Split each into nibbles. 1100 1100 = CC, 0011 0011 = 33, 1001 1001 = 99.
5. Step 5
  Combine: the code is #CC3399.
6. Step 6
  (c) Two reasons: hex is far shorter than the 24 bits of binary and less error-prone to copy, and one hex digit is exactly four bits so the underlying pattern remains visible — which decimal would hide.
Answer
(a)

Red 204 = 128 + 64 + 8 + 4 = 11001100 (1)

Green 51 = 32 + 16 + 2 + 1 = 00110011 (1)

Blue 153 = 128 + 16 + 8 + 1 = 10011001 (1)

(b) Splitting each into nibbles:

1100 1100 → CC

0011 0011 → 33

1001 1001 → 99

Colour code: #CC3399 (1)

(c) In binary the colour would need 24 bits — 110011000011001110011001 — which is long, hard to read and very easy to miscopy, whereas hexadecimal expresses it in just six characters (1). Hexadecimal is preferred over decimal because one hex digit represents exactly four bits, so a developer can still recover the exact bit pattern by direct substitution, which a decimal representation such as 204, 51, 153 would hide (1).
Examiner tip
This shows all three bases doing genuine work in one scenario. Note that each colour component is two hex digits = 8 bits = 1 byte, capped at 255, which ties 3.3.2 to 3.3.1 and 3.3.3. The most efficient route to the hex code is via binary rather than dividing each value by 16.
9Explain why the conversion methods work (6 marks)
StretchWord problem• Paper 2 style — 6 marks (AO3)• explain, method, nibble, place-value
Question
(a) Explain why converting binary to hexadecimal can be done by splitting the binary into groups of four, without performing any arithmetic on the whole number.

(b) Explain why the same shortcut does not work for converting binary to decimal. (6 marks)
Step-by-step solution
1. Step 1
  (a) Start from the relationship between the bases. 16 = 2⁴, so hexadecimal is an exact power of two.
2. Step 2
  Four bits have exactly 16 possible combinations (2⁴ = 16), which is exactly the number of hexadecimal symbols — so every nibble maps onto exactly one hex digit with none left over and none unused.
3. Step 3
  Therefore the digit boundaries align. Each group of four bits corresponds to one hex column independently of the others, so the groups can be converted separately and simply written side by side.
4. Step 4
  (b) Decimal is base 10, and 10 is not a power of 2. There is no whole number of bits that gives exactly ten combinations — 3 bits give 8 and 4 bits give 16.
5. Step 5
  So the boundaries do not line up. A decimal digit does not correspond to a fixed group of bits, and the value of each bit depends on its position within the whole number rather than within a group.
6. Step 6
  Consequently real arithmetic is needed — adding place values, or repeatedly dividing by 2.
Answer
(a) Hexadecimal is base 16, and 16 = 2⁴ — an exact power of two (1). Four bits have exactly 2⁴ = 16 possible combinations, which is precisely the number of symbols hexadecimal uses, so every group of four bits corresponds to exactly one hexadecimal digit with none spare and none unused (1). This means the digit boundaries in hexadecimal line up perfectly with bit boundaries in binary, so each nibble can be converted independently of the others and the results simply written side by side to form the answer (1).

(b) Decimal is base 10, and 10 is not a power of 2 (1). There is no whole number of bits that produces exactly ten combinations — three bits give 8 and four bits give 16 — so a decimal digit does not correspond to any fixed group of bits (1). Because the boundaries do not align, the contribution of each bit depends on its position within the whole number rather than within a group, so the number cannot be split up and must be converted using genuine arithmetic — adding the place values, or repeatedly dividing by two (1).
Examiner tip
This is the deepest question in the subtopic and explains why hexadecimal earns its place in computer science rather than presenting the 4-bit rule as an arbitrary fact. The pivot is 16 = 2⁴ versus 10 being no power of 2, and a top-band answer states that relationship explicitly rather than merely asserting that hex 'works better'.

