Encoding Schemes and Number System Class 11: ASCII, Unicode and Base Conversions

What Encoding Schemes and Number System is about

Weightage: This chapter is part of Unit I, Computer Systems and Organisation, which carries 10 of the 70 theory marks in the school annual exam together with Computer System and Boolean Logic.Source: CBSE Computer Science (083) syllabus 2026-27, Class XI distribution of marks, Unit I Computer Systems and Organisation

Key concepts

Encoding and ASCII

NCERT section NCERT 2.1, 2.1.1

A computer understands only 0s and 1s. When a key is pressed it is mapped to a code value, which is stored in binary. Encoding is converting data into such codes using a fixed scheme. Because everyone uses the same standard scheme, the letter A gives the same code on every keyboard.

ASCII (American Standard Code for Information Interchange) was built to standardise character codes. It uses 7 bits, so it can encode 2^7 = 128 characters: English letters, digits, punctuation and control codes. Some values to remember: space = 32, '0' = 48, 'A' = 65, 'a' = 97; the codes of letters and digits run in order, and a small letter's code is the capital's code + 32. The weakness of ASCII is that it covers English only.

Worked example

Encode the word CODE in ASCII and in 7-bit binary.

Try it first, then show the solutionSolution

1. C = 67, O = 79, D = 68, E = 69 (A = 65, so count on).
2. 67 = 1000011, 79 = 1001111, 68 = 1000100, 69 = 1000101.

Answer: ASCII: 67 79 68 69. Binary: 1000011 1001111 1000100 1000101.

ISCII and Unicode

NCERT section NCERT 2.1.2, 2.1.3

ISCII (Indian Script Code for Information Interchange) was developed in India in the mid-1980s for Indian languages. It is an 8-bit code, so it has 2^8 = 256 codes: the first 128 are the ASCII codes, and the upper codes (160-255) are used for the aksharas of Indian scripts.

Many separate schemes could not understand each other's text. Unicode solves this by giving a unique number (code point) to every character of every written language, whatever the device, operating system or software. Its first 128 values match ASCII, so Unicode is a superset of ASCII. Code points are written in hexadecimal, e.g. 'अ' is U+0905 and 'A' is U+0041.

A code point is stored using an encoding:

• UTF-8: variable length, 1 to 4 bytes per character. ASCII characters take 1 byte, so English text stays small; Devanagari letters take 3 bytes. It is the usual encoding of web pages.
• UTF-32: fixed length, 4 bytes (32 bits) for every character; simple to index but takes the most space.

(UTF-16, using 2 or 4 bytes, also exists.)

Worked example

How many bytes does the word 'India' take in UTF-8 and in UTF-32? Why the difference?

Try it first, then show the solutionSolution

1. 'India' has 5 ASCII characters; each takes 1 byte in UTF-8: 5 bytes.
2. In UTF-32 each takes 4 bytes: 5 × 4 = 20 bytes.
3. UTF-8 is variable length and uses the fewest bytes for ASCII characters.

Answer: 5 bytes in UTF-8 and 20 bytes in UTF-32.

Positional number systems

NCERT section NCERT 2.2

A number system is a way of writing numbers. The count of distinct symbols it uses is its base or radix:

System	Base	Symbols
Binary	2	0, 1
Octal	8	0-7
Decimal	10	0-9
Hexadecimal	16	0-9, A(10), B(11), C(12), D(13), E(14), F(15)

All four are positional: a digit's value depends on its position. In the integer part, the rightmost digit is at position 0, and positions rise by 1 to the left; in the fraction part, the first digit after the point is at position -1, then -2, and so on. The positional value of position p in base b is b^p. The number's value = sum of (digit × positional value). Example: (123.45)₁₀ = 1×10² + 2×10¹ + 3×10⁰ + 4×10⁻¹ + 5×10⁻². The base is written as a subscript, e.g. (1011)₂, (17)₈, (2F)₁₆.

Worked example

In (5307)₈, find the positional value of the digit 3 and the value it contributes.

Try it first, then show the solutionSolution

1. Positions from the right: 7 at 0, 0 at 1, 3 at 2, 5 at 3.
2. Positional value of position 2 in base 8 = 8² = 64.
3. Contribution = 3 × 64 = 192.

Answer: Positional value 64; the digit 3 contributes 192.

Decimal to binary, octal or hexadecimal

NCERT section NCERT 2.3.1, 2.3.4 (A)

For the integer part, divide repeatedly by the new base and note each remainder until the quotient is 0. Read the remainders from bottom to top (the last remainder is the leftmost digit). In hexadecimal write remainders 10-15 as A-F.

For a fractional part, multiply repeatedly by the new base; each time, write down the integer part of the product and carry on with the fraction left. Read the integer parts from top to bottom. Stop when the fraction becomes 0, when it starts repeating, or after a fixed number of steps (the answer is then approximate).

