AQA A-Level Computer Science Paper 2, June 2025: Question 12
9 marks · Medium difficulty · Calculation
Identify normalised floating point values, state reasons for normalisation, convert between two's complement floating point and decimal, and calculate relative error.
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Question text
12 A computer represents numbers using a normalised floating point representation
with an 8-bit mantissa and a 6-bit exponent, both stored using two’s complement,
as shown in Figure 7.
Five different bit patterns that are stored in this computer’s memory are listed in
Figure 7 and are labelled with the letters A to E. Four of the bit patterns are valid
normalised floating point numbers and one is not.
Figure 7
12.1 Shade one lozenge to indicate which bit pattern (A–E) in Figure 7 does not represent
a normalised value.
[1 mark]
A B C D E
12.2 Shade one lozenge to indicate which bit pattern (A–E) in Figure 7 shows the most
negative normalised value that can be represented.
[1 mark]
A B 31C D E
12.3 State one reason why values stored using a floating point representation are
usually normalised.
[1 mark]
Question parts 12.4 and 12.5 use a normalised floating point representation with an
8-bit mantissa and a 6-bit exponent, both stored using two’s complement.
12.4 Figure 8 shows a floating point representation of a number:
Figure 8
Calculate the decimal equivalent of the number in Figure 8.
Express your answer to at least four decimal places or as a fraction.
You should show your working.
[2 marks]
Answer
Question parts 12.4 and 12.5 use a normalised floating point representation with
an 8-bit mantissa and a 6-bit exponent, both stored using two’s complement.
12.5 The decimal value 43 057 152 cannot be represented exactly using this floating point
system.
Write the closest possible normalised floating point representation of the decimal
value 43 057 152 in the boxes below.
You should show your working.
[3 marks]
12.6 Calculate the relative error that occurred when you represented 43 057152 as
closely as possible in question part 12.5.
You should show your working.
Express your answer as a percentage to at least four decimal places.
[1 mark]
Answer
Mark scheme
Show the mark scheme
Total
Qu Pt Marking guidance
marks
12 1 Mark is AO1 (understanding) 1
C;
R. if more than one lozenge shaded
Total
Qu Pt Marking guidance
marks
12 2 Mark is AO1 (understanding) 1
D;
R. if more than one lozenge shaded
Total
Qu Pt Marking guidance
marks
12 3 Mark is AO1 (understanding) 1
Maximises precision / accuracy for given number of bits;
Note: Must have concept of given number of bits or an example of this
eg word length.
Unique representation of each number // simpler to test for equality of numbers;–A-LEVELCOMPUTER SCIENCE––
Max 1
Total
Qu Pt Marking guidance
marks
12 4 Marks are AO2 (apply) 2
10 1 0 1 0 0 0 1 1 1 1 0 0
Mantissa Exponent
2 marks for correct final answer written to 4dp: –0.0430 or as a fraction: –11/256
A. written to more decimal places (rounded or truncated), the exact answer is
–0.04296875
If answer is incorrect then award 1 method mark for either:
• showing correct value of both mantissa and exponent in decimal
(Mantissa = –0.6875 / –11/16 Exponent = –4)
• showing binary point shifted 4 places to left in binary number
• indicating that final answer has been calculated using
answer = mantissa × 2exponent and used either the correct mantissa, the correct
exponent, or both in this calculation.
