In binary number system, the left-most bit is called the * most* significant bit.

Base of a number is also known as * radix* of a number.

The decimal system is composed of * 10* symbols.

* ASCII* code is the most widely used alphanumeric code which is used in computers to translate text (letters, numbers and symbols).

Unicode is a new universal coding standard adopted by all new * Platforms*.

The * AND* operation is Boolean multiplication and the

*operation is Boolean addition.*

**OR**A statement is said to be a * Boolean/Logical Statement* if it has a definite value, which is either true or false.

* NOT* gate has only one input and it complements an input signal.

The values which are stored in binary variables are known as * Boolean Constants*.

A * Truth table* is a representation of a Boolean function or expression containing all possible combinations of input values and their result in a tabular format.

Binary data representation has only two symbols, 0 and 1.

**Answer**

True

**Reason** — Binary data representation uses only two symbols, 0 and 1, because it is based on base 2. In base 2, each digit (bit) can only be one of two values: 0 or 1. These digits are combined to represent numbers and data, with each position representing a power of 2, similar to how base 10 uses powers of 10.

ASCII, ISCII and UNICODE are the three internal storage representations.

**Answer**

True

**Reason** — ASCII, ISCII, and UNICODE are three types of character encoding schemes used for internal storage representations.

(128)_{8} is the representation of an octal number system.

**Answer**

False

**Reason** — The digit '8' is not valid in the octal system because the octal number system only includes digits from 0 to 7. Therefore, (128)_{8} is not a valid octal number.

The first 128 characters are the same in both the types of ASCII, i.e., ASCII-7 and ASCII-8.

**Answer**

True

**Reason** — The first 128 characters (ranging from 0 to 127) in both ASCII-7 and ASCII-8 are identical. ASCII-7 uses 7 bits to represent characters, covering the first 128 characters. ASCII-8 uses 8 bits, allowing for 256 characters in total. However, the first 128 characters remain consistent between both versions.

512 characters can be represented in ASCII-8.

**Answer**

False

**Reason** — ASCII-8 uses 8 bits, allowing for a total of 256 characters to be represented.

Unicode provides a consistent way of encoding multilingual plain text.

**Answer**

True

**Reason** — Unicode provides a unique number for every character irrespective of the platform, program and the language.

ABC is a valid Hexadecimal number.

**Answer**

True

**Reason** — A hexadecimal number system has sixteen (16) alphanumeric values from 0-9 and A-F. Therefore, "ABC" is a valid hexadecimal number.

A number with both integer and a fractional part has digits raised to both positive and negative powers of 2 in a decimal number system.

**Answer**

False

**Reason** — In the decimal number system, a number with both integer and fractional parts has digits raised to positive powers of 10 for the integer part and negative powers of 10 for the fractional part.

(52)_{16} is equivalent to 90 in decimal number system.

**Answer**

False

**Reason** — To convert (52)_{16} from hexadecimal to decimal, we calculate:

5×16^{1} + 2×16^{0} = 5×16 + 2×1 = 80 + 2 = 82

Thus, (52)_{16} is equivalent to 82 in the decimal number system.

A number system that uses eight digits-0, 1, 2, 3, 4, 5, 6, and 7—is called an octal number system.

**Answer**

True

**Reason** — Octal number system has only eight digits from 0 to 7. Every number can be represented with 0, 1, 2, 3, 4, 5, 6, 7 in this number system.

NAND and NOR are called universal gates.

**Answer**

True

**Reason** — The NAND and NOR gates are called universal gates because any digital circuits can be implemented using these gates.

What is information?

- Processed data
- Collection of programs
- Collection of instructions
- None of these

**Answer**

Processed data

**Reason** — Information is defined as processed data organized in a particular manner to generate meaningful piece of data.

What is the base of binary number system?

- 8
- 10
- 2
- 16

**Answer**

2

**Reason** — The base of binary number system is 2 because it has only two digits.

What is the base of decimal number system?

- 2
- 10
- 8
- 16

**Answer**

10

**Reason** — The decimal number system uses 10 digits (from 0 to 9) hence its has a base of 10.

What is the base of octal number system?

- 2
- 10
- 8
- 16

**Answer**

8

**Reason** — The base of octal number system is 8 because it has only 8 digits.

Which of the following is binary equivalent of (43)_{10}?

- (101011)
_{2} - (110011)
_{2} - (110101)
_{2} - (110010)
_{2}

**Answer**

(101011)_{2}

**Reason** — To convert (43)_{10} from decimal to binary, we perform the following calculation:

2 | 43 | Remainder |
---|---|---|

2 | 21 | 1 (LSB) |

2 | 10 | 1 |

2 | 5 | 0 |

2 | 2 | 1 |

2 | 1 | 0 |

0 | 1 (MSB) |

Therefore, (43)_{10} = (101011)_{2}

Which of the following is decimal equivalent of (10010)_{2}?

