Number-Systems
Question 1 |
In 16-bit 2's complement representation, the decimal number -28 is:
1111 1111 1110 0100 | |
1111 1111 0001 1100 | |
0000 0000 1110 0100 | |
1000 0000 1110 0100 |
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 ______.
0.502 | |
0.461 | |
0.402 | |
0.561 |
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:
n bits | |
n + 2 bits | |
n - 1 bits | |
n + 1 bits |
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?
None of (i), (ii), (iii), (iv) can be exactly represented
| |
Only (ii) cannot be exactly represented | |
Only (iii) and (iv) cannot be exactly represented | |
Only (i) and (ii) cannot be exactly represented |
= 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
2-f to 2i | |
2-f to (2i - 2-f) | |
0 to 2i | |
0 to (2i - 2-f ) |
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
136251 | |
736251 | |
571247 | |
136252 |
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:
The decimal value closest to this floating-point number is
1.45 × 101 | |
1.45 × 10-1 | |
2.27 × 10-1 | |
2.27 × 101 |

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=________.
8 | |
9 | |
10 | |
11 |
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:
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
p=3 and q=1 | |
p=3 and q=2 | |
p=4 and q=1 | |
p=4 and q=2 |
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 __________.
-11 | |
-12 | |
-13 | |
-14 |
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 _________.
1 | |
2 | |
3 | |
4 |
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
5 | |
6 | |
7 | |
8 |
(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 __________.
3 | |
5 | |
6 | |
7 |
(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
C1640000H | |
416C0000H | |
41640000H | |
C16C0000H |
(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:
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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
P+Q | |
![]() | |
P⨁Q | |
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⊕ 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
-256 | |
-128 | |
-127 | |
0 |
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?
xy+x'y' | |
x⊕y' | |
x'⊕y | |
x'⊕y' |
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
fraction bits of 000…000 and exponent value of 0 | |
fraction bits of 000…000 and exponent value of −1 | |
fraction bits of 100…000 and exponent value of 0 | |
no exact representation |
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
(C3D8)16 | |
(187B)16 | |
(F878)16 | |
(987B)16 |
(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
(1217)16
| |
(028F)16 | |
(2297)10 | |
(0B17)16 |
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
the normalized value 2 - 127 | |
the normalized value 2 - 126 | |
the normalized value + 0 | |
the special value + 0 |

Question 23 |
Let r denote number system radix. The only value(s) of r that satisfy the equation is/are
decimal 10 | |
decimal 11 | |
decimal 10 and 11 | |
any value > 2 |
(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?
g0(h3h2h1h0) = Σ(1,2,3,6,10,13,14,15) | |
g1(h3h2h1h0) = Σ(4,9,10,11,12,13,14,15) | |
g2(h3h2h1h0) = Σ(2,4,5,6,7,12,13,15) | |
g3(h3h2h1h0) = Σ(0,1,6,7,10,11,12,13)
|

