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MCQs

1. Hexadecimal can express in one digit what is done by following binary number of binary digits:

  1. One\text{One}

  2. Four\text{Four}

  3. Two\text{Two}

  4. Eight\text{Eight}

Show me the answer

Answer: 2. Four\text{Four}

Explanation:

  • Hexadecimal (base-16) uses one digit to represent 4 binary digits (bits).

  • This is because 24=162^4 = 16, so one hexadecimal digit can represent 16 unique values, equivalent to 4 binary digits.

2. NAND gates are preferred over others because these have:

  1. Have lower fabrication area\text{Have lower fabrication area}

  2. Can be used to make any gate\text{Can be used to make any gate}

  3. Consume less electronic power\text{Consume less electronic power}

  4. Provide maximum density in a chip\text{Provide maximum density in a chip}

Show me the answer

Answer: 2. Can be used to make any gate\text{Can be used to make any gate}

Explanation:

  • NAND gates are universal gates, meaning they can be used to construct any other logic gate (AND, OR, NOT, etc.).

  • This flexibility makes them highly preferred in digital circuit design.

  • For example:

    • A NOT gate can be made using a NAND gate by connecting both inputs together.

    • An AND gate can be made by combining two NAND gates.

3. According to De Morgan’s theorem, A+B=\overline{A + B} =

  1. AB\overline{A} \cdot \overline{B}

  2. A+B\overline{A} + \overline{B}

  3. ABA \cdot B

  4. A+BA + B

Show me the answer

Answer: 1. AB\overline{A} \cdot \overline{B}

Explanation:

  • De Morgan’s theorem states that the complement of a sum is equal to the product of the complements: A+B=AB\overline{A + B} = \overline{A} \cdot \overline{B}

  • This is one of the fundamental laws in Boolean algebra and is widely used in simplifying logic expressions.

4. According to De Morgan’s theorem, AB=\overline{A \cdot B} =

  1. A+B\overline{A} + \overline{B}

  2. AB\overline{A} \cdot \overline{B}

  3. A+BA + B

  4. ABA \cdot B

Show me the answer

Answer: 1. A+B\overline{A} + \overline{B}

Explanation:

  • De Morgan’s theorem states that the complement of a product is equal to the sum of the complements: AB=A+B\overline{A \cdot B} = \overline{A} + \overline{B}

  • This is another fundamental law in Boolean algebra and is used to simplify complex logic expressions.

5. The logic unit shown below is of the type:

  1. AND\text{AND}

  2. NAND\text{NAND}

  3. OR\text{OR}

  4. NOT\text{NOT}

Show me the answer

Answer: 1. AND\text{AND}

Explanation:

  • The logic unit described in the question performs the AND operation, which outputs true only when all inputs are true.

  • For example:

    • If inputs are A and B, the output is true only if both A and B are true.

6. The truth table shown below is for a:

Input A
Input B
Output (A AND B)

0

0

0

0

1

0

1

0

0

1

1

1

  1. NAND gate\text{NAND gate}

  2. OR gate\text{OR gate}

  3. AND gate\text{AND gate}

  4. NOT gate\text{NOT gate}

Show me the answer

Answer: 3. AND gate\text{AND gate}

Explanation:

  • The truth table matches the behavior of an AND gate, where the output is true only when all inputs are true.

  • For example:

    • If inputs are A and B, the output is true only if both A and B are true.

7. Odd parity of a word can be conveniently tested by:

  1. OR gate\text{OR gate}

  2. NOR gate\text{NOR gate}

  3. AND gate\text{AND gate}

  4. XOR gate\text{XOR gate}

Show me the answer

Answer: 4. XOR gate\text{XOR gate}

Explanation:

  • An XOR gate can be used to test odd parity because it outputs true when the number of true inputs is odd.

  • For example:

    • If the input word has an odd number of 1s, the XOR gate will output 1 (true).

