[1987: 1 M] The subtraction of a binary number Y from another binary number X, done by adding the 2’s complement of Y to X, results in a binary number without overflow. This implies that the result is:

(a) negative and is in normal form

(b) negative and is in 2’s complement form

(d) positive and is in 2’s complement form

Ans: (b) and (c)

Explanation

The negative is in the form of 2’s complement and positive is in the normal form .

[1993: 1 M] 2’s complement representation of a 16-bit number (one sign bit and 15 magnitude bits) if FFFF. Its magnitude in decimal representation is

(a) 0

(b) 1

(d) 65, 535

Ans: (b)

Explanation:- Given that 1 sign bits and 15 magnitude bits.

F F F F

1 1 1 1 = F

Taking 1’s complement of all F ,

0 0 0 0

0 0 0 1

= 0 0 0 1

= 1

So, option (b) is the correct answer.

[1997: 1 M] A signed integer has been stored in a byte using the 2’s complement format. We wish to store the same integer in a 16 bit word. We should

(a) copy the original byte to the less significant byte of the word and fill the more significant byte with zeros.

(b) copy the original byte to the more significant byte of the word and fill the less significant byte with zeros.

(c) copy the original byte to the less significant byte of the word and make each bit of the more significant byte equal to the most significant bit of the original byte.

(d) copy the original byte to the less significant bytes well as the more significant byte of the word.

Ans: (c)

Explanation

The MSB of the integer in 8 bit format should be repeated to expand the representation of 2’s complement form to 16 bit.

[1998: 1 M] An equivalent 2’s complement representation of the 2’s complement number 1101 is

(a) 110100

(b) 001111

(d) 111101

Ans: (d) 111101

Explanation:- The 2’s complement number is given 1101 and it is represented in 4 bit . Rest of the options are represented in 6 bit form. The 2’s complement of 1101 is 0011 i.e. (+3) .

Therefore, 0011 = 000011 (As per MSB if 0, zeroes are added if 1 is added, then the bits are extended) .

So, Taking 2’s complement of 000011 we get 111101

According to sign bit extension the sign bit can be extended towards left. So, option (d) is the correct answer.

[2001: 1 M] The 2’s complement representation of -17 is

(a) 101110

(b) 101111

(d) 110001

Ans: (b)

Explanation The 17 is represented by,

17 = 010001

Taking 2,s compliment we get

-17 = 101111

So, option (b) is the correct answer.

[2002: 1 M] 4-bit 2’s complement representation of a decimal number is 1000. The number is

(a) +8

(b) 0

(d) -8

Ans: (d) -8

Explanation:- Given decimal number is 1000.

MSB is 1 So, -ve number

Take 2’s complement for magnitude.

0111 + 0001 = 1000 (8)

So, option (d) is the correct answer.

[2004: 1 M] The range of signed decimal numbers that can be represented by 6-bit 1’s complement number is

(a) -31 to +31

(b) -63 to +63

(d) -32 to +31

Ans: (a)

Explanation

Range = -(2^n-1-1) to +(2^n-1-1)

= -(2^6-1-1) to +(2^6-1-1)

= -31 to +31

[2004: 2 M] 11001, 1001 and 111001 correspond to the 2’s complement representation of which one of the following sets of number?

(a) 25, 9 and 57 respectively

(b) -6, -6 and -6 respectively

(d) -25, -9 and -57 respectively

Ans: (c) -7, -7 and -7 respectively

Explanation:- Let us take the 2’s complement of the given numbers 11001, 1001 and 111001

11001 ⟶ 00111(+7), the 11001 is 2’s complement is representation is (-7)

1001 ⟶ 0111(+7)is in decimal form, so 1001 is 2’s complement representation of -7.

111001 ⟶ 000111(+7)

000111 is 7 in decimal form and 111001 is 2’s complement representation of -7.

∴ Numbers given in question in 2’s complement correspond to -7.

So, option (c) is the correct answer.

[2005: 1 M] Decimal 43 in Hexadecimal and BCD number system is respectively

(a) B2, 0100 0011

(b) 2B, 0100 0011

(d) B2, 0100, 0100

Ans: (b)

Explanation:- For finding the hexadecimal equivalent we will divide the decimal number 43 by 16.

∴ (2B)_H

By dividing we get, (2B) = Hexadecimal representation.

The 43 no. is represented as,

4 = 0100

3 = 0011

(43)₁₀= (01000011)_BCD

So, option (b) is the correct answer.

[2006: 2 M] A new binary coded Pentary (BCP) number system is proposed in which every digit of a base-5 number is represented by its corresponding 3-bit binary code. For example, the base-5 number 24 will be represented by its BCP code 010100. In this numbering system, the BCP code 100010011001 corresponds to the following number in base-5 system

(a) 423

(b) 1324

(d) 4231

Ans: (d)

Explanation:- Given BCP code is 100010011001

Now, grouping in binary of 3 bit

100010011001 → 4231

The first three bits i.e. 100 = 4

010 = 2 , 011=3 , 001 = 1

So, option (d) is the correct answer.

[2007: 1 M] X = 01110 and Y = 11001 are two 5-bit binary numbers represented in two’s complement format. The sum of X and Y represented in two’s complement format using 6 bits is

(a) 100111

(b) 001000

(d) 101001

Ans: (c)

Explanation:- Given that X = 01110 and Y = 11001 are two 5 bit binary numbers.

Adding X and Y

X + Y

X	=	0	1	1	1	0
Y	=	1	1	0	0	1
X+Y	=	0	0	1	1	1

Carry is discarded in the addition of numbers represented in 2’s complement form. X + Y in 6 bits is 000111.Therefore, option (c) is the correct answer.

[2008: 2 M] The two numbers represented in signed 2’s complement form are P = 11101101 and Q = 11100110. If Q is subtracted from P, the value obtained in signed 2’s complement form is

(a) 100000111

(b) 00000111

(d) 111111001

Ans: (b)00000111

Explanation:- Given that P=11101101 and Q = 11100110

We can notice that 1 is the most significant value in both P and Q . So both represents negative value.

Now taking 2’s complement of P and Q

2’s complement of P = 11101101 = (-19)₁₀

∵ 2’s complement of Q = 11100110 = (-26)₁₀

P – Q = -19-(-26) = -19+26 = 7 = (0000 0111)₂

So, option (b) is the correct answer.

[2014: 1 M, Set-2] The number of bytes required to represent the decimal number 1856357 in packed BCD (Binary Coded Decimal) form is ____.

Ans: 4

Explanation

For representing decimal number into BCD form, each decimal number is represented in 4 bits, while converting into BCD numbers.

1 ⟶ 0001

8 ⟶ 1000

5 ⟶ 0101

6 ⟶ 0110

3 ⟶ 0011

5 ⟶ 0101

7 ⟶ 0111

1		8	5	6	3	5	7
0000	0001	1000	0101	0110	0011	0101	0111
1 byte		1 byte	1 byte	1 byte	1 byte	1 byte	1 byte

In total 4 bytes are required for converting in BCD numbers.

Back to Course

Next Lesson