Number-Systems

Question 1

In 16-bit 2's complement representation, the decimal number -28 is:

A
1111 1111 1110 0100
B
1111 1111 0001 1100
C
0000 0000 1110 0100
D
1000 0000 1110 0100
       Digital-Logic-Design       Number-Systems       GATE 2019
Question 1 Explanation: 
+28 = 0000 0000 0001 1100

1’s complement = 1111 1111 1110 0011
2’s complement = 1’s complement + 1

2’s complement = 1111 1111 1110 0100 = (-28)
Question 2

Two numbers are chosen independently and uniformly at random from the set {1, 2, ..., 13}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is ______.

A
0.502
B
0.461
C
0.402
D
0.561
       Digital-Logic-Design       Number-Systems       GATE 2019
Question 2 Explanation: 
Correct answer is 0.502
1 - 0001
2 - 0010
3 - 0011
4 - 0100
5 - 0101
6 - 0110
7 - 0111
8 - 1000
9 - 1001
10 - 1010
11 - 1011
12 - 1100
13 - 1101
The probability that their 4-bit binary representations have the same most significant bit is
= P(MSB is 0) + P(MSB is 1)
= (7×7)/(13×13) + (6×6)/(13×13)
= (49+36)/169
= 85/169
= 0.502
Question 3

Consider Z = X - Y, where X, Y and Z are all in sign-magnitude form. X and Y are each represented in n bits. To avoid overflow, the representation of Z would require a minimum of:

A
n bits
B
n + 2 bits
C
n - 1 bits
D
n + 1 bits
       Digital-Logic-Design       Number-Systems       GATE 2019
Question 3 Explanation: 
In case of addition of two numbers with the same sign, there is a chance of overflow.
To store overflow/carry bit there should be extra space to accommodate it.
Hence, Z should be n+1 bits.
Question 4

Consider the unsigned 8-bit fixed point binary number representation below,

    b7b6b5b4b3 ⋅ b2b1b0

where the position of the binary point is between b3 and b2 . Assume b7 is the most significant bit. Some of the decimal numbers listed below cannot be represented exactly in the above representation:

    (i) 31.500    (ii) 0.875    (iii) 12.100    (iv) 3.001

Which one of the following statements is true?

A
None of (i), (ii), (iii), (iv) can be exactly represented
B
Only (ii) cannot be exactly represented
C
Only (iii) and (iv) cannot be exactly represented
D
Only (i) and (ii) cannot be exactly represented
       Digital-Logic-Design       Number-Systems       GATE 2018
Question 4 Explanation: 
(i) (31.5)10 = (11111.100)2 = 24 + 23 + 22 + 21 + 20 + 2-1
= 16 + 8 + 4 + 2 + 1 + 0.5
= (31.5)10
(ii) (0.875)10 = (00000.111)2
= 2-1 + 2-2 + 2-3
= 0.5 + 0.25 + 0.125
= (0.875)10
(iii) (12.100)10
It is not possible to represent (12.100)10
(iv) (3.001)10 It is not possible to represent (3.001)10
Question 5

The n-bit fixed-point representation of an unsigned real number X uses f bits for the fraction part. Let i = n-f. The range of decimal values for X in this representation is

A
2-f to 2i
B
2-f to (2i - 2-f)
C
0 to 2i
D
0 to (2i - 2-f )
       Digital-Logic-Design       Number-Systems       GATE 2017 [Set-1]
Question 5 Explanation: 
Size of the fixed point number → n-bits
Number of bits in fraction part → f-bits
Number of bits in integer part → (n – f) bits

Minimum value:
000…0.000…0 = 0
Maximum value:

= (2 n-f - 1) + (1 - 2 -f
= (2n-f - 2 -f)
= (2i - 2 -f )
Question 6

The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is

A
136251
B
736251
C
571247
D
136252
       Digital-Logic-Design       Number-Systems       GATE 2017 [Set-2]
Question 6 Explanation: 
X = (BCA9)16
Each hexadecimal digit is equal to a 4-bit binary number. So convert
X = (BCA9)16 to binary

Divide the binary data into groups 3 bits each because each octal digit is represented by 3-bit binary number.
X = (001 011 110 010 101 001)2
Note: Two zeroes added at host significant position to make number bits of a multiple of 3 (16 + 2 = 18)
X = (136251)8
Question 7

Given the following binary number in 32-bit (single precision) IEEE-754 format:

00111110011011010000000000000000

The decimal value closest to this floating-point number is

A
1.45 × 101
B
1.45 × 10-1
C
2.27 × 10-1
D
2.27 × 101
       Digital-Logic-Design       Number-Systems       GATE 2017 [Set-2]
Question 7 Explanation: 

For single-precision floating-point representation decimal value is equal to (-1)5 × 1.M × 2(E-127)
S = 0
E = (01111100)2 = (124).
So E – 127 = - 3
1.M = 1.11011010…0
= 20 + 2(-1) + 2(-1) + 2(-4) + 2(-5) + 2(-7)
= 1+0.5+0.25+0.06+0.03+0.007
≈ 1.847
(-1)5 × 1.M × 2(E-127)
= -10 × 1.847 × 2-3
≈ 0.231
≈ 2.3 × 10-1
Question 8

Consider a quadratic equation x2 - 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b=________.

A
8
B
9
C
10
D
11
       Digital-Logic-Design       Number-Systems       GATE 2017 [Set-2]
Question 8 Explanation: 
x2 - 13x + 36 = 0 ⇾(1)
Generally if a, b are roots.
(x - a)(x - b) = 0
x2 - (a + b)x + ab = 0
Given that x=5, x=6 are roots of (1)
So, a + b = 13
ab=36 (with same base ‘b’)
i.e., (5)b + (6)b = (13)b
Convert them into decimal value
5b = 510
610 = 610
13b = b+3
11 = b+3
b = 8
Now check with ab = 36
5b × 6b = 36b
Convert them into decimals
5b × 6b = (b×3) + 610
30 = b × 3 + 6
24 = b × 3
b = 8
∴ The required base = 8
Question 9

Consider a binary code that consists of only four valid code words as given below:

00000, 01011, 10101, 11110

Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are

A
p=3 and q=1
B
p=3 and q=2
C
p=4 and q=1
D
p=4 and q=2
       Digital-Logic-Design       Number-Systems       GATE 2017 [Set-2]
Question 9 Explanation: 
Hamming distance of a code is minimum distance between any two code words.
Minimum Distance = p = 3

Error bits that can be corrected = (p-1)/2 = (3-1)/2 = 1
∴ p=3 and q=1
Question 10

The 16-bit 2’s complement representation of an integer is 1111 1111 1111 0101; its decimal representation is __________.

A
-11
B
-12
C
-13
D
-14
       Digital-Logic-Design       Number-Systems       GATE 2016 [Set-1]
Question 10 Explanation: 
Given number is 1111 1111 1111 0101.
It is a negative number because MSB is 1.
Magnitude of 1111 1111 1111 0101 is 2’s complement of 1111 1111 1111 0101.
1111 1111 1111 0101
0000 0000 0000 1010 : 1’s Complement
0000 0000 0000 1011 : 2’s complement
= (11)10
Hence, 1111 1111 1111 0101 = -11
Question 11

Let X be the number of distinct 16-bit integers in 2’s complement representation. Let Y be the number of distinct 16-bit integers in sign magnitude representation.

Then X-Y is _________.

A
1
B
2
C
3
D
4
       Digital-Logic-Design       Number-Systems       GATE 2016 [Set-2]
Question 11 Explanation: 
X = 216
Since range is - 215 to 215 - 1
Y = 216 - 1
Here, +0 and -0 are represented separately.
X - Y = 216 - (216 - 1)
= 1
Question 12

The base (or radix) of the number system such that the following equation holds is_________.

        312/20 = 13.1     
A
5
B
6
C
7
D
8
       Digital-Logic-Design       Number-Systems       GATE 2014 [Set-1]
Question 12 Explanation: 
Let base of the number system is r.
(3r2 + r + 2) / 2r= (r+3+1/r)
(3r2 + r + 2) / 2r= (r2+3r+1) / r
(3r2 + r + 2) = (2r2+6r+2)
r2 -5r = 0
Therefor r = 5
Question 13

Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is __________.

A
3
B
5
C
6
D
7
       Digital-Logic-Design       Number-Systems       GATE 2014 [Set-2]
Question 13 Explanation: 
First we have to fullfill all the conditios,
(123)5 = (x8)y
In R.H.S. since y is base so y should be greater than x and 8, i.e.,
y > x
y > 8
Now, to solve let's change all the above bases number into base 10 number,
52 × 1 +2 × 5 + 3 = y × x + 8
38 = xy + 8
xy = 30
⇒ yx = 30
So the possible combinations are
(1,30), (2,15), (3,10), (5,6)
But we will reject (5,6) because it violates the condition (y > 8).
So, total solutions possible is 3.
Question 14

The value of a float type variable is represented using the single-precision 32-bit floating point format of IEEE-754 standard that uses 1 bit for sign, 8 bits for biased exponent and 23 bits for mantissa. A float type variable X is assigned the decimal value of −14.25. The representation of X in hexadecimal notation is

A
C1640000H
B
416C0000H
C
41640000H
D
C16C0000H
       Computer-Organization       Number-Systems       GATE 2014 [Set-2]
Question 14 Explanation: 
Given number is a negative number. So sign bit = 1
(14.25)10 = 1110.01000
= 1.11001000 x 23
23 bit Mantissa = 11001000000000000000000
Biased Exponent = exponent + bias
= 3 + 127 = 130 = 1000 0010
(-14.25) in 32-bit IEEE-754 floating point representation is
1 10000010 11001000000000000000000
= 1100 0001 0110 0100 0000 0000 000 0000
= (C 1 6 4 0 0 0 0)16
Question 15

Consider the following minterm expression for F:

  F(P,Q,R,S) = Σ0,2,5,7,8,10,13,15  

The minterms 2, 7, 8 and 13 are 'do not care' terms. The minimal sum-of-products form for F is:

A
B
C
D
       Digital-Logic-Design       Number-Systems       GATE 2014 [Set-3]
Question 15 Explanation: 
Question 16

Let ⊕ denote the Exclusive OR (XOR) operation. Let ‘1’ and ‘0’ denote the binary constants. Consider the following Boolean expression for F over two variables P and Q:

     F(P,Q) = ((1⊕P)⊕(P⊕Q))⊕((P⊕Q)⊕(Q⊕0)) 

The equivalent expression for F is

A
P+Q
B
C
P⨁Q
D
       Digital-Logic-Design       Number-Systems       GATE 2014 [Set-3]
Question 16 Explanation: 
((1 ⊕ P) ⊕ (P ⊕ Q)) ⊕ ((P ⊕ Q) ⊕ (Q ⊕ 0))
⊕ is associative i.e P ⊕ (Q ⊕ R) = (P⊕Q) ⊕ R.
P ⊕ P = 0, 1 ⊕ P = P’ and 0 ⊕ Q = Q
(1 ⊕ P) ⊕ ((P ⊕ Q) ⊕ (P ⊕ Q)) ⊕ (Q ⊕ 0)
= P’⊕ (0) ⊕ Q
= P’ ⊕ Q
= (P ⊕ Q)’
Question 17

The smallest integer that can be represented by an 8-bit number in 2’s complement form is

A
-256
B
-128
C
-127
D
0
       Digital-Logic-Design       Number-Systems       GATE 2013
Question 17 Explanation: 
The range of 8-bit signed numbers representable is – 2n-1 to 2n-1 -1.
The smallest 8-bit 2’s complement number is 1000 0000.
MSB is 1. So it is a negative number.
To know the magnitude again take 2’s complement of 1000 0000.
1000 0000
0111 1111 ← 1’s complement
1000 0000 ← 2’s complement (1’s complement +1)
= 128
-128 is 1000 0000 in 2’s complement representation.
Question 18

Which one of the following expressions does NOT represent exclusive NOR of x and y?