- 28
- 18
- 20
- 22

**Answer**

18

**Reason** —

Binary No | Power | Value | Result |
---|---|---|---|

0 (LSB) | 2^{0} | 1 | 0x1=0 |

1 | 2^{1} | 2 | 1x2=2 |

0 | 2^{2} | 4 | 0x4=0 |

0 | 2^{3} | 8 | 0x8=0 |

1 (MSB) | 2^{4} | 16 | 1x16=16 |

Equivalent decimal number = 2 + 16 = 18

Therefore, (10010)_{2} = (18)_{10}.

What is the full form of ISCII?

- International Standard Code for Information Interchange
- Indian Standard Code for Information Interchange
- International Script Code for Information Interchange
- None of these

**Answer**

Indian Standard Code for Information Interchange

**Reason** — The full form of ISCII is Indian Standard Code for Information Interchange.

Hexadecimal number system is composed of?

- 2 symbols
- 10 symbols
- 8 symbols
- 16 symbols

**Answer**

16 symbols

**Reason** — A hexadecimal number system has sixteen (16) alphanumeric values from 0 to 9 and A to F.

Which of the following is not a binary number?

- 1111
- 101
- 11E
- 000

**Answer**

11E

**Reason** — A binary number system only includes the digits 0 and 1. The number "11E" contains the letter 'E', which is not a valid digit in the binary system. Therefore, "11E" is not a binary number.

Which numbering system uses numbers and letters as symbols?

- Decimal
- Binary
- Octal
- Hexadecimal

**Answer**

Hexadecimal

**Reason** — A hexadecimal number system has sixteen (16) alphanumeric values from 0 to 9 and A to F.

Identify the logical/Boolean statement:

- 15+4=19
- I want to have coffee.
- Should I wear the mask or not?
- I love dogs.

**Answer**

Should I wear the mask or not?

**Reason** — A logical or Boolean statement is one that can be evaluated as either true or false. The statement "Should I wear the mask or not?" poses a question that requires a true or false answer, thus making it a logical or Boolean statement.

The boolean expression of the following logic circuit is:

- (A+B).C'+B
- A'B'+C+B'A
- A.B.C+(B'+C').A
- A+B+C'(A+B)

**Answer**

A.B.C+(B'+C').A

**Reason** — The above logic circuit is equivalent to the boolean expression `A.B.C + (B'+C').A`

.

The following is a ............... Logical Gate.

- XOR
- NAND
- AND
- OR

**Answer**

AND

**Reason** — The above logic gate is an AND operator.

**Assertion (A):** Data is stored inside a computer in the form of 0s and 1s.

**Reasoning (R):** Number system is a technique of representing numbers in different forms.

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

Both A and R are true but R is not the correct explanation of A.

**Explanation**

Data inside a computer gets stored and manipulated in digital (binary) form (0s and 1s). Number systems are the techniques to represent numbers in the computer system architecture. Every number system includes a set of unique characters or literals. It is a way to represent a number in different forms.

**Assertion (A):** The base/radix of binary number system is 2.

**Reasoning (R):** Binary number system has only two digits, i.e., 1 and 0.

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

Both A and R are true and R is the correct explanation of A.

**Explanation**

The base of binary number system is 2 because it has only two digits, i.e., 1 and 0. Every number can be represented with 0 and 1 in binary number system.

**Assertion (A):** The base/radix of Octal number system is 8 because it has 8 digits.

**Reasoning (R):** 342289 is an octal number.

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

A is true but R is false.

**Explanation**

The base of the octal number system is 8 because it has only 8 digits (0 to 7). The number 342289 is not an octal number because the octal number system only allows digits from 0 to 7, and 342289 contains digits outside this range (8 and 9).

**Assertion (A):** If A+B=1 then its dual will be A.B=0.

**Reasoning (R):** In dual, 1 changes to 0 and 0 changes to 1. Similarly, (+) changes to (.) and (.) to (+).

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

Both A and R are true and R is the correct explanation of A.

**Explanation**

The dual of the Boolean expression A+B=1 is A⋅B=0. This is derived by applying the duality principle, which states that the dual of any boolean expression is obtained by interchanging the operations (+) and (.), and swapping the constants 1 and 0.

**Assertion (A):** Unicode is used for multilingual encoding.

**Reasoning (R):** Unicode provides a unique number for all the characters of every written language of the world.

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

Both A and R are true and R is the correct explanation of A.

**Explanation**

Unicode represents a single encoding scheme for all languages and characters. Unicode provides a unique number for every character irrespective of the platform, program and the language.

**Assertion (A):** The number of symbols (digit and alphabet) used in a number system is called its radix/base.

**Reasoning (R):** Base value of a number system is used to distinguish a number in one number system from another number system.

- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false but R is true.

**Answer**

Both A and R are true and R is the correct explanation of A.

**Explanation**

The number of symbols (digits and alphabets) used in a number system defines its radix or base. The base value of a number system helps in distinguishing numbers from different number systems, such as binary, octal, decimal, and hexadecimal.

Convert decimal number into binary number.