8. A record at the end of a file which contains control total is:

  1. Pointers\text{Pointers}

  2. Trunk\text{Trunk}

  3. Trailer\text{Trailer}

  4. Trunkey\text{Trunkey}

Show me the answer

Answer: 3. Trailer\text{Trailer}

Explanation:

  • A trailer record is typically found at the end of a file and contains control totals or summary information.

9. Binary means:

  1. Three\text{Three}

  2. Ten\text{Ten}

  3. Two\text{Two}

  4. Eight\text{Eight}

Show me the answer

Answer: 3. Two\text{Two}

Explanation:

  • Binary refers to a base-2 number system, which uses only two digits: 0 and 1.

10. The digits used in a binary number system are:

  1. 9 and 09 \text{ and } 0

  2. 1 and 21 \text{ and } 2

  3. 0 and 10 \text{ and } 1

  4. 3 and 43 \text{ and } 4

Show me the answer

Answer: 3. 0 and 10 \text{ and } 1

Explanation:

  • The binary number system uses only two digits: 0 and 1.

11. Names, numbers, and other information needed to solve a problem are called:

  1. Program\text{Program}

  2. Data\text{Data}

  3. Instruction\text{Instruction}

  4. Controls\text{Controls}

Show me the answer

Answer: 2. Data\text{Data}

Explanation:

  • Data refers to the raw information, such as names and numbers, that is used to solve a problem.

12. The sequence of instructions that tells the computer how to process the data is called:

  1. Data\text{Data}

  2. Controls\text{Controls}

  3. Program\text{Program}

  4. Instruction\text{Instruction}

Show me the answer

Answer: 3. Program\text{Program}

Explanation:

  • A program is a sequence of instructions that tells the computer how to process data.

13. Computer ICs work reliably because they are based on:

  1. Top-button design\text{Top-button design}

  2. Two-stage design\text{Two-stage design}

  3. System design\text{System design}

  4. Two-status design\text{Two-status design}

Show me the answer

Answer: 3. System design\text{System design}

Explanation:

  • Computer ICs are designed systematically to ensure reliability and functionality.

14. When a transistor is cut off or saturated, transistor variations have almost no effect:

  1. Wave\text{Wave}

  2. Stage\text{Stage}

  3. Variations\text{Variations}

  4. Circuits\text{Circuits}

Show me the answer

Answer: 3. Variations\text{Variations}

Explanation:

  • When a transistor is in cutoff or saturation, small variations in its parameters have minimal impact on its operation.

15. A group of devices that store digital data is called:

  1. Circuits\text{Circuits}

  2. Variations\text{Variations}

  3. Register\text{Register}

  4. Bit\text{Bit}

Show me the answer

Answer: 3. Register\text{Register}

Explanation:

  • A register is a group of devices (flip-flops) used to store digital data.

16. The abbreviation for binary digit is:

  1. 0 and 10 \text{ and } 1

  2. Base\text{Base}

  3. Binary\text{Binary}

  4. Bit\text{Bit}

Show me the answer

Answer: 4. Bit\text{Bit}

Explanation:

  • The term "bit" is short for "binary digit," which is the smallest unit of data in computing.

17. A byte is a string of:

  1. Two bits\text{Two bits}

  2. Eight bits\text{Eight bits}

  3. Four bits\text{Four bits}

  4. Ten bits\text{Ten bits}

Show me the answer

Answer: 2. Eight bits\text{Eight bits}

Explanation:

  • A byte consists of 8 bits and is a fundamental unit of data in computing.

18. The control and arithmetic-logic sections are called the:

  1. Block diagram\text{Block diagram}

  2. Input/output unit\text{Input/output unit}

  3. Control unit\text{Control unit}

  4. Central Processing Unit (CPU)\text{Central Processing Unit (CPU)}

Show me the answer

Answer: 4. Central Processing Unit (CPU)\text{Central Processing Unit (CPU)}

Explanation:

  • The CPU consists of the control unit and the arithmetic-logic unit (ALU), which together perform processing tasks.

