(A) x’y’z’ + wx’y’ + w’y

(B) xyz + wx’y’ + w’y

(C) x’y’z’ + wy’ + w’yz + wz

(D) x’yz’ + x’y’ + w’y + xz

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**ANSWER: x’y’z’ + wx’y’ + w’y**

(A) 01, 10, 01, 10

(B) 11, 11, 00, 10

(C) 11, 10, 01, 00

(D) 00, 10, 10, 10

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**ANSWER: 11, 10, 01, 00**

(A) 1’s compliment of the input number is computed

(B) 2’s compliment of the input number is computed

(C) The input number is decremented

(D) The input number is increment

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**ANSWER: 2’s compliment of the input number is computed**

In the initial k bits of the input, number of 0’s is represented by zk and number of 1’s is represented by nk.

(zk + nk = k). The output of the circuit is 00 only if one of the conditions given below is correct.

Case 1. zk – nk = 2, the output at the kth and all subsequent clock ticks is 10.

Case 2. nk – zk = 2, the output at the kth and all subsequent clock ticks is 01.

In the state transition graph of the above circuit the minimum number of states required is

(A) 6

(B) 5

(C) 8

(D) 7

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**ANSWER: 7**

(A) DE5

(B) DBC

(C) 9D6

(D) ABC

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**ANSWER: DE5**

(A) (9, 10)

(B) (10, 9)

(C) (11, 11)

(D) (12, 13)

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**ANSWER: (9, 10)**

(A) 11000110

(B) 00101011

(C) 01011101

(D) 10010010

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**ANSWER: 00101011**

(A) f is independent of X.

(B) f is independent of Y.

(C) f is independent of Z.

(D) None of X, Y, Z is redundant.

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**ANSWER: None of X, Y, Z is redundant.**

(A) XY + YW

(B) (W + X)(W + Y)(X + Y)

(C) (W + Y) X

(D) None of the above

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**ANSWER: None of the above**

X: = A + B

Y: = A + C

X: = X + C

Y: = Y + B

What is the value of X and Y?

(A) X = Y = 0.0

(B) X = 0.0, Y = 1.0

(C) X = 1.0, Y = 0.0

(D) X = Y = 1.0

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**ANSWER: X = Y = 1.0 **

(A) (C + D)(C’ + D)(A + B)

(B) D(C + A)

(C) A’B + CD

(D) AD + A’B

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**ANSWER: D(C + A)**

(A) m2 + m3 + m0 + m6

(B) m2 + m0 + m3 + m1

(C) m2 + m4 + m6 + m1

(D) m0 + m4 + m6 + m1

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**ANSWER: m2 + m4 + m6 + m1**

(A) 1

(B) 2

(C) 3

(D) 4

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**ANSWER: 3**

(A) 0 – 100 + 100 - 1

(B) 0 – 100 + 1000

(C) 00 – 10 +100 - 1

(D) 0 – 1 + 100 - 1

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**ANSWER: 0 – 100 + 100 - 1**

If the input is valid then only V = 1. The truth table represents

(A) Demultiplexer

(B) Multiplexer

(C) Decoder

(D) Priority encoder

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**ANSWER: Priority encoder**

(A) 011111……..11110

(B) 101010………1010

(C) 100000…..001

(D) 111111…….1111

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**ANSWER: 101010………1010**

(A) 1’s compliment

(B) Sign magnitude

(C) Both A and B

(D) Neither A nor B

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**ANSWER: Both A and B**

(A) 0, 0

(B) 0, 1

(C) 1, 1

(D) 1, 0

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**ANSWER: 1, 0**

(A) AND gate and NOT gate

(B) AND gate and XOR gate

(C) 2 to 1 multiplexer

(D) XOR gate and NOT gate

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**ANSWER: AND gate and XOR gate**

(A) Fraction bits of 100….000 and exponent value of 0

(B) Fraction bits of 000….000 and exponent value of -1

(C) Fraction bits of 000….000 and exponent value of 0

(D) No exact representation

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**ANSWER: Fraction bits of 100….000 and exponent value of 0**

(A) f = x1 + x2

(B) f = x1x2 + x1’x2’

(C) f = x1 + x2’

(D) f = x1’ + x2

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**ANSWER: f = x1x2 + x1’x2’**

x + y + z = 1

xy = 0

xz + w = 1

xy + z’w’ = 0

The values of x, y, z and w are respectively

(A) 1 0 1 1

(B) 1 0 0 0

(C) 1 1 0 0

(D) 0 1 1 0

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**ANSWER: 1 0 1 1**

(A) x + y’

(B) x’ + z

(C) xyz

(D) None of the above

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**ANSWER: None of the above**