(a) 52

(b) 44

(c) 25.80

(d) 62.325

**Answer**

(a) 52

2 | 52 | Remainder |
---|---|---|

2 | 26 | 0 (LSB) |

2 | 13 | 0 |

2 | 6 | 1 |

2 | 3 | 0 |

2 | 1 | 1 |

0 | 1 (MSB) |

Therefore, (52)_{10} = (110100)_{2}.

(b) 44

2 | 44 | Remainder |
---|---|---|

2 | 22 | 0 (LSB) |

2 | 11 | 0 |

2 | 5 | 1 |

2 | 2 | 1 |

2 | 1 | 0 |

0 | 1 (MSB) |

Therefore, (44)_{10} = (101100)_{2}.

(c) 25.80

Let us first convert 25 into binary as shown below:

2 | 25 | Remainder |
---|---|---|

2 | 12 | 1 (LSB) |

2 | 6 | 0 |

2 | 3 | 0 |

2 | 1 | 1 |

0 | 1 (MSB) |

The binary equivalent of integer part (25)_{10} = (11001)_{2}.

Now let us convert (0.80)_{10} into binary as shown below:

Multiply | = | Integer | Fraction |
---|---|---|---|

0.80 x 2 | = | 1 | 0.6 (MSB) |

0.6 x 2 | = | 1 | 0.2 |

0.2 x 2 | = | 0 | 0.4 |

0.4 x 2 | = | 0 | 0.8 |

0.8 x 2 | = | 1 | 0.6 (LSB) |

The binary equivalent of fractional part (0.8)_{10} = (11001)_{2}.

Therefore, (25.80)_{10} = (11001.11001)_{2}.

(d) 62.325

Let us first convert 62 into binary as shown below:

2 | 62 | Remainder |
---|---|---|

2 | 31 | 0 (LSB) |

2 | 15 | 1 |

2 | 7 | 1 |

2 | 3 | 1 |

2 | 1 | 1 |

0 | 1 (MSB) |

The binary equivalent of integer part (62)_{10} = (111110)_{2}.

Now let us convert (0.325)_{10} into binary as shown below:

Multiply | = | Integer | Fraction |
---|---|---|---|

0.325 x 2 | = | 0 | 0.65 (MSB) |

0.65 x 2 | = | 1 | 0.3 |

0.3 x 2 | = | 0 | 0.6 |

0.6 x 2 | = | 1 | 0.2 |

0.2 x 2 | = | 0 | 0.4 |

0.4 x 2 | = | 0 | 0.8 |

0.8 x 2 | = | 1 | 0.6 (LSB) |

The binary equivalent of fractional part (0.325)_{10} = (0101001)_{2}.

Therefore, (62.325)_{10} = (111110.0101001)_{2}.

Convert decimal number into octal number.

(a) 911

(b) 540

(c) 61

(d) 132

**Answer**

(a) 911

8 | 911 | Remainder |
---|---|---|

8 | 113 | 7 (LSB) |

8 | 14 | 1 |

8 | 1 | 6 |

0 | 1 (MSB) |

Therefore, (911)_{10} = (1617)_{8}.

(b) 540

8 | 540 | Remainder |
---|---|---|

8 | 67 | 4 (LSB) |

8 | 8 | 3 |

8 | 1 | 0 |

0 | 1 (MSB) |

Therefore, (540)_{10} = (1034)_{8}.

(c) 61

8 | 61 | Remainder |
---|---|---|

8 | 7 | 5 (LSB) |

0 | 7 (MSB) |

Therefore, (61)_{10} = (75)_{8}.

(d) 132

8 | 132 | Remainder |
---|---|---|

8 | 16 | 4 (LSB) |

8 | 2 | 0 |

0 | 2 (MSB) |

Therefore, (132)_{10} = (204)_{8}.

Convert decimal number into hexadecimal number.

(a) 132

(b) 3619

(c) 206

(d) 752

**Answer**

(a) 132

16 | 132 | Remainder |
---|---|---|

16 | 8 | 4 (LSB) |

0 | 8 (MSB) |

Therefore, (132)_{10} = (84)_{16}.

(b) 3619

16 | 3619 | Remainder |
---|---|---|

16 | 226 | 3 (LSB) |

16 | 14 | 2 |

0 | E (14) (MSB) |

Therefore, (3619)_{10} = (E23)_{16}.

(c) 206

16 | 206 | Remainder |
---|---|---|

16 | 12 | E (14) (LSB) |

0 | C (12) (MSB) |

Therefore, (206)_{10} = (CE)_{16}.

(d) 752

16 | 752 | Remainder |
---|---|---|

16 | 47 | 0 (LSB) |

16 | 2 | F (15) |

0 | 2 (MSB) |

Therefore, (752)_{10} = (2F0)_{16}

Convert binary number into decimal number.

(a) 10111

(b) 111101

(c) 101010.011

(d) 101101

**Answer**

(a) 10111

Binary No | Power | Value | Result |
---|---|---|---|

1 (LSB) | 2^{0} | 1 | 1x1=1 |

1 | 2^{1} | 2 | 1x2=2 |

1 | 2^{2} | 4 | 1x4=4 |

0 | 2^{3} | 8 | 0x8=0 |

1 (MSB) | 2^{4} | 16 | 1x16=16 |

Equivalent decimal number = 1 + 2 + 4 + 16 = 23

Therefore, (10111)_{2} = (23)_{10}.