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:
- 2n-1 to (2n-1 - 1) | |
- (2n-1 - 1) to (2n-1 - 1) | |
- 2n-1 to 2n-1
| |
- (2n-1 + 1) 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
1AF | |
D78 | |
D71 | |
32F |
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:
0D 24 | |
0D 4D | |
4D 0D | |
4D 3D |
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
0A 20 | |
11 34 | |
4D D0
| |
4A E8 |
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
8, 16 | |
10, 12 | |
9, 13 | |
8, 11 |
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)
9.51 and 10.0 respectively
| |
10.0 and 9.51 respectively
| |
9.51 and 9.51 respectively | |
10.0 and 10.0 respectively |
= (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
1100 0100
| |
1001 1100
| |
1010 0101
| |
1101 0101 |
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?
11100111 | |
11100100 | |
11010111 | |
11011011 |
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?
2-40 | |
2-9 | |
222 | |
231 |
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
is equivalent to the binary value 0.1 | |
is equivalent to the binary value 0.01 | |
is equivalent to the binary value 0.00111… | |
cannot be represented precisely in binary |
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
1111 | |
11111 | |
111111 | |
10001 |
-15 = 11111
1's complement = 10000
2's complement = 10001
Question 36 |
Sign extension is a step in
floating point multiplication | |
signed 16 bit integer addition | |
arithmetic left shift | |
converting a signed integer from one size to another |
Question 37 |
In 2’s complement addition, overflow
is flagged whenever there is carry from sign bit addition | |
cannot occur when a positive value is added to a negative value | |
is flagged when the carries from sign bit and previous bit match | |
None of the above |
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.
Theory of Explanation is given below. |
Question 39 |
The 2’s complement representation of (-539)10 in hexadecimal is
ABE | |
DBC | |
DE5 | |
9E7 |
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?
f = x1' + x2 | |
f = x1'x2 + x1x2' | |
f = x1x2 + x1'x2' | |
f = x1 + x2' |
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
01010101 | |
11010101 | |
00101011 | |
10101011 |
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
X = 1.0, Y = 1.0 | |
X = 1.0, Y = 0.0 | |
X = 0.0, Y = 1.0 | |
X = 0.0, Y = 0.0 |
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
0 – 100 + 1000 | |
0 – 100 + 100 - 1 | |
0 – 1 + 100 – 10 + 1 | |
00 – 10 + 100 - 1 |

Question 44 |
Zero has two representations in
Sign magnitude | |
1’s complement | |
2’s complement | |
None of the above | |
Both A and B |
+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
226 | |
-98 | |
76 | |
-30 |
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?
80% | |
20% | |
60% | |
40% |
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:
10 | |
8 | |
5 | |
6 |
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
0 to 1 | |
0.5 to 1 | |
2-23 to 0.5 | |
0.5 to (1-2-23) |
Question 49 |
The number of 1’s in the binary representation of
(3*4096 + 15*256 + 5*16 + 3) are:
8 | |
8 | |
10 | |
12 |
= (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 ≤ _________
-2n-1 to 2n-1 - 1 |
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?
0x40800000 | |
0x83400000 | |
0xC8500000 | |
0xC0800000 |

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) 6.625, (b) (45E)H |
= 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 ________.
9 |
(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 _________
c1 = b1a0 ⊕ a1b0 |

⇒ c1 = b1a0 ⊕ a1b0
Question 55 |
The exponent of a floating-point number is represented in excess-N code so that:
The dynamic range is large. | |
The precision is high. | |
The smallest number is represented by all zeros. | |
Overflow is avoided. |
Question 56 |
The following bit pattern represents a floating point number in IEEE 754 single precision format
110000011101000000000000000000000The value of the number in decimal form is
-10 | |
-13 | |
-26 | |
None of these |
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:
1, 1, 0 | |
1, 0, 0 | |
0, 1, 0 | |
1, 0, 1 |

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 1101How many additions/Subtractions are required for the multiplication of the above two numbers?
6 | |
8 | |
10 | |
12 |
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 =
(135103.412)O | |
(564411.412)O | |
(564411.205)O | |
(135103.205)O |
= 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
0001 and an overflow | |
1001 and no overflow | |
0001 and no overflow | |
1001 and an overflow |
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
2Y and Y | |
-2Y and 2Y | |
-2Y and 0 | |
0 and Y |

⇒ -2Y and 0
Question 62 |
(34.4)8 × (23.4)8 evaluates to
(1053.6)8 | |
(1053.2)8 | |
(1024.2)8 | |
None of these |
(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
(A72E)16 and (22130232)4 | |
(A72E)16 and (22131122)4 | |
(A73E)16 and (22130232)4 | |
(A62E)16 and (22120232)4
|
= (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
1 and 3 | |
1, 2 and 3 | |
2 and 4 | |
1, 2, 3 and 4 |
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 |
3+n ternary digits | |
2n/3 ternary digits | |
n(log23) ternary digits | |
n(log32 ) ternary digits |
→ 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 |
subtract 0011 from the sum | |
add 0011 to the sum
| |
subtract 0110 from the sum
| |
add 0110 to the sum
|
Example:
x+3
y+3
-------
(x+y+6)
Here, sum is excess-6. Hence, subtract 0011 to make it excess-3.
Question 67 |
8 | |
9 | |
10 | |
12 |
= (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 |
0.60 | |
0.52 | |
0.54 | |
0.50 |
= (0.6)8
Option (A) is correct.
Question 69 |
1 × 10-128 and 215× 1015
| |
1 × 10-256 and 215× 10255 | |
1 × 10-128 and 215× 10127 | |
1 × 10-128and 215– 1 × 10127 |

