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NIMCET Previous Year Questions (PYQs)

NIMCET 1s Complement 2s Complement PYQ


NIMCET PYQ

Consider the following statements about the range of numbers in a $9$-bit $1$'s complement and $2$'s complement system.

I. In $9$-bit $1$'s complement, the range is $-255$ to $+255$, and there exist two representations of zero.

II. In $9$-bit $2$'s complement, the range is $-256$ to $+255$, and both $1$'s complement and $2$'s complement can represent exactly $512$ unique values.

III. The maximum positive number representable is $+255$ in both $1$'s complement and $2$'s complement $9$-bit systems.

Identify the CORRECT option.






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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

For $9$-bit $1$'s complement, the range is

$-(2^{8}-1)$ to $+(2^{8}-1)$

$=-255$ to $+255$

Also, in $1$'s complement, there are two representations of zero: positive zero and negative zero.

So, statement I is correct.

For $9$-bit $2$'s complement, the range is

$-2^8$ to $2^8-1$

$=-256$ to $+255$

But $1$'s complement does not represent exactly $512$ unique values because zero has two representations.

So, statement II is incorrect.

The maximum positive number in both systems is

$+255$

So, statement III is correct.

Therefore, statements I and III only are correct.


NIMCET PYQ
Consider a 9-bit representation. Which of the following correctly gives the smallest number that can be represented in: 
(i) 1's complement, 
(ii) 2's complement





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NIMCET Previous Year PYQ NIMCET NIMCET 2025 PYQ

Solution

For an n-bit system:

  • 1’s complement range: $-(2^{n-1}-1)$ to $+(2^{n-1}-1)$ ⇒ smallest $=-(2^{8}-1)={-255}$.
  • 2’s complement range: $-2^{n-1}$ to $+(2^{n-1}-1)$ ⇒ smallest $=-2^{8}={-256}$.

(9-bit means sign + 8 magnitude bits.)


NIMCET PYQ
Which of the following is the representation of decimal number (- 147) in 2's compliment notation on a 12-bit machine?





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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution


NIMCET PYQ
The smallest integer that can be represented by an 8 bit number in 2's complement form is





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NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ

Solution

✅ Given Information:

An 8-bit number in 2's complement form can represent values from \(-2^{n-1}\) to \(2^{n-1} - 1\).

✅ Step 1: Calculate the smallest integer:

Smallest value = \(-2^{8-1} = -2^7 = -128\)

✅ Final Answer:

The smallest integer that can be represented by an 8-bit number in 2's complement form is: -128


NIMCET PYQ
What is the 2's complement of 0011 0101 1001 1100?





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NIMCET Previous Year PYQ NIMCET NIMCET 2014 PYQ

Solution


NIMCET PYQ
What is the 8 bit 2's complement representation of the negative integer-93?





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Solution


NIMCET PYQ
The maximum and minimum value represented in signed 16-bit 2s compliment representation are





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NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ

Solution

 Maximum & Minimum in 16-bit 2's Complement

 Total Bits: 16

Format: 1 sign bit + 15 magnitude bits

  • Maximum (positive): 0111 1111 1111 1111(2) = +32,767
  • Minimum (negative): 1000 0000 0000 0000(2) = −32,768

✅ Final Answer:
Minimum = −32,768
Maximum = +32,767


NIMCET PYQ
If N is a 16-bit signed integer, then 2's complement representation of N is (F87B)16. The 2's complement representation of 8*N is





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NIMCET Previous Year PYQ NIMCET NIMCET 2019 PYQ

Solution


NIMCET PYQ
In an 8 bit representation of computer system the decimal number 47 has to be subtracted from 38 and the result in binary 2's complement is _________






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Solution


NIMCET PYQ
A computer with a 32 bit word size uses 2’s complement to represent numbers. The range of integers that can be represented by this computer is:





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NIMCET Previous Year PYQ NIMCET NIMCET 2009 PYQ

Solution

In $n$-bit 2’s complement, range is $-2^{n-1}$ to $2^{n-1}-1$.

NIMCET PYQ
When two binary numbers are added, then an overflow will never occur if





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NIMCET Previous Year PYQ NIMCET NIMCET 2011 PYQ

Solution

Overflow never occurs when carry into sign bit = carry out of sign bit.

NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

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


NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

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


NIMCET PYQ
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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NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ

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.


NIMCET PYQ
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


NIMCET PYQ
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


NIMCET PYQ
The maximum and minimum value represented in signed 16 bit 2's complement representations are





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NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ

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 $32767$

NIMCET PYQ
The 2's complement representation of the number (–100)10 in an 8 bit computer is





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Solution


NIMCET PYQ
Subtract (1010)2 from (1101)2 using first complement





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Solution


NIMCET PYQ
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:
$\therefore \boxed{-(2^{n-1}-1) \leq x \leq +(2^{n-1}-1)}$

NIMCET PYQ
An embedded computer uses $8$-bit $2$'s complement representation for signed integers. What is the minimum negative integer value which can be represented in this computer?





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NIMCET Previous Year PYQ NIMCET NIMCET 2026 PYQ

Solution

In $n$-bit $2$'s complement representation, the range of signed integers is:

$-2^{n-1}$ to $2^{n-1}-1$

Here,

$n=8$

So, range is:

$-2^{7}$ to $2^{7}-1$

$=-128$ to $127$

Therefore, the minimum negative integer value is $-128$.


NIMCET PYQ
The addition of 4 bit, 2’s complement binary numbers 1101 and 0100 results in





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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

1101 is a 4-bit 2’s complement number.
Its MSB is 1, so it is negative.

1101 = −3
0100 = +4

Adding:
1101
+0100
=10001

Taking only 4 bits → 0001

NIMCET PYQ
Let A = 11111010 and B = 00001010 be two 8 bit 2’s complement numbers. Their product in 2’s complement is





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NIMCET Previous Year PYQ NIMCET NIMCET 2008 PYQ

Solution

A = 11111010 → −6
B = 00001010 → +10

Product = −60

+60 in binary = 00111100
2’s complement of 00111100 = 11000100

NIMCET PYQ
Which one of the following is decimal equivalent of $8$-bit two's complement number $11010011$?





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Solution

Given $8$-bit two's complement number is:

$11010011$

Since the leftmost bit is $1$, the number is negative.

To find its magnitude, take two's complement.

First, invert all bits:

$11010011\to 00101100$

Now add $1$:

$00101100+1=00101101$

Now convert $00101101$ to decimal:

$00101101=32+8+4+1$

$=45$

Since the original number was negative, the decimal equivalent is:

$-45$



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