(b) 111101

Binary No | Power | Value | Result |
---|---|---|---|

1 (LSB) | 2^{0} | 1 | 1x1=1 |

0 | 2^{1} | 2 | 0x2=0 |

1 | 2^{2} | 4 | 1x4=4 |

1 | 2^{3} | 8 | 1x8=8 |

1 | 2^{4} | 16 | 1x16=16 |

1 (MSB) | 2^{5} | 32 | 1x32=32 |

Equivalent decimal number = 1 + 4 + 8 + 16 + 32 = 61

Therefore, (111101)_{2} = (61)_{10}.

(c) 101010.011

For the integer part (101010):

Binary No | Power | Value | Result |
---|---|---|---|

0 (LSB) | 2^{0} | 1 | 0x1=0 |

1 | 2^{1} | 2 | 1x2=2 |

0 | 2^{2} | 4 | 0x4=0 |

1 | 2^{3} | 8 | 1x8=8 |

0 | 2^{4} | 16 | 0x16=0 |

1 (MSB) | 2^{5} | 32 | 1x32=32 |

Equivalent decimal number = 2 + 8 + 32 = 42

For the fractional part (0.011):

Binary No | Power | Value | Result |
---|---|---|---|

0 (LSB) | 2^{-1} | 0.5 | 0x0.5=0 |

1 | 2^{-2} | 0.25 | 1x0.25=0.25 |

1 | 2^{-3} | 0.125 | 1x0.125=0.125 |

Equivalent decimal number = 0.25 + 0.125 = 0.375

Therefore, (101010.011)_{2} = (42.375)_{10}.

(d) 101101

Binary No | Power | Value | Result |
---|---|---|---|

1 (LSB) | 2^{0} | 1 | 1x1=1 |

0 | 2^{1} | 2 | 0x2=0 |

1 | 2^{2} | 4 | 1x4=4 |

1 | 2^{3} | 8 | 1x8=8 |

0 | 2^{4} | 16 | 0x16=0 |

1 (MSB) | 2^{5} | 32 | 1x32=32 |

Equivalent decimal number = 1 + 4 + 8 + 32 = 45

Therefore, (101101)_{2} = (45)_{10}.

Convert octal number into decimal number.

(a) 75

(b) 321

(c) 142

(d) 205

**Answer**

(a) 75

Octal No | Power | Value | Result |
---|---|---|---|

5 (LSB) | 8^{0} | 1 | 5x1=5 |

7 (MSB) | 8^{1} | 8 | 7x8=56 |

Equivalent decimal number = 5 + 56 = 61

Therefore, (75)_{8} = (61)_{10}.

(b) 321

Octal No | Power | Value | Result |
---|---|---|---|

1 (LSB) | 8^{0} | 1 | 1x1=1 |

2 | 8^{1} | 8 | 2x8=16 |

3 (MSB) | 8^{2} | 64 | 3x64=192 |

Equivalent decimal number = 1 + 16 + 192 = 209

Therefore, (321)_{8} = (209)_{10}.

(c) 142

Octal No | Power | Value | Result |
---|---|---|---|

2 (LSB) | 8^{0} | 1 | 2x1=2 |

4 | 8^{1} | 8 | 4x8=32 |

1 (MSB) | 8^{2} | 64 | 1x64=64 |

Equivalent decimal number = 2 + 32 + 64 = 98

Therefore, (142)_{8} = (98)_{10}.

(d) 205

Octal No | Power | Value | Result |
---|---|---|---|

5 (LSB) | 8^{0} | 1 | 5x1=5 |

0 | 8^{1} | 8 | 0x8=56 |

2 (MSB) | 8^{2} | 64 | 2x64=128 |

Equivalent decimal number = 5 + 128 = 133

Therefore, (205)_{8} = (133)_{10}.

Convert hexadecimal number into decimal number.

(a) A2

(b) 13B

(c) 271

(d) 132

**Answer**

(a) A2

Hexadecimal Number | Power | Value | Result |
---|---|---|---|

2 | 16^{0} | 1 | 2x1=2 |

A (10) | 16^{1} | 16 | 10x16=160 |

Equivalent decimal number = 2 + 160 = 162

Therefore, (A2)_{16} = (162)_{10}.

(b) 13B

Hexadecimal Number | Power | Value | Result |
---|---|---|---|

B (11) | 16^{0} | 1 | 11x1=11 |

3 | 16^{1} | 16 | 3x16=48 |

1 | 16^{2} | 256 | 1x256=256 |

Equivalent decimal number = 11 + 48 + 256 = 315

Therefore, (13B)_{16} = (315)_{10}.

