> For the complete documentation index, see [llms.txt](https://nec-license.gitbook.io/books/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://nec-license.gitbook.io/books/2.-digital-logic-and-microprocessor/mcqs/mcqs-on-digital-logic/set-10.md).
# set-10
#### 451. $$( 2 \times 10^1 + 8 \times 10^0 )$$ equal to
1. 10
2. 2.8
3. 280
4. 28
Show me the answer
**Answer:** 4. 28
**Explanation:**
* The expression evaluates to:\
$$2 \times 10 + 8 \times 1 = 20 + 8 = 28$$
* Therefore, the correct answer is **28**.
#### 452. The binary number 1101 is equal to the decimal number
1. 13
2. 11
3. 49
4. 3
Show me the answer
**Answer:** 1. 13
**Explanation:**
* The binary number **1101** converts to decimal as:\
$$1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 13$$
* Therefore, the correct answer is **13**.
#### 453. The binary number 11011101 is equal to the decimal number
1. 121
2. 441
3. 221
4. 256
Show me the answer
**Answer:** 3. 221
**Explanation:**
* The binary number **11011101** converts to decimal as:\
$$1 \times 2^7 + 1 \times 2^6 + 0 \times 2^5 + 1 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 128 + 64 + 0 + 16 + 8 + 4 + 0 + 1 = 221$$
* Therefore, the correct answer is **221**.
#### 454. The decimal number 17 is equal to the binary number
1. 10010
2. 10001
3. 11000
4. 01001
Show me the answer
**Answer:** 2. 10001
**Explanation:**
* The decimal number **17** converts to binary as:\
$$17 = 16 + 1 = 2^4 + 2^0 = 10001$$
* Therefore, the correct answer is **10001**.
#### 455. The decimal number 175 is equal to the binary number
1. 11001111
2. 10101111
3. 10101110
4. 11101111
Show me the answer
**Answer:** 2. 10101111
**Explanation:**
* The decimal number **175** converts to binary as:\
$$175 = 128 + 32 + 8 + 4 + 2 + 1 = 10101111$$
* Therefore, the correct answer is **10101111**.
#### 456. The sum of 11010 + 01111 equals
1. 101001
2. 110101
3. 101010
4. 101000
Show me the answer
**Answer:** 1. 101001
**Explanation:**
* The sum of **11010** and **01111** is:\
$$11010 + 01111 = 101001$$
* Therefore, the correct answer is **101001**.
#### 457. The difference of 110 - 010 equals
1. 001
2. 101
3. 010
4. 100
Show me the answer
**Answer:** 4. 100
**Explanation:**
* The difference of **110** and **010** is:\
$$110 - 010 = 100$$
* Therefore, the correct answer is **100**.
#### 458. The 1's complement of 10111001 is
1. 01000111
2. 11000110
3. 01000110
4. 10101010
Show me the answer
**Answer:** 3. 01000110
**Explanation:**
* The 1's complement of **10111001** is obtained by flipping all the bits:\
$$10111001 \rightarrow 01000110$$
* Therefore, the correct answer is **01000110**.
#### 459. The 2's complement of 11001000 is
1. 00110111
2. 01001000
3. 00110001
4. 00111000
Show me the answer
**Answer:** 4. 00111000
**Explanation:**
* The 2's complement of **11001000** is obtained by flipping all the bits and adding 1:\
$$11001000 \rightarrow 00110111 + 1 = 00111000$$
* Therefore, the correct answer is **00111000**.
#### 460. The decimal number -34 is expressed in the 2's complement form as
1. 01011110
2. 11011110
3. 10100010
4. 01011101
Show me the answer
**Answer:** 2. 11011110
**Explanation:**
* The 2's complement representation of **-34** is obtained by taking the 2's complement of the positive value:\
$$34 = 00100010 \rightarrow 11011101 + 1 = 11011110$$
* Therefore, the correct answer is **11011110**.
#### 461. The decimal number +122 is expressed in the 2's complement form as
1. 01111010
2. 01000101
3. 11111010
4. 10000101
Show me the answer
**Answer:** 1. 01111010
**Explanation:**
* The 2's complement representation of **+122** is the same as its binary representation:\
$$122 = 01111010$$
* Therefore, the correct answer is **01111010**.
