Encoding Schemes and Number System

The base (radix) of a number system equals the total number of unique digits (symbols) it uses.

| Number System | Symbols Used | Base Value |
|---|---|---|
| Binary | 0, 1 | 2 |
| Octal | 0, 1, 2, 3, 4, 5, 6, 7 | 8 |
| Hexadecimal | 0–9, A, B, C, D, E, F | 16 |

- Binary base = 2
- Octal base = 8
- Hexadecimal base = 16

ASCII — American Standard Code for Information Interchange

ISCII — Indian Script Code for Information Interchange

ASCII is a 7-bit (or 8-bit extended) encoding scheme used to represent English characters and control characters. ISCII is an 8-bit encoding scheme developed to represent characters of Indian scripts.

(i) $(514)_8 = (?)_{10}$

Formula: Multiply each digit by its positional power of 8.
$$5 \times 8^2 + 1 \times 8^1 + 4 \times 8^0$$
$$= 5 \times 64 + 1 \times 8 + 4 \times 1$$
$$= 320 + 8 + 4 = \boxed{332}$$
$$\therefore (514)_8 = (332)_{10}$$

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(ii) $(220)_8 = (?)_2$

Method: Replace each octal digit with its 3-bit binary equivalent.
$$2 \rightarrow 010, \quad 2 \rightarrow 010, \quad 0 \rightarrow 000$$
$$\therefore (220)_8 = (010\,010\,000)_2 = \boxed{(010010000)_2}$$

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(iii) $(76F)_{16} = (?)_{10}$

Note: $F = 15$
$$7 \times 16^2 + 6 \times 16^1 + 15 \times 16^0$$
$$= 7 \times 256 + 6 \times 16 + 15 \times 1$$
$$= 1792 + 96 + 15 = \boxed{1903}$$
$$\therefore (76F)_{16} = (1903)_{10}$$

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(iv) $(4D9)_{16} = (?)_{10}$

Note: $D = 13$
$$4 \times 16^2 + 13 \times 16^1 + 9 \times 16^0$$
$$= 4 \times 256 + 13 \times 16 + 9 \times 1$$
$$= 1024 + 208 + 9 = \boxed{1241}$$
$$\therefore (4D9)_{16} = (1241)_{10}$$

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(v) $(11001010)_2 = (?)_{10}$

Assign positional values (right to left, starting from $2^0$):
$$1\times2^7 + 1\times2^6 + 0\times2^5 + 0\times2^4 + 1\times2^3 + 0\times2^2 + 1\times2^1 + 0\times2^0$$
$$= 128 + 64 + 0 + 0 + 8 + 0 + 2 + 0 = \boxed{202}$$
$$\therefore (11001010)_2 = (202)_{10}$$

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(vi) $(1010111)_2 = (?)_{10}$

$$1\times2^6 + 0\times2^5 + 1\times2^4 + 0\times2^3 + 1\times2^2 + 1\times2^1 + 1\times2^0$$
$$= 64 + 0 + 16 + 0 + 4 + 2 + 1 = \boxed{87}$$
$$\therefore (1010111)_2 = (87)_{10}$$

(i) $(54)_{10} = (?)_2$

Method: Repeated division by 2, read remainders bottom to top.
$$54 \div 2 = 27 \text{ R } 0$$
$$27 \div 2 = 13 \text{ R } 1$$
$$13 \div 2 = 6 \text{ R } 1$$
$$6 \div 2 = 3 \text{ R } 0$$
$$3 \div 2 = 1 \text{ R } 1$$
$$1 \div 2 = 0 \text{ R } 1$$
Reading remainders from bottom to top: $\boxed{(110110)_2}$

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(ii) $(120)_{10} = (?)_2$

$$120 \div 2 = 60 \text{ R } 0$$
$$60 \div 2 = 30 \text{ R } 0$$
$$30 \div 2 = 15 \text{ R } 0$$
$$15 \div 2 = 7 \text{ R } 1$$
$$7 \div 2 = 3 \text{ R } 1$$
$$3 \div 2 = 1 \text{ R } 1$$
$$1 \div 2 = 0 \text{ R } 1$$
Reading bottom to top: $\boxed{(1111000)_2}$