(c) 271

Hexadecimal Number | Power | Value | Result |
---|---|---|---|

1 | 16^{0} | 1 | 1x1=1 |

7 | 16^{1} | 16 | 7x16=112 |

2 | 16^{2} | 256 | 2x256=512 |

Equivalent decimal number = 1 + 112 + 512 = 625

Therefore, (271)_{16} = (625)_{10}.

(d) 132

Hexadecimal Number | Power | Value | Result |
---|---|---|---|

2 | 16^{0} | 1 | 2x1=2 |

3 | 16^{1} | 16 | 3x16=48 |

1 | 16^{2} | 256 | 1x256=256 |

Equivalent decimal number = 2 + 48 + 256 = 306

Therefore, (132)_{16} = (306)_{10}.

Discuss UTF-8 encoding scheme. How is it different from UTF-32 encoding scheme?

**Answer**

UTF-8 is a variable-width encoding that can represent every character in the Unicode character set. The code unit of UTF-8 is 8 bits (an octet). It uses 1 to 4 octets to represent code points, depending on their size. For example, it uses 1 byte for ASCII characters and up to 4 bytes for others. This makes UTF-8 a multi-byte encoding that efficiently handles different character sets.

UTF-32, in contrast, is a fixed-length encoding scheme that always uses exactly 4 bytes (32 bits) to represent all Unicode code points, regardless of their size, making it less space-efficient than UTF-8.

What are Boolean constants?

**Answer**

The values which are stored in binary variables are known as Boolean Constants. Therefore, the values true/false, yes/no or 1/0 are Boolean constants.

What are Boolean operators? Explain AND, OR, NOT operators.

**Answer**

Operators used in Boolean algebra are known as Boolean/Logical operators.

1. **AND Operator** — AND operator is a binary operator that operates on two variables and the operation performed by AND operator is known as logical multiplication. The symbol used for logical multiplication is dot(.) operator. The truth table for AND operator is as follows:

**Truth Table**

A | B | Output A.B |
---|---|---|

0 | 0 | 0 |

0 | 1 | 0 |

1 | 0 | 0 |

1 | 1 | 1 |

The AND operation will result in true value (1) when both inputs are 1 (true/high) and for all other values it results in 0 (false/low).

2. **OR Operator** — The OR operator is a binary operator that operates on two variables and the operation performed by OR operator is known as logical addition. The symbol used for logical addition is plus (+) operator. The truth table for OR operator is as follows:

**Truth Table**

A | B | Output A + B |
---|---|---|

0 | 0 | 0 |

0 | 1 | 1 |

1 | 0 | 1 |

1 | 1 | 1 |

The OR operation results in true value (1) when either of the inputs is 1 (true) or both the inputs are 1 (true/high), and for all other values of inputs it results in 0 (false/low).

3. **NOT Operator** — The NOT operator is a unary operator that operates on one variable and the operation performed by NOT operator is known as negation or complementation. The truth table for NOT operator is as follows:

**Truth Table**

A | Output A' |
---|---|

0 | 1 |

1 | 0 |

It means that the logical statements A and A' are opposite to each other. If the value of a variable A is 0, its complement would be 1, and if the value of the variable A is 1, its complement would be 0.

What is the use of Hexadecimal number system in computers?

**Answer**

The use of the hexadecimal number system in computers includes the following:

- The Hexadecimal number system is used in computers to specify memory addresses (which are 16-bit or 32-bit long). For example, a memory address
`1101011010101111`

is a big binary address but with hex it is`D6AF`

which is easier to remember. - The Hexadecimal number system is used to represent color codes. For example,
`FFFFFF`

represents White,`FF0000`

represents Red, etc.

Verify the following using truth table:

(i) a.(a+b) = a

(ii) X.(Y+Z) = X.Y + X.Z

**Answer**

(i) a.(a+b) = a

Truth Table

a | b | a + b | a.(a+b) | a |
---|---|---|---|---|

0 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 0 | 0 |

1 | 0 | 1 | 1 | 1 |

1 | 1 | 1 | 1 | 1 |

As columns "a.(a+b)" and "a" have same values, hence the expression is proved.

(ii) X.(Y+Z) = X.Y + X.Z

Truth Table

X | Y | Z | Y+Z | X.(Y+Z) | X.Y | X.Z | X.Y+X.Z |
---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

As columns X.(Y+Z) and X.Y+X.Z have same values, hence the expression is proved.

Obtain the Boolean expression for the logic circuit shown below:

**Answer**

The boolean expression for the above logic circuit is $\overline{\overline{x^n.y} + \overline{x .y^n}}$.

State De Morgan's law and prove it with a truth table.

**Answer**

The first law states that when two (or more) input variables are OR'ed and negated, they are equivalent to the AND of the complements of the individual variables.

$\overline{A+B}$ = $\overline{A}.\overline{B}$

Proof using truth table:

A | B | A' | B' | A + B | (A+B)' | A'.B' |
---|---|---|---|---|---|---|

0 | 0 | 1 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 1 | 0 | 0 |

1 | 0 | 0 | 1 | 1 | 0 | 0 |

1 | 1 | 0 | 0 | 1 | 0 | 0 |

The second law states that when two (or more) input variables are AND'ed and negated, they are equivalent to the OR of the complements of the individual variables.