#### 462. A single-precision floating-point binary number has a total of
1. 8 bits
2. 24 bits
3. 16 bits
4. 32 bits
Show me the answer
**Answer:** 4. 32 bits
**Explanation:**
* A single-precision floating-point binary number has a total of **32 bits**.
* Therefore, the correct answer is **32 bits**.
#### 463. In the 2's complement form, the binary number 10010011 is equal to the decimal number
1. -19
2. +91
3. +109
4. -109
Show me the answer
**Answer:** 4. -109
**Explanation:**
* The binary number **10010011** in 2's complement form represents a negative number.
* The decimal value is calculated as:\
$$-128 + 16 + 2 + 1 = -109$$
* Therefore, the correct answer is **-109**.
#### 464. The binary number 101100111001010100001 can be written in hexadecimal as
1. 5471238
2. 2634521s
3. 5471241s
4. 23162501s
Show me the answer
**Answer:** 3. 5471241s
**Explanation:**
* The binary number **101100111001010100001** can be grouped into 4-bit segments and converted to hexadecimal:\
$$1011 \ 0011 \ 1001 \ 0101 \ 0001 = B \ 3 \ 9 \ 5 \ 1 = 5471241s$$
* Therefore, the correct answer is **5471241s**.
#### 465. The binary number 10001101010001101111 can be written in hexadecimal as
1. $$AD467(\_{16})$$
2. $$8D46F(\_{16})$$
3. $$8C46F(\_{16})$$
4. $$AE46F(\_{16})$$
Show me the answer
**Answer:** 2. $$8D46F(\_{16})$$
**Explanation:**
* The binary number **10001101010001101111** can be grouped into 4-bit segments and converted to hexadecimal:\
$$1000 \ 1101 \ 0100 \ 0110 \ 1111 = 8 \ D \ 4 \ 6 \ F = 8D46F\_{16}$$
* Therefore, the correct answer is $$8D46F(\_{16})$$.
#### 466. The binary number for $$F7A9(\_{16})$$ is
1. 1111011110101001
2. 1111111010110001
3. 1110111110101001
4. 1111011010101001
Show me the answer
**Answer:** 1. 1111011110101001
**Explanation:**
* The hexadecimal number **F7A9** converts to binary as:\
$$F = 1111, 7 = 0111, A = 1010, 9 = 1001$$\
$$F7A9 = 1111011110101001$$
* Therefore, the correct answer is **1111011110101001**.
#### 467. The BCD number for decimal 473 is
1. 111011010
2. 010001110011
3. 110001110011
4. 010011110011
Show me the answer
**Answer:** 2. 010001110011
**Explanation:**
* The BCD (Binary-Coded Decimal) representation of **473** is:\
$$4 = 0100, 7 = 0111, 3 = 0011$$\
$$473 = 010001110011$$
* Therefore, the correct answer is **010001110011**.
#### 468. Refer to Table 2-7. The word STOP in ASCII is
1. 1010011101010010011111010000
2. 1001010110110110011101010001
3. 1010010100110010011101010000
4. 1010011101010010011101100100
Show me the answer
**Answer:** 1. 1010011101010010011111010000
**Explanation:**
* The ASCII representation of the word **STOP** is:\
$$S = 1010011, T = 1010100, O = 1001111, P = 1010000$$\
$$STOP = 1010011101010010011111010000$$
* Therefore, the correct answer is **1010011101010010011111010000**.
#### 469. The number of parity bits to be added o an 8-bit word for constructing Hamming code for detection
1. 1
2. 3
3. 2
4. 4
Show me the answer
**Answer:** 4. 4
**Explanation:**
* For an 8-bit word, the number of parity bits required for Hamming code is:\
$$2^r \geq m + r + 1$$\
Where ( m = 8 ), so ( r = 4 ).
* Therefore, the correct answer is **4**.
#### 470. A 7-bit Hamming code (even parity) 001001 for a BCD digit is known to have single error the encoded BCD digit is
1. 9
2. 3
3. 5
4. 0
Show me the answer
**Answer:** 1. 9
**Explanation:**
* The 7-bit Hamming code **001001** with even parity and a single error corresponds to the BCD digit **9**.
* Therefore, the correct answer is **9**.
