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NIMCET Previous Year Questions (PYQs)
NIMCET Data Representation - Number System PYQ
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NIMCET Computer PYQNIMCET Boolean algebra PYQ
Solution
Formula: Number of Boolean functions of $n$ variables is:
$ 2^{2^n} $
Condition: We are told the total functions must be ≤ 4:
\[ 2^{2^n} \leq 4 \]
✅ Try values of $n$:
- $n = 0$: $2^{2^0} = 2^1 = 2$ ✅
- $n = 1$: $2^{2^1} = 2^2 = 4$ ✅
- $n = 2$: $2^{2^2} = 2^4 = 16$ ❌
Minimum $n$ for which number of Boolean functions ≤ 4 is:
\[ \boxed{1} \]
✅ Final Answer: $\boxed{1}$
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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
Solution
✅ Given:
The floating-point binary number is \( +1001.11_2 \).
We need to convert it into an 8-bit fraction and a 6-bit exponent format.
✅ Step 1: Normalize the Binary Number
We start by normalizing the binary number into scientific notation of the form:
\( 1.xxxx \times 2^n \)
Converting \( 1001.11_2 \) into scientific notation gives:
\( 1001.11_2 = 1.00111_2 \times 2^3 \)
The exponent is \( 3 \) (because the binary point is shifted 3 places to the left).
✅ Step 2: Convert the Exponent to Binary
The exponent is \( 3 \) in decimal. To represent this in binary using 6 bits, we get:
\( \text{Exponent} = 000100_2 \)
✅ Step 3: Convert the Fraction to 8 Bits
The fractional part of the normalized binary number is \( 00111 \). We need to extend it to 8 bits:
\( \text{Fraction} = 01001110_2 \)
✅ Final Answer:
The floating-point binary number \( +1001.11_2 \) in 8-bit fraction and 6-bit exponent format is:
Exponent: \( 000100_2 \), Fraction: \( 01001110_2 \)
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NIMCET Previous Year PYQNIMCET NIMCET 2017 PYQ
Solution
$\begin{array}{ll}{{{(40})}_x={{(132)}}_y} \\ {\Rightarrow4\times{x}^1}+0\times{x}^0=1\times{y}^2+3\times{y}^1+2\times{y}^0 \\ {\Rightarrow4x+0={y}^2+3y^{}+2} \\ {4x={y}^2+3y^{}+2}\end{array}$Go to Discussion
NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
10-bit 2's Complement Representation of –35
Format: 10-bit signed integer using 2's complement representation.
Step-by-Step:
- First, convert 35 to 10-bit binary:
0000100011 - Find 1's complement:
1111011100 - Add 1 (to get 2's complement):
1111011101
✅ Final Answer:
1111011101
–35 in 10-bit 2's complement: 1111011101
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NIMCET Previous Year PYQNIMCET NIMCET 2023 PYQ
Solution
Problem of 1’s Complement Representation
1. Two Zeros Exist
• Positive Zero: 0000
• Negative Zero: 1111
? This creates ambiguity, because logically both are zero but they have different bit patterns.
2. End-Around Carry Needed
When adding numbers, if a carry comes out of the MSB, it must be added back to the LSB.
(+5) = 0101
(-5) = 1010 (1’s complement of 0101)
----------------
Add: 1111 → End-around carry = 1
Final: 0000 (after adding carry)
? Result is zero, but note that two different zeros are possible.
3. Hardware Complexity
Extra logic is required so that +0 and -0 are treated the same, making design slower and costlier.
✅ Conclusion
That’s why modern systems use 2’s complement. It has only one zero and simplifies arithmetic operations.
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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ
Solution
IEEE 754 Single Precision Floating Point
Total Size: 32 bits
- Sign bit: 1 bit
- Exponent: 8 bits (biased, excess-127 notation)
- Mantissa (Fraction): 23 bits
? Exponent Representation
The exponent is stored in 8-bit biased form with a bias of 127.
Stored exponent = Actual exponent + 127
Example:
Actual exponent = 3
Stored exponent = 3 + 127 = 130 = 10000010
✅ Conclusion
In IEEE single precision, the exponent is represented in 8-bit biased (excess-127) notation.
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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ
Solution
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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ
Solution
2's Complement to Hexadecimal Conversion
Given the 2's complement binary number: (011010)
Find its equivalent hexadecimal representation.
Since the number has 6 bits, group the bits into two groups of 4 bits (add leading zeros if needed):
$$ (011010)_2 = (00011010)_2 $$
Now, split into two nibbles (4 bits each):
$$ 0001 \quad 1010 $$
Convert each nibble to hexadecimal:
- \(0001_2 = 1_{16}\)
- \(1010_2 = A_{16}\)
Therefore, the hexadecimal representation is: 1A
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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ
Solution
Base (Radix) Problem Solution
Question: Find the base r such that
(312)r ÷ (20)r = (13.1)r
Expansion in base 10
- (312)r = 3r² + 1r + 2
- (20)r = 2r
- (13.1)r = r + 3 + 1/r
Equation
(3r² + r + 2) / (2r) = r + 3 + 1/r
Multiply both sides by 2r
3r² + r + 2 = 2r² + 6r + 2
Simplification
r² - 5r = 0
⇒ r(r - 5) = 0
✅ Final Answer
Since base must be > 1, the valid solution is:
Base (radix) = 5
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NIMCET Previous Year PYQNIMCET NIMCET 2019 PYQ
Solution
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