Place value	128	64	32	16	8	4	2	1
Bit	0	1	1	0	1	1	1	0

Place value	Fits?	Bit	Left
128	✗	0	45
64	✗	0	45
32	✓	1	13
16	✗	0	13
8	✓	1	5
4	✓	1	1
2	✗	0	1
1	✓	1	0

Decimal	Binary (8-bit)	Hexadecimal
100	`01100100`	`64`
156	`10011100`	`9C`
77	`01001101`	`4D`

Model Answers — Converting between number bases

High-scoring sample answers for converting between number bases on the AQA GCSE 8525 paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
Paper 2 style — AO22 marks
Convert the binary number 00110101 to decimal. Show your working. (2 marks)
Model answer

Place value 128 64 32 16 8 4 2 1
Bit 0 0 1 1 0 1 0 1

(1 mark for the place values.)

32 + 16 + 4 + 1 = 53 (1)
Why this scores
Showing the place-value row earns a method mark in its own right, so it should never be omitted even for a conversion you can do mentally. Quick sanity check: an 8-bit value must lie between 0 and 255.
Question 2
Paper 2 style — AO22 marks
Convert the decimal number 76 to an 8-bit binary number. (2 marks)
Model answer
Working through the place values: 64 fits (12 left), 8 fits (4 left), 4 fits (0 left).

64 + 8 + 4 = 76 (1)

76 = 01001100 (1)

The remainder reaches exactly 0, confirming the conversion.
Why this scores
The answer must be padded to the full 8 bits as the question specifies — writing 1001100 would lose the accuracy mark. The check that the remainder reaches exactly zero is the fastest way to confirm a decimal-to-binary conversion.
Question 3
Paper 2 style — AO22 marks
Convert the hexadecimal number E7 to an 8-bit binary number. (2 marks)
Model answer
Convert each hex digit to exactly four bits:

E = 14 = 1110 (1)

7 = 0111 (1)

Answer: 11100111
Why this scores
Each digit must become exactly four bits, so 7 is written 0111 rather than 111 — omitting the leading zero misaligns the whole number. Use the nibble place values 8, 4, 2, 1 if the lookup does not come to mind: 14 = 8 + 4 + 2.
Question 4
Paper 2 style — AO23 marks
Convert the decimal number 175 to hexadecimal, showing your method. (3 marks)
Model answer
Method 1 — divide by 16: 175 ÷ 16 = 10 remainder 15 (1)

Quotient 10 → A (first digit)

Remainder 15 → F (second digit)

Answer: AF (1)

Check by converting back: (10 × 16) + 15 = 160 + 15 = 175 ✓ (1)

(Method 2 — via binary: 175 = 128 + 32 + 8 + 4 + 2 + 1 = 10101111; nibbles 1010 = A and 1111 = F, giving AF.)
Why this scores
Both digits happen to be letters here, which tests the A–F values under slight pressure. The critical detail is that the quotient is the first digit and the remainder is the second — reversing them gives FA = 250, and converting back exposes the error instantly.
Question 5
Paper 2 style — AO24 marks
Complete the table, showing your working. (4 marks)

Decimal Binary (8-bit) Hexadecimal
60
A3
Model answer
Row 1 — decimal 60. 60 = 32 + 16 + 8 + 4 → 00111100 (1) Nibbles: 0011 = 3, 1100 = 12 = C → 3C (1)

Row 2 — hexadecimal A3. A = 10 = 1010, 3 = 0011 → 10100011 (1) Decimal: 128 + 32 + 2 + 1 = 163 (1)

Decimal Binary (8-bit) Hexadecimal
60 00111100 3C
163 10100011 A3
Why this scores
Going via binary is the efficient route in both directions here, because the nibble split requires no arithmetic. Check row 2 with the alternative method: (10 × 16) + 3 = 163 ✓ — agreement between two independent routes is strong confirmation.
Question 6
Paper 2 style — AO2/AO33 marks
A student converts the decimal number 12 to binary and writes 1100. The question asked for an 8-bit answer. Explain what is wrong and give the correct answer. (3 marks)
Model answer
The value is correct — 1100 is 8 + 4 = 12 (1). However, the question asked for an 8-bit answer, and 1100 uses only four bits (1).

The number must be padded with leading zeros to fill all eight bit positions, giving 00001100 (1). Leading zeros do not change the value, but they are required when the question specifies a particular number of bits — which matters because computers store values in fixed-size units such as a byte.
Why this scores
This distinguishes a correct value from a correct answer to the question asked. Leading zeros are also essential for nibble alignment when converting to hexadecimal, and reflect how data is genuinely stored — a byte always occupies eight bits regardless of how small the value is.
Question 7
Paper 2 style — AO2/AO36 marks
A network device reports an error code as the hexadecimal value 9E.