#### 471. When the input to an inverter is HIGH (I), the output is
1. HIGH or 1
2. HIGH or 0
3. LOW or 1
4. LOW or 0
Show me the answer
**Answer:** 4. LOW or 0
**Explanation:**
* An inverter outputs **LOW (0)** when the input is **HIGH (1)**.
* Therefore, the correct answer is **LOW or 0**.
#### 472. An inverter performs an operation known as
1. Complementation
2. Inversion
3. Assertion
4. Both answers (A) and (B)
Show me the answer
**Answer:** 4. Both answers (A) and (B)
**Explanation:**
* An inverter performs **complementation** and **inversion** of the input signal.
* Therefore, the correct answer is **Both answers (A) and (B)**.
#### 473. The output of an AND gate with inputs A, B, and C is a 1 (HIGH) when
1. A = 1, B = 1, C = 1
2. A = 0, B = 0, C = 0
3. A = 1, B = 0, C = 1
4. Only answers (A) and (C)
Show me the answer
**Answer:** 4. Only answers (A) and (C)
**Explanation:**
* The output of an AND gate is **1 (HIGH)** only when **all inputs are 1**.
* Therefore, the correct answer is **Only answers (A) and (C)**.
#### 474. A pulse is applied to each input of a 2-input NAND gate. One pulse goes HIGH at t = 0 and goes back LOW at t = 1ms. The other pulse goes HIGH at t = 0.8 ms and goes back LOW at t = 3ms. The output pulse can be described as follows:
1. It goes LOW at t = 0 and back HIGH at t = 3ms
2. It goes LOW at t = 0.8 ms and back HIGH at t = 3 ms
3. It goes LOW at t = 0.8 ms and back HIGH at t = 1 ms
4. It goes LOW at t = 0.8 ms and back LOW at t = 1 ms
Show me the answer
**Answer:** 3. It goes LOW at t = 0.8 ms and back HIGH at t = 1 ms
**Explanation:**
* The output of a NAND gate goes **LOW** when **both inputs are HIGH**.
* Therefore, the output goes **LOW at t = 0.8 ms** and back **HIGH at t = 1 ms**.
* The correct answer is **It goes LOW at t = 0.8 ms and back HIGH at t = 1 ms**.
#### 475. A pulse is applied to each input of a 2-input NOR gate, one pulse goes HIGH at t = 0 and goes back LOW at t = 1ms. The other pulse goes HIGH at t = 0.8 ms and goes back LOW at t = 3 ms. The output pulse can be described as follows:
1. It goes LOW at t = 0 and back HIGH at t = 3 ms.
2. It goes LOW at t = 0.8 ms and back HIGH at t = 3ms
3. It goes LOW at t = 0.8 ms and back HIGH at t = 1 ms
4. It goes HIGH at t = 0.8 ms and back LOW at t = 1 ms
Show me the answer
**Answer:** 1. It goes LOW at t = 0 and back HIGH at t = 3 ms.
**Explanation:**
* The output of a NOR gate goes **LOW** when **any input is HIGH**.
* Therefore, the output goes **LOW at t = 0** and back **HIGH at t = 3 ms**.
* The correct answer is **It goes LOW at t = 0 and back HIGH at t = 3 ms**.
#### 476. A pulse is applied to each input of an exclusive-OR gate. One pulse goes HIGH at t = 0 and goes back LOW at t = 1 ms. The other pulse goes HIGH at t = 0.8 ms and goes back LOW at t = 3 ms. The output pulse can be described as follows:
1. It goes HIGH at t = 0 and back LOW at t = 3 ms
2. It goes HIGH at t = 0 and back LOW at t = 0.8ms
3. It goes HIGH at t = .1 ms and back LOW at t = 3ms
4. Both answers (B) and (C)
Show me the answer
**Answer:** 4. Both answers (B) and (C)
**Explanation:**
* The output of an XOR gate goes **HIGH** when **inputs are different**.
* Therefore, the output goes **HIGH at t = 0** and back **LOW at t = 0.8 ms**, and again **HIGH at t = 1 ms** and back **LOW at t = 3 ms**.
* The correct answer is **Both answers (B) and (C)**.
#### 477. For and AND gate
1. All LOW input produce a HIGH output
2. Output is HIGH if and only if all inputs are HIGH
3. Output is LOW if and only if all inputs are HIGH
4. Output is LOW if and only if all inputs are LOW
Show me the answer
**Answer:** 2. Output is HIGH if and only if all inputs are HIGH
**Explanation:**
* The output of an AND gate is **HIGH** only when **all inputs are HIGH**.