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(iii) $(76)_{10} = (?)_8$

Method: Repeated division by 8.
$$76 \div 8 = 9 \text{ R } 4$$
$$9 \div 8 = 1 \text{ R } 1$$
$$1 \div 8 = 0 \text{ R } 1$$
Reading bottom to top: $\boxed{(114)_8}$

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(iv) $(889)_{10} = (?)_8$

$$889 \div 8 = 111 \text{ R } 1$$
$$111 \div 8 = 13 \text{ R } 7$$
$$13 \div 8 = 1 \text{ R } 5$$
$$1 \div 8 = 0 \text{ R } 1$$
Reading bottom to top: $\boxed{(1571)_8}$

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(v) $(789)_{10} = (?)_{16}$

Method: Repeated division by 16.
$$789 \div 16 = 49 \text{ R } 5$$
$$49 \div 16 = 3 \text{ R } 1$$
$$3 \div 16 = 0 \text{ R } 3$$
Reading bottom to top: $\boxed{(315)_{16}}$

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(vi) $(108)_{10} = (?)_{16}$

$$108 \div 16 = 6 \text{ R } 12 \quad (12 = C)$$
$$6 \div 16 = 0 \text{ R } 6$$
Reading bottom to top: $\boxed{(6C)_{16}}$

(i) $(145)_8 = (?)_{10}$

$$1 \times 8^2 + 4 \times 8^1 + 5 \times 8^0$$
$$= 64 + 32 + 5 = \boxed{101}$$

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(ii) $(6760)_8 = (?)_{10}$

$$6 \times 8^3 + 7 \times 8^2 + 6 \times 8^1 + 0 \times 8^0$$
$$= 6 \times 512 + 7 \times 64 + 6 \times 8 + 0$$
$$= 3072 + 448 + 48 + 0 = \boxed{3568}$$

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(iii) $(455)_8 = (?)_{10}$

$$4 \times 8^2 + 5 \times 8^1 + 5 \times 8^0$$
$$= 256 + 40 + 5 = \boxed{301}$$

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(iv) $(10.75)_8 = (?)_{10}$

Integer part: $1 \times 8^1 + 0 \times 8^0 = 8$

Fractional part: $7 \times 8^{-1} + 5 \times 8^{-2} = \dfrac{7}{8} + \dfrac{5}{64} = 0.875 + 0.078125 = 0.953125$

$$\therefore (10.75)_8 = \boxed{(8.953125)_{10}}$$

(i) $(548)_{10} = (?)_{16}$

$$548 \div 16 = 34 \text{ R } 4$$
$$34 \div 16 = 2 \text{ R } 2$$
$$2 \div 16 = 0 \text{ R } 2$$
Reading bottom to top: $\boxed{(224)_{16}}$

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(ii) $(4052)_{10} = (?)_{16}$

$$4052 \div 16 = 253 \text{ R } 4$$
$$253 \div 16 = 15 \text{ R } 13 \quad (13 = D)$$
$$15 \div 16 = 0 \text{ R } 15 \quad (15 = F)$$
Reading bottom to top: $\boxed{(FD4)_{16}}$

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(iii) $(58)_{10} = (?)_{16}$

$$58 \div 16 = 3 \text{ R } 10 \quad (10 = A)$$
$$3 \div 16 = 0 \text{ R } 3$$
Reading bottom to top: $\boxed{(3A)_{16}}$

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(iv) $(100.25)_{10} = (?)_{16}$

Integer part (100):
$$100 \div 16 = 6 \text{ R } 4$$
$$6 \div 16 = 0 \text{ R } 6$$
Integer part in hex = $64$

Fractional part (0.25):
$$0.25 \times 16 = 4.00 \rightarrow \text{digit} = 4$$
Fractional part in hex = $.4$

$$\therefore (100.25)_{10} = \boxed{(64.4)_{16}}$$

(i) $(4A2)_{16} = (?)_{10}$

$A = 10$
$$4 \times 16^2 + 10 \times 16^1 + 2 \times 16^0$$
$$= 1024 + 160 + 2 = \boxed{1186}$$