A
xy+x'y'
B
x⊕y'
C
x'⊕y
D
x'⊕y'
       Digital-Logic-Design       Number-Systems       GATE 2013
Question 18 Explanation: 
x ⊕ y = x’y + xy’
x’ ⊕ y’ = xy’ + x’y = x⊕y. Hence option D is correct.
Question 19

The decimal value 0.5 in IEEE single precision floating point representation has

A
fraction bits of 000…000 and exponent value of 0
B
fraction bits of 000…000 and exponent value of −1
C
fraction bits of 100…000 and exponent value of 0
D
no exact representation
       Digital-Logic-Design       Number-Systems       GATE 2012
Question 19 Explanation: 
(0.5)10 = (1.0)2 × 2–1
So, value of the exponent = -1
and
fraction is 000…000 (Implicit representation)
Question 20

P is a 16-bit signed integer. The 2's complement representation of P is (F87B)16. The 2's complement representation of 8*P is

A
(C3D8)16
B
(187B)16
C
(F878)16
D
(987B)16
       Digital-Logic-Design       Number-Systems       GATE 2010
Question 20 Explanation: 
(F87B)16 is 2's complement representation of P.
(F87B)16=(1111 1000 0111 1011)2. (It is a negative number which is in 2's complement form)
P = 1111 1000 0111 1011 (2's complement form)
8 * P = 23* P = 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2k)
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)16.
Question 21

(1217)8 is equivalent to

A
(1217)16
B
(028F)16
C
(2297)10
D
(0B17)16
       Digital-Logic-Design       Number-Systems       GATE 2009
Question 21 Explanation: 
(1217)8 = (001 010 001 111)2
Divide the bits into groups, each containing 4 bits.
= (0010 1000 1111)2
= (28F)16
Question 22

In the IEEE floating point representation, the hexadecimal value 0×00000000 corresponds to

A
the normalized value 2 - 127
B
the normalized value 2 - 126
C
the normalized value + 0
D
the special value + 0
       Digital-Logic-Design       Number-Systems       GATE 2008
Question 22 Explanation: 
Value is ±0 if M=0 and E=0.
Question 23

Let r denote number system radix. The only value(s) of r that satisfy the equation is/are

A
decimal 10
B
decimal 11
C
decimal 10 and 11
D
any value > 2
       Digital-Logic-Design       Number-Systems       GATE 2008
Question 23 Explanation: 
√121r = 11r
(r2 + 2r + 1)1/2 = r + 1
(r + 1)2 * 1/2 = r + 1
r + 1 = r + 1 Any value of r will satisfy the above equation. But the radix should be greater than 2 because the 121 has 2. So r > 2 is correct.
Question 24

Consider numbers represented in 4-bit gray code. Let h3h2h1h0 be the gray code representation of a number n and let g3g2g1g0 be the gray code of (n+1)(modulo 16) value of the number. Which one of the following functions is correct?

A
g0(h3h2h1h0) = Σ(1,2,3,6,10,13,14,15)
B
g1(h3h2h1h0) = Σ(4,9,10,11,12,13,14,15)
C
g2(h3h2h1h0) = Σ(2,4,5,6,7,12,13,15)
D
g3(h3h2h1h0) = Σ(0,1,6,7,10,11,12,13)
       Digital-Logic-Design       Number-Systems       GATE 2006
Question 24 Explanation: 

g2(h3h2h1h0) = Σ(2,4,5,6,7,12,13,15)
Question 25

The range of integers that can be represented by an n bit 2's complement number system is:

A
- 2n-1 to (2n-1 - 1)
B
- (2n-1 - 1) to (2n-1 - 1)
C
- 2n-1 to 2n-1
D
- (2n-1 + 1) to (2n-1 - 1)
       Digital-Logic-Design       Number-Systems       GATE 2005
Question 25 Explanation: 
The maximum (positive) n bit number is 011….1 (i.e., 0 followed by n-1 ones) which is equal to 2n-1 - 1.
The smallest (negative) n bit number is 100..0 (i.e., 1 followed by n-1 zeros) which is equal to - 2n-1.
1000...00
0111...11 <- 1’s complement
1000..00 <- 2’s complement
= - 2n-1
Question 26

The hexadecimal representation of 6578 is

A
1AF
B
D78
C
D71
D
32F
       Digital-Logic-Design       Number-Systems       GATE 2005
Question 26 Explanation: 
(657)8 = (110 101 111)2
Make 3 zeros on the left side so that the number of bits is multiple of 4.
= (0001 1010 1111)2
= (1 A F)16
Question 27

Consider the following floating point format.

Mantissa is a pure fraction in sign-magnitude form.

The decimal number 0.239 × 213 has the following hexadecimal representation (without normalization and rounding off:

A
0D 24
B
0D 4D
C
4D 0D
D
4D 3D
       Digital-Logic-Design       Number-Systems       GATE 2005
Question 27 Explanation: 
Sign Bit = 0
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)2
Bias = 64. So biased exponent is 13+64 = 77= (1001101)2
0.239 × 213 = 0 1001101 00111101
= 0100 1101 0011 1101
= 4 D 3 D
Question 28

Consider the following floating point format.

Mantissa is a pure fraction in sign-magnitude form.

The normalized representation for the above format is specified as follows. The mantissa has an implicit 1 preceding the binary (radix) point. Assume that only 0's are padded in while shifting a field. The normalized representation of the above number (0.239 × 213) is

A
0A 20
B
11 34
C
4D D0
D
4A E8
       Digital-Logic-Design       Number-Systems       GATE 2005
Question 28 Explanation: 
Sign Bit = 0
Convert 0.239 to binary
0.239 * 2 = 0.478
0.478 * 2 = 0.956
0.956 * 2 = 1.912
0.912 * 2 = 1.824
0.824 * 2 = 1.648
0.648 * 2 = 1.296
0.296 * 2 = 0.512
0.512 * 2 = 1.024
Mantissa = (0. 00111101)2
0.239 × 213 = 1.11101000 x 210 <- Normalized Mantissa
Bias = 64. So biased exponent is 10+64 = 74 = (1001010)2
0.239 × 213 = 0 1001010 11101000
= 0100 1010 1110 1000
= (4 A E 8)16
Question 29

If 73x (in base-x number system) is equal to 54y (in base-y number system), the possible values of x and y are

A
8, 16
B
10, 12
C
9, 13
D
8, 11
       Digital-Logic-Design       Number-Systems       GATE 2004
Question 29 Explanation: 
(73)x = (54)y
7x+3 = 5y+4
7x-5y = 1
Only option (D) satisfies above equation.
Question 30

What is the result of evaluating the following two expressions using three-digit floating point arithmetic with rounding?

   (113. + -111.) + 7.51
   113. + (-111. + 7.51) 
A
9.51 and 10.0 respectively
B
10.0 and 9.51 respectively
C
9.51 and 9.51 respectively
D
10.0 and 10.0 respectively
       Digital-Logic-Design       Number-Systems       GATE 2004
Question 30 Explanation: 
(113. + -111.) + 7.51
= (2) + 7.51
= 9.51 (✔️)
113. + (-111. + 7.51)
= 113. + (-103.51)
= 113. + -103
= 10 (✔️)
Question 31

Let A = 1111 1010 and B = 0000 1010 be two 8-bit 2's complement numbers. Their product in 2's complement is

A
1100 0100
B
1001 1100
C
1010 0101
D
1101 0101
       Digital-Logic-Design       Number-Systems       GATE 2004
Question 31 Explanation: 
A = 1111 1010 = -610 [2's complement number]
B = 0000 1010 = 1010 [2's complement number]
A×B = -6×10 = - 6010
⇒ -6010 = 101111002
= 110000112 (1's complement)
= 110001002 (2's complement)
Question 32

Assuming all numbers are in 2's complement representation, which of the following numbers is divisible by 11111011?

A
11100111
B
11100100
C
11010111
D
11011011
       Digital-Logic-Design       Number-Systems       GATE 2003
Question 32 Explanation: 
Given: Binary numbers = 11111011
MSB bit is '1' then all numbers are negative
1's complement = 00000100
2's complement = 00000100 + 00000001 = 00000101 = -5
(A) 11100111 - (-25)10
(B) 11100100 - (-28)10
(C) 11010111 - (-41)10
(D) 11011011 - (-37)10
Answer: Option A (-25 is divisible by -5)
Question 33

The following is a scheme for floating point number representation using 16 bits.

Let s, e, and m be the numbers represented in binary in the sign, exponent, and mantissa fields respectively. Then the floating point number represented is:

What is the maximum difference between two successive real numbers representable in this system?

A
2-40
B
2-9
C
222
D
231
       Digital-Logic-Design       Number-Systems       GATE 2003
Question 33 Explanation: 
Largest gap will be in between two most largest numbers.
The largest number is 1.111111111× 262-31 = (2−2−9)×231
Second largest number is 1.111111110×262-31 = (2−2-8)×231
Difference = (2−2−9)×231 - (2−2-8)×231
= (2-8−2−9) ×231
= 2−9×231
= 222
Question 34

The decimal value 0.25

A
is equivalent to the binary value 0.1
B
is equivalent to the binary value 0.01
C
is equivalent to the binary value 0.00111…
D
cannot be represented precisely in binary
       Digital-Logic-Design       Number-Systems       GATE 2002
Question 34 Explanation: 
1st Multiplication iteration:
Multiply 0.25 by 2.
0.25×2 = 0.50 (product)
Fractional part = 0.50
Carry = 0
2nd Multiplication iteration:
Multiply 0.50 by 2.
0.50×2 = 1.00 (product)
Fractional part = 0.00
Carry = 1
The fractional part in the 2nd iteration becomes zero and so we stop the multiplication iteration.
Carry from 1st multiplication iteration becomes MSB and carry from 2nd iteration becomes LSB. So the result is 0.01.
Question 35

The 2’s complement representation of the decimal value -15 is

A
1111
B
11111
C
111111
D
10001
       Digital-Logic-Design       Number-Systems       GATE 2002
Question 35 Explanation: 
15 = 1111
-15 = 11111
1's complement = 10000
2's complement = 10001
Question 36

Sign extension is a step in

A
floating point multiplication
B
signed 16 bit integer addition
C
arithmetic left shift
D
converting a signed integer from one size to another
       Digital-Logic-Design       Number-Systems       GATE 2002
Question 36 Explanation: 
Sign extension is a step in converting a signed integer from on size to another.
Question 37

In 2’s complement addition, overflow

A
is flagged whenever there is carry from sign bit addition
B
cannot occur when a positive value is added to a negative value
C
is flagged when the carries from sign bit and previous bit match
D
None of the above
       Digital-Logic-Design       Number-Systems       GATE 2002
Question 37 Explanation: 
The left most bit of positive value is zero. And left most bit for negative value is one. The value of 0+1 becomes 1. Then overflow never occurs.
Question 38

Consider the following 32-bit floating-point representation scheme as shown in the formal below. A value is specified by 3 fields, a one bit sign field (with 0 for positive and 1 for negative values), a 24 bit fraction field (with the binary point being at the left end of the fraction bits), and a 7 bit exponent field (in excess-64 signed integer representation, with 16 being the base of exponentiation). The sign bit is the most significant bit.

(a) It is required to represent the decimal value –7.5 as a normalized floating point number in the given format. Derive the values of the various fields. Express your final answer in the hexadecimal.

(b) What is the largest values that can be represented using this format? Express your answer as the nearest power of 10.

A
Theory of Explanation is given below.
       Digital-Logic-Design       Number-Systems       GATE 2002
Question 39

The 2’s complement representation of (-539)10 in hexadecimal is

A
ABE
B
DBC
C
DE5
D
9E7
       Digital-Logic-Design       Number-Systems       GATE 2001
Question 39 Explanation: 
(539)10 = (0010 0001 1011)2
For (-539)10 = (1101 1110 0100)2
1's complement = (1101 1110 0100)2
2's complement = (1101 1110 0101)2
= (DE5)16
Question 40

Consider the circuit shown below. The output of a 2:1 Mux is given by the function (ac' + bc).

Which of the following is true?

A
f = x1' + x2
B
f = x1'x2 + x1x2'
C
f = x1x2 + x1'x2'
D
f = x1 + x2'
       Digital-Logic-Design       Number-Systems       GATE 2001
Question 40 Explanation: 
g = (a and x1′) or (b and x1)
g = (1 and x1’) or (0 and x1)
g = x1’
f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2
Question 41

The number 43 in 2’s complement representation is

A
01010101
B
11010101
C
00101011
D
10101011
       Digital-Logic-Design       Number-Systems       GATE 2000
Question 41 Explanation: 
Positive integers are represented in its normal binary form while negative numbers are represented in its 2′s complement form. Binary representation of 43 is 00101011.
Question 42

Consider the values A = 2.0 x 1030, B = -2.0 x 1030, C = 1.0, and the sequence

             X: = A + B    Y: = A + C
             X: = X + C    Y: = Y + B  

executed on a computer where floating-point numbers are represented with 32 bits. The values for X and Y will be

A
X = 1.0, Y = 1.0
B
X = 1.0, Y = 0.0
C
X = 0.0, Y = 1.0
D
X = 0.0, Y = 0.0
       Digital-Logic-Design       Number-Systems       GATE 2000
Question 42 Explanation: 
Given: 32 bits representation. So, the maximum precision can be 32 bits (In 32-bit IEEE representation, maximum precision is 24 bits but we take best case here). This means approximately 10 digits.
A = 2.0 * 1030, C = 1.0
So, A + C should make the 31st digit to 1, which is surely outside the precision level of A (it is 31st digit and not 31st bit). So, this addition will just return the value of A which will be assigned to Y.
So, Y + B will return 0.0 while X + C will return 1.0.
Question 43

Booth’s coding in 8 bits for the decimal number –57 is

A
0 – 100 + 1000
B
0 – 100 + 100 - 1
C
0 – 1 + 100 – 10 + 1
D
00 – 10 + 100 - 1
       Digital-Logic-Design       Number-Systems       GATE 1999
Question 43 Explanation: 
Option-B:
Question 44

Zero has two representations in

A
Sign magnitude
B
1’s complement
C
2’s complement
D
None of the above
E
Both A and B
       Digital-Logic-Design       Number-Systems       GATE 1999
Question 44 Explanation: 
Sign magnitude:
+0 = 0000
-0 = 1000
1's complement:
+0 = 0000
-0 = 1111
Question 45

The octal representation of an integer is (342)8. If this were to be treated as an eight-bit integer is an 8085 based computer, its decimal equivalent is

A
226
B
-98
C
76
D
-30
       Digital-Logic-Design       Number-Systems       GATE 1998
Question 45 Explanation: 
(342)8 = (011 100 010)2 = (1110 0010)2
If this can be treated as 8 bit integer, then the first becomes sign bit i.e., '1' then the number is negative.
8085 uses 2's complement then

⇒ -30
Question 46

Suppose the domain set of an attribute consists of signed four digit numbers. What is the percentage of reduction in storage space of this attribute if  it is stored as an integer rather than in character form?