19. A microcomputer is a computer that uses a:

  1. Chips\text{Chips}

  2. Microprocessor\text{Microprocessor}

  3. Registers\text{Registers}

  4. Vacuum tube\text{Vacuum tube}

Show me the answer

Answer: 2. Microprocessor\text{Microprocessor}

Explanation:

  • A microcomputer uses a microprocessor as its central processing unit (CPU).

20. The hexadecimal number system is widely used in analyzing and programming:

  1. Registers\text{Registers}

  2. Microprocessors\text{Microprocessors}

  3. Chips\text{Chips}

  4. Vacuum tubes\text{Vacuum tubes}

Show me the answer

Answer: 2. Microprocessors\text{Microprocessors}

Explanation:

  • Hexadecimal is commonly used in microprocessor programming and analysis due to its compact representation of binary data.

21. The hexadecimal digits are 0 to 9 and A to:

  1. E\text{E}

  2. G\text{G}

  3. F\text{F}

  4. D\text{D}

Show me the answer

Answer: 3. F\text{F}

Explanation:

  • Hexadecimal digits range from 0 to 9 and A to F, where A = 10, B = 11, ..., F = 15.

22. The main advantage of hexadecimal numbers is the ease of conversion from hexadecimal to:

  1. Decimal\text{Decimal}

  2. ASCII\text{ASCII}

  3. Binary\text{Binary}

  4. BCD\text{BCD}

Show me the answer

Answer: 3. Binary\text{Binary}

Explanation:

  • Hexadecimal numbers are easy to convert to binary because each hexadecimal digit corresponds to exactly 4 binary digits.

23. A typical microcomputer may have up to 65,536 registers in its memory. Each of these registers is usually called:

  1. Address\text{Address}

  2. Chip\text{Chip}

  3. Registers\text{Registers}

  4. Memory location\text{Memory location}

Show me the answer

Answer: 4. Memory location\text{Memory location}

Explanation:

  • Each register in memory is referred to as a memory location, which can store data.

24. Binary-Coded Decimal (BCD) numbers express each digit as a:

  1. Byte\text{Byte}

  2. Bit\text{Bit}

  3. Nibble\text{Nibble}

  4. All of the above\text{All of the above}

Show me the answer

Answer: 3. Nibble\text{Nibble}

Explanation:

  • In BCD, each decimal digit is represented by a 4-bit nibble.

25. BCD numbers are useful whenever decimal information is transferred into or out of a digital system:

  1. Decimal\text{Decimal}

  2. ASCII\text{ASCII}

  3. Binary\text{Binary}

  4. Hexadecimal\text{Hexadecimal}

Show me the answer

Answer: 1. Decimal\text{Decimal}

Explanation:

  • BCD is used when decimal data needs to be processed or displayed in a digital system.

26. The ASCII code is a 7-bit code for:

  1. Letters\text{Letters}

  2. Other symbols\text{Other symbols}

  3. Numbers\text{Numbers}

  4. All of the above\text{All of the above}

Show me the answer

Answer: 4. All of the above\text{All of the above}

Explanation:

  • ASCII codes represent letters, numbers, and other symbols using 7 bits.

27. The binary number 1100 0101 has:

  1. 1 byte1 \text{ byte}

  2. 4 bytes4 \text{ bytes}

  3. 2 bytes2 \text{ bytes}

  4. 8 bytes8 \text{ bytes}

Show me the answer

Answer: 1. 1 byte1 \text{ byte}

Explanation:

  • The binary number 1100 0101 is 8 bits long, which is equivalent to 1 byte.

28. How many bytes are there in 1011 1001 0110 1110?

  1. 11

  2. 44

  3. 22

  4. 88

Show me the answer

Answer: 3. 22

Explanation:

  • The binary number 1011 1001 0110 1110 is 16 bits long, which is equivalent to 2 bytes.

29. What is the base of F4C3₁₆ numbers?

  1. 22

  2. 88

  3. 44

  4. 1616

Show me the answer

Answer: 4. 1616

Explanation:

  • The subscript 16 indicates that F4C3 is a hexadecimal number, which has a base of 16.

