Qus : 2 NIMCET PYQ 2 The Process when processor fetch or decode another instruction during the execution of current instruction is called
1 Super computing 2 Pipelining 3 Cloud Computing 4 Accumulators Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution Pipelining is the process of accumulating instruction from the processor through a pipeline. It allows storing and executing instructions in an orderly process. It is also known as pipeline processing. Pipelining is a technique where multiple instructions are overlapped during execution.
Qus : 3 NIMCET PYQ 4 Which of the following is used by ALU to store the intermediate results?
1 Stack 2 Heap 3 Registers 4 Accumulators Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution An accumulator is a type of register included in a CPU. It acts as a temporary storage location which holds an intermediate value in mathematical and logical calculations. Intermediate results of an operation are progressively written to the accumulator, overwriting the previous value. For example, in the operation "3 + 4 + 5," the accumulator would hold the value 3, then the value 7, then the value 12. The benefit of an accumulator is that it does not need to be explicitly referenced, which conserves data in the operation statement.
Qus : 4 NIMCET PYQ 4 One TeraByte(TB)=_________GB and One ExaByte(EB)=_______GB
1 ${2}^{10}\, GB,\, {2}^{16}\, GB$ 2 ${2}^{10}\, GB,\, {2}^{20}\, GB$ 3 ${2}^{10}\, GB,\, {2}^{24}\, GB$ 4 ${2}^{10}\, GB,\, {2}^{30}\, GB$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution Unit Shortened Capacity Bit b 1 or 0 (on or off) Byte B 8 bits Kilobyte KB 1024 bytes Megabyte MB 1024 kilobytes Gigabyte GB 1024 megabytes Terabyte TB 1024 gigabytes Petabyte PB 1024 terabytes Exabyte EB 1024 petabytes Zettabyte ZB 1024 exabytes Yottabyte YB 1024 zettabytes
Qus : 5 NIMCET PYQ 2 The Cache Memory is more effective because of
1 Memory localization 2 Locality of reference 3 Memory Size 4 None of the mentioned Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution Locality of reference refers to a phenomenon in which a computer program tends to access same set of memory locations for a particular time period. In other words, Locality of Reference refers to the tendency of the computer program to access instructions whose addresses are near one another.
Qus : 6 NIMCET PYQ 1 Which of the following is the fastest means of memory access for CPU?
1 Registers 2 Cache 3 Main Memory 4 Stack Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution Registers are a type of computer memory used to quickly accept, store, and transfer data and instructions that are being used immediately by the CPU. The registers used by the CPU are often termed as Processor registers.
Qus : 8 NIMCET PYQ 1 To fetch data from secondary memory which one of the following register is used
1 MAR 2 PC 3 IR 4 MBR Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution MAR register is used to access data and instructions from memory during the execution phase of instruction. MAR holds the memory location of data that needs to be accessed. When reading from memory, data addressed by MAR is fed into the MDR (memory data register) and then used by the CPU. When writing to memory, the CPU writes data from MDR to the memory location whose address is stored in MAR. MAR, which is found inside the CPU, goes either to the RAM (random-access memory) or cache.
Qus : 10 NIMCET PYQ 4 Consider a computer system with speed of 106 instructions per second. A program P, having 2n2 steps is run on this system, where n is the input size. If n = 10000, what is the execution time for P?
1 1.2 seconds 2 20 seconds 3 100 seconds 4 200 seconds Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution $\text{Speed of computer} = 10^6 \text{per second} $
$\text{For n} =10000=10^4$
$$Time = \frac{\text{No of tasks}}{\text{Speed of computer}}$$
$$Time = \frac{2n^2}{10^6}$$
$$= \frac{2\times (10^4)^2}{10^6}$$
$$= \frac{2\times 10^8}{10^6}$$
$$=2 \times 10^2$$
$$= 200sec$$
Qus : 11 NIMCET PYQ 1 To access the I/O devices the status flags is continuously checked in
1 Program controlled I/O 2 Memory mapped I/O 3 I/O Mapped 4 None of the above Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2021 PYQ Solution Programmed I/O: In program-controlled I/O, the processor program controls the complete data transfer. So only when an I/O transfer instruction is executed, the transfer could take place. It is required to check that device is ready/not for the data transfer in most cases. Usually, the transfer is to & from a CPU register & peripheral. Here, CPU constantly monitors the peripheral. Here, until the I/O unit indicates that it is ready for transfer, the CPU wait & stays in a loop. It is time-consuming as it keeps the CPU busy needlessly.