A
80%
B
20%
C
60%
D
40%
       Digital-Logic-Design       Number-Systems       GATE 1998
Question 46 Explanation: 
We assume byte addressable memory - nothing smaller than a byte can be used.
We have four digits. So to represent signed 4 digit numbers we need 5 bytes, 4 bytes for four digits and 1 for the sign.
So required memory = 5 bytes.
Now, if we use integer, the largest no. needed to represent is 9999 and this requires 2 bytes of memory for signed representation.
9999 in binary requires 14 bits. So, 2 bits remaining and 1 we can use for sign bit.
So, memory savings,
= 5 - 2/5 × 100
= 60%
Question 47

Given √224)r = 13)r.
The value of the radix r is:

A
10
B
8
C
5
D
6
       Digital-Logic-Design       Number-Systems       GATE 1997
Question 47 Explanation: 
(√224)r = (13)r
Convert r base to decimal.
√2r2 + 25 + 4 = r + 3
Take square both sides,
2r2 + 2r + 4 = r2 + 6r + 9
r2 - 4r - 5 = 0
r2 - 5r + r - 5 = 0
r(r - 5) + (r - 5) = 0
r = -1, 5
r cannot be -1,
So r = 5 is correct answer.
Question 48

Consider the following floating point number representation

The exponent is in 2's complement representation and mantissa is in the sign magnitude representation. The range of the magnitude of the normalized numbers in this representation is

A
0 to 1
B
0.5 to 1
C
2-23 to 0.5
D
0.5 to (1-2-23)
       Digital-Logic-Design       Number-Systems       GATE 1996
Question 48 Explanation: 
Maximum value of mantissa will be 23, is where a decimal point is assumed before first 1. So the value is 1 - 2-23.
Question 49

The number of 1’s in the binary representation of
(3*4096 + 15*256 + 5*16 + 3) are:

A
8
B
8
C
10
D
12
       Digital-Logic-Design       Number-Systems       GATE 1995
Question 49 Explanation: 
3 × 4096 = 3 × 212
= (11000000000000)2
15 × 256 = 15 × 28
= (111100000000)2
5 × 16 = 5 × 24
= (1010000)2
3 = (11)2
Hence, all binary numbers,

∴ 101's
Question 50

Consider n-bit (including sign bit) 2’s complement representation of integer number. The range of integer values, N, that can be represented is _________ ≤ N ≤ _________

A
-2n-1 to 2n-1 - 1
       Digital-Logic-Design       Number-Systems       GATE 1994
Question 51

Consider three registers R1, R2 and R3 that store numbers in IEEE-754 single precision floating point format. Assume that R1 and R2 contain the values (in hexadecimal notation) 0x42200000 and 0xC1200000, respectively.

If R3 = R1/R2, what is the value stored in R3?

A
0x40800000
B
0x83400000
C
0xC8500000
D
0xC0800000
       Digital-Logic-Design       Number-Systems       GATE 2020
Question 51 Explanation: 
Given numbers are 0x42200000 and 0xC1200000 which are stored in the registers R1 and R2, respectively.

R1 = 1.0100..0 X 2132-127
= 1.0100..0 X 25
= 101.0 X 23
= 5 X 8
= 40

R2 = (-1) x 1.0100..0 X 2130-127
= (-1) x 1.0100..0 X 23
= (-1) x 101.0 X 21
= (-1) x5 X 2
= -10
R3 = R1/R2
= -4
= (-1)x 1.0 x 22
Sign = 1
Mantissa = 000..0
Exponent = 2+127 = 129

R3 = 1100 0000 1000 000..0
= 0x C 0 8 0 0 0 0 0
Question 52

Convert the following numbers in the given bases into their equivalents in the desired bases.
(a) 110.101)2 = x)10
(b) 1118)10 = y)H

A
(a) 6.625, (b) (45E)H
       Digital-Logic-Design       Number-Systems       GATE 1993
Question 52 Explanation: 
(a) 1*22 + 1*21 + 0*20 + 1*2-1 + 0*2-2 + 1*2-3
= 4 + 2 + 0 + 0.5 + 0 + 0.125
= 6.625
(b) 1118 mod 16 = E, quotient = 69
69 mod 16 = 5, quotient = 4
4 mod 16 = 4
Writing the mods result in reverse order gives (45E)H.
Question 53

Consider the number given by the decimal expression:

  163 * 9 + 162 * 7 + 16 * 5 + 3

The number of 1’s in the unsigned binary representation of the number is ________.

A
9
       Digital-Logic-Design       Number-Systems       GATE 1991
Question 53 Explanation: 
Hexadecimal representation of a given no. is,
(9753)16
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
Question 54

When two 4-bit binary number A = a3a2a1a0 and B = b3b2b1b0 are multiplied, the digit c1 of the product C is given by _________

A
c1 = b1a0 ⊕ a1b0
       Digital-Logic-Design       Number-Systems       GATE 1991
Question 54 Explanation: 

⇒ c1 = b1a0 ⊕ a1b0
Question 55

The exponent of a floating-point number is represented in excess-N code so that:

A
The dynamic range is large.
B
The precision is high.
C
The smallest number is represented by all zeros.
D
Overflow is avoided.
       Digital-Logic-Design       Number-Systems       GATE 1987
Question 55 Explanation: 
To avoid extra work, excess-N code is used so that all exponent can be represented in positive numbers, starting with 0.
Question 56

The following bit pattern represents a floating point number in IEEE 754 single precision format

 110000011101000000000000000000000
The value of the number in decimal form is
A
-10
B
-13
C
-26
D
None of these
       Digital-Logic-Design       Number-Systems       GATE 2008-IT
Question 56 Explanation: 
Sign bit is 1 then given number is negative.
Exponent bits - 10000011
Exponent can be added with 127 bias in IEEE single precision format then outval exponent
= 10000011 - 127
= 131 - 127
= 4
→ In IEEE format, an implied 1 is before mantissa, and hence the outval number is
→ 1.101 × 24 = -(11010)2 = -26
Question 57

A processor that has carry, overflow and sign flag bits as part of its program status word (PSW) performs addition of the following two 2's complement numbers 01001101 and 11101001. After the execution of this addition operation, the status of the carry, overflow and sign flags, respectively will be:

A
1, 1, 0
B
1, 0, 0
C
0, 1, 0
D
1, 0, 1
       Digital-Logic-Design       Number-Systems       GATE 2008-IT
Question 57 Explanation: 

Carry flag = 1
Overflow flag = 0
Sign bit = 0 (MSB bit is 0)
Overflow flag:
In computer processors, the overflow flag is usually a single bit in a system status register used to indicate when an arithmetic overflow has occurred in an operation.
Question 58

The two numbers given below are multiplied using the Booth's algorithm.

Multiplicand : 0101 1010 1110 1110 
Multiplier: 0111 0111 1011 1101
How many additions/Subtractions are required for the multiplication of the above two numbers?
A
6
B
8
C
10
D
12
       Digital-Logic-Design       Number-Systems       GATE 2008-IT
Question 58 Explanation: 
Take the multiples and add 0 to the LSB.
Now we have some values defined for pair of bits in Booth’s Algorithm,
00 → 0
11 → 0
01 → -1
10 → 1
Now after adding 0 to the LSB of the multiplier, start traversing from left to right and accordingly put the values defined above.

Hence, total 8 additions / subtractions required.
Question 59

(C012.25)H – (10111001110.101)B =

A
(135103.412)O
B
(564411.412)O
C
(564411.205)O
D
(135103.205)O
       Digital-Logic-Design       Number-Systems       GATE 2007-IT
Question 59 Explanation: 
(C012.25)H – (10111001110.101)B
= 1100000000010010.00100101 - 0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)O
Question 60

The addition of 4-bit, two’s complement, binary numbers 1101 and 0100 results in

A
0001 and an overflow
B
1001 and no overflow
C
0001 and no overflow
D
1001 and an overflow
       Digital-Logic-Design       Number-Systems       GATE 2006-IT
Question 60 Explanation: 
2's complement of 1101 = 0011
2's complement of 1100 = 1100
Add = 1111
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
Question 61

When multiplicand Y is multiplied by multiplier X = xn-1xn-2 ...x0 using bit-pair recoding in Booth's algorithm, partial products are generated according to the following table.

The partial products for rows 5 and 8 are

A
2Y and Y
B
-2Y and 2Y
C
-2Y and 0
D
0 and Y
       Digital-Logic-Design       Number-Systems       GATE 2006-IT
Question 61 Explanation: 

⇒ -2Y and 0
Question 62

(34.4)8 × (23.4)8 evaluates to

A
(1053.6)8
B
(1053.2)8
C
(1024.2)8
D
None of these
       Digital-Logic-Design       Number-Systems       GATE 2005-IT
Question 62 Explanation: 
First convert (34.4)8 and (23.4)8 to decimal.
(34.4)8 = 3×81 + 4×80 + 4×8-1
= 24 + 4 + 0.5
= (28.5)10
(23.4)8 = 2×81 + 3×80 + 4×8-1
= 16 + 3 + 0.5
= (19.5)10
Now,
(28.5)10 × (19.5)01 = (555.75)10
Now,
(555.75)10 = ( ? )8
To convert the integer part,

We get, 1053.
To convert the fractional part, keep multiplying by 8 till decimal part becomes 0,

∴ (555.75)10 = (1053.6)8
Question 63

The number (123456)8 is equivalent to

A
(A72E)16 and (22130232)4
B
(A72E)16 and (22131122)4
C
(A73E)16 and (22130232)4
D
(A62E)16 and (22120232)4
       Digital-Logic-Design       Number-Systems       GATE 2004-IT
Question 63 Explanation: 
(123456)8 = (001 010 011 100 101 110)2
= (00 1010 0111 0010 1110)2
= (A72E)16
Also,
(001 010 011 100 101 110)2
= (00 10 10 01 11 00 10 11 10)2
= (22130232)4
Question 64

Consider a parity check code with three data bits and four parity check bits. Three of the code words are 0101011, 1001101 and 1110001. Which of the following are also code words?

1. 0010111             
2. 0110110         
3. 1011010             
4. 0111010 
A
1 and 3
B
1, 2 and 3
C
2 and 4
D
1, 2, 3 and 4
       Digital-Logic-Design       Number-Systems       GATE 2004-IT
Question 64 Explanation: 
Let x1, x2, x3 are data bits, and c1, c2, c3 and c4 are parity check bits.
Given transmitted codewords are

By inspection we can find the rule for generating each of the parity bits,

Now from above we can see that (I) and (III) are only codewords.
Question 65
A computer uses ternary system instead of the traditional binary system. An n bit string in the binary system will occupy
A
3+n ternary digits
B
2n/3 ternary digits
C
n(log23) ternary digits
D
n(log32 ) ternary digits
       Digital-Logic-Design       Number-Systems       ISRO-2018
Question 65 Explanation: 
→ Binary numbers are maximum 2n-1.
→ But in question they are given ternary numbers, it means 3x-1.
→ Both will take different no. of bits to represent the same number.
3x -1 = 2n -1
3x = 2n
Apply log on both side
x= log3( 2n)
x=n*log32 .
Question 66
When two numbers are added in excess-3 code and the sum is less than 9, then in order to get the correct answer it is necessary to
A
subtract 0011 from the sum
B
add 0011 to the sum
C
subtract 0110 from the sum
D
add 0110 to the sum
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 66 Explanation: 
Subtract 0011 if there is no carry otherwise add 0011.
Example:
x+3
y+3
-------
(x+y+6)
Here, sum is excess-6. Hence, subtract 0011 to make it excess-3.
Question 67
The number of digit 1 present in the binary representation of 3 × 512 + 7 × 64 + 5 × 8 + 3
A
8
B
9
C
10
D
12
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 67 Explanation: 
3 × 512 + 7 × 64 + 5 × 8 + 3
= (2 + 1)× 512 + (4 + 2 + 1)× 64 + (4 + 1)× 8 + 2 + 1
= 1024 + 512 + 64 x 4 + 64 x 2 + 64 + 32 + 8 + 2 + 1
= 1024 + 512 + 256 + 128 + 64 + 32 + 8 + 2 + 1
As 1024 has ten 0’s followed by 1, 512 has nine 0’s followed by 1 and so on..
So, the expression will contain total nine 1’s and will be be represented as 11111101011.
Question 68
0.75 decimal system is equivalent to ____ in octal system
A
0.60
B
0.52
C
0.54
D
0.50
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 68 Explanation: 
0.75 = (0.110)2
= (0.6)8
Option (A) is correct.
Question 69
Consider a computer system that stores a floating-point numbers with 16-bit mantissa and an 8-bit exponent, each in two’s complement. The smallest and largest positive values which can be stored are
A
1 × 10-128 and 215× 1015
B
1 × 10-256 and 215× 10255
C
1 × 10-128 and 215× 10127
D
1 × 10-128and 215– 1 × 10127
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 69 Explanation: 