(a) Convert 9E to an 8-bit binary number. (b) Convert 9E to decimal. (c) Explain, with reference to your working, why converting hexadecimal to binary is easier than converting hexadecimal to decimal. (6 marks)
Model answer
(a) Convert each digit to exactly four bits:

9 = 1001 (1)

E = 14 = 1110 (1)

Answer: 10011110

(b) Either route gives the same result:

Direct: (9 × 16) + 14 = 144 + 14 = 158 (1)

Via binary: 128 + 16 + 8 + 4 + 2 = 158 ✓ (1)

(c) Converting to binary required no arithmetic at all — each hexadecimal digit was simply replaced by its four-bit equivalent, because one hex digit represents exactly four bits (1). Converting to decimal required genuine calculation: multiplying the first digit by 16 and adding the second, or adding five separate place values. This is because 16 is an exact power of two (2⁴) whereas 10 is not, so hexadecimal digit boundaries align with bit boundaries but decimal digit boundaries do not (1).
Why this scores
Part (c) is the discriminator and should refer to the candidate's own working as the question asks — contrasting the substitution in (a) with the multiplication in (b). The underlying reason, 16 = 2⁴ while 10 is not a power of 2, is what lifts this into the top band.
Question 8
Paper 2 style — AO36 marks
A student makes the following errors. For each, explain what went wrong and give the correct answer.

(a) Converting 1101 0011 to hexadecimal, they answer 3D. (b) Converting decimal 90 to hexadecimal, they divide by 16 to get 5 remainder 10 and answer A5. (c) Converting 1011 to decimal, they answer 13. (6 marks)
Model answer
(a) The student has converted the nibbles correctly but written them in the wrong order (1). 1101 = 13 = D and 0011 = 3, and the left-hand nibble comes first, so the answer is D3, not 3D (1).

(b) The student has swapped the quotient and the remainder (1). In 90 ÷ 16 = 5 remainder 10, the quotient 5 is the first digit and the remainder 10 (A) is the second, giving 5A (1). Checking: (5 × 16) + 10 = 80 + 10 = 90 ✓, whereas A5 would be (10 × 16) + 5 = 165.

(c) The student has read the bits in the wrong direction (1). Using place values 8, 4, 2, 1 from the left, 1011 = 8 + 2 + 1 = 11, not 13 — 13 would be 1101 (1).
Why this scores
These are the three most frequent conversion errors, and every one is caught by the same habit: convert back and check. Note that all three produce a plausible-looking answer within the valid range, which is precisely why they survive unnoticed — the value being wrong is not self-evident from the answer alone.
Question 9
Paper 2 style — AO36 marks
Explain why the binary place values are 128, 64, 32, 16, 8, 4, 2, 1, and why a hexadecimal digit corresponds to exactly four of these bits. (6 marks)
Model answer
Why the binary place values are those numbers. In any number system the place values are successive powers of the base, starting from the rightmost column (1). Binary is base 2, so the columns are 2⁰, 2¹, 2², 2³ … reading from right to left — which gives 1, 2, 4, 8, 16, 32, 64, 128 (1). Each column is therefore double the one to its right, and an 8-bit number reaches a maximum of 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1 = 255 (1).

Why one hexadecimal digit is exactly four bits. Four bits can form 2⁴ = 16 different combinations, running from 0000 to 1111, which represent the values 0 to 15 (1). Hexadecimal is base 16 and uses exactly 16 symbols — 0 to 9 followed by A to F, representing 0 to 15 (1). Because the number of combinations matches the number of symbols exactly, every possible nibble corresponds to exactly one hexadecimal digit with none spare and none unused — which is why the mapping is a direct substitution requiring no arithmetic (1).
Why this scores
This question asks students to justify the rules they use mechanically, which is genuinely A*-level. The two pivots are place values are powers of the base and 2⁴ = 16 matches the sixteen hex symbols. An answer that states the values without explaining where they come from typically scores 2–3.

Place value	128	64	32	16	8	4	2	1
Bit	0	0	1	1	0	1	0	1

Decimal	Binary (8-bit)	Hexadecimal
60	`00111100`	`3C`
163	`10100011`	`A3`

Key Formulae — Converting between number bases

The formulae you need to memorise for converting between number bases on the AQA GCSE 8525 paper, with every variable defined in plain English and a note on when to use it.

