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       Video-Explanation
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       Video-Explanation
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       Video-Explanation
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,
b7 b6 b5 b4 b3 b2 b1 b0
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       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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]       Video-Explanation
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: 
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

(a) An asynchronous serial communication controller that uses a start stop scheme for controlling the serial I/O of a system is programmed for a string of length seven bits, one parity bit (odd parity) and one step bit. The transmission rate is 1200 bits/second.
(i) What is the complete bit stream that is transmitted for the string ‘0110101’?
(ii) How many such strings can be transmitted per second?

(b) Consider a CRT display that has a text mode display format of 80 × 25 characters with a 9 × 12 character cell. What is the size of the video buffer RAM for the display to be used in monochrome (1 bit per pixel) graphics mode?

A
Theory Explanation.
       Digital-Logic-Design       Number-Systems       GATE 1995
Question 51

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 52

Write a program in 8085 Assembly language to Add two 16-bit unsigned BCD(8-4-2-1 Binary Coded Decimal) number. Assume the two input operands are in BC and DE Register pairs. The result should be placed in the register pair BC. (Higher order register in the register pair contains higher order digits of operand)

A
Theory Explanation.
       Digital-Logic-Design       Number-Systems       GATE 1994
Question 53

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 53 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 54

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 54 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 55

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       Video-Explanation
Question 55 Explanation: 
Hexadecimal representation of a given no. is,
(9753)16
It's binary representation is,
1001011101010011
∴ The no. of 1's is 9.
There are 55 questions to complete.

Access quiz wise question and answers by becoming as a solutions adda PRO SUBSCRIBER with Ad-Free content

Register Now

If you have registered and made your payment please contact solutionsadda.in@gmail.com to get access