30 Total
Qu Pt Marking guidance
marks
12 5 Marks are AO2 (apply) 3
3 marks for correct final answer:
01 0 1 0 0 1 0 0 1 1 0 1 0
Mantissa Exponent
If answer is incorrect then award up to 2 method marks, one for each point from
this list:
• correct (unsigned) fixed point representation of 43 057 152 in binary:
10 1001 0001 0000 00000000 0000 A. leading 0s
• identifying need to represent 225 223 and 220 // adding 225 223 and 220 //
calculating closest value that can be represented is 42 991 616
A. inclusion of 216 // 65 536
A. values given in decimal: 33 554 432 and 8 388608 and 1 048 576
R. other powers of 2 also given (apart from 16)
• showing the correct value of the exponent in decimal (26) or binary (11010) in
the working space or in binary in the final answer box // showing the binary
point being shifted 26 places left
• showing the correct value of the mantissa in binary in the working space or final
answer box – A-LEVEL COMPUTER SCIENCE – –
• giving mantissa as 1.0100100 and exponent as 011001
• giving mantissa as 0.0101001 and exponent as 011011
Total
Qu Pt Marking guidance
marks
12 6 Mark is AO2 (apply) 1 31
1 mark: 0.1522 (%)
A. written to more decimal places (rounded or truncated): 0.1522070015 to 10dp
R. –0.1522 %
How to answer it
Two’s Complement Floating Point Representation
This question evaluates your complete operational mastery of two’s complement normalised floating point numbers:
- Identifying Normalisation: Recognising valid positive ( 01... ) and negative ( 10... ) patterns.
- Extreme Values: Identifying the most negative representable value.
- Purpose of Normalisation: Maximising precision and providing a unique representation.
- Binary to Decimal Conversion: Calculating the value using mantissa × 2^exponent .
- Decimal to Floating Point Conversion: Approximating large integers within bit constraints and normalising.
- Relative Error: Quantifying approximation error as a percentage of the original value.
Question 12.1
Identify the Bit Pattern that is NOT Normalised
✅ Correct Answer
Shade lozenge C.
Mantissa for C is: 0.0 1 1 1 1 1 1
💡 Key Knowledge
In two’s complement normalised floating point:
- A positive number starts with 01 (sign bit 0 , first fractional bit 1 ).
- A negative number starts with 10 (sign bit 1 , first fractional bit 0 ).
🧠 Exam Technique
Inspect the first two bits of each mantissa immediately:
- A: 01... (Valid positive normalised)
- B: 10... (Valid negative normalised)
- C: 00... ❌ Not normalised! The first two bits are identical.
- D: 10... (Valid negative normalised)
- E: 01... (Valid positive normalised)
❌ Common Errors
Looking at the exponent instead of the mantissa. Normalisation rules apply only to the mantissa.
Question 12.2
Most Negative Normalised Value
✅ Correct Answer
Shade lozenge D.
📐 Analysis of Options B and D
To find the most negative value, we need a negative mantissa with the largest positive exponent:
- Mantissa: 1.0000000 represents the value -1.0 (the minimum possible mantissa).
- Option B exponent: 100000 = -32 (most negative exponent, making the number tiny: -1.0 × 2⁻³²).
- Option D exponent: 011111 = +31 (largest positive exponent: -1.0 × 2³¹ = -2,147,483,648).
❌ Common Errors
Confusing "most negative" with "smallest magnitude". Candidates often mistakenly select B because its exponent has the leading 1 (negative exponent), not realising that makes the number closest to zero, not most negative.
Question 12.3
Reason for Normalising Floating Point Values
✅ Acceptable Answers (Any One)
- Maximises precision / accuracy for a given number of bits (or given word length).
- Provides a unique representation for each number (making it simpler/faster to test for equality).
🧠 Examiner Trap: "Given Number of Bits"
Simply stating "it makes numbers more accurate" is often rejected! The mark scheme specifically notes:
"Must have concept of given number of bits or an example of this e.g. word length."
Always write: "Maximises precision for a given number of bits."
Question 12.4
Convert Floating Point to Decimal Equivalent
📐 Step-by-Step Calculation
1. Decode the Mantissa: 1.0 1 0 1 0 0 0
| -1 | 0.5 | 0.25 | 0.125 | 0.0625 | 0.03125 | 0.015625 | 0.0078125 |
|---|---|---|---|---|---|---|---|
| 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
Value = -1 + 0.25 + 0.0625 = -0.6875 (or -11/16).
2. Decode the Exponent: 1 1 1 1 0 0 (Two’s complement)
Weights: -32 + 16 + 8 + 4 + 0 + 0 = -32 + 28 = -4.