According to question 16 bit mantissa and 8 bit Exponent.
Since the mantissa is always 1.xxxxxxxxx in the normalised form, no need to represent the leading 1.
Single Precision: mantissa ===> 1 bit + 15 bits
The largest mantissa value value is 215-1 (one bit meant for sign)
The largest exponent value is 27-1=127
The smallest mantissa value is 0000 0000 0000 0000(one bit is always 1) =1
The Smallest (largest negative) exponent value is 1111 1111 (which is 2’s complement form) 2-8=-128
Question 70
The Hexadecimal equivalent of 01111100110111100011 is
A
CD73E
B
ABD3F
C
7CDE3
D
FA4CD
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 70 Explanation: 
Binary number = 0111 1100 1101 1110 0011
7 C D E 3

(7CDE3)16
Question 71
One approach to handling fuzzy logic data might be to design a computer using ternary (base-3) logic so that data could be stored as “true,” “false,” and “unknown.” If each ternary logic element is called a flit, how many blits are required to represent at least 256 different values?
A
4
B
5
C
6
D
7
       Digital-Logic-Design       Number-Systems       ISRO-2007
Question 71 Explanation: 
In binary representation, to represent 256 different values, you need log_2 (256) = 8 bits. Similarly in ternary representation, you would require log_3 (256) which is 5.something. Now rounding off to the upper integer (since number of bits is an integer) and we get 6
Question 72
If a variable can take only integral values from 0 to n, where n is an integer, then the variable can be represented as a bit field whose width is (the log in the Solutions are to the base 2, and [log n] means the floor of log n)
A
[log(n)] + 1 bits
B
[log (n-1)) + 1 bits
C
[log (n+1)] + 1 bits
D
None of the above
       Digital-Logic-Design       Number-Systems       ISRO-2018
Question 72 Explanation: 
Question 73
Given √(224)r = 13r the value of radix r is
A
10
B
8
C
6
D
5
       Digital-Logic-Design       Number-Systems       ISRO-2018
Question 73 Explanation: 
√(224)r = 13r
For f(x) to be maximum
f'(x) = 4x - 2 = 0
⇒ x = 1/2
So at x = 1/2, f(x) is an extremum (either maximum or minimum).
f(2) = 2(2)2 - 2(2) + 6 = 10
f(1/2) = 2 × (1/2)2 - 2 × 1/2 + 6 = 5.5
f(0) = 6
So, the maximum value is at x=2 which is 10 as there are no other extremum for the given function.
Question 74
When two n-bit binary numbers are added the sum will contain at the most
A
n bits
B
(n+3) bits
C
(n+2) bits
D
(n+1) bits
       Digital-Logic-Design       Number-Systems       ISRO-2017 May
Question 74 Explanation: 
→ When two n-bit binary numbers are added the sum will contain at the most (n+1) bits
Example = 2 Decimal numbers are (7)10 and (7)10
= Equivalent binary numbers are (111)2 + (111)2
= Adding two binary numbers, the final result will be n+1 number (1110)2
Question 75
(1217)8 is equivalent to
A
(1217)16
B
(028F)16
C
(2297)1o
D
(0B17)16
       Digital-Logic-Design       Number-Systems       ISRO-2017 May
Question 75 Explanation: 
(1217)8=(001 010 001 111)2
=(0010 1000 1111)2
=(2 8 F)16
Question 76
The logic circuit given below converts a binary code y1,y2,y3 into
A
Excess-3 code
B
Gray code
C
BCD code
D
Hamming Code
       Digital-Logic-Design       Number-Systems       ISRO-2016
Question 76 Explanation: 
X1= Y1
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
Question 77
If 12A7C16 = X8, then the value of X is
A
224174
B
425174
C
6173
D
225174
       Digital-Logic-Design       Number-Systems       ISRO-2016
Question 77 Explanation: 
Given, (12A7C)16 = (0001 0010 1010 0111 1100)2
MAke blocks of 3 bits each from LSB to MSB.
(Note: In the last block append zeros (as MSBs) if number bits is not three)
(000 010 010 101 001 111 100)
Each of the above blocks represents a digit in base 8 and they can be converted to base 8 as shown below.
= (0 2 2 5 1 7 4)8
Question 78
The Excess-3 code is also called
A
Cyclic Redundancy Code
B
Weighted Code
C
Self-Complementing Code
D
Algebraic Code
       Digital-Logic-Design       Number-Systems       ISRO-2016
Question 78 Explanation: 
Excess-3 code is also called Self-Complementing Code. Because 1’s complement of an excess-3 number is equivalent to 9’s complement of the corresponding decimal digit.
→ In excess-3 code, each of the 4-bit numbers represents decimal digit which is 3 less than the actual decimal digit. So the bits have no fixed weight.
Excess-3 code is neither CRC nor Algebraic Code which is used for error detection and/or correction.
Question 79
Which of the following binary number is the same as its 2’s complement?
A
1010
B
0101
C
1000
D
1001
       Digital-Logic-Design       Number-Systems       ISRO-2016
Question 79 Explanation: 
Hint: Number of bits=4
(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8
Question 80
The addition of 4-bit, two’s complement, binary numbers 1101 and 0100 results in
A
0001 and an overflow
B
1001 and no overflow
C
0001 and no overflow
D
1001 and an overflow
       Digital-Logic-Design       Number-Systems       ISRO CS 2009
Question 80 Explanation: 
Number one is 0100 (4-Decimal value)
Another number is 1101(-3 is decimal value)
Adding of -3 and 4, the result is 1 and there is no overflow
Question 81
What is the decimal value of the floating-point number C1D00000 (hexadecimal notation)? (Assume 32-bit, single precision floating point IEEE representation)
A
28
B
-15
C
-26
D
-28
       Digital-Logic-Design       Number-Systems       ISRO CS 2011
Question 81 Explanation: 
Floating Point number in Hexadecimal = C1D00000
Floating Point number in Binary = 1100 0001 1101 0000 0000 0000 0000 0000
In 32-bit, single precision floating point IEEE representation, first MSB represents sign of mantissa: 1 is used to represent a negative mantissa and 0 for a positive value of mantissa, next 8 bits are for exponent value and then 23 bits represents mantissa.
Value of exponent = 131-127 = 4
Mantissa = -1.1010000 0000 0000 0000 0000
Floating point number = -1.1010000 0000 0000 0000 0000
Converting the above one into decimal no -(1*20+1*2-1*0*2-2+1*2-2+0* 2-3 +.....)
= -(1+½+⅛)=-13/8
Decimal value =sign*Exponent*mantissa=1*4*-13/8

= -26
Question 82
In the standard IEEE 754 single precision floating point representation, there is 1 bit for sign, 23 bits for fraction and 8 bits for exponent. What is the precision in terms of the number of decimal digits?
A
5
B
6
C
7
D
8
       Digital-Logic-Design       Number-Systems       ISRO CS 2014
Question 82 Explanation: 
A floating-point variable can represent a wider range of numbers than a fixed-point variable of the same bit width at the cost of precision. A signed 32-bit integer variable has a maximum value of 231 − 1 = 2,147,483,647, whereas an IEEE 754 32-bit base-2 floating-point variable has a maximum value of (2 − 2−23) × 2127 ≈ 3.402823 × 1038.
In the IEEE 754-2008 standard, the 32-bit base-2 format is officially referred to as binary32; it was called single in IEEE 754-1985. IEEE 754 specifies additional floating-point types, such as 64-bit base-2 double precision and, more recently, base-10 representations.
We can convert the binary into decimal representation by using the following steps
let the number of digits in decimal digits be ‘x’
2-23 = 10-x
After taking log on both sides
log210-x =log2 2-23
-x log210 = -23log22 (The value of log22=1)
-3.322 x = -23 (The value of log210 = 3.321928)
x = 6.92
Question 83
How many different BCD numbers can be stored in 12 switches? (Assume two position or on-off switches)
A
212
B
212-1
C
1012
D
103
       Digital-Logic-Design       Number-Systems       ISRO CS 2014
Question 83 Explanation: 
Step-1: A binary-coded decimal (BCD) is a class of binary encodings of decimal numbers where each decimal digit is represented by a fixed number of bits, usually four (or) eight.
Step-2: Decimal number 0 can be represented 0000 and 9 can be represented by using 1001.
Step-3: A switch can store maximum 1 bit data that may be either 0 (or) 1. In switch terminology, 0 means “off” and 1 means “on”. With 4 bit we can represent 10 BCD numbers.
Step-4: A BCD digit can be from 0 to 9 (total 10 possibility).
Step-5: Different possible BCD numbers in 12 switches are = 10*10*10
= 1000
= 103
Question 84

To guarantee correction of upto t errors, the minimum Hamming distance dmin in a block code must be

A
t+1
B
t−2
C
2t−1
D
2t+1
       Digital-Logic-Design       Number-Systems       UGC-NET CS 2018 JUNE Paper-2
Question 84 Explanation: 
For detect t-bit errors a hamming distance of t+1 bit is needed but to correct the t-bit error the hamming distance of 2t+1 bit is needed.
Question 85

The hexadecimal equivalent of the binary integer number 110101101 is :

A
D 2 4
B
1 B D
C
1 A E
D
1 A D
       Digital-Logic-Design       Number-Systems       UGC-NET CS 2018 JUNE Paper-2
Question 85 Explanation: 
(110101101)2 = ( ? )16
24 = 16
So, 4-bits in binary will represent one integer in Hexadecimal.
So,
Question 86
The range of the numbers which can be stored in an eight bit register is
A
-128 to +127
B
-128 to +128
C
-999999 + +999999
D
none of these
       Digital-Logic-Design       Number-Systems       Nielit Scientist-C 2016 march
Question 86 Explanation: 
There are 2​ 8​ (256) different possible values for 8 bits. When unsigned, it has possible values ranging from 0 to 255; when signed, it has -128 to 127.
Question 87
The excess 3 code is also called
A
Cyclomatic redundancy code
B
Weighted code
C
Self complementing code
D
algebraic code
       Digital-Logic-Design       Number-Systems       Nielit Scientist-C 2016 march
Question 87 Explanation: 
Excess-3 is a self-complementing code. This is because in Excess-3 code we get the 9's complement of a number by just complementing each bit that means by replacing a '0' by '1' and '1' by '0'
Question 88

