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S3 ICT | Data Representation: Binary, Hexadecimal and Encoding | Tutor Preparation Guide
Students can say a computer uses ones and zeros, but ask what 10110110 is worth and they read it aloud digit by digit. This lesson turns reading into place values: from the right, each column doubles.
Objectives — Students can: (1) convert the eight-bit 10110110 to decimal 182 as 128 + 32 + 16 + 4 + 2, and reverse it; (2) split it into 1011 and 0110 to get hex B6, explaining that one hex digit holds four bits; (3) use bits to count values, so eight bits give 2^8 = 256 combinations, seven-bit ASCII gives 128 characters and 24-bit colour gives 2^24 = 16,777,216 colours.
01 Prerequisites (3-minute check)
If any of these fails, fix it before the lesson starts.
| Skill | Quick check | Pass mark |
|---|---|---|
| Powers | What is 2 to the power of 10? | 1,024, which is where one kibibyte comes from |
| Place value | What power of ten is the thousands column? | 10 to the power of 3, three zeros |
| Hex digits | What follows 9 in hexadecimal? | A to F, standing for 10 to 15 |
02 Lesson flow (45 minutes)
| Time | Stage | Tutor does | Student does |
|---|---|---|---|
| 0-6 min | Place values first | Draw eight boxes and write 1, 2, 4, 8, 16, 32, 64, 128 from the right | State that each box doubles the last |
| 6-16 min | Binary to decimal | Line 10110110 against the boxes to get 182, then 11111111 to get 255 | Both answers computed unprompted |
| 16-24 min | Decimal to binary | Subtract place values to rebuild 182 from 128, 32, 16, 4 and 2 | Five ones placed correctly |
| 24-32 min | Four bits per hex digit | 1011 gives B and 0110 gives 6, so B6; then 11111111 gives FF | Any eight bits become two hex digits |
| 32-39 min | Bits set the count | Use 2^8, 2^7 and 2^24 for ASCII and 24-bit colour | Each extra bit doubles the values |
| 39-45 min | Exit check | Collect three answers and mark on the spot | Hand in three answers |
03 Teaching script (read it out)
Scene 1: Act 1: number from the right, doubling each box
“Do not read 1, 0, 1, 1 from the left. Stick a price tag on the rightmost box first: it is worth 1, and each box to the left doubles, so 1, 2, 4, 8, 16, 32, 64, 128 for eight boxes. In 10110110 the ones stand at 128, 32, 16, 4 and 2, which adds to 182. The rightmost bit is the least significant, worth 1; the leftmost is worth 128. With eight ones, 128 plus 64 plus 32 plus 16 plus 8 plus 4 plus 2 plus 1 gives 255, so eight bits hold 0 to 255, which is 256 values.”
Tag from the right and double; eight bits top out at 255.
Scene 2: Act 2: four bits per hex digit, cut from the right
“Hexadecimal exists so nobody has to read a long string of ones and zeros. Cut every four bits from the right: 1011 and 0110. The left group is 8 plus 2 plus 1, which is 11, written B. The right group is 4 plus 2, which is 6. So 10110110 is B6, written 0xB6 to mark the base. Why four? Because 2 to the power of 4 is 16, so one hex digit holds exactly four bits. Eight bits therefore give two hex digits, twelve bits give three, never more and never fewer. 11111111 is FF, which is 255, a full byte.”
Cut from the right; B and 6 make B6.
Scene 3: Act 3: bits set the number of values, not the number of boxes
“A question asking how many values eight bits hold wants 2 to the power of 8, which is 256, covering 0 to 255. Seven-bit ASCII gives 2 to the power of 7, which is 128 characters. Twenty-four-bit colour gives 2 to the power of 24, which is 16,777,216, over sixteen million. The pattern is that one more bit doubles the count, which is why memory costs double per extra bit of width. Keep bits and bytes apart too: one byte is eight bits, and one kibibyte is 1,024 bytes, not 1,000.”
One more bit doubles the count; a byte is eight bits.
