NIMCET 2027 1s Complement 2s Complement Previous Year Questions (PYQs) – Page 2 of 3
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When two binary numbers are added, then an overflow will never occur if
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Solution
Overflow never occurs when carry into sign bit = carry out of sign bit.
Consider
the following 4- bit binary numbers represented in the 2’s complement form :
1101 and 0100 What would be the result when we add them?
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Solution
2's Complement Addition (4-bit)
Given: 1101 and 0100 (in 2’s complement)
Step-by-step:
- 1101 = −3 (in decimal)
- 0100 = +4 (in decimal)
- Sum = −3 + 4 = +1
- +1 in 4-bit 2’s complement = 0001
✅ Final Answer: 0001
Given
that numbers A and B are two 8 bit 2’s complement numbers with A = 11111111,
B = 11111111. Then sum A + B is _________
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Solution
2's Complement Addition (8-bit)
Given:
- A = 11111111 → (−1)
- B = 11111111 → (−1)
Sum: −1 + (−1) = −2
Convert −2 to 8-bit 2's complement:
- +2 = 00000010
- Invert = 11111101
- Add 1 = 11111110
✅ Final Answer: 11111110
Let
the given number 11001, 1001 and 111001 be correspond to the 2’s complement representation.
Then with which one of the following decimal number, the given numbers match
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Solution
Binary to Decimal: 2's Complement Conversion
Given binary numbers:
- 11001 (5-bit)
- 1001 (4-bit)
- 111001 (6-bit)
Step-by-step (2's complement):
- Each starts with 1 → negative number
- Convert by inverting and adding 1
- All result in binary 0111 → decimal 7
- So final value = −7
✅ Final Answer: Each binary number corresponds to the decimal number −7.
Let x = 11111010 and y = 00001010 be two 8-bit 2’s complement numbers. Their product in 2’s complement notation is
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Solution
The range of numbers that can be stored in 8 bits, if negative numbers are stored in 2’s complement form is
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Solution
The maximum and minimum value represented in signed
16 bit 2's complement representations are
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Solution
Range of 2's complement $-2^{n-1}$ to $2^{n-1}+1$
Range for 16 bits = $-2^{16-1}$ to $2^{16-1}+1$
Range for 16 bits = $-2^{15}$ to $2^{25}+1$
Range for 16 bits = $-32768$ to $32767$
The 2's complement representation of the number (–100)10 in an 8 bit computer is
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Solution
The range of n-bit signed magnitude representation is
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Solution
For an $n$-bit Signed Magnitude Representation:
The leftmost bit (MSB) is the sign bit:
$0 \rightarrow$ Positive
$1 \rightarrow$ Negative
Remaining $(n-1)$ bits represent the magnitude
Maximum positive value:
Sign bit $= 0$, remaining $(n-1)$ bits all $1$s
$\Rightarrow +(2^{n-1} - 1)$
Maximum negative value:
Sign bit $= 1$, remaining $(n-1)$ bits all $1$s
$\Rightarrow -(2^{n-1} - 1)$
Note: $+0$ and $-0$ both exist in signed magnitude representation
$\therefore$ Range of $n$-bit Signed Magnitude Representation is:
$\therefore \boxed{-(2^{n-1}-1) \leq x \leq +(2^{n-1}-1)}$
The leftmost bit (MSB) is the sign bit:
$0 \rightarrow$ Positive
$1 \rightarrow$ Negative
Remaining $(n-1)$ bits represent the magnitude
Maximum positive value:
Sign bit $= 0$, remaining $(n-1)$ bits all $1$s
$\Rightarrow +(2^{n-1} - 1)$
Maximum negative value:
Sign bit $= 1$, remaining $(n-1)$ bits all $1$s
$\Rightarrow -(2^{n-1} - 1)$
Note: $+0$ and $-0$ both exist in signed magnitude representation
$\therefore$ Range of $n$-bit Signed Magnitude Representation is:
$\therefore \boxed{-(2^{n-1}-1) \leq x \leq +(2^{n-1}-1)}$
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