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 |
CD73E | |
ABD3F | |
7CDE3 | |
FA4CD |
7 C D E 3
(7CDE3)16
Question 71 |
4 | |
5 | |
6 | |
7 |
Question 72 |
[log(n)] + 1 bits | |
[log (n-1)) + 1 bits | |
[log (n+1)] + 1 bits | |
None of the above |

Question 73 |
10 | |
8 | |
6 | |
5 |
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 |
n bits | |
(n+3) bits | |
(n+2) bits | |
(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)16 | |
(028F)16 | |
(2297)1o | |
(0B17)16 |
=(0010 1000 1111)2
=(2 8 F)16
Question 76 |

Excess-3 code | |
Gray code | |
BCD code | |
Hamming Code |
X2= Y1⊕ Y2
X3= Y2 ⊕ Y3
Question 77 |
224174 | |
425174 | |
6173 | |
225174 |
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 |
Cyclic Redundancy Code | |
Weighted Code | |
Self-Complementing Code | |
Algebraic Code |
→ 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 |
1010 | |
0101 | |
1000 | |
1001 |

(Decimal value of maximum 4-bit number +1 )/2= (15+1)/2=8
Question 80 |
0001 and an overflow
| |
1001 and no overflow
| |
0001 and no overflow | |
1001 and an overflow
|
Another number is 1101(-3 is decimal value)
Adding of -3 and 4, the result is 1 and there is no overflow
Question 81 |
28 | |
-15 | |
-26 | |
-28 |
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 |
5 | |
6 | |
7 | |
8 |
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 |
212 | |
212-1 | |
1012 | |
103 |
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
t+1 | |
t−2
| |
2t−1 | |
2t+1 |
Question 85 |
The hexadecimal equivalent of the binary integer number 110101101 is :
D 2 4
| |
1 B D
| |
1 A E | |
1 A D |
24 = 16
So, 4-bits in binary will represent one integer in Hexadecimal.
So,

Question 86 |
-128 to +127 | |
-128 to +128 | |
-999999 + +999999 | |
none of these |
Question 87 |
Cyclomatic redundancy code | |
Weighted code | |
Self complementing code | |
algebraic code |
Question 88 |
In computers, subtraction is generally carried out by
1’s complement
| |
10’s complement
| |
2’s complement
| |
9’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 |
30 | |
60 | |
90 | |
120 |
Therefore, it takes approximately above 90 binary numbers. So, 120 is correct answer.
Question 90 |
75 16 | |
65 16 | |
5E 16 | |
10 16 |
(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 |
2 -128 to (1 – 2 –23 ) * 2 127 | |
(1 – 2 –23 ) * 2 -127 to 2 128 | |
(1 – 2 –23 ) * 2 –127 to 2 23> | |
2 –129 to (1 – 2 –23 ) * 2 127 |
Question 92 |
1001 | |
1010 | |
1011 | |
1100 |
12 equivalent binary number is 1100.
Question 93 |
31845 | |
41117 | |
41052 | |
32546 |
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 |