Worked example

Convert (156)₁₀ to binary, octal and hexadecimal.

Try it first, then show the solutionSolution

1. Binary: 156/2 = 78 r0, 78/2 = 39 r0, 39/2 = 19 r1, 19/2 = 9 r1, 9/2 = 4 r1, 4/2 = 2 r0, 2/2 = 1 r0, 1/2 = 0 r1. Bottom to top: 10011100.
2. Octal: 156/8 = 19 r4, 19/8 = 2 r3, 2/8 = 0 r2: 234.
3. Hex: 156/16 = 9 r12 (C), 9/16 = 0 r9: 9C.

Answer: (10011100)₂ = (234)₈ = (9C)₁₆.

Binary, octal or hexadecimal to decimal

NCERT section NCERT 2.3.2, 2.3.4 (B)

Write the position number under each digit, multiply each digit by base^position, and add. In hexadecimal replace letters by their values (A = 10 ... F = 15). For digits after the point use negative powers: base^-1, base^-2, ...

Memorise the binary positional values 1, 2, 4, 8, 16, 32, 64, 128, 256 and the fractions 0.5, 0.25, 0.125: then a binary number can be converted by adding the values under its 1s.

Worked example

Convert (2AF)₁₆ and (101.101)₂ to decimal.

Try it first, then show the solutionSolution

1. 2AF: 2×16² + 10×16¹ + 15×16⁰ = 512 + 160 + 15 = 687.
2. 101.101: 1×4 + 0×2 + 1×1 + 1×0.5 + 0×0.25 + 1×0.125 = 5.625.

Answer: (2AF)₁₆ = (687)₁₀; (101.101)₂ = (5.625)₁₀.

Binary to octal and hexadecimal (and back)

NCERT section NCERT 2.3.3

Since 8 = 2³ and 16 = 2⁴, each octal digit stands for exactly 3 bits and each hexadecimal digit for exactly 4 bits.

• Binary to octal: group the bits in threes from the binary point outwards (right to left for the integer part, left to right for the fraction), padding with 0s at the ends; replace each group by its octal digit.
• Binary to hexadecimal: the same with groups of four.
• Octal or hex to binary: replace each digit by its 3-bit or 4-bit group.
• Octal to hexadecimal (or the reverse): go through binary.

Hex	Binary	Hex	Binary
0	0000	8	1000
1	0001	9	1001
2	0010	A	1010
3	0011	B	1011
4	0100	C	1100
5	0101	D	1101
6	0110	E	1110
7	0111	F	1111

Worked example

Convert (1011011101)₂ to octal and to hexadecimal.

Try it first, then show the solutionSolution

1. Octal groups from the right: 1 | 011 | 011 | 101 → 001 011 011 101 → 1 3 3 5.
2. Hex groups from the right: 10 | 1101 | 1101 → 0010 1101 1101 → 2 D D.

Answer: (1335)₈ and (2DD)₁₆ (both equal 733 in decimal).

Why hexadecimal is used

NCERT section NCERT 2.2.5

Long binary strings are hard to read and copy, so programmers use hexadecimal (or octal) as a short form: one hex digit replaces four bits.

• Memory addresses: a 16-bit address such as 1100 0000 1111 0001 is written C0F1.
• Colour codes: a screen colour is made of red, green and blue parts of 8 bits each (24 bits in all, about 16 million colours). Each part is written as two hex digits: red is (FF, 00, 00), white (FF, FF, FF), black (00, 00, 00), grey (80, 80, 80). Web pages write these as #FF0000 and so on.
• Unicode code points are also written in hex (U+0905).

Worked example

A web colour is #2D9CFF. Find the decimal values of its red, green and blue parts.

Try it first, then show the solutionSolution

1. Red 2D = 2×16 + 13 = 45.
2. Green 9C = 9×16 + 12 = 156.
3. Blue FF = 15×16 + 15 = 255.

Answer: (45, 156, 255): a bright sky blue.

Encoding Schemes and Number System in real life

Each concept, met in an everyday situation. Read the situation, answer the question in your own words, then compare with the model answer. The one-line hook is what to remember.

Topics in this chapter

• ASCII, ISCII and Unicode: Encoding Text with Calculations
• Number System Conversions: Decimal, Binary, Octal and Hexadecimal

Encoding Schemes and Number System: syntax and key facts

Value of a number in base b

N = Σ digit × b^position (positions ... 2, 1, 0 . -1, -2 ...)

Used for any base to decimal.

Decimal to base b

Integer: divide by b, remainders bottom to top; fraction: multiply by b, integer parts top to bottom

Stop the fraction when it becomes 0 or repeats.