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(ii) $(9E1A)_{16} = (?)_{10}$

$E = 14,\ A = 10$
$$9 \times 16^3 + 14 \times 16^2 + 1 \times 16^1 + 10 \times 16^0$$
$$= 9 \times 4096 + 14 \times 256 + 1 \times 16 + 10$$
$$= 36864 + 3584 + 16 + 10 = \boxed{40474}$$

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(iii) $(6BD)_{16} = (?)_{10}$

$B = 11,\ D = 13$
$$6 \times 16^2 + 11 \times 16^1 + 13 \times 16^0$$
$$= 1536 + 176 + 13 = \boxed{1725}$$

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(iv) $(6C.34)_{16} = (?)_{10}$

$C = 12,\ 3 = 3,\ 4 = 4$

Integer part:
$$6 \times 16^1 + 12 \times 16^0 = 96 + 12 = 108$$

Fractional part:
$$3 \times 16^{-1} + 4 \times 16^{-2} = \frac{3}{16} + \frac{4}{256} = 0.1875 + 0.015625 = 0.203125$$

$$\therefore (6C.34)_{16} = \boxed{(108.203125)_{10}}$$

Method for Octal: Group bits in sets of 3 from right (for integer part) and from left (for fractional part). Replace each group with its octal digit.

Method for Hexadecimal: Group bits in sets of 4 from right (for integer part) and from left (for fractional part). Replace each group with its hex digit.

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(i) $(1110001000)_2$

To Octal (groups of 3):
$$001\; 110\; 001\; 000$$
$$= 1\quad 6\quad 1\quad 0$$
$$\therefore \boxed{(1610)_8}$$

To Hexadecimal (groups of 4):
$$0011\; 1000\; 1000$$
$$= 3\quad 8\quad 8$$
$$\therefore \boxed{(388)_{16}}$$

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(ii) $(110110101)_2$

To Octal (groups of 3):
$$110\; 110\; 101$$
$$= 6\quad 6\quad 5$$
$$\therefore \boxed{(665)_8}$$

To Hexadecimal (groups of 4):
$$0001\; 1011\; 0101$$
$$= 1\quad B\quad 5$$
$$\therefore \boxed{(1B5)_{16}}$$

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(iii) $(1010100)_2$

To Octal (groups of 3):
$$001\; 010\; 100$$
$$= 1\quad 2\quad 4$$
$$\therefore \boxed{(124)_8}$$

To Hexadecimal (groups of 4):
$$0101\; 0100$$
$$= 5\quad 4$$
$$\therefore \boxed{(54)_{16}}$$

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(iv) $(1010.1001)_2$

To Octal:
Integer: $001\; 010 \rightarrow 1\; 2$
Fractional: $100\; 1\mathbf{00} \rightarrow 100\; 100 \rightarrow 4\; 4$ (pad right with zeros)
$$\therefore \boxed{(12.44)_8}$$

To Hexadecimal:
Integer: $1010 \rightarrow A$
Fractional: $1001 \rightarrow 9$
$$\therefore \boxed{(A.9)_{16}}$$

Method: Replace each octal digit with its 3-bit binary equivalent.

Octal–Binary table:
| Octal | Binary |
|---|---|
| 0 | 000 |
| 1 | 001 |
| 2 | 010 |
| 3 | 011 |
| 4 | 100 |
| 5 | 101 |
| 6 | 110 |
| 7 | 111 |

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(i) $(2306)_8$
$$2 \to 010,\quad 3 \to 011,\quad 0 \to 000,\quad 6 \to 110$$
$$\therefore \boxed{(010011000110)_2}$$

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(ii) $(5610)_8$
$$5 \to 101,\quad 6 \to 110,\quad 1 \to 001,\quad 0 \to 000$$
$$\therefore \boxed{(101110001000)_2}$$

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(iii) $(742)_8$
$$7 \to 111,\quad 4 \to 100,\quad 2 \to 010$$
$$\therefore \boxed{(111100010)_2}$$

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(iv) $(65.203)_8$
$$6 \to 110,\quad 5 \to 101 \;.\; 2 \to 010,\quad 0 \to 000,\quad 3 \to 011$$
$$\therefore \boxed{(110101.010000011)_2}$$

Method: Replace each hexadecimal digit with its 4-bit binary equivalent.