2.5 | |
3.0 | |
3.5 | |
4.5 |
→ 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 |
– 2n – 1 to 2n – 1 – 1 | |
– (2n – 1 – 1) to (2n – 1 – 1) | |
– 2n – 1 to 2n – 1 | |
– (2n – 1 + 1) to (2n – 1 – 1) |
Question 96 |
ASCII | |
BCD | |
EBCDIC | |
Gray |
→ 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 |
8 | |
9 | |
10 | |
12 |
=(12,288+3840+80+3)
=(16211)10
=(0011111101010011)2
Total number of 1’s in binary representation is 10.
Question 98 |
200 | |
213 | |
246 | |
277 |
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
20.33335
| |
24.45453
| |
16.35353
| |
20.53333 |
= 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 |
50 | |
74 | |
40 | |
60 | |
None of these |
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 2n -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 = 13.18 | |
4 | |
5 | |
6 |
(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?
(2059)16
| |
(2159)16
| |
(2195)16
| |
(2157)16
|
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 |
Division by 2 | |
Addition by 2
| |
Multiplication by 2 | |
Subtraction by 2 |
→ 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 |
1111 1111 1110 1011 | |
1111 1111 1110 1001 | |
1111 1111 1110 0111 | |
1111 1111 1110 1000 |
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 |
50 | |
74 | |
40 | |
None of the above |
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 |
Octal code | |
Binary Code | |
Gray code | |
Excess-3 Code |
→ Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.
Question 107 |
32,64 and 80 | |
32,64 and 128 | |
16,32 and 64 | |
16,32 and 80 |

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 |
26 | |
36 | |
46 | |
56 |
→ 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?
![]() | |
![]() | |
![]() | |
![]() |
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 |
1 | |
2 | |
3 | |
4 |
→ In the question, the unique digits are 0,1 and X(possible x value is 2) then number system is 3;
Question 111 |
0000000 | |
1111111 | |
2222222 | |
12121212 |
● The data is 0000 and Hamming(7,4) transmitted is 0000000
Question 112 |
16 | |
10 | |
12 | |
13 |
So, total 13 bits required to represent 4096 decimal number.
Question 113 |
The signed 2’s complement representation of -33 is:
11011111
| |
00100001 | |
01011111
| |
10100001
|
Step-2: Perform 2’s complement.

Question 114 |
Using signed 2’s complement subtraction the result of 11111010-11110011 is:
10000111 | |
00000111 | |
10001101 | |
00001101 |
-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 |
32 bits | |
8 bits | |
16 bits | |
24 bits |
→ Sign bit: 1 bit
→ Exponent width: 8 bits
→ Significand precision: 24 bits (23 explicitly stored)
Question 116 |
(10208)10 | |
(1276)10 | |
(2374)10 | |
(1272)10 |
Question 117 |
0|10001101|11111011100000000000000 | |
0|11001101|11111011100000000000000 | |
1|11001101|11111011100000000000000 | |
0|10001110|11111011100000000000000 |
3.248x104 =32480= 1111110111
= 1.111110111 x 214
Mantissa = 11111011100...00
Biased exponent = 14 +127= 141 = 10001101
Question 118 |
46252 | |
11450 | |
11452 | |
45250 |
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 |
7353 | |
1353 | |
5651 | |
5657 |
1 011 101 011
1 3 5 3
(1011101011) 2 = (1353) 8
Question 120 |
4AC8 | |
4CA8 | |
CCA8 | |
4CA4 |
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 |
(635) 4 | |
(32312) 4 | |
(21323) 4 | |
(1301) 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.

Question 122 |
0.5100098 | |
0.2096 | |
0.52 | |
0.4192 |

Question 123 |
2EE | |
2FF | |
4EF | |
4FE |
(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 |
11....11 | |
00.....00 | |
100.....0 | |
000......1 |
→ 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 |
(214.2) 10 and (D6.8)) 16 | |
(212.5) 10 and (D6.8)) 16 | |
(214.5) 10 and (D6.8)) 16 | |
(214.5) 10 and (D6.4)) 16 |
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 |
Sign-magnitude | |
1’s complement | |
2’s complement | |
9’s complement |
Question 127 |
16 | |
32 | |
48 | |
64 |
→ 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 |
1100 | |
1001 | |
1000 | |
1111 |
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 |
100101100 | |
1110001110001 | |
11110011 | |
10101010101010 |
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 |
4 | |
8 | |
10 | |
16 |
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 |
1010 | |
0101 | |
1000 | |
1001 |
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 |
150 | |
241 | |
142 | |
101011110 |
(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 |
8421 code | |
Gray code | |
Excess-3 code | |
7421 code |
→ 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 |
8 | |
9 | |
10 | |
11 |
(2027) 10 = (111 1110 1011) 2
Here, total number of 1’s are 9.
Question 135 |
0101 | |
0111 | |
1011 | |
1000 |
= 6*1 + 4*0 + 2*1 + -3*1
= 6+2-3
=5
Question 136 |
6 full adders and 2 half adders | |
5 full adders and 3 half adders | |
4 full adders and 3 half adders | |
5 full adders and 2 half adders |
→ 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 binary sum of a code and its 9’s complement is equal to 9. | |
It is a weighted code | |
Complement can be generated by inverting each bit pattern | |
The binary sum of a code and its 10’s complement is equal to 9 |