$\overline{A.B}$ = $\overline{A} + \overline{B}$

Proof using truth table:

A | B | A' | B' | A.B | (A.B)' | A' + B' |
---|---|---|---|---|---|---|

0 | 0 | 1 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 1 | 1 |

1 | 1 | 0 | 0 | 1 | 0 | 0 |

What is the advantage of preparing a digital content in Indian language using UNICODE font?

**Answer**

Following are the advantages of preparing a digital content in Indian language using UNICODE font:

- Unicode provides a universal character encoding standard, ensuring consistent display and compatibility across different platforms, devices, and applications.
- It supports a wide range of Indian scripts, allowing seamless text exchange and readability without font issues.
- Content prepared using Unicode can be easily indexed by search engines, improving accessibility and searchability.

Encode the word 'COMPUTER' using ASCII and convert the encoded value into binary values.

**Answer**

Representing the word 'COMPUTER' in ASCII values of its characters:

COMPUTER → 67 79 77 80 85 84 69 82

Binary Equivalent → 01000011 01001111 01001101 01010000 01010101 01010100 01000101 01010010

**Explanation**

ASCII value of C is 67 and its equivalent 7-bit binary code = 01000011

ASCII value of O is 79 and its equivalent 7-bit binary code = 01001111

ASCII value of M is 77 and its equivalent 7-bit binary code = 01001101

ASCII value of P is 80 and its equivalent 7-bit binary code = 01010000

ASCII value of U is 85 and its equivalent 7-bit binary code = 01010101

ASCII value of T is 84 and its equivalent 7-bit binary code = 01010100

ASCII value of E is 69 and its equivalent 7-bit binary code = 01000101

ASCII value of R is 82 and its equivalent 7-bit binary code = 01010010

Hence, binary value for the word 'COMPUTER' is 01000011 01001111 01001101 01010000 01010101 01010100 01000101 01010010.

Give full form of ASCII and ISCII.

**Answer**

ASCII — American Standard Code for Information Interchange

ISCII — Indian Standard Code for Information Interchange

Draw a logic circuit of the following Boolean expression:

(i) (A.$\overline{B}$) + (C + $\overline{D}$) + (B.$\overline{D}$)

(ii) (A+B).(BC+$\overline{D}$)

**Answer**

(i)

(ii)

Write binary equivalent of the following octal numbers:

(a) 2306

(b) 5610

(c) 742

(d) 65.203

**Answer**

(a) 2306

Octal Number | Binary Equivalent |
---|---|

6 | 110 |

0 | 000 |

3 | 011 |

2 | 010 |

Therefore, (2306)_{8} = (**$\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{110}}$**)_{2}.

(b) 5610

Octal Number | Binary Equivalent |
---|---|

0 | 000 |

1 | 001 |

6 | 110 |

5 | 101 |

Therefore, (5610)_{8} = (**$\bold{\underlinesegment{101}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{001}}\medspace\bold{\underlinesegment{000}}$**)_{2}.

(c) 742

Octal Number | Binary Equivalent |
---|---|

2 | 010 |

4 | 100 |

7 | 111 |

Therefore, (742)_{8} = (**$\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{100}}\medspace\bold{\underlinesegment{010}}$**)_{2}.

(d) 65.203

Octal Number | Binary Equivalent |
---|---|

3 | 011 |

0 | 000 |

2 | 010 |

5 | 101 |

6 | 110 |

Therefore, (65.203)_{8} = (**$\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{101}}.\medspace\bold{\underlinesegment{010}}\medspace\bold{\underlinesegment{000}}\medspace\bold{\underlinesegment{011}}$**)_{2}.

Write binary representation of the following hexadecimal numbers:

(a) 4026

(b) BCA1

(c) 98E

(d) 132.45

**Answer**

(a) 4026

Hexadecimal Number | Binary Equivalent |
---|---|

6 | 0110 |

2 | 0010 |

0 | 0000 |

4 | 0100 |

Therefore, (4026)_{16} = (0100000000100110)_{2}.

(b) BCA1

Hexadecimal Number | Binary Equivalent |
---|---|

1 | 0001 |

A (10) | 1010 |

C (12) | 1100 |

B (11) | 1011 |

B → 1011

C → 1100

A → 1010

1 → 0001

Therefore, (BCA1)_{16} = (1011110010100001)_{2}.

(c) 98E

Hexadecimal Number | Binary Equivalent |
---|---|

E | 1110 |

8 | 1000 |

9 | 1001 |

Therefore, (98E)_{16} = (100110001110)_{2}.

(d) 132.45

Hexadecimal Number | Binary Equivalent |
---|---|

5 | 0101 |

4 | 0100 |

2 | 0010 |

3 | 0011 |

1 | 0001 |

Therefore, (132.45)_{16} = (000100110010.01000101)_{2}.