04 Question types and worked steps
| Type | Example | Steps | Answer |
|---|---|---|---|
| Binary to decimal | Convert 10110110 to decimal. | (1) Write place values 1 to 128 from the right (2) circle the ones (3) add (4) check it is under 255 | 128 + 32 + 16 + 4 + 2 = 182 |
| Decimal to binary | Convert 182 to eight-bit binary. | (1) Take the largest place value not above 182 (2) remainder 54 (3) take 32, 16, 4, 2 (4) pad with zeros to eight places | 128 + 32 + 16 + 4 + 2 gives 10110110 |
| To hexadecimal | Write 10110110 in hexadecimal and explain why it takes two digits. | (1) Cut four bits from the right (2) 1011 is 11, written B (3) 0110 is 6 (4) prefix 0x | 0xB6; eight divided by four is two digits |
| Bits and counts | How many colours does 24-bit colour give? And eight-bit? | (1) Write two to the power of the bit count (2) compute (3) compare (4) note the doubling rule | 2^24 = 16,777,216 and 2^8 = 256, which is 65,536 times smaller |
Sanity check (teach this): Sanity check: any eight-bit conversion must land between 0 and 255, so 300 means a wrong addition; a hex answer must be exactly two characters, and three or one means the cut was misplaced; converting the answer back to binary and adding again must return the original; 2^24 is over sixteen million, not one point six million, so count the digits before the decimal point; bits and bytes always differ by eight, which is where capacity questions usually fail.
05 Independent practice (3 questions)
| Question | Answer | What to watch |
|---|---|---|
| Convert 11001010 to decimal and to hexadecimal. | 202 from 128+64+8+2; hex CA because 1100 is 12 (C) and 1010 is 10 (A) | Check both against 202 |
| How many values do twelve bits hold, and how many hex digits write them? | 2^12 = 4,096 values; twelve divided by four is three hex digits | Divide by four, not eight |
| Roughly how many bytes is an uncompressed 1,024 by 768 image at 24 bits? | 1,024 times 768 is 786,432 pixels, times three bytes is 2,359,296 bytes, about 2.25 MiB | 24 bits is three bytes |
06 Common mistakes and what to say
| Mistake | Student writes | Say this |
|---|---|---|
| Reading binary from the left | “10110110 means ten million one hundred and one thousand.” | “Binary is read by place, not as a number. Tag 1, 2, 4, 8, 16, 32, 64, 128 from the right and add: 182.” |
| Cutting hex groups from the left | “Just cut four bits from the left and it works.” | “Always cut from the right, padding the left with zeros: 101101 becomes 0010 1101, which is 2D.” |
| Confusing bits with values | “Eight bits means eight values.” | “Eight boxes with two options each give 2 to the power of 8, so 256 values from 0 to 255.” |
| Mixing bytes and bits | “One kilobyte is 1,000 bits.” | “One kilobyte is 1,024 bytes, and each byte is eight bits, so 8,192 bits.” |
07 Homework (paper, section and question numbers)
- Chapter practice: convert five decimal numbers to eight-bit binary and then to hex — Data representation chapter, practice question 1
- Chapter practice: compute representable counts for ASCII, Unicode and 24-bit colour — Data representation chapter, practice question 2
- Extension: size an uncompressed 1,920 by 1,080 24-bit image, then explain why JPEG is far smaller — Data representation chapter, practice question 3
08 Exit check (last 5 minutes)
| Question | Answer | Decision |
|---|---|---|
| What is 10110110 in decimal? | 182 | Wrong means reteach Act 1 |
| Why do hex digits hold four bits? | Because 2^4 is 16, so one digit covers four bits | Wrong means reteach Act 2 |
| How many colours in 24-bit colour? | 16,777,216 | Answering 256 means Act 3 again |
09 Tutor FAQ
Is octal needed?
The S3 syllabus does not require it; drilling binary and hex is enough, and octal only adds confusion.
Must the ASCII table be memorised?
Not in full, but know A is 65, a is 97, 0 is 48, and that upper and lower case differ by 32.
Should sampling be taught with colour depth?
Keep the image size fixed here and teach depth only; sampling and compression belong to the multimedia chapter.
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A TutorZone lesson-prep guide. Written by the TutorZone team; questions and lesson flow are original and are not taken from any school or publisher material.