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451. (2×101+8×100)( 2 \times 10^1 + 8 \times 10^0 )(2×101+8×100) equal to

  1. 10

  2. 2.8

  3. 280

  4. 28

Show me the answer

Answer: 4. 28

Explanation:

  • The expression evaluates to: 2×10+8×1=20+8=282 \times 10 + 8 \times 1 = 20 + 8 = 282×10+8×1=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: 1×23+1×22+0×21+1×20=8+4+0+1=131 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 131×23+1×22+0×21+1×20=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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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.

  • 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:

  • Therefore, the correct answer is 5471241s.

465. The binary number 10001101010001101111 can be written in hexadecimal as

Show me the answer

Explanation:

  1. 1111011110101001

  2. 1111111010110001

  3. 1110111110101001

  4. 1111011010101001

Show me the answer

Answer: 1. 1111011110101001

Explanation:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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:

  • 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.

The binary number 11011101 converts to decimal as: 1×27+1×26+0×25+1×24+1×23+1×22+0×21+1×20=128+64+0+16+8+4+0+1=2211 \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 = 2211×27+1×26+0×25+1×24+1×23+1×22+0×21+1×20=128+64+0+16+8+4+0+1=221

The decimal number 17 converts to binary as: 17=16+1=24+20=1000117 = 16 + 1 = 2^4 + 2^0 = 1000117=16+1=24+20=10001

The decimal number 175 converts to binary as: 175=128+32+8+4+2+1=10101111175 = 128 + 32 + 8 + 4 + 2 + 1 = 10101111175=128+32+8+4+2+1=10101111

The sum of 11010 and 01111 is: 11010+01111=10100111010 + 01111 = 10100111010+01111=101001

The difference of 110 and 010 is: 110−010=100110 - 010 = 100110−010=100

The 1's complement of 10111001 is obtained by flipping all the bits: 10111001→0100011010111001 \rightarrow 0100011010111001→01000110

The 2's complement of 11001000 is obtained by flipping all the bits and adding 1: 11001000→00110111+1=0011100011001000 \rightarrow 00110111 + 1 = 0011100011001000→00110111+1=00111000

The 2's complement representation of -34 is obtained by taking the 2's complement of the positive value: 34=00100010→11011101+1=1101111034 = 00100010 \rightarrow 11011101 + 1 = 1101111034=00100010→11011101+1=11011110

The 2's complement representation of +122 is the same as its binary representation: 122=01111010122 = 01111010122=01111010

The decimal value is calculated as: −128+16+2+1=−109-128 + 16 + 2 + 1 = -109−128+16+2+1=−109

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=5471241s1011 \ 0011 \ 1001 \ 0101 \ 0001 = B \ 3 \ 9 \ 5 \ 1 = 5471241s1011 0011 1001 0101 0001=B 3 9 5 1=5471241s

AD467(16)AD467(_{16})AD467(16​)

8D46F(16)8D46F(_{16})8D46F(16​)

8C46F(16)8C46F(_{16})8C46F(16​)

AE46F(16)AE46F(_{16})AE46F(16​)

Answer: 2. 8D46F(16)8D46F(_{16})8D46F(16​)

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=8D46F161000 \ 1101 \ 0100 \ 0110 \ 1111 = 8 \ D \ 4 \ 6 \ F = 8D46F_{16}1000 1101 0100 0110 1111=8 D 4 6 F=8D46F16​

Therefore, the correct answer is 8D46F(16)8D46F(_{16})8D46F(16​).

466. The binary number for F7A9(16)F7A9(_{16})F7A9(16​) is

The hexadecimal number F7A9 converts to binary as: F=1111,7=0111,A=1010,9=1001F = 1111, 7 = 0111, A = 1010, 9 = 1001F=1111,7=0111,A=1010,9=1001 F7A9=1111011110101001F7A9 = 1111011110101001F7A9=1111011110101001

The BCD (Binary-Coded Decimal) representation of 473 is: 4=0100,7=0111,3=00114 = 0100, 7 = 0111, 3 = 00114=0100,7=0111,3=0011 473=010001110011473 = 010001110011473=010001110011

The ASCII representation of the word STOP is: S=1010011,T=1010100,O=1001111,P=1010000S = 1010011, T = 1010100, O = 1001111, P = 1010000S=1010011,T=1010100,O=1001111,P=1010000 STOP=1010011101010010011111010000STOP = 1010011101010010011111010000STOP=1010011101010010011111010000

For an 8-bit word, the number of parity bits required for Hamming code is: 2r≥m+r+12^r \geq m + r + 12r≥m+r+1 Where ( m = 8 ), so ( r = 4 ).

The frequency of a pulse occurring every 10 ms is: f=1T=110×10−3=100 Hzf = \frac{1}{T} = \frac{1}{10 \times 10^{-3}} = 100 \text{ Hz}f=T1​=10×10−31​=100 Hz

If the period is twice the pulse width, the duty cycle is: Duty Cycle=Pulse WidthPeriod×100=12×100=50%\text{Duty Cycle} = \frac{\text{Pulse Width}}{\text{Period}} \times 100 = \frac{1}{2} \times 100 = 50\%Duty Cycle=PeriodPulse Width​×100=21​×100=50%