Question 138 |
0xC2206000 | |
0xC2006666 | |
0xC2006000 | |
0xC2206666 |
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 |
8 bits and 16 bits | |
16 bits and 32 bits | |
32 bits and 64 bits
| |
64 bits and 128 bits |
Question 140 |
2 | |
8 | |
10 | |
None of the above |
Question 141 |
216 – 1 | |
215 – 1 | |
216 | |
215 |
= 215 -1
Question 142 |
216 – 1 | |
215 – 1 | |
216 | |
215 |
= 215 -1
Question 143 |
signed magnitude form | |
1’s complement form | |
2’s complement form | |
none of the above |
1. Signed magnitude form
2. 1’s complement form
3. 2’s complement form
Question 144 |
0101 | |
0101 | |
0110 | |
1001 |
Question 145 |
2 16 − 1 | |
2 15 − 1 | |
2 16 | |
2 15 |
Question 146 |
3F2.89 | |
3F2.98 | |
2F3.89 | |
2F3.98 |
→ 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 |
01001101.0100 | |
01001101.0010 | |
10110010.1011 | |
10110010.1101 |


Question 148 |
ASCII | |
Gray Code | |
Excess-3 Code | |
BCD |
Question 149 |
111 | |
10 | |
110 | |
11 |
Question 150 |
-42 | |
-22 | |
-21 | |
-106 |
Question 151 |
8-4-2-1 | |
Excess-3 | |
2-4-2-1 | |
3-2-1-0 |
Question 152 |
the number of one’s is divisible by 3 | |
the number of one’s is divisible by 3, but not by 9 | |
the number of zeroes is divisible by 3 | |
the difference of alternate sum, i.e., ( a 0 + a 2 + . ..) − ( a 1 + a 2 + . ..) is divisible by 3 |
Question 153 |
2095.75 | |
2095.075 | |
2095.937 | |
2095.0937 |

Question 154 |
01001001.1100 | |
11001001.1100 | |
10110110.0100 | |
10110110.1100 |


Question 155 |
0011110 | |
0101110 | |
0010110 | |
0011100 |
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 |
22.0 | |
22.2 | |
20.2 | |
21.2 | |
None of the above |
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 |
t | |
2t-1 | |
2t | |
2t+1 |

Question 158 |
47.21 | |
12.74 | |
12.71 | |
17.21 | |
None of the above |

Question 159 |
150 | |
241 | |
142 | |
101011110 |

Question 160 |
8 | |
9 | |
10 | |
11 |
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 |
t-1 | |
t | |
t+1 | |
2t+1 |

Question 162 |
2t+1 | |
2t | |
2t-1 | |
t-1 |

Question 163 |
100 | |
150 | |
377 | |
737 |
(FF)16 = (1111 1111)2
Step-2: Divide binary number into 3 segments from LSB(Least significant bit).