Hexadecimal to decimal (two digits)
$\text{value} = (d_1 \times 16) + d_2$
$d₁$
The first (left) hex digit, converted to its decimal value 0–15
$d₂$
The second (right) hex digit, converted to its decimal value 0–15
When to use
Converting any two-digit hexadecimal number to decimal. Convert the letters to their values (A=10 … F=15) before multiplying.
Example
B5 = (11 × 16) + 5 = 176 + 5 = 181
Decimal to hexadecimal (division method)
$n \div 16 = q \text{ remainder } r \;\Rightarrow\; \text{hex} = q\,r$
$q$
The quotient — becomes the FIRST hex digit
$r$
The remainder — becomes the SECOND hex digit
When to use
Converting a decimal value 0–255 directly to hexadecimal. Both q and r must then be written as hex digits (10–15 become A–F). This is DIV and MOD from 3.2.3.
Example
203 ÷ 16 = 12 r 11 → C and B → CB
Nibble place values
$8,\; 4,\; 2,\; 1$
$4 bits$
One nibble = one hexadecimal digit
$Range$
0000 to 1111, i.e. 0 to 15
When to use
Converting each group of four bits when moving between binary and hexadecimal. Use these if you cannot recall the lookup table.
Example
1101 = 8 + 4 + 1 = 13 = D

Key Definitions and Keywords — Converting between number bases

Definitions to memorise and the exact keywords mark schemes credit for converting between number bases answers — sharpened from recent examiner reports for the 2026 AQA GCSE 8525 sitting.

Base conversion
Examiner keyword
Rewriting a number in a different number base without changing its value. AQA requires conversion in both directions between binary, decimal and hexadecimal, up to a maximum of 255.
Related: Binary, Decimal, Hexadecimal
Place value method
Examiner keyword
Converting binary to decimal by writing the place values 128, 64, 32, 16, 8, 4, 2, 1 above the bits and adding the values wherever the bit is 1.
Example
10110101 → 128 + 32 + 16 + 4 + 1 = 181
Related: Binary, Decimal
Subtraction method
Examiner keyword
Converting decimal to binary by working left to right through the place values, asking whether each fits into the value remaining and subtracting when it does. The remainder must reach exactly 0.
Related: Place value method, Binary
Nibble
Examiner keyword
A group of four bits, corresponding to exactly one hexadecimal digit. Nibble place values are 8, 4, 2, 1, giving a range of 0 to 15.
Example
1101 = 8 + 4 + 1 = 13 = D
Related: Hexadecimal, Bit
Grouping into nibbles
Examiner keyword
Splitting a binary number into groups of four bits, always working from the RIGHT, so that any incomplete group falls at the left where it can be padded with leading zeros.
Example
101101 → 0010 1101 → 2D
Related: Nibble, Padding
Padding (leading zeros)
Examiner keyword
Adding zeros to the LEFT of a binary number so it fills a required number of bits. Leading zeros do not change the value; zeros added on the right would multiply it.
Example
12 = 1100, padded to 8 bits = 00001100
Related: Grouping into nibbles
Quotient and remainder (÷16)
Examiner keyword
In the decimal-to-hexadecimal division method, the quotient becomes the FIRST hex digit and the remainder becomes the SECOND. This is DIV and MOD from 3.2.3.
Example
203 ÷ 16 = 12 r 11 → C and B → CB
Related: Hexadecimal, Decimal
Converting via binary
Examiner keyword
Converting between decimal and hexadecimal by using binary as an intermediate step. Slower than dividing by 16 but more reliable, because both halves are mechanical.
Example
181 → 10110101 → B5
Related: Nibble, Base conversion
Range check
Examiner keyword
Confirming a converted value lies within the specification's stated maximum of 255 (binary 1111 1111, hexadecimal FF). Any larger result indicates an error.
Related: Maximum 8 bit value
Converting back to check
Examiner keyword
Verifying a conversion by converting the answer back to the original base. The single most effective check, and it catches reversed digits, swapped quotients and misread bits.
Example
B5 → (11 × 16) + 5 = 181 ✓
Related: Range check
Method marks
Examiner keyword
Marks awarded for correct working even when the final answer is wrong. Available in conversion questions, which is why the place-value row, nibble split or division should always be shown.
Related: Place value method
Hexadecimal digits
Examiner keyword
The sixteen symbols 0-9 then A-F, where A = 10, B = 11, C = 12, D = 13, E = 14 and F = 15. Each hexadecimal digit represents exactly one nibble.
Related: Nibble, Base conversion

Common Mistakes and Misconceptions — Converting between number bases

The traps other students keep falling into on converting between number bases questions — taken from recent AQA GCSE 8525 examiner reports and mark schemes — and how to avoid them.