Key hex–binary values: $A=1010,\ B=1011,\ C=1100,\ D=1101,\ E=1110,\ F=1111$

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(i) $(4026)_{16}$
$$4 \to 0100,\quad 0 \to 0000,\quad 2 \to 0010,\quad 6 \to 0110$$
$$\therefore \boxed{(0100000000100110)_2}$$

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(ii) $(BCA1)_{16}$
$$B \to 1011,\quad C \to 1100,\quad A \to 1010,\quad 1 \to 0001$$
$$\therefore \boxed{(1011110010100001)_2}$$

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(iii) $(98E)_{16}$
$$9 \to 1001,\quad 8 \to 1000,\quad E \to 1110$$
$$\therefore \boxed{(100110001110)_2}$$

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(iv) $(132.45)_{16}$
$$1 \to 0001,\quad 3 \to 0011,\quad 2 \to 0010 \;.\; 4 \to 0100,\quad 5 \to 0101$$
$$\therefore \boxed{(000100110010.01000101)_2}$$

The computer converts each character to its 7-bit ASCII code. The decimal ASCII values and their 7-bit binary equivalents are given below.

Standard ASCII decimal values (selected):
- A=65, B=66, C=67, D=68, E=69, F=70, G=71, H=72, I=73, J=74, K=75, L=76, M=77, N=78, O=79, P=80, Q=81, R=82, S=83, T=84, U=85, V=86, W=87, X=88, Y=89, Z=90
- a=97, e=101, i=105, m=109, n=110, s=115

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(i) HOTS

| Character | ASCII (Decimal) | 7-bit Binary |
|---|---|---|
| H | 72 | 1001000 |
| O | 79 | 1001111 |
| T | 84 | 1010100 |
| S | 83 | 1010011 |

The computer understands HOTS as: `1001000 1001111 1010100 1010011`

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(ii) Main

| Character | ASCII (Decimal) | 7-bit Binary |
|---|---|---|
| M | 77 | 1001101 |
| a | 97 | 1100001 |
| i | 105 | 1101001 |
| n | 110 | 1101110 |

The computer understands Main as: `1001101 1100001 1101001 1101110`

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(iii) CaSe

| Character | ASCII (Decimal) | 7-bit Binary |
|---|---|---|
| C | 67 | 1000011 |
| a | 97 | 1100001 |
| S | 83 | 1010011 |
| e | 101 | 1100101 |

The computer understands CaSe as: `1000011 1100001 1010011 1100101`

Note: Uppercase and lowercase letters have different ASCII codes, so the computer treats 'C' and 'c' as different characters.

Given: The hexadecimal number system uses 16 unique symbols/literals: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F.

Concept: The base (radix) of a number system = total number of unique symbols used.

$$\text{Base of Hexadecimal} = 16$$

Therefore, the base value of the hexadecimal number system is $\boxed{16}$.

Given: A number system X uses exactly $B$ unique symbols.

Concept: The base (radix) of any number system is equal to the total count of distinct symbols (digits/literals) it uses.

$$\text{Base of number system X} = B$$

For example:
- If $B = 2$ (symbols: 0, 1) → Binary system with base 2
- If $B = 8$ (symbols: 0–7) → Octal system with base 8
- If $B = 16$ (symbols: 0–9, A–F) → Hexadecimal system with base 16

Therefore, the base value of number system X is $\boxed{B}$.

The phrase "हम सब एक" is in Hindi (Devanagari script). It is encoded using Unicode (specifically UTF-16 or Unicode code points).

The Unicode (hexadecimal) code points for Devanagari characters are in the range U+0900 to U+097F.

| Character | Unicode Code Point (Hex) | Binary (16-bit) |
|---|---|---|
| ह | U+0939 | 0000 1001 0011 1001 |
| म | U+092E | 0000 1001 0010 1110 |
| (space) | U+0020 | 0000 0000 0010 0000 |
| स | U+0938 | 0000 1001 0011 1000 |
| ब | U+092C | 0000 1001 0010 1100 |
| (space) | U+0020 | 0000 0000 0010 0000 |
| ए | U+090F | 0000 1001 0000 1111 |
| क | U+0915 | 0000 1001 0001 0101 |

Each Devanagari character is represented by its Unicode code point in hexadecimal, and the corresponding 16-bit binary value is obtained by converting that hex value to binary.

