Digital Logic Test 5

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Digital Logic Test 5

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Weekly Quiz Competition

Question 1
1 / -0

The representation of the value of a 16 bit unsigned integer X in hexadecimal number system is BF62. The representation of the value of X in octal number system is

A
123674

B
137542

C
145236

D
135742

Solution

X = (BF62)₁₆ = (1011 1111 0110 0010)₂
Clubbing it in the group of 3 bits:
(001 011 111 101 100 010)₂ = (137542)₈
Question 2
1 / -0

Let x be a 16-bit number in 2’s complement form where x is given as
1
1
1
1
1
1
1
1
1
1
0
0
1
1
1
1
What is the equivalent decimal number for x?

A
-49--49

B
-494--491

C
-497--496

D
-499--492

Solution

Concept:
In 2’s complement number system leading 1’s (only continuous 1’s) apart from the last can be discarded.
Calculation:
(1111111111001111)2 = (1001111)2
(1001111)2 = - 1×26 + 1 × 23 + 1 × 22 + 1 × 21 + 1 × 20 = - 49
Example:
(11101)2 = -1×24 + 1× 23 + 1 × 22 + 1× 20 = -3
(101)2 = -1×22 + 1 × 20 = - 3
Question 3
1 / -0

Consider a decimal number equal to x = -6410. If x has to be in 2's complement form and in the signed form then which of the following is/are true from the below-given options?

A
x will take minimum of 7 bits in 2's complement form

B
x will take minimum of 8 bits in 2's complement form

C
x will take minimum of 9 bits in the signed form

D
x will take minimum of 8 bits in the signed form

Solution

A range of numbers in 2’s complement form:
\( \text{0 to } {2^{n - 1}} - 1 \text{ ...for positive values}\)
\(- {2^{n - 1}} \text{ to } -1\text{ ...for negative values}\)
If n = 7, then a range of values by 7-bit signed 2's complement is from -64 to 63 for negative values.
For numbers in 2's complement form,
-6410 = (1000000)2 \(\Rightarrow\) (-64)
For numbers in signed magnitude,
-6410 = (1100 0000)2 \(\Rightarrow\) (-64) // 1's bit represents negative
Question 4
1 / -0

What is the value of y if the given expression is (y)b + (145)b = (412)b and b = 11?

A
278-278

B
2784-2781

C
2787-2786

D
2789-2782

Solution

Calculation:
(145)_b= (145)₁₁ = 1 × 11²+ 4 × 11 + 5 = (170)₁₀
(412)b = (412)11 = 4 × 112 + 1 × 11 + 2 = (497)10
(y)b = (278)11 = 2 × 112 + 7 × 11 + 8 = (327)₁₀
170 + 327 = 497
LHS = RHS
Approach:
(170)10 + (y)b = (497)10
(y)b= 327
11
327

11
29
8
9
2
7

0
2
Read from bottom to up:
(y) = (278)
Question 5
1 / -0

The floating point number represented by 0xC1700000 in IEEE 754 single precision is ____________

A
-15--15

B
-154--151

C
-157--156

D
-159--152

Solution

Concept:
32-bit floating-point representation of a binary number in IEEE- 754 is
Sign (1 bit)
Exponent (8 bit)
Mantissa bit (23 bits)

Calculation:
0xC1700000 is equal to
1100 0001 0111 0000 0000 0000 0000 0000
Here, the sign bit is 1. So, the number is negative
1 10000010 11100000000000000000000
Mantissa bits M = 11100000000000000000000
Exponent bits = E = 10000010 = 130 (in decimal)
In IEEE-754 format, 32-bit (single precision)
(-1)s × 1.M × 2130 – 127
= (-1)1 × 1.111 × 2130 – 127
= -1111 × 2^-3 × 23
= -15
Question 6
1 / -0

The Booth technique for recording multiply of -15 and +10 [110001 and 01010] is:

A
1101 1100 10

B
1011 1010 10

C
1101 1010 10

D
1101 1011 00

Solution

Concept:
(-15)10 → (111001)2
(01010)10 → (11010)2
The given number is in 2’s complement
Reason
(110001)2 = 2^-5 + 24 + 20 = -32 + 16 + 1 = 15
Multiply
-15 × 10 = -150
Convert 78 in binary:

Remainder
2
150

2
75
0
2
37
1
2
18
1
2
9
0
2
4
1
2
2
0
2
1
0
0 1

(150) = (1001110) → (0010 0101 10)
2’s complement of (-150) = 1101 1010 10
Tips and Tricks:
By taking the option and converting into decimal
1101 1010 10 = - 29 + 28 + 26 + 25 + 23 + 21 = -150
Question 7
1 / -0

Consider a given number system in which P = 10, Q = 11, R = 12, …, U = 15 is given by expression
16⁵ × U + 16⁴ × P + 16² × R + 16 × 7 + T
The number of 0’s in the unsigned binary representation of the number is

A
10-10

B
104-101

C
107-106

D
109-102

Solution

(UP0R7T)₁₆= 16⁵ × U + 16⁴ × P + 16³ × 0 + 16² × R + 16 × 7 + T
2⁴ = 16 (4 bits needed to represent a digit)
U → 15 → 1111
P → 10 → 1010
0 → 0000
R → 12 → 1100
7 → 0111
T → 14 → 1110
Number of 0’s = 0 + 2 + 4 + 2 + 1 + 1 = 10
Question 8
1 / -0

Consider the unsigned 8-bit point binary number representation below:
X7X6X5X4 .X3X2X1X0
where the position of the binary point is between X4 and X3. X7 is the MSB. Which of the decimal numbers can be represented exactly in the below given representation:

A
30.1875

B
15.6875

C
4.5125

D
9.8125

Solution

Option 1: 30.1875
30 = 11110
Since 5 bits is consumed by 30 it is out of range ∴ it cannot be represented.
Option 2: 15.6875
15.6875 = 1111. 1101
∴ it can be represented.
Option 3: 4.5125
15.6875 = 1111. 1011
.5125 cannot be represented
Option 4: 9.8125
1001 = 1111. 1011
∴ it can be represented.
Question 9
1 / -0

Consider the equation (146)₇ = (x6)_y with x and y as unknown. What is the number of possible solutions?

A
2-2

B
24-21

C
27-26

D
29-22

Solution

1*7² + 4*7 + 6 = xy + 6
49 + 28 + 6 = xy + 6
xy = 77
Possible combinations:
x = 1, y = 77
x = 77, y = 1
x = 11, y = 7
x = 7, y = 11
Since the radix y should be greater than x, the possible solutions are:
x = 1, y = 77
x = 7, y = 11
Hence, number of possible solutions = 2

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11	327
11	29	8
9	2	7
	0	2

		Remainder
2	150
2	75	0
2	37	1
2	18	1
2	9	0
2	4	1
2	2	0
2	1	0
	0	1

Digital Logic Test 5

Result

Digital Logic Test 5

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Question 1

123674

137542

145236

135742

Solution

Question 2

-49--49

-494--491

-497--496

-499--492

Solution

Question 3

x will take minimum of 7 bits in 2's complement form

x will take minimum of 8 bits in 2's complement form

x will take minimum of 9 bits ​in the signed form

x will take minimum of 8 bits ​in the signed form

Solution

Question 4

278-278

2784-2781

2787-2786

2789-2782

Solution

Question 5

-15--15

-154--151

-157--156

-159--152

Solution

Question 6

1101 1100 10

1011 1010 10

1101 1010 10

1101 1011 00

Solution

Question 7

10-10

104-101

107-106

109-102

Solution

Question 8

30.1875

15.6875

4.5125

9.8125

Solution

Question 9

2-2

24-21

27-26

29-22

Solution

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x will take minimum of 9 bits in the signed form

x will take minimum of 8 bits in the signed form