* Therefore, the correct answer is **Output is HIGH if and only if all inputs are HIGH**.
#### 478. The output of a gate is LOW when atleast one of its inputs is HIGH. This is true for
1. AND
2. OR
3. NAND
4. NOR
Show me the answer
**Answer:** 4. NOR
**Explanation:**
* The output of a **NOR gate** is **LOW** when **at least one input is HIGH**.
* Therefore, the correct answer is **NOR**.
#### 479. The output of a gate is LOW when atleast one of its inputs is LOW. It is true for
1. AND
2. NAND
3. OR
4. NOR
Show me the answer
**Answer:** 1. AND
**Explanation:**
* The output of an **AND gate** is **LOW** when **at least one input is LOW**.
* Therefore, the correct answer is **AND**.
#### 480. The output of a gate is HIGH when atleast one of its inputs is LOW. It is true for
1. XOR
2. NOR
3. NAND
4. OR
Show me the answer
**Answer:** 3. NAND
**Explanation:**
* The output of a **NAND gate** is **HIGH** when **at least one input is LOW**.
* Therefore, the correct answer is **NAND**.
#### 481. The output of a gate is HIGH when atleast one of its inputs is HIGH. It is true for
1. NAND
2. OR
3. AND
4. XOR
Show me the answer
**Answer:** 2. OR
**Explanation:**
* The output of an **OR gate** is **HIGH** when **at least one input is HIGH**.
* Therefore, the correct answer is **OR**.
#### 482. The output of a gate is HIGH if and only if all its inputs are HIGH. It is true for
1. XOR
2. OR
3. AND
4. NAND
Show me the answer
**Answer:** 3. AND
**Explanation:**
* The output of an **AND gate** is **HIGH** only when **all inputs are HIGH**.
* Therefore, the correct answer is **AND**.
#### 483. The output of a gate is LOW if and only if all its inputs are HIGH. It is true for
1. AND
2. NOR
3. XNOR
4. NAND
Show me the answer
**Answer:** 4. NAND
**Explanation:**
* The output of a **NAND gate** is **LOW** only when **all inputs are HIGH**.
* Therefore, the correct answer is **NAND**.
#### 484. The output of a gate is HIGH if and only if all its inputs are LOW. It is true for
1. NOR
2. NAND
3. XOR
4. XNOR
Show me the answer
**Answer:** 1. NOR
**Explanation:**
* The output of a **NOR gate** is **HIGH** only when **all inputs are LOW**.
* Therefore, the correct answer is **NOR**.
#### 485. The output of a gate is LOW if and only if all its inputs are LOW. It is true for
1. XOR
2. OR
3. AND
4. NOR
Show me the answer
**Answer:** 2. OR
**Explanation:**
* The output of an **OR gate** is **LOW** only when **all inputs are LOW**.
* Therefore, the correct answer is **OR**.
#### 486. The output of a 2-input gates is 1 if and only if its inputs are unequal. It is true for
1. OR
2. XNOR
3. XOR
4. NOR
Show me the answer
**Answer:** 3. XOR
**Explanation:**
* The output of an **XOR gate** is **1** only when **inputs are unequal**.
* Therefore, the correct answer is **XOR**.
#### 487. The output of a 2-input gates is 0 if and only if its inputs are unequal. It is true for
1. XNOR
2. NOR
3. AND
4. NAND
Show me the answer
**Answer:** 1. XNOR
**Explanation:**
* The output of an **XNOR gate** is **0** only when **inputs are unequal**.
* Therefore, the correct answer is **XNOR**.
#### 488. The output of a 2-input gates is 1 if and only if its inputs are equal. It is true for
1. AND
2. OR
3. XOR
4. XNOR
Show me the answer
**Answer:** 4. XNOR
**Explanation:**
* The output of an **XNOR gate** is **1** only when **inputs are equal**.
* Therefore, the correct answer is **XNOR**.
#### 489. The output of a 2-input gates is 0 if and only if its inputs are unequal. It is true for
1. AND
2. OR
3. XOR
4. NOR
Show me the answer
**Answer:** 3. XOR
**Explanation:**
* The output of an **XOR gate** is **0** only when **inputs are unequal**.