Perform the following conversions from decimal number to other number systems:

(a) (54)_{10} = (?)_{2}

(b) (120)_{10} = (?)_{2}

(c) (76)_{10} = (?)_{8}

(d) (889)_{10} = (?)_{8}

(e) (789)_{10} = (?)_{16}

(f) (108)_{10} = (?)_{16}

**Answer**

(a) (54)_{10}

2 | 54 | Remainder |
---|---|---|

2 | 27 | 0 (LSB) |

2 | 13 | 1 |

2 | 6 | 1 |

2 | 3 | 0 |

2 | 1 | 1 |

0 | 1 (MSB) |

Therefore, (54)_{10} = (110110)_{2}.

(b) (120)_{10}

2 | 120 | Remainder |
---|---|---|

2 | 60 | 0 (LSB) |

2 | 30 | 0 |

2 | 15 | 0 |

2 | 7 | 1 |

2 | 3 | 1 |

2 | 1 | 1 |

0 | 1 (MSB) |

Therefore, (120)_{10} = (1111000)_{2}.

(c) (76)_{10}

8 | 76 | Remainder |
---|---|---|

8 | 9 | 4 (LSB) |

8 | 1 | 1 |

0 | 1 (MSB) |

Therefore, (76)_{10} = (114)_{8}.

(d) (889)_{10}

8 | 889 | Remainder |
---|---|---|

8 | 111 | 1 (LSB) |

8 | 13 | 7 |

8 | 1 | 5 |

0 | 1 (MSB) |

Therefore, (889)_{10} = (1571)_{8}.

(e) (789)_{10}

16 | 789 | Remainder |
---|---|---|

16 | 49 | 5 (LSB) |

16 | 3 | 1 |

0 | 3 (MSB) |

Therefore, (789)_{10} = (315)_{16}.

(f) (108)_{10}

16 | 108 | Remainder |
---|---|---|

16 | 6 | 12 (C) (LSB) |

0 | 6 (MSB) |

Therefore, (108)_{10} = (6C)_{16}.

Express the following binary numbers into their equivalent octal numbers:

(a) 100101111

(b) 111011010

(c) 1010011

(d) 10011101

**Answer**

(a) 100101111

Grouping in bits of 3:

$\underlinesegment{100} \quad \underlinesegment{101} \quad \underlinesegment{111}$

Binary Number | Equivalent Octal |
---|---|

100 | 4 |

101 | 5 |

111 | 7 |

Therefore, (100101111)_{2} = (457)_{8}

(b) 111011010

Grouping in bits of 3:

$\underlinesegment{111} \quad \underlinesegment{011} \quad \underlinesegment{010}$

Binary Number | Equivalent Octal |
---|---|

111 | 7 |

011 | 3 |

010 | 2 |

Therefore, (111011010)_{2} = (732)_{8}

(c) 1010011

Grouping in bits of 3:

$\underlinesegment{001} \quad \underlinesegment{010} \quad \underlinesegment{011}$

Binary Number | Equivalent Octal |
---|---|

001 | 1 |

010 | 2 |

011 | 3 |

Therefore, (1010011)_{2} = (123)_{8}

(d) 10011101

Grouping in bits of 3:

$\underlinesegment{010} \quad \underlinesegment{011} \quad \underlinesegment{101}$

Binary Number | Equivalent Octal |
---|---|

010 | 2 |

011 | 3 |

101 | 5 |

Therefore, (10011101)_{2} = (235)_{8}

Express the following binary numbers into their equivalent hexadecimal numbers:

(a) 1111101101100011

(b) 100000101011100

(c) 1000111010100011

(d) 111011111

**Answer**

(a) 1111101101100011

Grouping in bits of 4:

$\underlinesegment{1111} \quad \underlinesegment{1011} \quad \underlinesegment{0110} \quad \underlinesegment{0011}$

Binary Number | Equivalent Hexadecimal |
---|---|

1111 | F |

1011 | B |

0110 | 6 |

0011 | 3 |

Therefore, (1111101101100011)_{2} = (FB63)_{16}.

(b) 100000101011100

Grouping in bits of 4:

$\underlinesegment{0100} \quad \underlinesegment{0001} \quad \underlinesegment{0101} \quad \underlinesegment{1100}$

Binary Number | Equivalent Hexadecimal |
---|---|

0100 | 4 |

0001 | 1 |

0101 | 5 |

1100 | C |

Therefore, (100000101011100)_{2} = (415C)_{16}.

(c) 1000111010100011

Grouping in bits of 4:

$\underlinesegment{1000} \quad \underlinesegment{1110} \quad \underlinesegment{1010} \quad \underlinesegment{0011}$

Binary Number | Equivalent Hexadecimal |
---|---|

1000 | 8 |

1110 | E |

1010 | A |

0011 | 3 |

Therefore, (1000111010100011)_{2} = (8EA3)_{16}.

(d) 111011111

Grouping in bits of 4:

$\underlinesegment{0001} \quad \underlinesegment{1101} \quad \underlinesegment{1111}$

Binary Number | Equivalent Hexadecimal |
---|---|

0001 | 1 |

1101 | D |

1111 | F |

Therefore, (111011111)_{2} = (1DF)_{16}.