3. Calculate the Number:
Value = Mantissa × 2^(Exponent) = -0.6875 × 2⁻⁴ = -0.6875 / 16
Value = -0.04296875
To 4 decimal places: -0.0430 (or as fraction: -11/256).
✅ Final Answer
-0.0430 (or -11/256)
Exact value: -0.04296875
🧠 Method Mark Criteria (If final answer incorrect)
1 method mark awarded for:
- Showing mantissa = -0.6875 (-11/16) and exponent = -4 in decimal.
- OR showing the binary point shifted 4 places to the left.
- OR showing formula: answer = mantissa × 2^exponent and applying it to either correct part.
❌ Common Errors
- Forgetting that the mantissa is negative (missing the leading minus sign).
- Misinterpreting the 6-bit exponent 111100 as +60 instead of -4.
- Failing to round to at least 4 decimal places as instructed (e.g. writing -0.04).
Question 12.5
Convert Large Decimal to Closest Normalised Floating Point
📐 Step-by-Step Working
Step 1: Express 43 057 152 in powers of 2
- 2²⁵ = 33 554 432 → Remainder: 43 057 152 - 33 554 432 = 9 502 720
- 2²⁴ = 16 777 216 → Too large (0)
- 2²³ = 8 388 608 → Remainder: 9 502 720 - 8 388 608 = 1 114 112
- 2²² = 4 194 304 (0)
- 2²¹ = 2 097 152 (0)
- 2²⁰ = 1 048 576 → Remainder: 1 114 112 - 1 048 576 = 65 536
- 65 536 = 2¹⁶
Full Binary: 10 1001 0001 0000 0000 0000 0000
Step 2: Normalise for an 8-bit mantissa
A positive number must start with 01 :
0.1 0 1 0 0 1 0 0 0 1 ... × 2²⁶
Since we only have an 8-bit mantissa:
| Sign | Bit 1 | Bit 2 | Bit 3 | Bit 4 | Bit 5 | Bit 6 | Bit 7 |
|---|---|---|---|---|---|---|---|
| 0 | 1 | 0 | 1 | 0 | 0 | 1 | 0 |
The next bit was 0 ( ...001... ), so round down to: 01010010 .
This represents: 2²⁵ + 2²³ + 2²⁰ = 33 554 432 + 8 388 608 + 1 048 576 = 42 991 616.
Step 3: Convert the exponent (26) to 6-bit two's complement
26 in 6-bit binary: 16 + 8 + 2 = 0 1 1 0 1 0
✅ Final Answer
Mantissa (8 bits):
Exponent (6 bits):
🧠 Method Marks (Up to 2 if final answer incorrect)
- Correct unsigned binary for 43 057 152: 10 1001 0001 0000 0000 0000 0000
- Identifying need to represent 2²⁵, 2²³, and 2²⁰ (or calculating closest value: 42 991 616).
- Correct exponent of 26 (in binary 011010 or decimal 26).
- Correct mantissa 01010010 shown in working.
Question 12.6
Calculate Relative Error
📐 Relative Error Formula & Calculation
Formula:
Relative Error = |Actual Value - Stored Value| / Actual Value × 100%
- Actual Value = 43 057 152
- Stored Value = 42 991 616 (from 12.5)
- Absolute Error = 43 057 152 - 42 991 616 = 65 536 (which was 2¹⁶!)
Calculation:
Relative Error = (65 536 / 43 057 152) × 100%
Relative Error = 0.00152207... × 100% = 0.1522%
🧠 Exam Technique & Pitfalls
Answer: 0.1522% (at least 4 d.p.)
- Always divide by the ORIGINAL actual value (43 057 152), not the stored value.
- Sign: Error is typically an absolute magnitude. The mark scheme explicitly states: Reject -0.1522% . Never write a negative percentage!
- Precision: Question specifies "to at least four decimal places". Writing 0.15% loses the mark.
Topics
4.5 Fundamentals of data representation · 4.5.4 Binary number system
Question and mark scheme from the AQA A-Level Computer Science examination, Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.