Question 164 |
00011100 | |
10011101 | |
10011100 | |
11100100 |

Question 165 |
8 | |
9 | |
10 | |
11 |
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 |
t + 1 | |
t | |
t - 2 | |
t / 2 |
Question 167 |
8 | |
7 | |
10 | |
16 |
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 |
320 | |
480 | |
640 | |
768 |
→ 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 |
10011001 | |
00011001 | |
00111000 | |
11000110 |

So data bits are:
00011001
Question 170 |
FADED | |
AEOBE | |
ADOBE | |
ACABE |

Question 171 |
2 r - 1 | |
2 r - r - 1 | |
2 r - r + 1 | |
2 r + r - 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 |
(a) 8421 code
(b) 2421 code
(c) excess-3 code
(d) excess-3 gray code
Choose the correct option:
(a) and (b) | |
(b) and (c) | |
(c) and (d) | |
(d) and (a) |

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 ___________
24 | |
576 | |
14.4 | |
49.2 |
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.

A1B4 | |
81B4 | |
A184 | |
8184 |




Question 175 |
Given following equation:
(142)b + (112)b-2 = (75)8. Find base b.
3 | |
6 | |
7 | |
5 |
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 |
12; 9 | |
7; 12 | |
9; 12 | |
12; 7 |
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?

1 | |
2 | |
3 | |
4 |
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 |
Biquinary code | |
BCD code | |
2421 code | |
Excess-3 code |
The term bi-quinary indicates that the code comprises both a two-state (bi) and a five-state (quinary) component.
Question 179 |
Address Data
0 X 104 78
0 X 103 56
0 X 102 34
0 X 101 12
0 X 104 12 0 X 103 56 0 X 102 34 0 X 101 12 | |
0 X 104 12 0 X 103 34 0 X 102 56 0 X 101 78 | |
0 X 104 56 0 X 103 78 0 X 102 12 0 X 101 34 | |
0 X 104 56 0 X 103 12 0 X 102 78 0 X 101 34 |
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 |
Gates required for parity generation differ from those required for parity checking | |
Exclusive-OR gates | |
RS-flipflop | |
Counters |
Question 181 |
2 | |
7 | |
3 | |
5 |


Question 182 |
Logical shift right | |
Logical shift left | |
Arithmetic shift right | |
Arithmetic shift left |
Question 183 |
The gray code for decimal number 6 is equivalent
1100 | |
1001 | |
0101 | |
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:
0101 | |
1010
| |
1100
| |
0011 |
2+3 = 5
Binary form of 5 is 0101.
Question 185 |
MSB stands for
Middle significant bit
| |
Measured Significant bit | |
Maximum significant bit
| |
Most significant bit
|
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?
The minuend is added to the 10's complement of the subtrahend and carry is dropped | |
The 10's complement of a decimal number is equal to 9's complement of given number minus one
| |
The subtrahend is added to the 10's complement of the minuend and carry is dropped | |
The 10's complement of a decimal number is equal to 9's complement of given number plus one |
Question 187 |
The multiplicand register and multiplier register of hardware circuit implementation of booth's algorithm have (11101) and (1100). the result shall be:
(812)10 | |
(-812)10
| |
(-12)10
| |
(12)10
|
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
(140)5 | |
(410)5
| |
(042)5
| |
(321)5 |

∴ Answer is (410)5
Question 189 |
Detect d bit errors and correct (d – 1) bit errors. | |
Detect (d – 1) bit errors only with no error correction. | |
Detect (d – 1) bit errors and correct (d – 1)/2 bit errors. | |
Detect and correct all d bit errors. |
So using above solution most appropriate answer is option C.
Question 190 |
1110 0011 01 | |
1110 1100 10 | |
1110 1010 10 | |
1110 0011 00 |
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 |
0.1 | |
0.2 | |
0.4 | |
0.5 | |
All the above |
Question 192 |
the number of possible Gray codes is even | |
the number of possible Gray codes is odd | |
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 | |
in any Gray code, if two strings are separated by k other strings in the ordering, then they must differ in exactly k bits | |
none of the above |
Question 193 |
-3.5 | |
-14 | |
-7/8 | |
-2 |
Question 194 |
00000000 | |
10101110 | |
01000000 | |
10000000 | |
11000000 |
Question 195 |
{n ∈ N : n ≤ 182 and n has exactly four ones in its binary representation}
91 | |
70 | |
54 | |
35 | |
27 |