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Test: Number Systems & Binary Codes - 1 - Electronics and Communication Engineering (ECE) MCQ


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Test: Number Systems & Binary Codes - 1 - Question 1

Find the Octal equivalent of hexa decimal number (FB2)16?

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 1

Concept:
A number system includes the number of independent digits used in the number system (the base), the place values of the different digits constituting the number, and the maximum numbers that can be written with the given number of digits.
Octal numbers: These numbers use digits from 0 to 7, total of 8 digits, and hence they are called octal number system. Octal numbers have base 8.
Hexadecimal numbers: The numbers which have base 16. It uses 16 different digits to represent the numbers. It is denoted as h16, where h is a hexadecimal number. It may be a combination of alphabets and numbers. Thus, it includes numbers from 0 to 9 and alphabets A to F

Calculation:
Hexadecimal to Octal Conversion:
Given, (FB2)16 is a hexadecimal number.
F → 1111, B → 1011, 2 → 0010
1111 1011 0010
Now group them from right to left, each having 3 digits.
111, 110, 110, 010
111 → 7, 110 → 6, 110 → 6, 010 → 2
Hence, FB216 = 76628

*Answer can only contain numeric values
Test: Number Systems & Binary Codes - 1 - Question 2

If (154)b / (14)b = (8)10, then what is the radix b is _______?


Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 2

Convert the given b-radix into a decimal system.
(154)b / (14)b = (8)10
(1 x b2 + 5 x b1 + 4 x b0)10 / (1 x b1 + 4 x b0)10 = (8)10 (b2 + 5 b + 4) / (b + 4) = 8
(b2 + 5 b + 4) = 8 (b + 4) 
b2 + 5b + 4 = 8b + 32
b2 - 3b - 28 = 0
b2 - 7b + 4b - 28 = 0
b(b-7) + 4(b - 7) = 0
b = -4 or b = 7
b = 7 Here b value should be greater than 5 because it maximum digit in b radix.
Hence the correct answer is 7.

Test: Number Systems & Binary Codes - 1 - Question 3

Convert (23)8 into its decimal number.

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 3

Octal to Decimal Conversion:

  • Step 1: Since an octal number only uses digits from 0 to 7, we first arrange the octal number with the power of 8.
  • Step 2: We evaluate all the power of 8 values such as 80 is 1, 81 is 8, etc., and write down the value of each octal number.
  • Step 3: Final step is to add the product of all the numbers to obtain the decimal number.

 
Application:
Step 1: Write 23 with the power of 8. Start from the right-hand side.
2 × 81 + 3 × 80
Step 2: Evaluate the power of 8 values for each octal number.
16 + 3 = 19 (Decimal Number)

Test: Number Systems & Binary Codes - 1 - Question 4

Which of the following is not an invalid BCD Code?

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 4

Binary Coded Decimal (BCD) code:

  • BCD is a way to express each of the decimal digits with a binary code.
  • In this code, each decimal digit is represented by its 4-bit binary equivalent.
  • Also, with four bits we can represent sixteen numbers (0000 to 1111)
  • But as there are 10 decimal digits from 0 to 9, BCD code uses only the first ten of these (0000 to 1001). The remaining six code combinations i.e. 1010 to 1111 are invalid in BCD.

Test: Number Systems & Binary Codes - 1 - Question 5

In Binary-coded Decimal (BCD) systems, the decimal number 81 is represented as

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 5
  • In BCD each decimal digit is represented by a 4-bit binary number.
  • The binary representation of 8 → 1000
  • The binary representation of 1 → 0001
  • (81)10 = 10000001
Test: Number Systems & Binary Codes - 1 - Question 6

If (110)x = (132)4, then x =

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 6

Concept:
Another number system to Decimal

In each and every representation of numbers with different bases, the maximum value in a number system with the base ‘r’ is r – 1. Since numbers vary from 0 to r – 1.
To convert any number which is in the different base to decimal number system we use binary-weighted representation.
Eg: let the number be (abc⋯ ⋯ yz)r
Now to convert the above number into the decimal number system
a × rn-1 + b × rn-2 + ⋯ ⋯ + y × r1 + z × r0
If we convert all numbers into decimal then we can perform normal addition and subtraction etc.
Application:
Given:

(110)x = (132)4,
The decimal equivalent of this number will be:
1 x x2 + 1 x x1 + 0 x x0 = 1 x 42 + 3 x 4 + 2 x 40
x2 + x + 0 = 16 + 12 + 2
x2 + x - 30 = 0 
On solving this quadratic equation we'll get:
x = 5, -6
Base can't be a negative so;
x = 5

Test: Number Systems & Binary Codes - 1 - Question 7

The given logic circuit represents:

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 7

The circuit is redrawn as:

y1 = x1
y2 = x1 ⊕ x2
y3 = x2 ⊕ x3
y4 = x3 ⊕ x4   
Let the input to the circuit be 1010
∴ For an input 1010, we get the output as 1111.
Similarly, let the input be 0110.
The output for 0110 input is 0101
Observations:

∴ The given circuit converts 4 bit binary to Grey code converter.

Test: Number Systems & Binary Codes - 1 - Question 8

2's complement of -7 is

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 8

Concept:

1. Signed magnitude representation uses the most significant bit (MSB) a sign bit.

  • If the sign bit is ‘0’ then the number is positive.
  • If the sign bit is ‘1’ then the number is negative.

The remaining bits represent the magnitude of the binary number.
2. 1’s complement representation:
It is a representation of a binary number obtained by toggling all bits in it i.e. transforming the 0 bit to 1 and the 1 bit to 0.
3. 2’s complement representation:
It is obtained by simply adding 1 to the 1’s complement of that binary number.
Calculation:
The binary form of 7 ↔ 0111
1's complement of -7 = 1(for sign) (000)(1's complement of 7)
= 1000
2's complement of -7 = 1's complement + 1 = 1000 + 1 = 1001

Test: Number Systems & Binary Codes - 1 - Question 9

Which of the following is an invalid state in 8-4-2-1 Binary Coded Decimal counter

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 9

Concept:

  • 8421 is also known as BCD code
  • BCD is a weighted code.
  • In weighted codes, each successive digit from right to left represents weights equal to some specified value, and to get the equivalent decimal number to add the products of the weights by the corresponding binary digit.

The following represents the 4-bit binary representation of decimal values: 

  • A number with an 'n' decimal digit will require 4k bits in BCD.
  • Ex- Decimal 396 is represented in BCD with 12 bits as 0011 1001 0110, with each group of 4 bits representing one decimal digit.
  • As the decimal digits relating to 1010, 1011, 1100, 1101, 1110, and 1111 do not exist, these are invalid.

In BCD valid states are from 0 to 9. So 1100 is an invalid state.

Test: Number Systems & Binary Codes - 1 - Question 10

State the octal equivalent of hexa decimal number (B34)16.

Detailed Solution for Test: Number Systems & Binary Codes - 1 - Question 10

Concept:
A number system includes the number of independent digits used in the number system (the base), the place values of the different digits constituting the number, and the maximum numbers that can be written with the given number of digits.
Octal numbers: These numbers use digits from 0 to 7, total of 8 digits, and hence they are called octal number system. Octal numbers have base 8.
Hexadecimal numbers: The numbers which have base 16. It uses 16 different digits to represent the numbers. It is denoted as h16, where h is a hexadecimal number. It may be a combination of alphabets and numbers. Thus, it includes numbers from 0 to 9 and alphabets A to F.
Calculation:
Hexadecimal to Octal Conversion:
Given, B3416 is a hexadecimal number.
B → 1011, 3 → 0011, 4 → 0100
1011 0011 0100
Now group them from right to left, each having 3 digits.
101, 100, 110, 100
101 → 5, 100 → 4, 110 → 6, 100 → 4
Hence, B3416 = 54648

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