Grouping

1 octal digit = 3 bits; 1 hex digit = 4 bits

Group outwards from the binary point; pad with zeros at the ends.

Code sizes

1. 1.ASCII 7 bits
2. 2.128 codes; ISCII 8 bits
3. 3.256 codes; UTF-8: 1-4 bytes; UTF-32: 4 bytes

n bits give 2^n different codes.

Key ASCII values

space 32, '0' 48, 'A' 65, 'a' 97

Lowercase = uppercase + 32.

Python checks

bin(n), oct(n), hex(n); int('101', 2), int('17', 8), int('1F', 16); ord('A'), chr(65)

Use them to check hand conversions.

How the school exam asks Encoding Schemes and Number System

The question patterns to expect, and the method that works for each.

Tricks to remember Encoding Schemes and Number System

Short memory aids. Say each one aloud once, then use it on a question the same day.

• Divide down, read up

  Decimal integer to another base: divide down the page, read the remainders up.

  Use it when: Integer conversions.

• Multiply down, read down

  Decimal fraction: multiply down the page and read the integer parts down.

  Use it when: Fraction conversions.

• Oct 3, Hex 4

  8 = 2³ and 16 = 2⁴, so octal digits take groups of 3 bits and hex digits groups of 4 bits.

  Use it when: Binary-octal-hex shortcuts.

• 65 and 97

  Capital A is 65; small a is 97 = 65 + 32; digit 0 is 48.

  Use it when: Encoding words in ASCII.

• UTF-8 is elastic, UTF-32 is fixed

  8 stretches from 1 to 4 bytes; 32 bits is always 4 bytes.

  Use it when: Comparing Unicode encodings.

Encoding Schemes and Number System: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

n = 156
print(bin(n), oct(n), hex(n))
print(int('2AF', 16), int('745', 8), int('101101', 2))
print(ord('A') + 32, chr(ord('a') - 32))

Case-study practice (Section E style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 4-mark question split into 1 + 1 + 1 + 1 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Encoding Schemes and Number System

The slips examiners see most often, and what to write instead.

• ✗Mistake: Reading remainders top to bottom

  The first remainder is the least significant (rightmost) digit.

  Wrong working

  13 to binary: remainders 1, 0, 1, 1 → 1011

  Right working

  Read bottom to top → 1101

  ✓Do this: Draw an arrow upwards beside the remainder column.

  Usually costs: 1 mark

• ✗Mistake: Grouping bits from the left

  Groups must start at the binary point; starting from the left gives a different number.

  Wrong working

  11010 to octal: 110 | 10 → 6 2

  Right working

  11 | 010 → 011 010 → 3 2, so (32)₈

  ✓Do this: Group integer bits from the right, fraction bits from the left.

  Usually costs: 1 mark

• ✗Mistake: Using 10 as a hex digit

  Each hex digit is one symbol; 10 to 15 are A to F.

  Wrong working

  (122)₁₀ = (7 10)₁₆

  Right working

  (122)₁₀ = (7A)₁₆

  ✓Do this: Change remainders 10-15 to letters before writing the answer.

  Usually costs: 1 mark

• ✗Mistake: Positive powers after the point

  Fraction digits are at positions -1, -2, ...

  Wrong working

  (0.11)₂ = 1×2¹ + 1×2² = 6

  Right working

  (0.11)₂ = 1×2⁻¹ + 1×2⁻² = 0.75

  ✓Do this: Write position numbers under every digit, negative after the point.

  Usually costs: 1 mark

• ✗Mistake: Saying UTF-8 always uses 8 bytes, or Unicode uses 16 bits only

  UTF-8 uses 8-bit units and 1 to 4 of them; UTF-32 uses a fixed 4 bytes.

  Wrong working

  UTF-8 stores every character in 8 bytes.

  Right working

  UTF-8 stores a character in 1 to 4 bytes; ASCII characters need only 1.

  ✓Do this: Link the number in the name to bits per unit, not to bytes per character.

  Usually costs: 1 mark

NCERT exercise guide: Encoding Schemes and Number System

What each exercise in the NCERT chapter tests, how to approach it, and a question of the same kind solved step by step. Questions here are written in our own words.

Practice questions on Encoding Schemes and Number System

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

1. Q1 · 1 mark · MCQExam level

   Which is the base of the hexadecimal system? (a) 2 (b) 8 (c) 10 (d) 16

   Hint

   Hexa = 6, deca = 10.

   Answer

   (d) 16

2. Q2 · 2 marks · Short answerExam level

   Convert (45)₁₀ to binary and octal.

   Hint

   45 = 32 + 8 + 4 + 1.

   Answer

   (101101)₂ and (55)₈.

3. Q3 · 2 marks · Short answerHarder exam level

   Convert (9C4)₁₆ to binary and octal.

   Hint

   Four bits per hex digit, then regroup in threes.