Advantages of preparing digital content in Indian languages using Unicode:

1. Universal Compatibility: Unicode is a universal encoding standard. Content created using Unicode can be read and displayed correctly on any device, operating system, or platform (Windows, Mac, Linux, Android, iOS) without any special font installation.

2. Web and Internet Support: Unicode-encoded Indian language content is fully supported by web browsers, search engines, and the internet. It can be indexed and searched by search engines like Google.

3. Interoperability: Unicode allows seamless exchange of Indian language documents between different applications (MS Word, Google Docs, email clients, etc.) without encoding errors or garbled text.

4. Single Standard for All Scripts: Unicode supports all Indian scripts (Devanagari, Tamil, Telugu, Bengali, Gujarati, etc.) under one standard, making it easy to create multilingual documents.

5. Long-term Preservation: Since Unicode is an internationally accepted standard, digital content created in Unicode will remain accessible and readable in the future.

6. No Dependency on Specific Fonts: Unlike older encoding schemes (like ISCII-based fonts), Unicode content does not require the recipient to have a specific font installed to read the text correctly.

7. Software Support: All modern software applications, databases, and programming languages support Unicode, making it easy to process, store, and retrieve Indian language content.

Steps to type in an Indian language (e.g., Hindi) using Unicode on a Windows computer:

Step 1: Add the Indian Language Input Method
- Go to Control Panel → Clock, Language, and Region → Language.
- Click Add a language and select the desired Indian language (e.g., Hindi).
- Click Add to install the language pack.

Step 2: Enable the Keyboard Layout
- After adding the language, click on Options next to the language.
- Add an Input Method (keyboard layout), such as Devanagari – INSCRIPT or Hindi Phonetic keyboard.

Step 3: Switch the Input Language
- Use the Language Bar on the taskbar (or press Windows key + Space / Alt + Shift) to switch between English and the Indian language input.

Step 4: Open a Unicode-Compatible Application
- Open any Unicode-compatible application such as Notepad, MS Word, or a web browser.
- Ensure the font selected supports the Indian script (e.g., Mangal, Nirmala UI for Hindi/Devanagari).

Step 5: Start Typing
- With the Indian language input method active, start typing. The keys will produce the corresponding Indian script characters encoded in Unicode.
- For phonetic keyboards, type the Roman transliteration and the software converts it to the Indian script.

Step 6: Using Online Tools (Alternative)
- Alternatively, use online Unicode typing tools such as Google Input Tools or Quillpad, which allow phonetic typing and produce Unicode output.

Note: On modern systems (Windows 10/11, macOS, Linux), Unicode support for Indian languages is built-in and the above steps may vary slightly depending on the OS version.

Given: Word = COMPUTER

Concept: Each character is assigned a unique 7-bit ASCII decimal value. We then convert each decimal value to its 7-bit binary equivalent.

| Character | ASCII Decimal Value | 7-bit Binary |
|---|---|---|
| C | 67 | 1000011 |
| O | 79 | 1001111 |
| M | 77 | 1001101 |
| P | 80 | 1010000 |
| U | 85 | 1010101 |
| T | 84 | 1010100 |
| E | 69 | 1000101 |
| R | 82 | 1010010 |

Verification of a few conversions:
- $C = 67$: $64+2+1 = 2^6+2^1+2^0 \Rightarrow 1000011$
- $O = 79$: $64+8+4+2+1 = 2^6+2^3+2^2+2^1+2^0 \Rightarrow 1001111$
- $M = 77$: $64+8+4+1 = 2^6+2^3+2^2+2^0 \Rightarrow 1001101$

Result:
The word COMPUTER is encoded in ASCII and binary as:
$$\text{C O M P U T E R}$$
$$\text{67 79 77 80 85 84 69 82}$$
$$\text{1000011 1001111 1001101 1010000 1010101 1010100 1000101 1010010}$$