In computers, subtraction is generally carried out by

A
1’s complement
B
10’s complement
C
2’s complement
D
9’s complement
       Digital-Logic-Design       Number-Systems       UGC-NET CS 2018 DEC Paper-2
Question 88 Explanation: 
• In computers, subtraction is generally carried out by 2’s complement.
• In two's-complement representation, positive numbers are simply represented as themselves, and negative numbers are represented by the two's complement of their absolute value.
• In the subtraction there may possibility of negative number as a result.
Question 89
A decimal number has 30 digits. Approximately, how many digits would the binary representation have?
A
30
B
60
C
90
D
120
       Digital-Logic-Design       Number-Systems       Nielit Scientist-B CS 2016 march
Question 89 Explanation: 
Here, 30 digits numbers means 123....30. 10​30​ -1=1000000000000000000000000000000-1=999999999999999999999999999999
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Question 90
The result of the subtraction FD​16 - ​8816​ is
A
75​ 16
B
65 16
C
5E 16
D
10 16
       Digital-Logic-Design       Number-Systems       Nielit Scientist-B CS 2016 march
Question 90 Explanation: 
Step-1: Convert Hexadecimal numbers into decimal numbers.
(FD)​ 16<​/sub> = (253)​ 10
(88)​ 16<​/sub> = (136)​ 10
Step-2: Perform subtraction 253-136=(117)​ 10
Step-3: Convert (117)​ 10 ​ =(75)​ 16
Question 91
The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is
A
2​ -128​ to (1 – 2​ –23​ ) * 2​ 127
B
(1 – 2​ –23​ ) * 2​ ​ -127 ​ to 2​ 128
C
(1 – 2​ –23​ ) * 2​ –127​ to 2​ 23>
D
2​ –129​ to (1 – 2​ –23​ ) * 2​ 127
       Digital-Logic-Design       Number-Systems       UGC NET CS 2014 Dec-Paper-2
Question 91 Explanation: 
The range of representable normalized numbers in the floating point binary fractional representation in a 32-bit word with 1-bit sign, 8-bit excess 128 biased exponent and 23-bit mantissa is​ ​ 2 –129 to (1 – 2​ –23​ ) * 2​ 127
Question 92
What will be the Excess-3 code for 1001?
A
1001
B
1010
C
1011
D
1100
       Digital-Logic-Design       Number-Systems       Nielit Scientist-B CS 4-12-2016
Question 92 Explanation: 
Excess-3 number starts with 3. Here, 1001 menas decimal number-9. So, we have to add +3.
12 equivalent binary number is 1100.
Question 93
The Decimal equivalent of the Hexadecimal number (A09D)16 is
A
31845
B
41117
C
41052
D
32546
       Digital-Logic-Design       Number-Systems       Nielit Scientist-B CS 4-12-2016
Question 93 Explanation: 
Given Hexadecimal number is ​ (A09D)16
A decimal number is the sum of the digits multiplied with its power of 10.
(A09D)16​ is equal to each digit multiplied with its corresponding power of 16:
Ax16​ 3​ +0x16​ 2​ +9x16​ 1​ +Dx16​ 0​ =(10x4096+144+13x1) [ Where A=10,D=13]
=40960+144+13=41117
Question 94
Which of the given number has its IEEE-754 32 bit floating point representation as
A
2.5
B
3.0
C
3.5
D
4.5
       Digital-Logic-Design       Number-Systems       ISRO CS 2015
Question 94 Explanation: 
→ Sign bit S = 0 (It means positive number)
→ E=1000 0000B = 128D (in normalized form)
→ Fraction is 1.11B (with an implicit leading 1) = 1 + 1×2-1 + 1×2-2
= 1.75D
→ The number is +1.75 × 2(128-127)
= +3.5D
Question 95
The range of integers that can be represented by an n bit 2’s complement number system is:
A
– 2n – 1 to 2n – 1 – 1
B
– (2n – 1 – 1) to (2n – 1 – 1)
C
– 2n – 1 to 2n – 1
D
– (2n – 1 + 1) to (2n – 1 – 1)
       Digital-Logic-Design       Number-Systems       ISRO CS 2015
Question 95 Explanation: 
In 2’s complement numbers, the range of integers are from -2n-1 to 2n-1 – 1
Question 96
The code which uses 7 bits to represent a character is
A
ASCII
B
BCD
C
EBCDIC
D
Gray
       Digital-Logic-Design       Number-Systems       ISRO CS 2015
Question 96 Explanation: 
→ ISO/IEC 646, like ASCII, is a 7-bit character set. It does not make any additional codes available, so the same code points encoded different characters in different countries. Escape codes were defined to indicate which national variant applied to a piece of text, but they were rarely used, so it was often impossible to know what variant to work with and, therefore, which character a code represented, and in general, text-processing systems could cope with only one variant anyway.
→ Extended Binary Coded Decimal Interchange Code (EBCDIC) is an 8-bit binary code for numeric and alphanumeric characters.
→ BCD encoding uses 4 bits to represent each digit from the range 0 to 9 in its binary form.
→ In case of Gray codes, any number of bits can be used to represent a character, according to the requirement.
Question 97
The number of 1’s in the binary representation of (3*4096 + 15*256 + 5*16 + 3) are:
A
8
B
9
C
10
D
12
       Digital-Logic-Design       Number-Systems       ISRO CS 2015
Question 97 Explanation: 
Binary expression of (3*4096 + 15*256 + 5*16 + 3)
=(12,288+3840+80+3)
=(16211)10
=‭(0011111101010011‬)2
Total number of 1’s in binary representation is 10.
Question 98
The decimal number has 64 digits.The number of bits needed for its equivalent binary representation is?
A
200
B
213
C
246
D
277
       Digital-Logic-Design       Number-Systems       ISRO CS 2015
Question 98 Explanation: 
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 103-1 which is 999.
Then, Decimal number has 64 digits, so maximum number is 1064-1
Similarly, in the binary representation with “n” bits the maximum number is 2n-1
So we can write 1064 –1 = 2n – 1 --->1064 = 2n
After taking log2 on both sides
log22n=log21064
n log22=64 log 210
n=64*(3.322) [ log22=1 & log210 =3.322]
n=212.608
n=213
Question 99

Suppose x and y are floating point variables that have been assigned the values x = 8.8 and y = 3.5. What will be the value of the following arithmetic expression?

2 * x / 3 * y

A
20.33335
B
24.45453
C
16.35353
D
20.53333
       Digital-Logic-Design       Number-Systems       JT(IT) 2018 PART-B Computer Science
Question 99 Explanation: 
x = 8.8 y=3.5
= 1 (equal priority
Associativity (left to Right)
(((2 * x)/3) * y)
((2 x *) / 3) * y)
(2x * 31) * y
2x * 31) y*
(2x * 31 y*
Put the 2 * 8.8 * 3 / 3.5 *
= 20.53333
Question 100
A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
A
50
B
74
C
40
D
60
E
None of these
       Digital-Logic-Design       Number-Systems       Nielit Scientific Assistance IT 15-10-2017
Question 100 Explanation: 
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10​ 3​ -1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10​ 25​ -1
Similarly, in the binary representation with “n” bits the maximum number is 2​n​ -1
So we can write 10​ 25​ –1 = 2​ n​ – 1 --->10​ 25​ = 2​ n After taking log​ 2​ on both sides
log​ 2​ 2​ n​ =log ​ 2​ 10​ 25
n log​ 2​ 2=25 log ​ 2​ 10
n=25*(3.322) [ log​ 2​ 2=1 & log​ 2​ 10 =3.322]
n=83
Question 101

What is the base(radix) of the number system whose numbers 312, 20 and 13.1 satisfy the following equation?

312/20 = 13.1
A
8
B
4
C
5
D
6
       Digital-Logic-Design       Number-Systems       JT(IT) 2018 PART-B Computer Science
Question 101 Explanation: 
Let base of the number system is r.
(3r2 + r + 2) / 2r = (r + 3 + 1/r)
(3r2 + r + 2) / 2r = (r2 + 3r + 1) / r
(3r2 + r + 2) = (2r2 + 6r + 2)
r2 - 5r = 0
Therefore, r = 5
Question 102

What is the hexadecimal representation of the decimal number 8537?

A
(2059)16
B
(2159)16
C
(2195)16
D
(2157)16
       Digital-Logic-Design       Number-Systems       JT(IT) 2018 PART-B Computer Science
Question 102 Explanation: 
Step-1: First convert decimal number into binary number
Step-2: (8537)10 = (0010000101011001)2
Step-3: Divide binary number into 4 segments (0010 0001 0101 1001)2
Step-4: Write equivalent number of hexadecimal (2159)16
Question 103
Shifting a register content to left by one bit is equivalent to____
A
Division by 2
B
Addition by 2
C
Multiplication by 2
D
Subtraction by 2
       Digital-Logic-Design       Number-Systems       KVS 22-12-2018 Part-B
Question 103 Explanation: 
→ The left-shift operator (<<), which moves the bits of shift-expression to the left.
→ The bit positions that have been vacated by the shift operation are zero-filled.
→ For example a=5 and equivalent binary value is 101 and shifting one bit left side means the result binary value is 1010 whose decimal value is 10
Question 104
What shall be the 2’s complement represented of -24 in a 16 bit computer?
A
1111 1111 1110 1011
B
1111 1111 1110 1001
C
1111 1111 1110 0111
D
1111 1111 1110 1000
       Digital-Logic-Design       Number-Systems       KVS 22-12-2018 Part-B
Question 104 Explanation: 
Negative numbers are represented in 2’s complement form.
The binary equivalent of 24 is 0000 0000 0001 1000
One’s complement is 1111 1111 1110 0111 (Flipping the bits 1 by 0 and 0 by 1)
Two’s complement is 1111 1111 1110 1000 (adding 1 to the LSB bit)
Question 105
A decimal has 25 digits. the number of bits needed for its equivalent binary representation is approximately
A
50
B
74
C
40
D
None of the above
       Digital-Logic-Design       Number-Systems       Nielit Scientific Assistance CS 15-10-2017
Question 105 Explanation: 
Consider three digits(1,2,3) of decimal numbers.Maximum number, we can generate by that three digits are 10​ 3​ -1 which is 999.
Then, Decimal number has 25 digits, so maximum number is 10​ 25​ -1
Similarly, in the binary representation with “n” bits the maximum number is 2​ 25​ -1
So we can write 10​ 25​ –1 = 2​ n​ – 1 → 10​ 25​ = 2​ n
After taking log​ 2​ on both sides
log​ 2​ 2​ n​ =log ​ 2​ 10​ 25
n log​ 2​ 2=25 log ​ 2​ 10
n = 25 log ​ 2​ 10
n = 25 x 3.3 [ log​ 2​ 2=1 & log​ 2​ 10 =3.322]
n = 82.5
Note: Original question paper given option D is 60. But actual answer is 82.5.
Question 106
Which of the following is minimum error code?
A
Octal code
B
Binary Code
C
Gray code
D
Excess-3 Code
       Digital-Logic-Design       Number-Systems       Nielit Scientific Assistance CS 15-10-2017
Question 106 Explanation: 
→ "Gray code" as an alternative name is "reflected binary code". one of those also lists "minimum error code" and "cyclic permutation code" among the names.
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Question 107
In single-precision, double-precision and extended-precision representation of floating point numbers, as defined by ANSI/IEEE standard 754-1985, the no.of bits used are____ respectively.
A
32,64 and 80
B
32,64 and 128
C
16,32 and 64
D
16,32 and 80
       Digital-Logic-Design       Number-Systems       KVS 22-12-2018 Part-B
Question 107 Explanation: 

Extended precision, the third format, is usually an 80-bit word, with 1 bit sign, 15 bit exponent and 64 bit significand, with leading bit of a normalized number not hidden
Question 108
If the original size of data is 40 then after adding error detection redundancy bit the size of data length is
A
26
B
36
C
46
D
56
       Digital-Logic-Design       Number-Systems       Nielit Scientific Assistance CS 15-10-2017
Question 108 Explanation: 
→ Imagine that we want to design a code with m message bits and r check bits that will allow all single errors to be corrected.
→ Each of the 2​ m​ legal messages has n illegal codewords at a distance of 11 from it.
→ These are formed by systematically inverting each of the n bits in the n-bit codeword formed from it. Thus, each of the 2​ m​ legal messages requires n+1 bit patterns dedicated to it.
→ Since the total number of bit patterns is 2​ n​ ,​ We must have (n+1)2​ m​ ≤ 2​ n​ .
→ Using n=m+r, this requirement becomes
= (m+r+1) ≤ 2​ r
= 40+6+1 ≤ 26
= 47 ≤ 2
r=6
Message size will be 6+40=46
Question 109

Which of the following is a recursive algorithm to convert a positive decimal integers into equivalent binary integers?

A
B
C
D
       Digital-Logic-Design       Number-Systems       JT(IT) 2018 PART-B Computer Science
Question 109 Explanation: 
Algorithm:
Take decimal number is 12.
Step-1 → 12%2 which is equal to 0+10*(⌊12/2⌋)%2
Step-2 → 6%2 which is equal to 0+10*(⌊6/2⌋)%2
Step-3 → 3%2 which is equal to 1+10*(⌊3/2⌋)%2
Step-4 → 1%2 which is equal to 1+10*(⌊1/2⌋)%2
Question 110
Consider an arbitrary number system with the independent digits as 0,1 and X. What is the radix of this number system?
A
1
B
2
C
3
D
4
       Digital-Logic-Design       Number-Systems       KVS DEC-2013
Question 110 Explanation: 
→ The radix of a number system is the number of unique digits, including, zero, that are used to represent larger numbers. In the decimal system that would be 0 to 9.
→ In the question, the unique digits are 0,1 and X(possible x value is 2) then number system is 3;
Question 111
The hamming(7,4) code for 0000 using even parity is
A
0000000
B
1111111
C
2222222
D
12121212
       Digital-Logic-Design       Number-Systems       KVS DEC-2013
Question 111 Explanation: 
● Hamming(7,4) is a linear error-correcting code that encodes four bits of data into seven bits by adding three parity bits.
● The data is ​ 0000 and Hamming(7,4) transmitted is 0000000
Question 112
The number of bits required to represent decimal number 4096 in binary form is___
A
16
B
10
C
12
D
13
       Digital-Logic-Design       Number-Systems       KVS DEC-2017
Question 112 Explanation: 
(4098)​ 10​ =(1000000000010)​ 2
So, total 13 bits required to represent 4096 decimal number.
Question 113

The signed 2’s complement representation of -33 is:

A
11011111
B
00100001
C
01011111
D
10100001
       Digital-Logic-Design       Number-Systems       JT(IT) 2016 PART-B Computer Science
Question 113 Explanation: 
Step-1: -33 corresponding positive number 00100001
Step-2: Perform 2’s complement.
Question 114

Using signed 2’s complement subtraction the result of 11111010-11110011 is:

A
10000111
B
00000111
C
10001101
D
00001101
       Digital-Logic-Design       Number-Systems       JT(IT) 2016 PART-B Computer Science
Question 114 Explanation: 
11111010 → 250(Decimal)
-11110011 → -243(Decimal)
Step1: convert signed number into 2’s complement
11110011
00001100 → (1’s complement)
+1
-------------
00001101 → (2’s complement)
Step 2: Add 11111010 into 2’s complement number
11111010
00001101
--------------
00000111 → solution
Question 115
How many bits are used in the exponent part of IEEE single precision for the representation of floating point numbers?
A
32 bits
B
8 bits
C
16 bits
D
24 bits
       Digital-Logic-Design       Number-Systems       KVS DEC-2017
Question 115 Explanation: 
The IEEE 754 standard specifies a binary floating point format(binary32) as having:
→ Sign bit: 1 bit
→ Exponent width: 8 bits
→ Significand precision: 24 bits (23 explicitly stored)
Question 116
Convert the following octal number into its decimal equivalent: 2 3 7 4 in octal
A
(10208)10
B
(1276)10
C
(2374)10
D
(1272)10
       Digital-Logic-Design       Number-Systems       KVS 30-12-2018 Part B
Question 116 Explanation: 
(2374)8=2x83+3x82+7x81+4x80=1024+192+56+4=1276
Question 117
Represent the decimal number 3.248*104 into a single precision floating point binary number(using standard format).
A
0|10001101|11111011100000000000000
B
0|11001101|11111011100000000000000
C
1|11001101|11111011100000000000000
D
0|10001110|11111011100000000000000
       Digital-Logic-Design       Number-Systems       KVS 30-12-2018 Part B
Question 117 Explanation: 
Given number is in base 10. Convert it to base-10.
3.248x104 =32480= 1111110111
= 1.111110111 x 214
Mantissa = 11111011100...00
Biased exponent = 14 +127= 141 = 10001101
Question 118
What is the octal equivalent of the hexadecimal number 132A?
A
46252
B
11450
C
11452
D
45250
       Digital-Logic-Design       Number-Systems       KVS DEC-2017
Question 118 Explanation: 
Step-1: Convert hexadecimal number in binary format. It is nothing but representing 4 binary values of each character.
Step-2: (132A)​ 16​ =(0001 0011 0010 1010)​ 2 Step-3: Divide 4 binary numbers in to 3 binary numbers from right to left.
Step-4: (0 001 001 100 101 010)​ 2
(0 1 1 4 5 2)​ 8
Question 119
The Octal equivalent of the binary number 1011101011 is :
A
7353
B
1353
C
5651
D
5657
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Nov- paper-2
Question 119 Explanation: 
We have to divide binary number into 3 bit pairs from LSB.
1 011 101 011
1 3 5 3
(1011101011)​ 2​ = (1353)​ 8
Question 120
Which is the hexadecimal number equivalent to the octal number 46250
A
4AC8
B
4CA8
C
CCA8
D
4CA4
       Digital-Logic-Design       Number-Systems       KVS 30-12-2018 Part B
Question 120 Explanation: 
Given octal number is 46250 and the corresponding binary number we will get by writing each decimal digit into three binary digits form which is 100110010101000
Now group the four digits from LSB and write corresponding equivalent of Hexadecimal digit of binary digits.
The Hexadecimal number of 100(4) 1100(C) 1010(A) 1000(8) which is nothing but 4CA8
Question 121
Let m=(313)​ 4​ and n=(322)​ 4​ . Find the base 4 expansion of m+n.
A
(635)​ 4
B
(32312)​ 4
C
(21323)​ 4
D
(1301)​ 4
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Nov- paper-2
Question 121 Explanation: 
In this problem, they are asking to find (m+n)​ 4
→ We are using addition for decimal number system. But in this problem m and n values are base 4. So, we can’t add directly.
Step-1: We have to perform base 4 into decimal values of m and n.
m = 3*42 + 1*41 + 3*40
m = 48 + 4 + 3
m = 55
n= 3*42 + 2*41 +2*40
n= 48 + 8 +2
n= 58
Step-2: The resultant decimal values should perform addition.
m+n = 55 + 58
m+n = 113
Step-3: Finally we have to convert decimal value into base 4 value.
(1301)4
Question 122
Convert the octal number 0.4051 into its equivalent decimal number.
A
0.5100098
B
0.2096
C
0.52
D
0.4192
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Jan -paper-2
Question 122 Explanation: 
Question 123
The hexadecimal equivalent of the octal number 2357 is :
A
2EE
B
2FF
C
4EF
D
4FE
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Jan -paper-2
Question 123 Explanation: 
Step-1: Convert octal number into binary number
(2357)​ 8​ = (010 011 101 111)​ 2
Step-2: Divide 4 bits from LSB then will get hexadecimal number
0100 1110 1111
2 E F
(2EF)​ 16​ = (2357)​ 8
Question 124
If X is a binary number which is power of 2, then the value of X & (X – 1) is :
A
11....11
B
00.....00
C
100.....0
D
000......1
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Jan -paper-2
Question 124 Explanation: 
Given data,
→ X is binary number which is power of 2. It means, we have to take powers of 2 numbers only.
Ex: 1,2,4,8,16,32,..,
Let X=4
X=4 equivalent binary number is 100
X-1=3 equivalent binary number is 011
100
011
-----
000 (AND operation)
-----
Ex-2:
X=8 and X-1=7
8 binary value is
1000
7 binary number is 0111
--------
0000(AND operation)
--------
So, Option B is correct answer.
Question 125
The octal number 326.4 is equivalent to
A
(214.2)​ 10​ and (D6.8))​ 16
B
(212.5)​ 10​ ​ and (D6.8))​ ​ 16
C
(214.5)​ 10​ and (D6.8))​ ​ 16
D
(214.5)​ 10​ and (D6.4))​ ​ 16
       Digital-Logic-Design       Number-Systems       UGC NET CS 2016 Aug- paper-2
Question 125 Explanation: 
(326.4) 8 = ( ?)
Step1: First convert given octal no. to binary number because it will be easier to solve this way.

Step 2: Now convert above binary no. into decimal .
(011010110.100) 2
= ( 1 * 2 7 ) + ( 1 × 2 6 ) + ( 1 × 2 4 ) + ( 1 × 2 2 ) + ( 1 × 2 1 ) · [ 1 × ( 1/2) ]
= (214.5) 10
(326.4) 8 = (?) 16 Step 1: Convert given octal no. into binary no.

Step 2: Now convert above binary no. into decimal .
(011010110.100) 2
= ( 1 * 2 7 ) + ( 1 × 2 6 ) + ( 1 × 2 4 ) + ( 1 × 2 2 ) + ( 1 × 2 1 ) · [ 1 × ( 1/2) ]
= (214.5) 10
(326.4) 8 = (?) 16

Question 126
Which of the following is the most efficient to perform arithmetic operations on the numbers?
A
Sign-magnitude
B
1’s complement
C
2’s complement
D
9’s complement
       Digital-Logic-Design       Number-Systems       UGC NET CS 2016 Aug- paper-2
Question 126 Explanation: 
2’s complement has single representation for zero , but Sign-magnitude, 1’s complement and 9’s complement have two representations for 0 (i.e., both positive zero and negative zero). While doing arithmetic operations like addition or subtraction using 1's complement(or 9's complement), we have to add an extra carry bit, i.e 1 to the result to get the correct answer. 2's complement doesn't require such extra calculation.
Question 127
The IEEE-754 double-precision format to represent floating point numbers, has a length of _____ bits.
A
16
B
32
C
48
D
64
       Digital-Logic-Design       Number-Systems       UGC NET CS 2016 July- paper-2
Question 127 Explanation: 
→ The IEEE-754 double-precision format to represent floating point numbers has a length of 64 bits
→ In the IEEE 754-2008 standard, the 64-bit base-2 format is officially referred to as binary64 called double in IEEE 754-1985.
→ IEEE 754 specifies additional floating-point formats, including 32-bit base-2 single precision and, more recently, base-10 representations.
Question 128
An example of a binary number which is equal to its 2​ ’s complement is :
A
1100
B
1001
C
1000
D
1111
       Digital-Logic-Design       Number-Systems       UGC NET CS 2004 Dec-Paper-2
Question 128 Explanation: 
Option-A: 1100 convert into 2’s complement is
1100
1’s complement: 0011
2’s complement: 1
--------
0100
Option-B: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement:
1
--------
0111
---------
Option-C: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement:
1
--------
1000
---------
Option-D: 1111 convert into 2’s complement is
1111
1’s complement: 0000
2’s complement:
1
--------
0001
---------
So, Option-C is correct answer.
Question 129
Which of the following is divisible by 4 ?
A
100101100
B
1110001110001
C
11110011
D
10101010101010
       Digital-Logic-Design       Number-Systems       UGC NET CS 2005 Dec-Paper-2
Question 129 Explanation: 
Option-A: (100101100)​ 2​ = (300)​ 10
300 is divisible by 4
Option-B: (1110001110001)​ 2​ = (7281)​ 10
7281 is not divisible by 4
Option-C: (11110011)​ 2​ = (243)​ 10
243 is not divisible by 4
Option-D: (10101010101010)​ 2​ = (10,922)​ 10
10,922 is not divisible by 4.
Question 130
(101011)​ 2​ =(53)​ b​ , then ‘b’ is equal to :
A
4
B
8
C
10
D
16
       Digital-Logic-Design       Number-Systems       UGC NET CS 2005 june-paper-2
Question 130 Explanation: 
We are dividing binary number into 3 then
101 011
5 3
(101011)​ 2 = (53)​ 8
Suppose, we are dividing binary digits into 4 then
0010 1011
2 B
So, it is not hexadecimal number.
Question 131
Which of the following binary number is the same as its 2’s complement :
A
1010
B
0101
C
1000
D
1001
       Digital-Logic-Design       Number-Systems       UGC NET CS 2005 june-paper-2
Question 131 Explanation: 
Option-A: 1010 convert into 2’s complement is
1010
1’s complement: 0101
2’s complement: 1
--------
0110
--------- Option-B: 0101 convert into 2’s complement is
0101
1’s complement: 1010
2’s complement: 1
--------
1011
---------
Option-C: 1000 convert into 2’s complement is
1000
1’s complement: 0111
2’s complement: 1
--------
1000
---------
Option-D: 1001 convert into 2’s complement is
1001
1’s complement: 0110
2’s complement: 1
--------
0111
---------
So, Option-C is correct answer.
Question 132
The hexadecimal equivalent of (10111)​ 2​ ×(1110)​ 2 is :
A
150
B
241
C
142
D
101011110
       Digital-Logic-Design       Number-Systems       UGC NET CS 2006 Dec-paper-2
Question 132 Explanation: 
Step-1: First convert binary number into decimal number
(10111)​ 2​ = (23)​ 10
(1110)​ 2​ = (14)​ 10
Step-2: Perform multiplication 23*14=(322)​ 10
Step-3: Convert (322)​ 10​ into hexadecimal number (322)​ 10​ = (142)​ 16
Question 133
An example of a self complementing code is :
A
8421 code
B
Gray code
C
Excess-3 code
D
7421 code
       Digital-Logic-Design       Number-Systems       UGC NET CS 2006 Dec-paper-2
Question 133 Explanation: 
→ Excess-3 code is also called Self-Complementing Code. Because 1’s complement of excess-3 number is equivalent to 9’s complement of corresponding decimal digit.
→ In excess-3 code, each of the 4-bit number represents decimal digit which is 3 less than actual decimal digit. So the bits have no fixed weight.
→ Excess-3 code is ​ neither​ CRC ​ nor​ Algebraic Code which are used for error detection and/or correction.
Question 134
The number of 1​ ’s present in the binary representation of (3*512 + 7*64 +5*8 +3)​ 10​ is :
A
8
B
9
C
10
D
11
       Digital-Logic-Design       Number-Systems       UGC NET CS 2006 June-Paper-2
Question 134 Explanation: 
(3*512 + 7*64 +5*8 +3) =2027
(2027)​ 10​ = (111 1110 1011)​ 2
Here, total number of 1’s are 9.
Question 135
​ In a weighted code with weight 6, 4, 2, -3 the decimal 5 is represented by :
A
0101
B
0111
C
1011
D
1000
       Digital-Logic-Design       Number-Systems       UGC NET CS 2006 June-Paper-2
Question 135 Explanation: 
The decimal value 5 is represented by 1011.
= 6*1 + 4*0 + 2*1 + -3*1
= 6+2-3
=5
Question 136
The BCD adder to add two decimal digits needs minimum of
A
6 full adders and 2 half adders
B
5 full adders and 3 half adders
C
4 full adders and 3 half adders
D
5 full adders and 2 half adders
       Digital-Logic-Design       Number-Systems       UGC NET CS 2014 Dec-Paper-2
Question 136 Explanation: 
→ Each digit is represented by a 4-bit BCD code.
→ To add two 4-bit number, we need 1 Half Adder(to add LSBs) and 3 Full Adders(remaining three bits of both number along with carry bits).
→ To make the resultant Sum as valid BCD sum, we need to add 0110 to the sum.
→ This can be done with 1 Half adder and 2 Full Adder
(Note: LSB bit of 0110 is always zero. So there is no need of ADDER to add LSBs.)
→ Here, Half adder is used to add next significant bits.
Total 5 Full Adders and 2 Half Adders are needed
Question 137
The Excess-3 decimal code is a self-complementing code because
A
The binary sum of a code and its 9’s complement is equal to 9.
B
It is a weighted code
C
Complement can be generated by inverting each bit pattern
D
The binary sum of a code and its 10’s complement is equal to 9
       Digital-Logic-Design       Number-Systems       UGC NET CS 2014 Dec-Paper-2
Question 137 Explanation: 
The Excess-3 decimal code is a self-complementing code because complement can be generated by inverting each bit pattern.
Question 138
The decimal floating point number -40.1 represented using IEEE-754 32-bit representation and written in hexadecimal form is
A
0xC2206000
B
0xC2006666
C
0xC2006000
D
0xC2206666
       Digital-Logic-Design       Number-Systems       UGC NET CS 2018-DEC Paper-2
Question 138 Explanation: 
1. Fraction part can be converted into binary form by multiplying it with 2 and taking non fractional part of the product. Take fractional part and multiply again as explained above.
0.1 x 2= 0.2 → 0
0.2 x 2= 0.4 → 0
0.4 x 2= 0.8 → 0
0.8 x 2= 1.6 → 1
0.6 x 2= 1.2 → 1
(0.1)​ 10​ = (0.00011)​ 2
(40)​ 10​ = (101000)​ 2
101000.00011
Normalize the number
1.0100000011 x 2​ 5
Biased exponent= 5+127= 132=(1000 0100)​ 2
Mantissa= 01000000110000000000000
Sign= 1
Question 139
The IEEE single-precision and double-precision format to represent floating-point numbers, has a length of ______ and ______ respectively.
A
8 bits and 16 bits
B
16 bits and 32 bits
C
32 bits and 64 bits
D
64 bits and 128 bits
       Digital-Logic-Design       Number-Systems       UGC NET CS 2013 Sep-paper-2
Question 139 Explanation: 
The IEEE single-precision and double-precision format to represent floating-point numbers, has a length of 32 bits and 64 bits respectively.
Question 140
Given that (292)10 = (1204)x in some number system x. The base x of that number system is
A
2
B
8
C
10
D
None of the above
       Digital-Logic-Design       Number-Systems       UGC NET CS 2013 Dec-paper-2
Question 140 Explanation: 
Option-A: It is false because it is equivalent to (292)10 =(100100100)2 Option-B: It is false because it is equivalent to (292)10 =(444)8 Option-B: It is false because it is equivalent to (292)10 =(292)10 So, option-D is the correct answer. The actual x value is 6.
Question 141
If an integer needs two bytes of storage, then the maximum value of a signed integer is
A
216 – 1
B
215 – 1
C
216
D
215
       Digital-Logic-Design       Number-Systems       UGC NET CS 2012 June-Paper2
Question 141 Explanation: 
= If an integer needs two bytes of storage, then the maximum value of a signed integer is 2n-1 -1
= 215 -1
Question 142
If an integer needs two bytes of storage, then the maximum value of unsigned integer is
A
216 – 1
B
215 – 1
C
216
D
215
       Digital-Logic-Design       Number-Systems       UGC NET CS 2011 Dec-Paper-2
Question 142 Explanation: 
= If an integer needs two bytes of storage, then the maximum value of a signed integer is 2n-1 -1
= 215 -1
Question 143
Negative numbers cannot be represented in
A
signed magnitude form
B
1’s complement form
C
2’s complement form
D
none of the above
       Digital-Logic-Design       Number-Systems       UGC NET CS 2011 Dec-Paper-2
Question 143 Explanation: 
Negative numbers can be represented in
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
Question 144
Two’s complement of a binary number 1010 is
A
0101
B
0101
C
0110
D
1001
       Digital-Logic-Design       Number-Systems       NIELIT Junior Teachnical Assistant_2016_march
Question 145
If integer needs two by storage, then maximum value of an unsigned integer is
A
2 16 − 1
B
2 15 − 1
C
2 16
D
2 15
       Digital-Logic-Design       Number-Systems       NIELIT Junior Teachnical Assistant_2016_march
Question 146
The hexadecimal number equivalent to (1762.46)8 is
A
3F2.89
B
3F2.98
C
2F3.89
D
2F3.98
       Digital-Logic-Design       Number-Systems       UGC NET CS 2011 June-Paper-2
Question 146 Explanation: 
(1762.46)8
→ For making the conversion easy, first convert number into binary
(001 111 110 010. 100 110)2
→ Now convert above binary no. into hexadecimal, we need 4-bits to represent a no. into hexadecimal