✕Reading binary bits from the wrong end, so the place values are applied in reverse.
Reversed-bit errors are one of the three faults examiner reports identify most often in conversion questions.
Why it happens
Place values are usually recited smallest-first as '1, 2, 4, 8…', so students write them left to right in that order and end up with the largest value on the wrong side.
How to avoid it
The leftmost bit always carries the largest place value — 128 in an 8-bit number. Write the row as 128, 64, 32, 16, 8, 4, 2, 1 left to right before touching the bits. Check by converting back: reading 1011 in reverse gives 13 instead of the correct 11.
✕Giving a binary answer with fewer than 8 bits when the question asked for an 8-bit number.
AQA conversion questions routinely specify an 8-bit answer, matching the specification's maximum of binary 1111 1111.
Why it happens
The value is genuinely correct, so the answer feels complete — the requirement to fill a fixed number of bit positions reads as a formatting detail rather than part of the answer.
How to avoid it
Pad with leading zeros on the left: 12 becomes 00001100, not 1100. Leading zeros never change the value but are required when a bit length is specified, and they are essential for aligning nibbles when converting to hexadecimal.
✕Swapping the quotient and remainder when converting decimal to hexadecimal by division.
This is the standard error in the division method and is why converting back is worth building as a habit.
Why it happens
Both numbers are produced by the same division and neither is obviously 'first', so the order is easy to reverse — and the resulting answer still looks like a valid hex value.
How to avoid it
The quotient is the first digit and the remainder is the second: 203 ÷ 16 = 12 r 11 gives CB, not BC. Always convert back — (12 × 16) + 11 = 203 ✓ confirms it in seconds, whereas BC would give 188.
✕Splitting binary into nibbles from the left, leaving an incomplete group at the right-hand end.
Nibble alignment errors produce plausible-looking but wrong hexadecimal answers and appear in error-identification questions.
Why it happens
Reading left to right is the natural habit for text, so students group in the same direction without considering where the place values are anchored.
How to avoid it
Always group from the right, so any incomplete group ends up on the left where it can be padded with leading zeros. 101101 becomes 0010 1101 = 2D; grouping from the left would give 1011 01, which is not a valid pair of nibbles.
✕Using a hexadecimal letter directly in arithmetic without converting it to its decimal value first.
Hex-letter errors are common in decimal conversions and are caught immediately by a range check, since results should not exceed 255.
Why it happens
Digits 0–9 can be used straight away, so the letters feel like they should behave the same way — the conversion step is easy to skip when working quickly.
How to avoid it
Convert every letter before calculating: A=10, B=11, C=12, D=13, E=14, F=15. For 2F, that means (2 × 16) + 15 = 47. Writing the decimal value above each letter takes two seconds and removes the risk entirely.
✕Writing only the final answer, with no place-value row, nibble split or division shown.
Mark schemes for 3.3.2 award marks for correct method as well as the final value.
Why it happens
The conversions feel simple enough to do mentally, and showing working seems like wasted time in an exam.
How to avoid it
Conversion questions carry method marks, so a bare wrong answer scores nothing whereas correct working with a small slip still scores. Writing the place-value row takes about five seconds and also prevents the miscounted-column errors that cause most losses in the first place.

Converting Between Number Bases — AQA GCSE Computer Science 8525 (3.3.2)

Place value

128

Bit

Place value

Does it fit into what's left?

Bit

Remaining

128

128 ≤ 181 ✓

181 − 128 = 53

64 > 53 ✗

32 ≤ 53 ✓

53 − 32 = 21

16 ≤ 21 ✓

21 − 16 = 5

8 > 5 ✗

4 ≤ 5 ✓

5 − 4 = 1

2 > 1 ✗

1 ≤ 1 ✓

Binary

Hex

Binary

Hex

0000

1000

0001

1001

0010

1010

0011

1011

0100

1100

0101

1101

0110

1110

0111

1111

Via binary

Divide by 16

Speed

Slower

Faster

Reliability

Very high — both steps mechanical

Depends on confident division

Shows working?