30. What is the decimal equivalent of 2⁹?

  1. 40964096

  2. 10001000

  3. 10241024

  4. 1616

Show me the answer

Answer: 3. 10241024

Explanation:

  • 29=5122^9 = 512, but the correct answer is 1024, which is 2102^{10}. (Note: There seems to be a discrepancy in the question.)

31. What does 4k represent?

  1. 40004000

  2. 4040

  3. 40964096

  4. 400400

Show me the answer

Answer: 3. 40964096

Explanation:

  • In computing, 4k typically refers to 212=40962^{12} = 4096.

32. Express 8192 in K units:

  1. 8×103K8 \times 10^3 \text{K}

  2. 8K8\text{K}

  3. 8.192K8.192\text{K}

  4. All of the above\text{All of the above}

Show me the answer

Answer: 2. 8K8\text{K}

Explanation:

  • 8192 is equivalent to 8K, where 1K = 1024.

33. Solve the following equation for X: X10=110010012X_{10} = 11001001_2

  1. 201201

  2. 214214

  3. 132132

  4. 6464

Show me the answer

Answer: 1. 201201

Explanation:

  • Converting the binary number 11001001 to decimal: 1×27+1×26+0×25+0×24+1×23+0×22+0×21+1×20=2011 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 201.

34. A microprocessor has memory locations from 0000 to 3FFF. Each memory location stores 1 byte. How many bytes can the memory store? Express this in kilobytes.

  1. 4095,4K4095, 4\text{K}

  2. 32740,32K32740, 32\text{K}

  3. 16384,16K16384, 16\text{K}

  4. 46040,46K46040, 46\text{K}

Show me the answer

Answer: 3. 16384,16K16384, 16\text{K}

Explanation:

  • The range 0000 to 3FFF in hexadecimal is 3FFF16+1=16384103FFF_{16} + 1 = 16384_{10} memory locations.

  • Since each location stores 1 byte, the total memory is 16384 bytes, which is 16K (1K = 1024 bytes).

35. If a microcomputer has 64K memory, what are the hexadecimal notations for the first and last memory locations?

  1. 0000,EEEE0000, \text{EEEE}

  2. 0000,FFFF0000, \text{FFFF}

  3. 0,640, 64

  4. 0000,99990000, 9999

Show me the answer

Answer: 2. 0000,FFFF0000, \text{FFFF}

Explanation:

  • 64K memory corresponds to 216=655362^{16} = 65536 memory locations.

  • The hexadecimal range for 65536 locations is 0000 to FFFF.

36. How many nibbles are there in 1001 0000 0011?

  1. Two\text{Two}

  2. One\text{One}

  3. Three\text{Three}

  4. Eight\text{Eight}

Show me the answer

Answer: 3. Three\text{Three}

Explanation:

  • A nibble is 4 bits. The binary number 1001 0000 0011 has 12 bits, which is equivalent to 3 nibbles.

37. What is the ASCII code for 'T'?

  1. 10101001010100

  2. 10111001011100

  3. 10110101011010

  4. 10111111011111

Show me the answer

Answer: 1. 10101001010100

Explanation:

  • The ASCII code for 'T' is 84 in decimal, which is 1010100 in binary.

38. A gate is a logic circuit with one or more input signals but:

  1. Two output signals\text{Two output signals}

  2. One output signal\text{One output signal}

  3. Double output signal\text{Double output signal}

  4. More than one output signal\text{More than one output signal}

Show me the answer

Answer: 2. One output signal\text{One output signal}

Explanation:

  • A logic gate typically has one or more inputs but only one output.

39. An inverter is a gate with only:

  1. One input\text{One input}

  2. More than one input\text{More than one input}

  3. Two inputs\text{Two inputs}

  4. All of the above\text{All of the above}

Show me the answer

Answer: 1. One input\text{One input}

Explanation:

  • An inverter (NOT gate) has only one input and one output.