(3F2.98)16
Question 147
8-bit 1’s complement form of –77.25 is
A
01001101.0100
B
01001101.0010
C
10110010.1011
D
10110010.1101
       Digital-Logic-Design       Number-Systems       UGC NET CS 2011 June-Paper-2
Question 147 Explanation: 
(-77.25)10 Step 1: Convert (77)10into binary form.

Question 148
In which of the following codes the successive numbers differ in only one bit position ?
A
ASCII
B
Gray Code
C
Excess-3 Code
D
BCD
       Digital-Logic-Design       Number-Systems       NIELIT Technical Assistant_2016_march
Question 149
The binary addition of 1+1+1 is
A
111
B
10
C
110
D
11
       Digital-Logic-Design       Number-Systems       NIELIT Technical Assistant_2016_march
Question 150
Which one of the following is decimal value of a signed binary number 1101010, if it is in 2’s complement form ?
A
-42
B
-22
C
-21
D
-106
       Digital-Logic-Design       Number-Systems       UGC NET CS 2013 June-paper-2
Question 150 Explanation: 
Step-1: 2's complement number are weighted number. So, 1101010= -1*(0.26) + 1*25 + 1*23 + 1*21 = -64 + 32 +8 +2 = -22 Hence (1101010)=(-22) in 2's complement form
Question 151
Which of the following weights makes the complement operation easier in BCD form ?
A
8-4-2-1
B
Excess-3
C
2-4-2-1
D
3-2-1-0
       Digital-Logic-Design       Number-Systems       NIELIT Technical Assistant_2016_march
Question 152
Let a n a n−1 ...a 1 a 0 be the binary representation of an integer b . The integer b is divisible by 3 if
A
the number of one’s is divisible by 3
B
the number of one’s is divisible by 3, but not by 9
C
the number of zeroes is divisible by 3
D
the difference of alternate sum, i.e., ( a 0 + a 2 + . ..) − ( a 1 + a 2 + . ..) is divisible by 3
       Digital-Logic-Design       Number-Systems       NIELIT Technical Assistant_2016_march
Question 153
The decimal number equivalent of (4057.06)8 is
A
2095.75
B
2095.075
C
2095.937
D
2095.0937
       Digital-Logic-Design       Number-Systems       UGC NET CS 2010 Dec-Paper-2
Question 153 Explanation: 

Question 154
12-bit 2’s complement of –73.75 is
A
01001001.1100
B
11001001.1100
C
10110110.0100
D
10110110.1100
       Digital-Logic-Design       Number-Systems       UGC NET CS 2010 Dec-Paper-2
Question 154 Explanation: 


Question 155
Encoding of data bits 0011 into 7-bit even Parity Hamming Code is
A
0011110
B
0101110
C
0010110
D
0011100
       Digital-Logic-Design       Number-Systems       UGC NET CS 2010 Dec-Paper-2
Question 155 Explanation: 
(m) Data = 0011
No. of parity bits needed is decided using
2p≥ m+p+1
p=3
Bit pattern:

P1(check even parity at 1, 3, 5, 7 bit) = 0
P2(check even parity at 2, 3, 6, 7 bit) = 1
P4(check even parity at 4, 5, 6, 7 bit) = 1
So encoded data = 0011110
Question 156
What is decimal equivalent of BCD 11011.1100 ?
A
22.0
B
22.2
C
20.2
D
21.2
E
None of the above
       Digital-Logic-Design       Number-Systems       UGC NET CS 2010 June-Paper-2
Question 156 Explanation: 
The question is not properly framed.
In the question it is mentioned that 11011.1100 is BCD.
We know that BCD code is different from Binary number system.
Each of the decimal digit has a 4-bit binary code or in other words every block of 4 bits has a corresponding decimal digit.
0000- 0
0001- 1
:
:
1001- 9
1010 - 1111 are not valid codes.
If the given code 11011.1100 is BCD then divide it into blocks of 4 bits.
0001 1011. 1100 (Note: Appending zeros on the left side will not change value.)
But 1011 is not a valid BCD code.
Question 157
In order that a code is ‘t’ error correcting, the minimum Hamming distance should be :
A
t
B
2t-1
C
2t
D
2t+1
       Digital-Logic-Design       Number-Systems       UGC NET CS 2009-June-Paper-2
Question 157 Explanation: 

Question 158
The octal equivalent of hexadecimal (A.B)16 is:
A
47.21
B
12.74
C
12.71
D
17.21
E
None of the above
       Digital-Logic-Design       Number-Systems       UGC NET CS 2009-June-Paper-2
Question 158 Explanation: 

Question 159
The answer of the operation (10111)2* (1110)2 in hex equivalence is
A
150
B
241
C
142
D
101011110
       Digital-Logic-Design       Number-Systems       UGC NET CS 2009 Dec-Paper-2
Question 159 Explanation: 

Question 160
How many 1’s are present in the binary representation of 3 × 512 + 7 × 64 + 5 × 8 + 3
A
8
B
9
C
10
D
11
       Digital-Logic-Design       Number-Systems       UGC NET CS 2009 Dec-Paper-2
Question 160 Explanation: 
Given expression is 3 × 512 + 7 × 64 + 5 × 8 + 3
We can write above statement based on precedence is (3 * 512) + (7 * 64) + (5 * 8) + 3
Step-1: 1536+448+40+3
Step-2: (2027)10
Step-3: Equivalent binary number is (‭011111101011‬)2
Note: Total number of 1’s is 9.
Question 161
If a code is ‘t’ error detecting, the minimum hamming distance should be equal to :
A
t-1
B
t
C
t+1
D
2t+1
       Digital-Logic-Design       Number-Systems       UGC NET CS 2008 Dec-Paper-2
Question 161 Explanation: 

Question 162
If a code is t-error correcting, the minimum Hamming distance is equal to :
A
2t+1
B
2t
C
2t-1
D
t-1
       Digital-Logic-Design       Number-Systems       UGC NET CS 2008-june-Paper-2
Question 162 Explanation: 

Question 163
The octal equivalent of the hexadecimal number FF is :
A
100
B
150
C
377
D
737
       Digital-Logic-Design       Number-Systems       UGC NET CS 2008-june-Paper-2
Question 163 Explanation: 
Step-1: Convert FF into binary number
(FF)16 = (1111 1111)2
Step-2: Divide binary number into 3 segments from LSB(Least significant bit).

Question 164
2’s complement of -100 is :
A
00011100
B
10011101
C
10011100
D
11100100
       Digital-Logic-Design       Number-Systems       UGC NET CS 2007 June-Paper-2
Question 164 Explanation: 

Question 165
How many 1’s are present in the binary representation of 15*256 + 5*16 + 3 :
A
8
B
9
C
10
D
11
       Digital-Logic-Design       Number-Systems       UGC NET CS 2007 June-Paper-2
Question 165 Explanation: 
Step-1: The precedence to given decimal number is (15*256) + (5*16) + 3
Step-2: 3840+80+3=(3923)10
Step-3: Convert decimal number into binary
(3923)10 = (‭111101010011‬ )2
Step-4: Total number of 1’s are 8.
Question 166
Let C be a binary linear code with minimum distance 2t + 1 then it can correct upto _____ bits of error.
A
t + 1
B
t
C
t - 2
D
t / 2
       Digital-Logic-Design       Number-Systems       UGC NET CS 2017 Jan- paper-3
Question 166 Explanation: 
A binary linear code with minimum distance 2t + 1 then it can correct up to ‘T’ bits of error.
Question 167
Consider the equation (146)​ b​ + (313)​ b-2​ = (246)​ 8​ . Which of the following is the value of b?
A
8
B
7
C
10
D
16
       Digital-Logic-Design       Number-Systems       UGC NET June-2019 CS Paper-2
Question 167 Explanation: 
(146)7+(317)7-2=(246)8
Substitute 7 in b,
(146)7+(317)7-2=(246)8
(146)7+(317)5=(246)8
(146)7=1*72+4*71+6*70
=49+28+6
=83
(317)5=3*52+1*51+7*50
=75+5+7
= 87
(246)8=2*82+4*81+6*80
=128+32+6
= 166
∴ (146)7+(317)5=(246)8
=83+87
=166
166=166
LHS = RHS equal only if b is 7.
Question 168
How many bit strings of length ten either start with a 1 bit or end with two bits 00?
A
320
B
480
C
640
D
768
       Digital-Logic-Design       Number-Systems       UGC NET June-2019 CS Paper-2
Question 168 Explanation: 
→ Number of bit strings of length 10 that start with 1: 29 = 512.
→ Number of bit strings of length 10 that end with 00: 28 = 256
→ Number of bit strings of length 10 that start with 1 and end with 00: 27 = 128
→ Applying the subtraction rule, the number is 512+256-128 = 640
Question 169
For the 8 - bit word 00111001, the check bits stored with it would be 0111. Suppose when the wo​rd is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory?
A
10011001
B
00011001
C
00111000
D
11000110
       Digital-Logic-Design       Number-Systems       UGC NET CS 2015 June Paper-3
Question 169 Explanation: 
Here, the check bits stored along with data are 0111 and the check bits calculated at receiver side are 1101. Both check bits are not same is the indication of error bit in data whose position can be calculated by performing the EX-OR operation.