   Answer

   1001 1100 0100 → (100111000100)₂; groups of three 100 111 000 100 → (4704)₈.

4. Q4 · 2 marks · Short answerHarder exam level

   Convert (0.8125)₁₀ to binary and hexadecimal.

   Hint

   0.8125 × 16 = 13.

   Answer

   Binary: ×2 gives 1, 1, 0, 1 → (0.1101)₂. Hex: 0.8125 × 16 = 13.0 → (0.D)₁₆.

5. Q5 · 1 mark · Very short answerHarder exam level

   Find x if (1x01)₂ = (13)₁₀.

   Hint

   8 + 4x + 0 + 1 = 13.

   Answer

   x = 1.

6. Q6 · 2 marks · Short answerExam level

   Why does ISCII need 8 bits while ASCII needs only 7?

   Hint

   Number of characters.

   Answer

   7 bits give only 128 codes, all used by ASCII. ISCII keeps those 128 and needs extra codes for Indian aksharas, so it uses 8 bits, giving 256 codes.

7. Q7 · 3 marks · Short answerHarder exam level

   Predict the output:

   word = 'Ok'
   for ch in word:
       print(ch, ord(ch), bin(ord(ch))[2:])
   print(len('Ok'.encode('utf-8')), len('Ok'.encode('utf-32-be')))

   Hint

   O is 79, k is 107; [2:] drops '0b'.

   Answer

   O 79 1001111
   k 107 1101011
   2 8

8. Q8 · 2 marks · Short answerHarder exam level

   A 16-bit memory address is 1010 1111 0000 0011. Write it in hexadecimal and octal.

   Hint

   For octal, group in threes from the right: 1 010 111 100 000 011.

   Answer

   Hex AF03. Octal: 1 010 111 100 000 011 → (127403)₈.

9. Q9 · 1 mark · MCQHarder exam level

   Assertion (A): UTF-8 text in English takes less space than the same text in UTF-32. Reason (R): UTF-8 stores ASCII characters in one byte, while UTF-32 uses four bytes for every character. Choose (a) both true, R explains A (b) both true, R does not explain A (c) A true, R false (d) A false, R true.

   Hint

   Compare bytes per English letter.

   Answer

   (a)

10. Q10 · 3 marks · Short answerHOTS / competency

    The largest number that can be written with 3 hexadecimal digits is how many times the largest 3-digit octal number? Show working.

    Hint

    FFF and 777 in decimal.

    Answer

    (FFF)₁₆ = 4095; (777)₈ = 511; 4095 ÷ 511 ≈ 8.01, so about 8 times (exactly 4095/511).

11. Q11 · 2 marks · Short answerHOTS / competency

    A student says (0.1)₁₀ = (0.1)₂. Explain the error and give the first five binary places of 0.1.

    Hint

    (0.1)₂ = 0.5.

    Answer

    (0.1)₂ means 1 × 2⁻¹ = 0.5, not 0.1. 0.1 × 2: 0.2 (0), 0.4 (0), 0.8 (0), 1.6 (1), 1.2 (1) → (0.00011...)₂, a repeating fraction.

12. Q12 · 2 marks · Short answerExam level

    Convert (605.12)₈ to decimal.

    Hint

    6×64 + 0 + 5 + 1/8 + 2/64.

    Answer

    384 + 0 + 5 + 0.125 + 0.03125 = 389.15625.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. 1.How many characters can 7-bit ASCII represent?Show answer

   2^7 = 128.

2. 2.What is the ASCII code of 'a'?Show answer

   97.

3. 3.How many bits is ISCII and how many codes does it have?Show answer

   8 bits, 256 codes.

4. 4.How many bytes does UTF-32 use per character?Show answer

   4 bytes.

5. 5.What is the base of the octal system?Show answer

   8.

6. 6.Hexadecimal digit for 13?Show answer

   D.

7. 7.Positional value of the first digit after the binary point?Show answer

   2^-1 = 0.5.

8. 8.(F)₁₆ in binary?Show answer

   1111.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Encoding maps each character to a code, stored in binary.
• ☐ASCII: 7 bits, 128 characters, English only; 'A' 65, 'a' 97, '0' 48.
• ☐ISCII: 8 bits, 256 codes, ASCII plus Indian aksharas.
• ☐Unicode: one code point for every character; superset of ASCII; UTF-8 1-4 bytes, UTF-32 4 bytes.
• ☐Bases: binary 2, octal 8, decimal 10, hex 16 (A-F = 10-15).
• ☐To decimal: sum of digit × base^position.
• ☐From decimal: divide and read remainders upwards; fractions multiply and read downwards.
• ☐3 bits per octal digit, 4 bits per hex digit; group from the point.

Encoding Schemes and Number System: frequently asked questions