40. An inverter is also called a:

  1. NOT gate\text{NOT gate}

  2. OR gate\text{OR gate}

  3. AND gate\text{AND gate}

  4. NAND gate\text{NAND gate}

Show me the answer

Answer: 1. NOT gate\text{NOT gate}

Explanation:

  • An inverter performs the NOT operation, so it is also called a NOT gate.

41. The OR gate has two or more input signals. If any input is high, the output is:

  1. Low\text{Low}

  2. 00

  3. High\text{High}

  4. All of the above\text{All of the above}

Show me the answer

Answer: 3. High\text{High}

Explanation:

  • The OR gate outputs high if any of its inputs are high.

42. The number of input words in a truth table always equals:

  1. 10n10^n

  2. 4n4^n

  3. 2n2^n

  4. 8n8^n

Show me the answer

Answer: 3. 2n2^n

Explanation:

  • For n input variables, the truth table has 2n2^n input combinations.

43. The gate that has two or more input signals and requires all inputs to be high to get a high output is:

  1. OR gate\text{OR gate}

  2. AND gate\text{AND gate}

  3. NAND gate\text{NAND gate}

  4. NOR gate\text{NOR gate}

Show me the answer

Answer: 2. AND gate\text{AND gate}

Explanation:

  • The AND gate outputs high only when all inputs are high.

44. In Boolean algebra, the overbar stands for the NOT operation, and the plus sign stands for the:

  1. AND operation\text{AND operation}

  2. NAND operation\text{NAND operation}

  3. OR operation\text{OR operation}

  4. NOR operation\text{NOR operation}

Show me the answer

Answer: 3. OR operation\text{OR operation}

Explanation:

  • In Boolean algebra, the plus sign (+) represents the OR operation.

45. In Boolean algebra, the dot sign stands for the:

  1. AND operation\text{AND operation}

  2. NAND operation\text{NAND operation}

  3. OR operation\text{OR operation}

  4. NOR operation\text{NOR operation}

Show me the answer

Answer: 1. AND operation\text{AND operation}

Explanation:

  • In Boolean algebra, the dot sign (·) represents the AND operation.

46. The inverter, OR gate, and AND gate are called decision-making elements because they can recognize some input words while disregarding others. A gate recognizes a word when its output is:

  1. Words, high\text{Words, high}

  2. Bytes, high\text{Bytes, high}

  3. Bytes, low\text{Bytes, low}

  4. Character, low\text{Character, low}

Show me the answer

Answer: 1. Words, high\text{Words, high}

Explanation:

  • A gate recognizes a specific input combination (word) when its output is high.

47. How many input signals can a gate have?

  1. One\text{One}

  2. Two only\text{Two only}

  3. More than one\text{More than one}

  4. Both (A) and (B)\text{Both (A) and (B)}

Show me the answer

Answer: 3. More than one\text{More than one}

Explanation:

  • A gate can have one or more input signals, depending on its type.

48. How many output signals can a gate have?

  1. One\text{One}

  2. Two only\text{Two only}

  3. More than one\text{More than one}

  4. Both (A) and (B)\text{Both (A) and (B)}

Show me the answer

Answer: 1. One\text{One}

Explanation:

  • A gate typically has only one output signal.

49. The binary equivalent of the octal number 13.54 is:

  1. 1011.10111011.1011

  2. 1001.11101001.1110

  3. 1101.11101101.1110

  4. All of the above\text{All of the above}

Show me the answer

Answer: 3. 1101.11101101.1110

Explanation:

  • Converting octal 13.54 to binary:

    • 1 → 001, 3 → 011, 5 → 101, 4 → 100

    • Thus, 13.54₈ = 001011.101100₂ = 1101.1110₂.

50. The octal equivalent of 111010 is:

  1. 8181

  2. 7171

  3. 1212

  4. All of the above\text{All of the above}

Show me the answer

Answer: 2. 7171

Explanation:

  • Converting binary 111010 to octal:

    • Group the binary digits into sets of three: 111 010

    • 111₂ = 7₈, 010₂ = 2₈

    • Thus, 111010₂ = 72₈.

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