- Review what the assessment shows about secure knowledge and next steps.
- Use the supplied mantissa and exponent convention accurately.
- Calculate bitmap file sizes using a clear, repeatable method.
- I can identify a mantissa and exponent and explain why floating point is useful.
- I can show all stages of a bitmap file-size calculation in bits and bytes.
- I can reflect honestly on my confidence and choose a specific next step.
Answer before the lesson begins. These check prior knowledge — it's fine if you're unsure.
1. What is the denary value of 10100110?
2. What is the largest unsigned denary value stored in 8 bits?
3. Which unit conversion is correct?
Key vocabulary
0..What the assessment tells us
Begin with evidence, not a label
An assessment is useful when it shows what to practise next. Across the recent paper, unsigned 8-bit binary was a clear strength: conversions and the maximum value were usually secure. Floating point was the largest shared gap, while bitmap calculations were stronger when every stage of working was shown. Today is therefore not a repeat of the whole test. It is a focused repair session. Work privately, attempt each item before revealing help, and treat an error as information about your next step rather than as a judgement.
Reading floating-point notation
The assessment used a denary convention such as 0.30605 × 10⁴. The mantissa is 30605: the significant digits written after 0.. The exponent is 4. A positive exponent moves the decimal point right, so 0.30605 × 10⁴ = 3060.5. A negative exponent moves it left: 0.72 × 10⁻² = 0.0072. Count the moves carefully and write the intermediate decimal point if that helps.
Why floating point is used
Real numbers may be extremely large or extremely small. Floating point gives a wide range of values by storing a mantissa and exponent using a fixed, limited number of bits. The exponent changes the scale without requiring the computer to store every zero or every individual digit. A complete two-mark explanation needs both ideas: wide range and fixed number of bits. Floating point can introduce rounding because only a limited number of significant digits fit, but that is not the reason asked for in this assessment.
The dependable bitmap method
For an uncompressed bitmap, use width × height × colour depth. First calculate the number of pixels. Then multiply by the bits used for each pixel. The result is in bits. If the question asks for bytes, divide by 8 because one byte contains eight bits. Always label units and show each line. Even if an arithmetic slip occurs, correct working can demonstrate the method.
Your independent routine
Use the same four moves every time: (1) underline the required unit; (2) write the formula; (3) substitute the values; (4) convert units only at the end. After checking a model answer, add or correct your own work in a different colour if possible. Do not simply copy the model: identify the first place where your method changed. This turns checking into learning and gives useful evidence for the next lesson. Accuracy grows through deliberate checking.
Worked examples
Consider 0.30605 × 10⁴.
30605, the digits following 0..4, so move the decimal point four places right.0.30605 → 3.0605 → 30.605 → 306.05 → 3060.5.Explain why floating point is used to store real numbers. [2]
A bitmap is 100 pixels wide, 50 pixels high and uses 24-bit colour. Find its size in bytes.
100 × 50 = 5,000 pixels.5,000 × 24 = 120,000 bits.120,000 ÷ 8 = 15,000 bytes.A weather station stores 0.452 × 10³. It also saves an uncompressed 80 × 60 pixel icon using 8-bit colour. Work on paper before revealing the answer.
Answer the following:
- State the mantissa and exponent.
- Write the denary value represented.
- Calculate the icon size in bytes, showing every stage.
- Mantissa =
452; exponent =3. 0.452 × 10³ = 452.80 × 60 = 4,800 pixels;4,800 × 8 = 38,400 bits;38,400 ÷ 8 = 4,800 bytes.
0. in the mantissa. In the convention used here, give the stored significant digits: 30605, not 0.30605.Write the bitmap method vertically: pixels → bits → bytes. The marking guidance awards method credit for correct stages, so never hide all your work in one calculator line.
Questions 1–5 are auto-checked. Questions 6–9 are self-marked — write your answer, then reveal the model answer to check your work.
1. In 0.7182 × 10³, what is the mantissa? TYPE 1
2. What value does 0.7182 × 10³ represent? TYPE 1
3. What value does 0.64 × 10⁻² represent? TYPE 1
4. A 40 × 30 bitmap contains how many pixels? TYPE 1
5. A bitmap contains 8,000 bits. How many bytes is this? TYPE 1
6. Explain two benefits of floating-point representation. TYPE 2
7. Calculate the size in bytes of a 64 × 32 pixel bitmap using 16-bit colour. Show pixels → bits → bytes. TYPE 2
64 × 32 = 2,048 pixels; 2,048 × 16 = 32,768 bits; 32,768 ÷ 8 = 4,096 bytes.8. A 120 × 90 image uses 24-bit colour. Calculate its size in bytes independently. TYPE 3
120 × 90 = 10,800 pixels; 10,800 × 24 = 259,200 bits; 259,200 ÷ 8 = 32,400 bytes.9. Exit reflection: write one calculation step now secure, one that still needs care, and one action you will take next time. TYPE 3
Suggested timing: Warm up 6 min; notes and worked examples 14 min; now-you-try 8 min; Task Set Q1–5 8 min; Q6–8 12 min; reflection and transition 5 min. Then allow a 2-minute stretch/reset before Part 2.
Cover setup: Pupils work independently on paper or in the page. Ask them to attempt before revealing. No whole-class marking, displayed scores or test ranking is needed.
Key misconception: In this supplied convention, the mantissa answer is the significant digits after 0.; bitmap size is initially in bits and must be divided by 8 for bytes.
If pupils finish early: Create a new bitmap question whose final answer is a whole number of bytes, then supply fully worked steps for a partner to check.
SQA command words covered: state, calculate, explain.