(A) 3 time unit

(B) 15 time unit

(C) 10 time unit

(D) 6 time unit

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**ANSWER: 6 time unit**

(A) 2

(B) 3

(C) 4

(D) 5

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**ANSWER: 3**

(A) 1100 0100

(B) 1010 0010

(C) 1100 1100

(D) 1000 1111

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**ANSWER: 1100 0100**

(A) ac’ and bc’

(B) a’c only

(C) a’c and b’c

(D) a’c and ac’

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**ANSWER: a’c and ac’**

(A) Signed 16bit integer addition

(B) Arithmetic left shift

(C) Converting a signed integer from one size to another

(D) Floating point multiplication

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**ANSWER: Floating point multiplication**

For questions 29 and 30 refer to the data given below:

The floating-point format is given below whose mantissa is a pure fraction in sign-magnitude form:

(A) 4D0D

(B) 4D3D

(C) 0D24

(D) 0D4D

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**ANSWER: 4D3D**

The mantissa has an implicit preceding the binary (radix) point. Assuming that only 0’s are padded in while shifting a field the normalized representation of the above number 0.239 x

2

(A) D0D0

(B) 4AE8

(C) 49D0

(D) D048

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**ANSWER: 4AE8**

(A) b3 (a3a2a1a0) = ∑ (0, 1, 6, 7, 10, 13, 14, 15)

(B) b2 (a3a2a1a0) = ∑ (2, 4, 5, 6, 10, 13, 14, 15)

(C) b1 (a3a2a1a0) = ∑ (4, 9, 10, 11, 12, 13, 14, 15)

(D) b0 (a3a2a1a0) = ∑ (1, 2, 3, 6, 10, 13, 14, 15)

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**ANSWER: b1 (a3a2a1a0) = ∑ (4, 9, 10, 11, 12, 13, 14, 15)**

(A) 10001

(B) 11100

(C) 11010

(D) 11111

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**ANSWER: 10001**

(A) X = Y * Z

(B) Y = X * Z

(C) X * Y * Z = 1

(D) All of the above

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**ANSWER: All of the above **

(A) 2

(B) 2

(C) 2

(D) 2

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**ANSWER: 2 ^{n-1} line to 1 line**

(A) It cannot be represented precisely in binary

(B) It is equivalent to binary value 0.001

(C) It is equivalent to binary value 0.01

(D) It is equivalent to binary value 0.1

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**ANSWER: It is equivalent to binary value 0.01**

(A) 10, 12

(B) 8, 11

(C) 9, 13

(D) None of the above

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**ANSWER: 8, 11**

(A) BC’D’ + A’C’D + AB’D

(B) ABC’ + BCD’ + AB’D

(C) A’B’C’ + A’CD + AC’D’

(D) ACD’ + A’B’C + ABD

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**ANSWER: BC’D’ + A’C’D + AB’D**

(113 + - 111) + 7.51

113 + (-111 + 7.51)

The result will be

(A) 10.0 and 9.51

(B) 10.0 and 10.0

(C) 9.51 and 10.0

(D) 9.51 and 9.51

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**ANSWER: 9.51 and 10.0**

(A) Q = 1, Q’ = 1

(B) Q = 0, Q’ = 1

(C) Q = 1, Q’ = 0

(D) Indeterminate states

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**ANSWER: Indeterminate states**

(A) 11011011

(B) 11100111

(C) 11010111

(D) 11100100

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**ANSWER: 11100111**

(A) x’ + y’

(B) x + y’

(C) x’ + y

(D) x + y

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**ANSWER: x’ + y**

(A) Independent of two variables

(B) Independent of one variable

(C) Independent of three variables

(D) Dependent on all the variables

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**ANSWER: Independent of two variables**

(A) 108

(B) 32F

(C) D78

(D) 1AF

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**ANSWER: 1AF**

(A) (1217)

(B) (2424)

(C) (028F)

(D) (0B25)

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**ANSWER: (028F) _{16}**

(A) n

(B) 2n

(C) n

(D) None of the above

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**ANSWER: None of the above**

(A) -(2

(B) -2

(C) -(2

(D) -2

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**ANSWER: -2 ^{n-1} to (2^{n-1 }–1)**

(A) The special value +0

(B) The normalized value 2

(C) The normalized value 2

(D) The normalized value +0

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**ANSWER: The special value +0**

(A) 11

(B) 8

(C) 15

(D) 9

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**ANSWER: 9**

11

(A) Any value >2

(B) Any value <2

(C) Decimal 10

(D) None of the above

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**ANSWER: Any value >2**

(A) (1870)

(B) (C3D8)

(C) (ACD8)

(D) (C103)

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**ANSWER: (C3D8) _{16}**

(A) 5

(B) 4

(C) 3

(D) 2

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**ANSWER: 3**

(A) 121

(B) 5

(C) 258

(D) 9

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**ANSWER: 9**