So data bits are:
00011001
Question 170
The equivalent hexadecimal notation for octal number 2550276 is:
A
FADED
B
AEOBE
C
ADOBE
D
ACABE
       Digital-Logic-Design       Number-Systems       UGC NET CS 2015 June Paper-3
Question 170 Explanation: 
Question 171
In a binary Hamming Code the number of check digits is r then number of message digits is equal to:
A
2​ r​ - 1
B
2​ r​ - r - 1
C
2​ r​ - r + 1
D
2​ r​ + r - 1
       Digital-Logic-Design       Number-Systems       UGC NET CS 2015 June Paper-3
Question 171 Explanation: 
2 p = m + p + 1 ---(1)
where p is the no. of parity bits.
m = no. of msg. digits
No. of check bits in a msg = p
In question no. of check-bits are given as ‘r’
So r = p
Put it in equation (1)
2 r = m + r + 1
m = 2 r − r − 1
Question 172
Which of the following binary codes for decimal digits are self complementing?
(a) 8421 code
(b) 2421 code
(c) excess-3 code
(d) excess-3 gray code
Choose the correct option:
A
(a) and (b)
B
(b) and (c)
C
(c) and (d)
D
(d) and (a)
       Digital-Logic-Design       Number-Systems       UGC-NET DEC-2019 Part-2
Question 172 Explanation: 



Note: 8421 is not self complementing. Self complementing is nothing but reverse order from starting and last number.
Ex: 0 value is 0000
9 value is 1111
So, it is self complementing.
Note: Excess-3 and Gray code is different. So, it is not relevant.
Question 173

Let the population of chromosomes in genetic algorithm is represented in terms of binary number. The strength of fitness of a chromosome in decimal form, x, is given by



The population is given by P where:

P = {(01101, (11000), (01000), (10011)}

The strength of fitness of chromosome (11000) is ___________
A
24
B
576
C
14.4
D
49.2
       Digital-Logic-Design       Number-Systems       UGC-NET DEC-2019 Part-2
Question 174

The following program is stored in the memory unit of the basic computer. Give the content of accumulator register in hexadecimal after the execution of the program.


A
A1B4
B
81B4
C
A184
D
8184
       Digital-Logic-Design       Number-Systems       UGC-NET DEC-2019 Part-2
Question 174 Explanation: 








Question 175

Given following equation:

(142)b + (112)b-2 = (75)8. Find base b.
A
3
B
6
C
7
D
5
       Digital-Logic-Design       Number-Systems       UGC-NET DEC-2019 Part-2
Question 175 Explanation: 
Option-A: Definitely wrong. Because 142 having 4 in a number but base is 3 only. So,eliminate it.
Option-B:
Step-1: Converting (142)5 to base 10.
1*52=25
4*51=20
2*50=2
Adding all to get (47)10
Step-2: Converting (112)3 to base 10.
1*32=9
1*31=3
2*30=2
Adding all to get (14)10
Step-3: Convert (61)10 to (?)8 // 47+14=61. Both are in decimal so we can add directly.
8|_61
8|_7 → 5
8|_7 → 7
Ans:(75)8
No need to check Option-C & D.
Question 176
The decimal number 395, when converted into binary occupies______binary digits, whereas when it is represented using BCD codes using BCD codes, occupies_____ binary digits.
A
12; 9
B
7; 12
C
9; 12
D
12; 7
       Digital-Logic-Design       Number-Systems       CIL Part - B
Question 176 Explanation: 
Decimal Number= (395)10 =(110001011)2 =(613)8 = (18B)16
Total 9 digits of binary number required to convert decimal number
Question 177

What will be the minimum Hamming distance for the following coding scheme?


A
1
B
2
C
3
D
4
       Digital-Logic-Design       Number-Systems       CIL Part - B
Question 177 Explanation: 
The minimum Hamming distance is the smallest Hamming distance between all possible pairs.
The Hamming distance can easily be found if we apply the XOR operation on the two words and count the number of 1s in the result. Note that the Hamming distance is a value greater than zero
The minimum hamming distance is between the words 10 and 11
10101 exor 11100 = 01001 which is “2”.
Question 178
Which of the following is a seven-bit code?
A
Biquinary code
B
BCD code
C
2421 code
D
Excess-3 code
       Digital-Logic-Design       Number-Systems       CIL Part - B
Question 178 Explanation: 
Bi-quinary coded decimal is a numeral encoding scheme used in many abacuses and in some early computers, including the Colossus.
The term bi-quinary indicates that the code comprises both a two-state (bi) and a five-state (quinary) component.
Question 179
An array of two byte integers is stored in big endian machine in byte addresses as shown below. What will be its storage pattern in little endian machine?
Address Data
0 X 104 78
0 X 103 56
0 X 102 34
0 X 101 12
A
0 X 104 12
0 X 103 56
0 X 102 34
0 X 101 12
B
0 X 104 12
0 X 103 34
0 X 102 56
0 X 101 78
C
0 X 104 56
0 X 103 78
0 X 102 12
0 X 101 34
D
0 X 104 56
0 X 103 12
0 X 102 78
0 X 101 34
       Digital-Logic-Design       Number-Systems       ISRO CS 2020
Question 179 Explanation: 
In little endian the LSB parts of the data are stored first, whereas in big endian the MSB parts are stored first.
In the given question each integer is two bytes. 0x101, 0x102 are part of one word.
In the big endian 0x101 has 12 and 0x102 has 34. In the little endian 0x101 will have 34 and 0x102 will have 12.
Similarly in the big endian 0x103 has 56 and 0x104 has 78...in the little endian 0x103 will have 78 and 0x104 will have 56.
Question 180
Which of the following digital components are used for ‘odd’ parity bit generation and checking in the process of transmission error detection?
A
Gates required for parity generation differ from those required for parity checking
B
Exclusive-OR gates
C
RS-flipflop
D
Counters
       Digital-Logic-Design       Number-Systems       APPSC-2016-DL-CS
Question 180 Explanation: 
Ex-OR gates generate the output as ‘1’ when there are odd numbers of ‘1’ applied in input. Hence if we want to check odd parity then Ex-OR gates can be used.
Question 181
How many ‘1’s are there in the result produced by 8-bit 2’s complement addition of 29 and –46?
A
2
B
7
C
3
D
5
       Digital-Logic-Design       Number-Systems       APPSC-2016-DL-CS
Question 181 Explanation: 


Question 182
The micro-operation that performs division of a signed binary integer by 2 is ---------
A
Logical shift right
B
Logical shift left
C
Arithmetic shift right
D
Arithmetic shift left
       Digital-Logic-Design       Number-Systems       APPSC-2016-DL-CS
Question 182 Explanation: 
Division of signed binary integer by 2 is nothing but arithmetic shift right. Similarly ,multiplication of signed binary integer by 2 is nothing but arithmetic shift left.
Question 183

The gray code for decimal number 6 is equivalent

A
1100
B
1001
C
0101
D
0110
       Digital-Logic-Design       Number-Systems       APPSC-2016-DL-CA
Question 183 Explanation: 
The binary value of 6 is 0110.
The binary value b3, b2, b1, b0 and find the gray code g3, g2, g1, g0 based on the below concept, g3 = b3 = 0
g2 = b3 XOR b2 = 0 xor 1 = 1
g1 = b2 XOR b1 = 1 xor 1 = 0
g0 = b1 XOR b0 = 1 xor 0 = 1
Question 184

Excess-3 code of decimal number 2 is:

A
0101
B
1010
C
1100
D
0011
       Digital-Logic-Design       Number-Systems       CIL 2020
Question 184 Explanation: 
To find Excess-3 code, add 3 to the given decimal no. and then find the binary form of that no.
2+3 = 5
Binary form of 5 is 0101.
Question 185

MSB stands for

A
Middle significant bit
B
Measured Significant bit
C
Maximum significant bit
D
Most significant bit
       Digital-Logic-Design       Number-Systems       CIL 2020
Question 185 Explanation: 
MSB stands for Most Significant Bit. It is the left most bit of any binary string.
LSB stands for Least Significant Bit. It is the right most bit of any binary string.
Question 186

Which is the correct method to obtain 10's complement of a decimal number?

A
The minuend is added to the 10's complement of the subtrahend and carry is dropped
B
The 10's complement of a decimal number is equal to 9's complement of given number minus one
C
The subtrahend is added to the 10's complement of the minuend and carry is dropped
D
The 10's complement of a decimal number is equal to 9's complement of given number plus one
       Digital-Logic-Design       Number-Systems       CIL 2020
Question 186 Explanation: 
The 10’s complement of a decimal no. is equal to the 9’s complement of given number plus 1.
Question 187

The multiplicand register and multiplier register of hardware circuit implementation of booth's algorithm have (11101) and (1100). the result shall be:

A
(812)10
B
(-812)10
C
(-12)10
D
(12)10
       Digital-Logic-Design       Number-Systems       CIL 2020
Question 187 Explanation: 
In Booth’s multiplication the binary nos. are in 2’s complement form.
Hence the decimal value of multiplicand,
11101 = 101 = -3
The decimal value of multiplier,
1100 = 100 = -4
Hence the required result is,
-3 × -4 = 12
Question 188

The number 105 in decimal system is equal to

A
(140)5
B
(410)5
C
(042)5
D
(321)5
       Digital-Logic-Design       Number-Systems       APPSC-2012-DL CA
Question 188 Explanation: 

∴ Answer is (410)5
Question 189
A code with a Hamming distance d can :
A
Detect d bit errors and correct (d – 1) bit errors.
B
Detect (d – 1) bit errors only with no error correction.
C
Detect (d – 1) bit errors and correct (d – 1)/2 bit errors.
D
Detect and correct all d bit errors.
       Digital-Logic-Design       Number-Systems       TNPSC-2017-Polytechnic-CS
Question 189 Explanation: 
To detect t bit error hamming distance should be t+1. And to correct t bit error hamming distance should be 2t+1.
So using above solution most appropriate answer is option C.
Question 190
The Booth technique for recording multiply of +13 and –6 [01101 and 11010] is :
A
1110 0011 01
B
1110 1100 10
C
1110 1010 10
D
1110 0011 00
       Digital-Logic-Design       Number-Systems       TNPSC-2017-Polytechnic-CS
Question 190 Explanation: 
Firstly,
13 × -6 = -78
Let’s first find binary value of 78,

Now to get -78 let’s take 2’s complement of above binary no.,
10110010
which can be also written as,
1110110010
Question 191
Which of the following decimal numbers can be exactly represented in binary notation with a finite number of bits?
A
0.1
B
0.2
C
0.4
D
0.5
E
All the above
       Digital-Logic-Design       Number-Systems       TIFR PHD CS & SS 2019
Question 192
For any natural number n, an ordering of all binary strings of length n is a Gray code if it starts with 0n, and any successive strings in the ordering differ in exactly one bit (the first and last string must also differ by one bit). Thus, for n = 3, the ordering (000, 100, 101, 111, 110, 010, 011, 001) is a Gray code. Which of the following must be TRUE for all Gray codes over strings of length n?
A
the number of possible Gray codes is even
B
the number of possible Gray codes is odd
C
in any Gray code, if two strings are separated by k other strings in the ordering, then they must differ in exactly k +1 bits
D
in any Gray code, if two strings are separated by k other strings in the ordering, then they must differ in exactly k bits
E
none of the above
       Digital-Logic-Design       Number-Systems       TIFR PHD CS & SS 2017
Question 193
A certain computer represents floating point numbers by means of a signed magnitude fractional mantissa and an excess-16 base 4 exponent. The floating point format number is 110010111000. What is its decimal value?
A
-3.5
B
-14
C
-7/8
D
-2
       Digital-Logic-Design       Number-Systems       HCU PHD CS MAY 2019
Question 194
Let N be the sum of all numbers from 1 to 1023 except the five primes numbers: 2, 3, 11, 17, 31. Suppose all numbers are represented using two bytes (sixteen bits). What is the value of the least significant byte (the least significant eight bits) of N?
A
00000000
B
10101110
C
01000000
D
10000000
E
11000000
       Digital-Logic-Design       Number-Systems       TIFR PHD CS & SS 2012
Question 195
Consider the (decimal) number 182, whose binary representation is 10110110. How many positive integers are there in the following set?
{n ∈ N : n ≤ 182 and n has exactly four ones in its binary representation}
A
91
B
70
C
54
D
35
E
27
       Digital-Logic-Design       Number-Systems       TIFR PHD CS & SS 2020
There are 195 questions to complete.
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