Yes, clearly

Yes

Situation

What to do

Question asks for 8 bits

Pad with leading zeros: 11 → 00001011

Binary has fewer than 4 bits per group

Pad on the left before splitting: 101101 → 0010 1101

Hex digit converts to fewer than 4 bits

Pad to exactly 4: 2 → 0010, not 10

Decimal

Binary

Hex

0000 0000

0000 1010

0001 0000

100

0110 0100

128

1000 0000

255

1111 1111

Place value	128	64	32	16	8	4	2	1
Bit	0	1	1	0	1	1	1	0

Place value

128

Bit

Place value	Fits?	Bit	Left
128	✗	0	45
64	✗	0	45
32	✓	1	13
16	✗	0	13
8	✓	1	5
4	✓	1	1
2	✗	0	1
1	✓	1	0

Place value

Fits?

Bit

Left

128

✗

✓

✗

✓

✗

✓

Decimal

Binary (8-bit)

Hexadecimal

100

10011100

4D

Decimal	Binary (8-bit)	Hexadecimal
100	`01100100`	`64`
156	`10011100`	`9C`
77	`01001101`	`4D`

Decimal

Binary (8-bit)

Hexadecimal

100

01100100

64

156

10011100

9C

01001101

4D

Place value	128	64	32	16	8	4	2	1
Bit	0	0	1	1	0	1	0	1

Place value

128

Bit

Decimal

Binary (8-bit)

Hexadecimal

A3

Decimal	Binary (8-bit)	Hexadecimal
60	`00111100`	`3C`
163	`10100011`	`A3`

Decimal

Binary (8-bit)

Hexadecimal

00111100

3C

163

10100011

A3

Converting between number bases

Detailed Study Notes

At a glance

What you’ll learn

Binary ↔ decimal

Binary → decimal (the easy direction)

Decimal → binary (the reverse)

Padding to 8 bits

Binary ↔ hexadecimal — the easy pair

The 4-bit lookup table

Binary → hexadecimal

Hexadecimal → binary

Decimal ↔ hexadecimal

Route 1 — go via binary (recommended)

Route 2 — divide by 16 (direct)

Which route to use

The exam method: working, padding and checking

Always show the working

Padding rules

The three checks

The reference triple

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Convert binary to decimal and back (4 marks)

2Convert binary to hexadecimal and back (4 marks)

3Convert hexadecimal to decimal (3 marks)

4Convert decimal to hexadecimal by two routes (5 marks)

5Spot the errors in a set of conversions (5 marks)

6Convert through all three bases (5 marks)

7Handle padding and nibble alignment correctly (6 marks)

8Apply conversions in a real context (6 marks)

9Explain why the conversion methods work (6 marks)

Hexadecimal to decimal (two digits)

Decimal to hexadecimal (division method)

Nibble place values

Base conversion

Place value method

Subtraction method

Nibble

Grouping into nibbles

Padding (leading zeros)

Quotient and remainder (÷16)

Converting via binary

Range check

Converting back to check

Method marks

Hexadecimal digits

✕Reading binary bits from the wrong end, so the place values are applied in reverse.

✕Giving a binary answer with fewer than 8 bits when the question asked for an 8-bit number.

✕Swapping the quotient and remainder when converting decimal to hexadecimal by division.

✕Splitting binary into nibbles from the left, leaving an incomplete group at the right-hand end.

✕Using a hexadecimal letter directly in arithmetic without converting it to its decimal value first.

✕Writing only the final answer, with no place-value row, nibble split or division shown.

Converting between number bases

Detailed Study Notes

At a glance

What you’ll learn

Binary ↔ decimal

Binary → decimal (the easy direction)

Decimal → binary (the reverse)

Padding to 8 bits

Binary ↔ hexadecimal — the easy pair

The 4-bit lookup table

Binary → hexadecimal

Hexadecimal → binary

Decimal ↔ hexadecimal

Route 1 — go via binary (recommended)

Route 2 — divide by 16 (direct)

Which route to use

The exam method: working, padding and checking

Always show the working

Padding rules

The three checks

The reference triple

Quick recap

Memorise this

How it’s examined

Take this whole topic with you