* Therefore, the correct answer is **XOR**.
#### 490. The most suitable gate for comparing two bits is
1. AND
2. NAND
3. OR
4. XOR
Show me the answer
**Answer:** 4. XOR
**Explanation:**
* The **XOR gate** is the most suitable for comparing two bits, as it outputs **1** when the bits are different.
* Therefore, the correct answer is **XOR**.
#### 491. Which of the following gates can be used as an inverter?
1. AND
2. XOR
3. OR
4. None of the above
Show me the answer
**Answer:** 2. XOR
**Explanation:**
* An **XOR gate** can be used as an inverter by connecting one input to a constant HIGH (1).
* Therefore, the correct answer is **XOR**.
#### 492. Which of the following gates cannot be used as an inverter?
1. NAND
2. NOR
3. AND
4. XNOR
Show me the answer
**Answer:** 3. AND
**Explanation:**
* An **AND gate** cannot be used as an inverter because it does not invert the input signal.
* Therefore, the correct answer is **AND**.
#### 493. The maximum number of 3-inputs gates in a 16-pin IC will be
1. 2
2. 4
3. 3
4. 5
Show me the answer
**Answer:** 3. 3
**Explanation:**
* A 16-pin IC can accommodate a maximum of **3** 3-input gates, as each gate requires 3 input pins and 1 output pin.
* Therefore, the correct answer is **3**.
#### 494. A quality having continuous values is
1. A digital quantity
2. A binary quantity
3. An analog quantity
4. A natural quantity
Show me the answer
**Answer:** 3. An analog quantity
**Explanation:**
* A quantity having **continuous values** is referred to as an **analog quantity**.
* Therefore, the correct answer is **An analog quantity**.
#### 495. The term bit means
1. A small amount of data
2. Binary digit
3. A 1 or a 0
4. Both answers (B) and (C)
Show me the answer
**Answer:** 4. Both answers (B) and (C)
**Explanation:**
* The term **bit** stands for **binary digit**, which can be either **1 or 0**.
* Therefore, the correct answer is **Both answers (B) and (C)**.
#### 496. The time interval on the leading edge of a pulse between 10% and 90% of the amplitude is the
1. Rise time
2. Pulse width
3. Fall time
4. Period
Show me the answer
**Answer:** 1. Rise time
**Explanation:**
* The time interval between **10% and 90%** of the amplitude on the leading edge of a pulse is known as the **rise time**.
* Therefore, the correct answer is **Rise time**.
#### 497. A pulse in a certain waveform occurs every 10 ms. The frequency is
1. 1 kHz
2. 100Hz
3. 1 Hz
4. 10 Hz
Show me the answer
**Answer:** 2. 100Hz
**Explanation:**
* The frequency of a pulse occurring every **10 ms** is:\
$$f = \frac{1}{T} = \frac{1}{10 \times 10^{-3}} = 100 \text{ Hz}$$
* Therefore, the correct answer is **100Hz**.
#### 498. In a certain digital waveform, the period is twice the pulse width. The duty cycle is
1. 100%
2. 50%
3. 200%
4. 150%
Show me the answer
**Answer:** 2. 50%
**Explanation:**
* If the period is twice the pulse width, the duty cycle is:\
$$\text{Duty Cycle} = \frac{\text{Pulse Width}}{\text{Period}} \times 100 = \frac{1}{2} \times 100 = 50%$$
* Therefore, the correct answer is **50%**.
#### 499. An inverter
1. Performs the NOT operation
2. Changes a LOW to a HIGH
3. Changes a HIGH to a LOW
4. Does all of the above
Show me the answer
**Answer:** 4. Does all of the above
**Explanation:**
* An inverter **performs the NOT operation**, **changes a LOW to a HIGH**, and **changes a HIGH to a LOW**.
* Therefore, the correct answer is **Does all of the above**.
#### 500. The output of an AND gate is HIGH when
1. Any input is HIGH
2. No inputs are HIGH
3. All inputs are HIGH
4. Both answers (A) and (C)
Show me the answer
**Answer:** 3. All inputs are HIGH
**Explanation:**
* The output of an AND gate is **HIGH** only when **all inputs are HIGH**.
* Therefore, the correct answer is **All inputs are HIGH**.