Qus : 18 NIMCET PYQ 4 Consider the equation (40)x = (132)y is some bases x and y. Then a possible set of value of x and y are
1 8, 12 2 12, 8 3 6 and 12 4 14 and 6 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2017 PYQ
Solution $\begin{array}{ll}{{{(40})}_x={{(132)}}_y} \\ {\Rightarrow4\times{x}^1}+0\times{x}^0=1\times{y}^2+3\times{y}^1+2\times{y}^0 \\ {\Rightarrow4x+0={y}^2+3y^{}+2} \\ {4x={y}^2+3y^{}+2}\end{array}$Qus : 32 NIMCET PYQ 1
The maximum and minimum value represented in signed 16-bit 2s compliment representation are
1
-32768 and 32767
2
0 and 32767
3
0 and 65535
4
-16384 and 16383
Go to Discussion 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
Qus : 33 NIMCET PYQ 4
Which of the following is true about Von Neumann architecture?
1
It has separate memory for data and instructions
2
It has separate storage for input/output operations
3
It has a separate processing unit for data and instructions
4
It has a single memory unit for both data and instructions
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution Qus : 35 NIMCET PYQ 2
Consider the following minterm for F:F(P, Q, R, S) = Σ0, 2, 5, 7, 8, 10, 13, 15. The minterms 2, 7, 8, and 13 are don't care terms. The minimal sum of products form for F is
1 $$\overline{Q}S+Q\overline{S}$$ 2 $$\overline{Q}\, \overline{S}+QS$$ 3 $$\overline{Q}\, \overline{R}\, \overline{S}+\overline{Q}\, R\overline{S}+Q\overline{R}\, S+QRS$$ 4 $$\overline{P}\, \overline{Q}\, \overline{S}+\overline{P}\, QS+PQS+P\, \overline{Q}\, \overline{S}$$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution Qus : 36 NIMCET PYQ 3
Suppose we have a 10-bit computer that uses 10-bit int (2's complement representation). the number representation of - 35 is
1 0000100011 2 1100100011 3 1111011101 4 1111011111 Go to Discussion NIMCET Previous Year PYQ NIMCET 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
Qus : 37 NIMCET PYQ 4
A wrong sentence related to FAT 32 and NTFS file systems is
1
Read and write speeds of NTFS are faster than that of FAT 32
2
FAT 32 has lower disk utilisation compared to NTFS
3
NTFS stands for New Technology File System
4
FAT 32 store individual files of size up to 32 GB
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution Qus : 38 NIMCET PYQ 1
A bulb in the staircase has two switches, one switch is at the ground floor and the other one is at the first floor. The bulb can be turned ON and also can be turned OFF by any of the switches irrespective of the state of the other switch. The logic of the switching of the bulb resembles
1
XOR Gate
2
XNOR Gate
3
AND Gate
4
OR Gate
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution
Staircase Bulb Switch Logic
Question:
A staircase bulb is controlled by two switches — one at each floor. Each switch can independently turn the bulb ON or OFF, regardless of the other switch's position. What logic gate does this resemble?
✅ Correct Answer:
Exclusive OR (XOR) Gate
Explanation:
In XOR logic, the output is ON (1) when inputs are different, and OFF (0) when inputs are the same. Similarly, the bulb glows when the switches are in different states and turns OFF when both are in the same state.
Logic Used: XOR Gate
Qus : 39 NIMCET PYQ 4
Suppose we have a 10-bit computer that uses 10-bit floating point computational unit (Float number uses IEEE floating-point arithmetic where a floating point number has 1 sign bit, 5 exponent bits, and 4 fraction bits). The representation for +∞ (plus infinity) is
1 0 11111 1111 2 1 11111 0000 3 0 00000 1111 4 0 11111 0000 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution
10-bit Floating Point: +∞ Representation
Format: The 10-bit floating point is structured as follows:
1 bit for sign (S)
5 bits for exponent (E)
4 bits for fraction/mantissa (F)
IEEE Rule for +∞:
In IEEE floating-point format, +∞ is represented as:
Sign bit (S): 0
Exponent bits (E): all 1s → 11111
Fraction bits (F): all 0s → 0000
✅ Final 10-bit Representation: 0111110000
Qus : 40 NIMCET PYQ 1 Cosider the following Boolean Expression for F:
$F(P,Q,R,S)=PQ+\overline{P}QR+\overline{P}Q\overline{R}S$ .
