OCR A-Level Computer Science Computer systems (01), June 2025: Question 3

✅ Correct Answers

• (i): 1011 1101 [1 mark]
• (ii): BD [1 mark]
• (iii): 228 [1 mark]

📐 Step-by-Step Conversion

❌ Common Errors

• Miscounting hex values: confusing B (11) and D (13) with C (12) or E (14). Remember: A=10, B=11, C=12, D=13, E=14, F=15.
• Writing a 7-bit answer for (i) instead of an 8-bit binary byte.

🧠 Exam Technique Shortcut

Always convert denary to hex by going through binary first! Splitting an 8-bit binary number into two 4-bit nibbles is far less prone to arithmetic mistakes than performing division by 16.

✅ Correct Answer

A right shift of 3 places will:

• Divides the number [1 mark]
• ...by 8 (or 2³) [1 mark]

💡 Key Knowledge

Each single shift to the right divides an integer value by 2 (integer division / truncation if bits are discarded):

• 1 place right = ÷ 2 (2¹)
• 2 places right = ÷ 4 (2²)
• 3 places right = ÷ 8 (2³)
• n places right = ÷ 2ⁿ

❌ Common Errors & Examiner Insight

• Missing the power: Stating "divides by 3" or "divides by 6" instead of 2³ = 8. Shifts are exponential powers of 2, not linear multiplications!
• Vague answers: Saying "the number gets smaller" gains zero marks. You must state the exact mathematical operation ("divides") and the factor ("8" or "2³").

Step 1: Decode Exponents and Mantissas [MP1]
Both numbers have an assumed binary point between the 1st and 2nd bits of the mantissa:

• Number 1: Mantissa = 0.11010 , Exponent = 0011 = +3
• Number 2: Mantissa = 0.10010 , Exponent = 0010 = +2

Step 2: Equalise Exponents (Align Mantissas) [MP2 & MP3]
To perform subtraction directly, both numbers must share the same exponent. We scale Number 2 up to exponent 0011 (+3) by shifting its mantissa 1 place right:

• Number 1: 0.11010 × 2³
• Number 2: 0.01001 × 2³ (shifted right by 1 place; 0 shifted off the end)

Alternative method accepted by mark scheme: Shift binary points to absolute denary/fixed point ( 0110.10 and 0010.010 ).

Step 3: Binary Subtraction [MP4]
Subtract the aligned mantissas (or use Two's Complement addition):

	0 . 1 1 0 1 0	(Number 1)
−	0 . 0 1 0 0 1	(Number 2 aligned)
=	0 . 1 0 0 0 1	(Difference)

(Check via denary: Number 1 = 6.5, Number 2 = 2.25; 6.5 − 2.25 = 4.25. 0.10001₂ × 2³ = 4.25. Correct!)

Step 4: Check Normalisation & Final Format [MP5 & MP6]
A positive floating-point number is normalised if it begins with 01 .

• The resulting mantissa 010001 begins with 01 , so it is already normalised!
• No shifting needed, so the exponent stays 0011 (+3).

✅ Final Answer

Final 6-bit mantissa: 010001 [1 mark]

Final 4-bit exponent: 0011 [1 mark]

🧠 Mark Scheme & Method Warning

Strict Examiner Rule: Where denary working is clearly used to calculate the answer instead of showing binary arithmetic working, MP4 is NOT awarded.

Always show binary column subtraction or two's complement addition clearly in the working space!

📐 (i) Applying the AND Mask

An AND operation outputs 1 only when both bits are 1:

Byte:	1010	0011
AND Mask:	1010	1110
Result:	1010	0010

• Nibble 1: 1010 [1 mark]
• Nibble 2: 0010 [1 mark]

✅ (ii) Purpose of Bitwise AND Mask

Purpose:

• To extract, enable, disable, or clear specific bits in a register/value. [1 mark]

💡 How Bitwise Masks Work in Practice

• AND with 0: Clears / disables the bit (forces it to 0).
• AND with 1: Preserves / extracts the bit (leaves it unchanged: 0 AND 1 = 0; 1 AND 1 = 1).
• Common use cases: Checking if an interrupt flag is set, reading specific sensor pins, or extracting individual RGB colour channels.

❌ Common Misconceptions

• Confusing AND with OR or XOR: OR masks are used to set bits to 1; XOR masks are used to invert/flip bits.
• Mixing up bit positions when copying down nibbles. Always line the bits up vertically before applying the logic gate rules.