Stage 1 of 8 · Start here
Chapter 1 review: everything so far
I’m not in today, so this website is the first half of your lesson. It goes back over all five Chapter 1 lessons, one at a time. Each stage reminds you of the method, then gives you a few quick practice questions. Spend about 18 minutes here.
How this lesson works
- Type your name in the box below.
- Work through the stages in order. Read the Remind me box first, then practise.
- Each practice box has a few dots. Fill the dots, then press Next. Do not get stuck on one stage — the clock is running.
- Stuck? Press Show hint. It gives you the method, not the answer.
- At the end you get a completion code. Copy it into the top of your worksheet.
- Then complete the Chapter 1 Review Worksheet and hand it in.
Timings — keep an eye on the clock
| stage | topic | about |
|---|---|---|
| 2 | Lesson 1 — Binary and denary | 3 min |
| 3 | Lesson 2 — Hexadecimal | 4 min |
| 4 | Lesson 3 — Binary addition and overflow | 3 min |
| 5 | Lesson 4 — Logical shifts | 3 min |
| 6 | Lesson 5 — Two’s complement | 3 min |
| 7–8 | Final check and your code | 2 min |
| Website total | 18 min | |
| Then the worksheet (real exam questions) | 22 min |
Type your full name. It is used to make your completion code.
Stage 2 of 8 · Lesson 1 · about 3 minutes
Binary and denary
A computer is made of switches. Each switch is off (0) or on (1). One switch is a bit. Eight bits make a byte.
Remind me
The place values. Always write them above the bits, biggest on the left:
Binary → denary. Add the place values that have a 1 under them.
01001101 = 64 + 8 + 4 + 1 = 77
Denary → binary. Start at 128. Does it fit? Write 1 and subtract. If not, write 0. Move right.
77: 128 no · 64 yes (13 left) · 32 no · 16 no · 8 yes (5 left) · 4 yes (1 left) · 2 no · 1 yes → 01001101
Largest 8-bit number: 255 (all 1s). Number of different values: 256, because 0 counts too.
Optional: try it — click the bits
Click any bit to switch it on or off. Can you make 100? Can you make 255?
Stage 3 of 8 · Lesson 2 · about 4 minutes
Hexadecimal
Hexadecimal is base 16. It needs sixteen digits, so after 9 it uses letters.
Remind me
| hex | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | A | B | C | D | E | F |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| denary | 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 |
1 hex digit = 4 bits (a nibble). So one byte is exactly two hex digits.
Binary → hex. Split into nibbles from the right. Each nibble uses 8 4 2 1.
1011 0110 → 11 and 6 → B6
Hex → denary. Two columns: 16s and 1s.
B6 = (11 × 16) + 6 = 176 + 6 = 182
Denary → hex. Divide by 16. The answer is the first digit, the remainder is the second.
182 ÷ 16 = 11 remainder 6 → B6. (Or go through binary: 182 → 1011 0110 → B6.)
Why use hex? It is shorter than binary, so people make fewer mistakes reading and writing it, and it converts easily to and from binary. The computer still stores binary.
Where is it used? HTML colour codes (#FF8800), MAC addresses, IPv6 addresses, memory dumps, error codes, assembly code.
Optional: try it — see all three at once
Click the bits. Watch the denary and hexadecimal change. Notice each nibble makes one hex digit.
Stage 4 of 8 · Lesson 3 · about 3 minutes
Binary addition and overflow
Add from the right, one column at a time, just like denary. Carry into the next column on the left.
Remind me
Worked example: 45 + 23 = 68. The carries are shown in amber.
Always show your carries — they earn a mark on their own.
The result is too big for the register to hold (bigger than 255 in 8 bits),
so the carry out of the last column is lost.
Stage 5 of 8 · Lesson 4 · about 3 minutes
Logical shifts
A logical shift moves every bit in the register left or right.
Remind me
| places | left shift | right shift |
|---|---|---|
| 1 | × 2 | ÷ 2 |
| 2 | × 4 | ÷ 4 |
| 3 | × 8 | ÷ 8 |
When a 1 falls off, the rule breaks. Left shift: the most significant bits are lost, so the answer is wrong. Right shift: the least significant bits are lost, so the answer is rounded down.
Optional: try it — shift the register
Click bits to set a number, then shift. Watch what falls off.
Stage 6 of 8 · Lesson 5 · about 3 minutes
Two’s complement
Two’s complement stores negative numbers. Only one thing changes: the front column is worth −128.
Remind me
Reading one. Front bit 1 means negative. Add the columns, with the front one worth −128.
10110010 = −128 + 32 + 16 + 2 = −78
Writing a negative number. Three steps:
Positive numbers are written as ordinary binary, with a 0 at the front. No flip, no +1.
Range of 8-bit two’s complement: −128 to +127.
Optional: try it — the same bits, two readings
Click the bits. Try 11111111 — ordinary binary says 255, two’s complement says −1.
Stage 7 of 8 · Final check · about 2 minutes
Final check: all five lessons
Seven quick questions. Two are real Cambridge exam questions. The wrong answers are the mistakes students really make. You get one try at each, so think first.
Stage 8 of 8 · Finish
Your completion code
This shows how much of the website you completed. Go back and fill any missing dots to improve it.
Your progress
Answer the questions to generate your code.
Copy this code exactly into the box at the top of your worksheet.
What to do now
- Open the Chapter 1 Review Worksheet.
- Type your name, class, date and completion code at the top.
- Keep this tab open. You may look back at the Remind me boxes, but no calculator. You have about 22 minutes.
- Grey cells are the question. Type in the white cells. Press Tab to move between cells.
- Save and hand in the worksheet before the end of the lesson.
Teacher only
Check completion codes
Paste one pupil per line as Name, CODE. The name must be typed the way the pupil typed it (capitals and spaces do not matter).
Code = NNN-SS-AA-K. NNN matches the name. SS = correct answers and AA = attempts, both in hexadecimal. K = sections completed out of 6 (each practice target reached, plus all seven final-check questions answered).