The minimum sum of products form of F is
1 $$PQ+QR+QS$$ 2 $$P+Q+R+S$$ 3 $$\overline{P}+\overline{Q}+\overline{R}+\overline{S}$$ 4 $$\overline{P}\, R+\, \overline{P}\, \overline{R}\, S+P$$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution Qus : 41 NIMCET PYQ 2
What is the name of the storage device that compensates the difference in rates of flow of data from one device to another?
1
Concentrator
2
Buffer
3
Cache
4
Cache
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution
Data Flow Management
Question:
What is the name of the storage device that compensates the difference in rates of flow of data from one device to another?
✅ Correct Answer:
Buffer
Explanation:
A Buffer is a temporary storage area that helps in matching the speed of data transfer between a fast and a slow device, ensuring smooth data flow without loss.
Final Answer: Buffer
Qus : 43 NIMCET PYQ 4
A CPU generates 32-bit virtual addresses. The page size is 4 KB. The processor has a translation look-aside buffer (TLB) which can hold a total of 128 page table entries and is 4 -way set associative. The minimum size of the TLB tag is:
1 20 bits 2 11 bits 3 13 bits 4 15 bits Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution To determine the minimum size of the TLB tag, let's break down the problem step by step.
1. Understand the Virtual Address Structure
The CPU generates 32-bit virtual addresses.
The page size is 4 KB, which means the offset within a page is determined by the lower bits of the virtual address.
2. Calculate the Number of Offset Bits
Page size = 4 KB = $2^{12}$ bytes.
The offset within a page is determined by the lower 12 bits of the virtual address.
3. Calculate the Number of Virtual Page Number (VPN) Bits
Total virtual address bits = 32.
Offset bits = 12.
VPN bits = Total bits - Offset bits = 32−12=20 bits.
4. Understand the TLB StructureThe TLB is 4-way set associative , meaning there are 4 entries per set.
The TLB can hold a total of 128 page table entries.
Number of sets = Total entries / Associativity = 128/4=32 sets.
5. Calculate the Number of Set Index Bits
Number of sets = 32 = $2^5$.
The set index is determined by the lower 5 bits of the VPN.
6. Calculate the Number of Tag Bits
VPN bits = 20.
Set index bits = 5.
Tag bits = VPN bits - Set index bits = 20−5=15 bits.
7. Final Answer
The minimum size of the TLB tag is 15 bits.
Qus : 45 NIMCET PYQ 4 Let $\oplus$ and $\odot$ denote the Exclusive - OR and Exclusive-NOR operations respectively. Which of the following is not correct?
1 $$\, \overline{P}\oplus\overline{Q}=P\odot Q$$ 2 $$\, \overline{P}\oplus Q=P\odot Q$$ 3 $$\, \overline{P}\oplus\overline{Q}=P\oplus Q$$ 4 $$(P\oplus\overline{P})\oplus Q=(P\odot\overline{P})\overline{Q}$$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ
Solution Qus : 46 NIMCET PYQ 3
Which of the following registers is used to keep track of address of the memory location where the next instruction is located?
1
Memory Data Register
2
Memory Address Register
3
Program Counter
4
Instruction counters
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution
Register Responsible for Next Instruction
Question:
Which of the following registers is used to keep track of the address of the memory location where the next instruction is located?
✅ Correct Answer:
Program Counter (PC)
Explanation:
The Program Counter (PC) holds the address of the next instruction to be fetched from memory and executed by the CPU. After fetching the current instruction, it automatically updates to point to the next one.
Final Answer: Program Counter
Qus : 47 NIMCET PYQ 4
The time required for fetching and execution of one machine instruction is:
1
Seek time
2
Real time
3
Delay time
4
CPU cycle
Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2023 PYQ Solution
Time for Fetch & Execute
Definition:
The time required to fetch and execute one machine instruction is called a
Machine Cycle / CPU Cycle .
Includes:
Instruction fetch
Instruction decode
Operand fetch (if any)
Execution
Result store (if any)
✅ Final Answer: Machine Cycle / CPU Cycle.
Qus : 61 NIMCET PYQ 4 The quotient, if the binary number 11010111 is divided by 101, is___
1 101101
2 101010 3 111001
4 101011 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Binary Division
Question: What is the quotient when 11010111
is divided by 101
in binary?
Step 1: Convert to decimal:
Step 2: Divide: 215 ÷ 5 = 43
Step 3: Convert 43 to binary = 101011
✅ Final Answer: 101011
Qus : 63 NIMCET PYQ 1 Which of the following interfaces
perform the transfer of data between the memory and the I/O peripheral without
involving the CPU?
1 DMA 2 Serial Interface 3 branch
interface 4 DDA Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Data Transfer Interface in Computer Systems
Question: Which interface transfers data between memory and I/O peripheral without CPU involvement ?
✅ Correct Answer: Direct Memory Access (DMA)
Why DMA?
The CPU only initiates the process.
The DMA controller directly transfers data between I/O and memory.
Frees the CPU for other tasks, improving efficiency.
❌ Incorrect Options (if any): Programmed I/O, Interrupt-driven I/O – both require CPU involvement.
Qus : 64 NIMCET PYQ 3 Cache memory functions as an
intermediary between
1 CPU
and Hard Disk 2 None
of these 3 CPU
and RAM 4 RAM
and ROM Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Computer Architecture: Role of Cache Memory
Question: Cache memory functions as an intermediary between?
✅ Correct Answer: CPU and Main Memory (RAM)
Explanation:
CPU is very fast, but RAM is slower in comparison.
Cache memory holds frequently accessed data closer to the CPU.
This reduces data access time and improves overall system performance.
Hierarchy: CPU ↔ Cache ↔ Main Memory (RAM) ↔ Secondary Storage
Qus : 65 NIMCET PYQ 4 A CPU generates 32 bits virtual
addresses. The page size is 4 KB. The processor has a translation look-aside
buffer (TLB) which can hold a total of 128-page table entries and is 4- way set
associate. The minimum size of the TLB tag is
1 11 bits 2 20 bits 3 13 bits 4 15 bits Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
TLB Tag Size Calculation
Given:
Virtual Address = 32 bits
Page Size = 4 KB = \(2^{12}\) → Offset = 12 bits
VPN = 32 − 12 = 20 bits
TLB entries = 128, 4-way set associative → 32 sets
Set index bits: \( \log_2(32) = 5 \text{ bits} \)
TLB Tag = VPN − Set Index = 20 − 5 = 15 bits
✅ Final Answer: 15 bits
Qus : 66 NIMCET PYQ 2 The expression P+QR is the reduced
form of _____
1 (P
+ R)Q 2 (P+Q)
(P+R) 3 PQ
+ QR 4 (P+Q)
R Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Boolean Simplification
Given Expression: P + QR
We ask: This is the simplified (reduced) form of which expression?
Try expanding: (P + Q)(P + R)
Using distributive law:
(P + Q)(P + R) = P(P + R) + Q(P + R) = P + PR + PQ + QR
= P + PQ + PR + QR = P + QR (since P absorbs PQ and PR)
✅ Final Answer: (P + Q)(P + R)
Qus : 67 NIMCET PYQ 1 The
primary purpose of cache memory in a computer system is
1 to temporarily store frequently accessed data and instruction for faster access by the CPU. 2 To permanently store data and programs 3 to provide additional storage space when the main memory is full. 4 to manage input and output operations between the CPU and peripherals. Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Computer Architecture: Purpose of Cache Memory
Question: What is the primary purpose of cache memory in a computer system?
✅ Correct Answer: To increase the speed of data access by storing frequently used data closer to the CPU
Explanation:
Cache memory is much faster than main memory (RAM).
It holds frequently accessed data and instructions.
Helps reduce CPU waiting time and improves performance.
Hierarchy: CPU → Cache → RAM → Storage
Qus : 68 NIMCET PYQ 4 Consider the program below which uses
six temporary variables a, b, c, d, e, and f.
a = 10
b = 20
c = 30
d=a+c
f=c+c
b=c+e
e=b+f
d=5+e
return d+f
Assuming that all the above operations take their operands from registers, the minimum number of registers needed to execute this program without spilling is
Assuming that all the above operations
take their operands from registers, the minimum number of registers needed to execute
this program without spilling is
1 6 2 4 3 5 4 3 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Detailed Register Allocation Analysis
Objective: Determine the minimum number of registers needed to execute the program without spilling.
Live Range Analysis:
a: Line 1 → 4
b: Line 6 → 7
c: Line 3 → 6
d: Line 8 → 9
e: Line 7 → 8
f: Line 5 → 9
Max live variables: 3 (after lines 6 and 7)
✅ Final Answer: 3 registers are required to execute the program without spilling.
Qus : 69 NIMCET PYQ 1 Which of the following is the smallest
unit of data in a computer ?
1 Bit 2 Byte 3 KB 4 Nibble Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Basic Computer Knowledge: Smallest Unit of Data
Question: Which of the following is the smallest unit of data in a computer?
✅ Correct Answer: Bit (Binary Digit)
Explanation:
A bit is the most fundamental unit of data in computing.
It represents a binary value: 0
or 1
.
All other units (Byte, Kilobyte, etc.) are multiples of bits.
Common Units of Data:
1 Bit = Smallest unit
1 Byte = 8 Bits
1 Kilobyte (KB) = 1024 Bytes
1 Megabyte (MB) = 1024 KB
? Note: Every piece of data in a computer—from text and images to video and sound—ultimately breaks down into bits.
Qus : 70 NIMCET PYQ 1 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?
1 0001
and no overflow 2 1001
and an overflow 3 0001
and an overflow 4 1001
and no overflow Go to Discussion 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
Qus : 71 NIMCET PYQ 2 A
computer system has 16-bit wide address/ data bus that uses RAM chips of 4K $\times$ 8-bit capacity. The number of RAM chips are needed to provide a memory capacity
of 64 Kbytes memory is
1 32 2 16 3 64 4 8 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Memory Chip Calculation
Given:
RAM chip size = 4K × 8-bit = 4096 bytes = 4 KB
Required memory = 64 KB = 65536 bytes
Calculation:
Total chips needed = 65536 / 4096 = 16 chips
✅ Final Answer: 16 RAM chips are needed to build 64 KB of memory.
Qus : 72 NIMCET PYQ 3 Given
that numbers A and B are two 8 bit 2’s complement numbers with A = 11111111,
B = 11111111. Then sum A + B is _________
1 00000010
2 11111100 3 11111110
4 00000000 Go to Discussion 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
Qus : 74 NIMCET PYQ 3 Consider
an arbitrary number system with independent digits as 0,1 and A. If we generate
first few numbers in sequence as 00, 01, 0A, 10, 11, 1A and if this process is
continued to generate the numbers, then the position of 10A is ________
1 15 2 10 3 12 4 9 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Custom Number System: Position of 10A
Given digits: 0, 1, A (base-3)
Convert 10A to base-10:
1 → 1
0 → 0
A → 2
10A = 1×9 + 0×3 + 2 = 11
List of numbers in sequence:
00
01
0A
10
11
1A
A0
A1
AA
100
101
10A
✅ Final Answer: 12
Qus : 75 NIMCET PYQ 2 Which
of the following components is not a part of an instruction formation in CPU
processing?
1 Opcode
2 Register
file 3 Source
operand 4 Destination
operand Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2024 PYQ Solution
Instruction Formation in CPU Processing
Question: Which of the following is not a part of instruction formation?
Options:
Opcode
Register file
Source operand
Destination operand
✅ Correct Answer: Register File
Explanation:
Opcode, Source operand, Destination operand — all are part of the instruction format.
Register File — a hardware structure that stores registers, but it is not encoded into the instruction.
Qus : 79 NIMCET PYQ 2 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
1 -6,
-6 and -6, respectively 2 -7,
-7 and -7 respectively 3 -25,
-9 and -57 respectively 4 25,
9 and 57, respectively Go to Discussion 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 .
Qus : 80 NIMCET PYQ 4 The maximum and minimum value represented in signed
16 bit 2's complement representations are
1 -16384 and 16383 2 0 and 32767 3 0 and 65535 4 -32768 and 32767 Go to Discussion 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 $32767$
Qus : 82 NIMCET PYQ 1 Which of the following is equivalent to the Boolean expression:
$(X+Y).(X+\overline{Y}).(\overline{X}+Y)$
1 $XY$ 2 $X\overline{Y}$ 3 $\overline{X}Y$ 4 $\overline{X}\, \overline{Y}$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ Solution $(X+Y).(X+\overline{Y}).(\overline{X}+Y)$
=$(X+Y)(X+Y')(X'+Y)$
=$(XX+XY+YX+YY')(X'+Y)$
=$(X+XY)(X'+Y)$
=$X(1+Y)(X'+Y)$
=$X(X'+Y)$
=$XX'+XY$
=$XY$
Qus : 83 NIMCET PYQ 4 Suppose the largest n bit number requires ‘d’ digits in decimal representation. Which of the following relations between ‘n’ and ‘d’ is approximately correct
1 $$d=2^n$$ 2 $$n=2^d$$ 3 $$d{\lt}n\log _{10}2$$ 4 $$d{{\gt}}n\log _{10}2$$ Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ Solution n bits binary number required d -decimal digits.
So, ${10}^d{\gt}{2}^n$
Take on both side
$\log _{10}({10}^d)\gt{\log _{10}({2}^n)}^{}$
$d{\gt}n\log _{10}(2)$
Qus : 84 NIMCET PYQ 2 If a processor clock is rated as 2500 million cycles per second, then its clock period is:
1 $$2.50 \times 10^{-10} sec$$ 2 $$4.0 \times 10^{-10} sec$$ 3 $$1.00 \times 10^{-10} sec$$ 4 None of the above Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ Solution we know that Frequency is defined as the number of cycles in one second
Number of cycle in 1 sec = 2500 million
=> Frequency = 2500 Mhz
we know that time period is the inverse of frequency and is defined as the time taken by one cycle.
$T = \frac{1}{F}$
$T = \frac{1}{2500 \times 10^{-6}}$
$T=4 \times 10^{-10} $ sec
Qus : 86 NIMCET PYQ 2 FFFF will be the last memory location in a memory of size
1 1k 2 64k 3 32k 4 16k Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ Solution The Hexadecimal digits are 0-9 and A-F. The Hexadecimal system represents numbers in16 symbols, zero to nine and ten to fifteen is represented by the English alphabet A-F.
The Hexadecimal character represents 4 bits.
The last memory location in a memory of size 64K is FFFF.
64K is $2^{16}$ bytes, i.e.
$16^4$ bytes = 1000 bytes in hexadecimal code.
The last accessible address is 1000-1 = FFFF.
Qus : 89 NIMCET PYQ 4 If a signal passing through a gate is inhibited by sending a low into one of the inputs, and the output is HIGH, the gate is a(n):
1 NOR 2 AND 3 OR 4 NAND Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2022 PYQ Solution Output is high if any of the input is low. The truth table for NAND gate is:
A B Output 0 0 0 0 1 1 1 0 1 1 1 1
Table representing NAND Gate.
Qus : 93 NIMCET PYQ 1 Consider a hard disk with 16 recording surfaces (0 – 15) having 16384 cylinders (0 – 16383) and each cylinder contains 64 sectors (0 – 63). Data storage capacity in each sector is 512 bytes. Data are organized stored in the disk and the starting disk location of the file is < 1200, 9, 40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?
1 1284 2 1282 3 1286 4 1288 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
Solution Qus : 94 NIMCET PYQ 2 Consider the following min term expression for F.
F(P,Q,R,S) = ∑ (0, 2, 5, 7, 8, 10, 13, 15)
The minterms 2, 7, 8 and 13 are ‘do not care' terms.
The minimal sum of products form for F is
1 2 3 4 Go to Discussion NIMCET Previous Year PYQ NIMCET NIMCET 2016 PYQ
Solution [{"qus_id":"4189","year":"2019"},{"qus_id":"4190","year":"2019"},{"qus_id":"4191","year":"2019"},{"qus_id":"4192","year":"2019"},{"qus_id":"4193","year":"2019"},{"qus_id":"4195","year":"2019"},{"qus_id":"4196","year":"2019"},{"qus_id":"4197","year":"2019"},{"qus_id":"4198","year":"2019"},{"qus_id":"4194","year":"2019"},{"qus_id":"4376","year":"2017"},{"qus_id":"4377","year":"2017"},{"qus_id":"4378","year":"2017"},{"qus_id":"4379","year":"2017"},{"qus_id":"4380","year":"2017"},{"qus_id":"4381","year":"2017"},{"qus_id":"4382","year":"2017"},{"qus_id":"4383","year":"2017"},{"qus_id":"4384","year":"2017"},{"qus_id":"4385","year":"2017"},{"qus_id":"9519","year":"2020"},{"qus_id":"9520","year":"2020"},{"qus_id":"9521","year":"2020"},{"qus_id":"9522","year":"2020"},{"qus_id":"9523","year":"2020"},{"qus_id":"9524","year":"2020"},{"qus_id":"9525","year":"2020"},{"qus_id":"9526","year":"2020"},{"qus_id":"9527","year":"2020"},{"qus_id":"9528","year":"2020"},{"qus_id":"10752","year":"2021"},{"qus_id":"10753","year":"2021"},{"qus_id":"10754","year":"2021"},{"qus_id":"10755","year":"2021"},{"qus_id":"10756","year":"2021"},{"qus_id":"10757","year":"2021"},{"qus_id":"10758","year":"2021"},{"qus_id":"10759","year":"2021"},{"qus_id":"10760","year":"2021"},{"qus_id":"10761","year":"2021"},{"qus_id":"11199","year":"2022"},{"qus_id":"11200","year":"2022"},{"qus_id":"11201","year":"2022"},{"qus_id":"11202","year":"2022"},{"qus_id":"11203","year":"2022"},{"qus_id":"11204","year":"2022"},{"qus_id":"11205","year":"2022"},{"qus_id":"11206","year":"2022"},{"qus_id":"11207","year":"2022"},{"qus_id":"11208","year":"2022"},{"qus_id":"11591","year":"2023"},{"qus_id":"11592","year":"2023"},{"qus_id":"11593","year":"2023"},{"qus_id":"11594","year":"2023"},{"qus_id":"11595","year":"2023"},{"qus_id":"11596","year":"2023"},{"qus_id":"11597","year":"2023"},{"qus_id":"11598","year":"2023"},{"qus_id":"11599","year":"2023"},{"qus_id":"11600","year":"2023"},{"qus_id":"11601","year":"2023"},{"qus_id":"11602","year":"2023"},{"qus_id":"11603","year":"2023"},{"qus_id":"11604","year":"2023"},{"qus_id":"11605","year":"2023"},{"qus_id":"11606","year":"2023"},{"qus_id":"11607","year":"2023"},{"qus_id":"11608","year":"2023"},{"qus_id":"11609","year":"2023"},{"qus_id":"11610","year":"2023"},{"qus_id":"4583","year":"2016"},{"qus_id":"4584","year":"2016"},{"qus_id":"4586","year":"2016"},{"qus_id":"4587","year":"2016"},{"qus_id":"4588","year":"2016"},{"qus_id":"4589","year":"2016"},{"qus_id":"4590","year":"2016"},{"qus_id":"4591","year":"2016"},{"qus_id":"4592","year":"2016"},{"qus_id":"4585","year":"2016"},{"qus_id":"11909","year":"2024"},{"qus_id":"11910","year":"2024"},{"qus_id":"11911","year":"2024"},{"qus_id":"11912","year":"2024"},{"qus_id":"11913","year":"2024"},{"qus_id":"11914","year":"2024"},{"qus_id":"11915","year":"2024"},{"qus_id":"11916","year":"2024"},{"qus_id":"11917","year":"2024"},{"qus_id":"11918","year":"2024"},{"qus_id":"11928","year":"2024"},{"qus_id":"11925","year":"2024"},{"qus_id":"11924","year":"2024"},{"qus_id":"11923","year":"2024"},{"qus_id":"11922","year":"2024"},{"qus_id":"11921","year":"2024"},{"qus_id":"11920","year":"2024"},{"qus_id":"11919","year":"2024"},{"qus_id":"11927","year":"2024"}]