Express the following octal numbers into their equivalent decimal numbers.

(a) 145

(b) 6760

(c) 455

(d) 10.75

**Answer**

(a) 145

Octal No | Power | Value | Result |
---|---|---|---|

5 (LSB) | 8^{0} | 1 | 5x1=5 |

4 | 8^{1} | 8 | 4x8=32 |

1 (MSB) | 8^{2} | 64 | 1x64=64 |

Equivalent decimal number = 5 + 32 + 64 = 101

Therefore, (145)_{8} = (101)_{10}.

(b) 6760

Octal No | Power | Value | Result |
---|---|---|---|

0 (LSB) | 8^{0} | 1 | 0x1=0 |

6 | 8^{1} | 8 | 6x8=48 |

7 | 8^{2} | 64 | 7x64=448 |

6 (MSB) | 8^{3} | 512 | 6x512=3072 |

Equivalent decimal number = 48 + 448 + 3072 = 3568

Therefore, (6760)_{8} = (3568)_{10}.

(c) 455

Octal No | Power | Value | Result |
---|---|---|---|

5 (LSB) | 8^{0} | 1 | 5x1=5 |

5 | 8^{1} | 8 | 5x8=40 |

4 (MSB) | 8^{2} | 64 | 4x64=256 |

Equivalent decimal number = 5 + 40 + 256 = 301

Therefore, (455)_{8} = (301)_{10}.

(d) 10.75

Integer part:

Octal No | Power | Value | Result |
---|---|---|---|

0 (LSB) | 8^{0} | 1 | 0x1=0 |

1 (MSB) | 8^{1} | 8 | 1x8=8 |

Equivalent decimal number = 8

Fractional part:

Octal No | Power | Value | Result |
---|---|---|---|

7 (LSB) | 8^{-1} | 0.125 | 7x0.125=0.875 |

5 (MSB) | 8^{-2} | 0.015625 | 5x0.015625=0.078125 |

Equivalent decimal number = 0.875 + 0.078125 = 0.953125

Therefore, (10.75)_{8} = (8.953125)_{10}.

The Hexadecimal number system uses 16 literals (0 - 9, A - F). Write down its base value.

**Answer**

The base value of the hexadecimal number system is 16.

Explore and list the steps required to type in an Indian language using UNICODE.

**Answer**

To insert any Indian language character using Unicode in Microsoft Word, the steps are:

- Click Insert → Symbol → More Symbol
- From the Symbols dialog that appears, select a Unicode supporting font and Indian language's subset.
- Choose desired character and click Insert.

Convert the following base of number system:

(a) (1010100)_{10} = (...............)_{2}

(b) (3674)_{8} = (...............)_{2}

(c) (266)_{10} = (...............)_{8}

(d) (9F2)_{16} = (...............)_{2}

**Answer**

(a) (1010100)_{10}

2 | 1010100 | Remainder |
---|---|---|

2 | 505050 | 0 (LSB) |

2 | 252525 | 0 |

2 | 126262 | 1 |

2 | 63131 | 0 |

2 | 31565 | 1 |

2 | 15782 | 1 |

2 | 7891 | 0 |

2 | 3945 | 1 |

2 | 1972 | 1 |

2 | 986 | 0 |

2 | 493 | 0 |

2 | 246 | 1 |

2 | 123 | 0 |

2 | 61 | 1 |

2 | 30 | 1 |

2 | 15 | 0 |

2 | 7 | 1 |

2 | 3 | 1 |

2 | 1 | 1 |

0 | 1 (MSB) |

Therefore, (1010100)_{10} = (11110110100110110100)_{2}.

(b) (3674)_{8}

Octal Number | Binary Equivalent |
---|---|

4 | 100 |

7 | 111 |

6 | 110 |

3 | 011 |

Therefore, (3674)_{8} = (**$\bold{\underlinesegment{011}}\medspace\bold{\underlinesegment{110}}\medspace\bold{\underlinesegment{111}}\medspace\bold{\underlinesegment{100}}$**)_{2}.

(c) (266)_{10}

8 | 266 | Remainder |
---|---|---|

8 | 33 | 2 (LSB) |

8 | 4 | 1 |

0 | 4 (MSB) |

Therefore, (266)_{10} = (412)_{8}.

(c) (9F2)_{16}

Hexadecimal Number | Binary Equivalent |
---|---|

2 | 0010 |

F | 1111 |

9 | 1001 |

Therefore, (9F2)_{16} = (100111110010)_{2}.

Write the name of following law and verify using truth table:

a(b + c) = ab + ac

**Answer**

`a(b + c) = ab + ac`

is distributive law.

**Truth Table**

a | b | c | b+c | a.(b+c) | a.b | a.c | a.b + a.c |
---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 1 | 1 | 0 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 0 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

As columns "a(b + c)" and "ab + ac" have same values, hence the expression is proved.

Draw the logic circuit for the following Boolean expression:

(X'+Y).Z + W'

**Answer**

The logic circuit for the Boolean expression is shown below: