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Past Questions
Microprocessor with ‘n’ address lines is capable of addressing
A)
2n locations
B)
2(n + 1)locations
C)
2n locations
D)
n2 locations
In computer architecture, the address bus consists of 'n' parallel lines. Each line can carry a binary digit (0 or 1). The total number of unique combinations (unique addresses) that can be formed with 'n' binary digits is calculated as 2 raised to the power of n (2^n). For example, if a microprocessor has 16 address lines, it can address 2^16 = 65,536 locations. Note: In the provided options, '2n' is a common typographical representation for 2^n in many technical exam papers where superscripts are not properly formatted. Option 'a' and 'c' appear identical in the prompt, but the mathematical principle is 2^n. Option 'b' (2^(n+1)) and 'd' (n^2) are mathematically incorrect for binary addressing logic.
The ASCII code is for information interchange by a binary code for
A)
numbers only
B)
alphabets only
C)
alphanumeric and other common symbols
D)
none of these
ASCII stands for American Standard Code for Information Interchange. It is a character encoding standard used to represent text in computers and other devices.
1. **Scope of ASCII**: ASCII defines a set of characters that can be represented by a 7-bit binary number (providing 128 possible characters).
2. **What it includes**:
- **Numbers**: Digits 0-9.
- **Alphabets**: Both uppercase (A-Z) and lowercase (a-z) English letters.
- **Common Symbols**: Punctuation marks (e.g., !, @, #, $, %), mathematical symbols (+, -, =), and control characters (e.g., Carriage Return, Line Feed, Backspace).
Evaluating the options:
- Option (a) is incorrect because ASCII is not limited to numbers.
- Option (b) is incorrect because ASCII is not limited to alphabets.
- Option (c) is correct because it encompasses the full range of alphanumeric characters and symbols defined in the standard.
Swapping of two numbers is done by
A)
Direct interchanging of values with each other
B)
Storing the value of one in a temp variable and shifting the value of the other into the first and then copying the value of temp to the other
C)
Transfer one number to other number without storing any number in the temp variable
D)
None of these
In C programming, swapping two variables requires a way to preserve the value of one variable while it is being overwritten by another.
1. **Why Option (b) is correct:** This describes the standard 'Temporary Variable' algorithm.
- Step 1: `temp = a;` (Store value of 'a' in a temporary location)
- Step 2: `a = b;` (Move value of 'b' into 'a'; the original 'a' is now gone from its original variable but safe in 'temp')
- Step 3: `b = temp;` (Move the saved value from 'temp' into 'b')
This is the most common and readable method for swapping.
2. **Why Option (a) is incorrect:** In C, you cannot simply 'interchange' values (e.g., you cannot write a command like `swap(a, b)` as a built-in operator). Assignment is unidirectional (`a = b` only moves b to a).
3. **Why Option (c) is incorrect:** While it is technically possible to swap without a temp variable using arithmetic (e.g., `a = a + b; b = a - b; a = a - b;`) or XOR bitwise operations, the phrasing 'Transfer one number to other number without storing any number' is logically incomplete and misleading. To swap without a temp variable, you still have to store intermediate sums or XOR results within the existing variables, not simply 'transfer' them. Option (b) represents the fundamental, universal definition of the swapping process taught in introductory C programming.
In __________ testing, each component of a composite conditional expression assumes both true and false values.
A)
Branch coverage
B)
Condition coverage
C)
Path coverage
D)
Control flow
In software testing, specifically white-box testing, different coverage criteria are used to ensure the quality of the code.
1. **Condition Coverage (Option b):** This criterion requires that each individual boolean sub-expression (component) in a composite conditional statement (e.g., 'if (A || B)') evaluates to both True and False at least once. This ensures that every single condition is tested independently of the overall outcome of the composite expression.
2. **Branch Coverage (Option a):** This requires that every possible exit from a decision point (the entire 'if' statement) is executed at least once. It focuses on the overall result of the composite expression (True or False), not the individual components within it.
3. **Path Coverage (Option c):** This is a more rigorous criterion that requires every possible unique path through the code (from start to finish) to be executed. This includes all combinations of branches.
4. **Control Flow (Option d):** This is not a testing coverage metric, but rather a representation (Control Flow Graph) of the order in which statements are executed in a program.
Since the question specifically mentions that 'each component of a composite conditional expression' must assume both true and false values, the correct answer is Condition Coverage.
Deletion of a node from a singly linked list, in general, requires adjustment of pointer field of ______ node(s)
A)
1
B)
2
C)
3
D)
4
In a singly linked list, each node contains a data field and a pointer (next) to the subsequent node. To delete a node (let's call it Node B) that is located between Node A and Node C, the pointer of the preceding node (Node A) must be updated to point directly to the node following the deleted node (Node C).
Step-by-step process:
1. Locate the node to be deleted (Node B) and its predecessor (Node A).
2. Update the 'next' pointer of Node A to store the address of Node C (which is currently stored in Node B's 'next' pointer).
3. Deallocate the memory of Node B.
Since only the pointer field of the predecessor node (Node A) is modified to bypass the deleted node, only 1 node's pointer field requires adjustment.
Note: While the head pointer might be updated if the first node is deleted, the head pointer is a reference to the list, not a 'pointer field of a node' in the context of the list's internal structure. In general cases (deleting a middle or end node), only one node's pointer is changed.
In a _______ team, there is no formal team hierarchy.
A)
Chief programmer
B)
Democratic
C)
Mixed control
D)
All of these
In software engineering project management, different team organizational structures are used to manage personnel.
1. **Democratic Team (Option b):** In this structure, there is no formal hierarchy or single leader. Decisions are made through consensus, and responsibilities are shared equally among team members. This is the correct answer as it explicitly lacks a formal hierarchy.
2. **Chief Programmer Team (Option a):** This is a highly structured, hierarchical approach. A single 'Chief Programmer' is responsible for all major technical decisions and project coordination, while other members (backup programmers, librarians, etc.) support them in a strict chain of command.
3. **Mixed Control Team (Option c):** This is a hybrid approach that combines elements of both the chief programmer and democratic styles, typically maintaining some level of formal leadership while allowing for collaborative decision-making in specific areas. Therefore, it still possesses a degree of hierarchy.
Since the question asks for the team with 'no formal team hierarchy,' the Democratic team is the only correct choice.
The primary advantage of adhering to a life cycle model is
A)
It encourages development in systematic and disciplined manner
B)
It encourages development in transparent way.
C)
It offers good testing strategy.
D)
It offers good communication skill.
A Software Development Life Cycle (SDLC) model is a structured framework that defines the phases (such as requirements analysis, design, implementation, testing, and maintenance) that a software project must go through.
1. Why Option (a) is correct: The primary purpose of any life cycle model (whether it is Waterfall, Iterative, Spiral, or Agile) is to provide a roadmap for the development process. By defining specific stages, deliverables, and entry/exit criteria for each phase, it ensures that the development is carried out in a systematic, organized, and disciplined manner, reducing chaos and unpredictability in the project.
2. Why Option (b) is incorrect: While some models (like Agile) increase transparency through frequent demos and sprints, transparency is a characteristic of specific methodologies or management styles, not the primary goal of having a life cycle model in general.
3. Why Option (c) is incorrect: Testing is a phase within the life cycle model, but the model itself is not a 'testing strategy.' A testing strategy refers to the specific approach to verifying the software (e.g., unit testing, integration testing), whereas the life cycle model manages the overall project flow.
4. Why Option (d) is incorrect: Communication skills are human soft skills. While a model might facilitate communication through defined reporting lines or meetings, it cannot 'offer' or 'provide' a skill to the developers.
Binary numbers can be converted into equivalent octal numbers by making groups of three bits _____________.
A)
Starting from the MSB
B)
Starting from the LSB
C)
Ending at the MSB
D)
Ending at the LSB
To convert a binary number to an octal number, the binary digits are grouped into sets of three because 2^3 = 8. The grouping must start from the Least Significant Bit (LSB), which is the rightmost digit of the binary number. This ensures that if the total number of bits is not a multiple of three, the padding zeros are added to the left (the Most Significant Bit side), which does not change the value of the number. For example, to convert (11011)_2 to octal: 1. Group from LSB: (011) and (11). 2. Pad the leftmost group: (011) and (011). 3. Convert each group: 011=3, 011=3. Result: (33)_8. If one were to start from the MSB, the padding would occur at the LSB, which would fundamentally change the numerical value of the binary string.
A documented life cycle model prevents
A)
Misinterpretation
B)
Inconsistencies
C)
Omission
D)
All of these
A documented life cycle model provides a clear, written framework for the phases, activities, and deliverables of a project. Because it is explicitly documented, it helps avoid misinterpretation of requirements or process steps, reduces omissions by ensuring all stages are covered, and prevents inconsistencies by keeping the workflow uniform and traceable. Therefore, it prevents all of these.
Why the others are not correct individually: (a) Misinterpretation is prevented, but not the only benefit; (b) Inconsistencies are also prevented, but again not the only one; (c) Omission is likewise prevented, but the model addresses all three issues together. Hence, the best answer is (d) All of these.
Which of the following is used as storage locations both in the ALU and in the control section of a computer
A)
accumalator
B)
register
C)
adder
D)
decoder
Step-by-step reasoning:
1. **Understanding the Requirement**: The question asks for a storage location used in both the Arithmetic Logic Unit (ALU) and the Control Unit (CU) of a computer.
2. **Analyzing Option (b) Register**: A register is a small, high-speed storage location within the CPU. In the ALU, registers (like the accumulator or temporary registers) hold operands and results of calculations. In the Control Unit, registers (like the Program Counter (PC), Instruction Register (IR), and Memory Address Register (MAR)) are used to track the execution flow and hold current instructions. Therefore, registers are the fundamental storage units used in both sections.
3. **Analyzing Option (a) Accumulator**: While an accumulator is a type of register, it is specifically associated with the ALU to store intermediate arithmetic and logic results. It is not typically used as a general storage location within the control section.
4. **Analyzing Option (c) Adder**: An adder is a combinational logic circuit used to perform addition. It is a processing component, not a storage location.
5. **Analyzing Option (d) Decoder**: A decoder is a combinational circuit that converts binary information from n input lines to a maximum of 2^n unique output lines. It is used for instruction decoding in the control unit, but it is not a storage location.
The function fseek()
A)
Finds a given word or phrase in a file
B)
Finds the correct file
C)
Helps access records in the middle of a file
D)
None of these
The fseek() function in C is used for 'random access' in a file. It allows the program to move the file position pointer to a specific location without having to read through all the preceding data.
1. **Function Signature**: `int fseek(FILE *stream, long offset, int whence);`
2. **How it works**: It takes an offset (number of bytes to move) and a starting point (`SEEK_SET` for the beginning, `SEEK_CUR` for the current position, or `SEEK_END` for the end of the file). By manipulating these, a programmer can jump directly to a specific record or byte in the middle of a file.
**Analysis of other options:**
- **Option a**: Finding a word or phrase requires reading the file content and comparing strings (e.g., using `fgets()` and `strstr()`), not just moving the pointer.
- **Option b**: Finding or opening a file is handled by the `fopen()` function.
- **Option d**: Incorrect because option c accurately describes the purpose of fseek().
In OSI model, which of the following layer transforms information from machine format into that understandable by user?
A)
Application
B)
Session
C)
Presentation
D)
Physical
The OSI (Open Systems Interconnection) model consists of seven layers. The Presentation Layer (Layer 6) is specifically responsible for the 'translation' of data. Its primary roles include: 1. Translation: Converting data from the application-specific format (machine format) into a common network format and vice versa, ensuring that the receiving system can understand the information. 2. Encryption/Decryption: Securing data for transmission. 3. Compression: Reducing the size of the data.
Why other options are incorrect:
- Application Layer (a): This is the interface between the user and the network (e.g., HTTP, FTP), but it does not handle the internal data formatting/translation.
- Session Layer (b): This layer manages the establishment, maintenance, and termination of sessions between applications.
- Physical Layer (d): This layer deals with the actual hardware transmission of raw bits (0s and 1s) over a physical medium.
The fastest type of memory is
A)
tape
B)
semiconductor memory
C)
disk
D)
bubble memory
To determine the fastest type of memory among the given options, we must analyze the physical mechanism used to access data in each:
1. **Semiconductor Memory (Option b):** This refers to memory based on integrated circuits (ICs), such as RAM (Random Access Memory) and ROM. It is electronic in nature, meaning data is accessed via electrical signals. Because it does not require any moving mechanical parts, it has the lowest latency and the highest access speed. This category includes the fastest memories in a computer system, such as Cache (SRAM) and Main Memory (DRAM).
2. **Tape (Option a):** Magnetic tape is a sequential access storage medium. To reach a specific piece of data, the tape must be physically wound or unwound to that position, making it the slowest of all the listed options.
3. **Disk (Option c):** Magnetic disks (like HDDs) provide direct access, but they rely on mechanical components (spinning platters and a moving read/write head). The time taken for the head to move (seek time) and the disk to rotate (rotational latency) makes it significantly slower than semiconductor memory.
4. **Bubble Memory (Option d):** This is an older form of non-volatile memory that uses magnetic bubbles. While it was designed to be a middle ground between disks and RAM, it is still a physical/magnetic process and is far slower than the electronic switching of semiconductor memory.
Conclusion: Semiconductor memory is the fastest because it operates at electronic speeds without mechanical movement.
What is the name of routing type in which every incoming packet is sent to every neighbor router except the one from which the packet came?
A)
distribution
B)
multicast
C)
link-state
D)
flooding
The process described is the definition of 'Flooding' in computer networking. In flooding, every incoming packet is sent out on every outgoing line except the one it arrived on. This ensures that the packet reaches all possible destinations in the network, though it can lead to significant network congestion (broadcast storms) if not managed by mechanisms like sequence numbers or Time-to-Live (TTL) counters.
Why other options are incorrect:
- a. Distribution: This is a general term and not a specific routing algorithm or technique for packet forwarding.
- b. Multicast: This is a method of sending a packet to a specific group of destination computers, not to every single neighbor regardless of destination.
- c. Link-state: This is a type of routing protocol (like OSPF) where routers build a complete map of the network topology. While link-state protocols use flooding to distribute routing updates, the act of sending every packet to every neighbor is the mechanism of 'flooding' itself, not the definition of the routing type 'link-state'.
Which type of switching uses the entire capacity of a dedicated link?
A)
Circuit switching
B)
Datagram
C)
Virtual Circuit
D)
Message switching
Step-by-step reasoning:
1. **Circuit Switching**: In this method, a dedicated physical path (circuit) is established between the sender and the receiver before data transmission begins. Once the connection is established, the entire capacity (bandwidth) of that dedicated link is reserved for the duration of the session, regardless of whether data is being transmitted or not. This ensures a constant bit rate and no congestion once the call is connected (e.g., traditional PSTN telephone networks).
2. **Datagram (Packet Switching)**: In datagram switching, data is broken into packets. Each packet is routed independently based on the destination address. There is no dedicated link; packets share the network capacity dynamically, meaning the entire capacity of a link is not reserved for a single user.
3. **Virtual Circuit (Packet Switching)**: This is a hybrid approach. While it establishes a logical path before transmission, it still uses packet switching. The physical links are shared among multiple virtual circuits using multiplexing, so a single user does not have the entire capacity of the physical link dedicated to them.
4. **Message Switching**: This is a 'store-and-forward' mechanism where the entire message is sent to an intermediate node, stored, and then forwarded. It does not use dedicated links and is highly inefficient for real-time communication.
Therefore, only Circuit Switching utilizes the entire capacity of a dedicated link.
Good software design should possess
A)
Understandability
B)
Maintainability
C)
Both a and b
D)
None of these
In Software Engineering, a 'good' software design is one that adheres to quality attributes that ensure the software can be evolved, fixed, and understood over its lifecycle.
1. Understandability: This refers to the ease with which a developer (either the original author or a new maintainer) can comprehend the logic, structure, and purpose of the code. A design that is easy to understand reduces the risk of introducing bugs during updates.
2. Maintainability: This is the ease with which a software system can be modified to correct faults, improve performance, or adapt to a changed environment. High maintainability is achieved through low coupling and high cohesion.
Since both understandability and maintainability are fundamental pillars of high-quality software design, option 'c' is the correct choice.
By taking 2’s complement again of the 2’s complement of a binary, one gets
A)
the 1’s complement
B)
the 2’s complement
C)
the original number
D)
the sign magnitude form of the numbers
To find the 2's complement of a binary number, the standard procedure is to find the 1's complement (invert all bits) and then add 1 to the least significant bit (LSB). Mathematically, for an n-bit number X, the 2's complement is defined as (2^n - X).
If we take the 2's complement again of the result:
1. Let the original number be X.
2. First 2's complement: Y = 2^n - X.
3. Second 2's complement: Z = 2^n - Y.
4. Substituting Y: Z = 2^n - (2^n - X) = X.
Example:
Let the binary number be 1010 (Decimal 10).
1. 1's complement: 0101
2. 2's complement: 0101 + 1 = 0110
3. Now, take 2's complement of 0110:
- 1's complement: 1001
- 2's complement: 1001 + 1 = 1010 (Original number).
Therefore, the operation is self-inverting, and applying it twice returns the original binary number.
Insertion of a node at the beginning of a linked list requires adjustment of pointer field of _________ node(s)
A)
Head
B)
Tail
C)
Middle
D)
Top
In a singly linked list, the 'Head' pointer is a special pointer that stores the address of the first node of the list. To insert a new node at the beginning of the list, the following steps are performed: 1. Create a new node. 2. Set the 'next' pointer of the new node to point to the current first node (the node currently pointed to by the Head). 3. Update the 'Head' pointer to point to the new node. Since the Head pointer must be updated to reflect the new starting point of the list, the adjustment occurs at the Head.
Why other options are incorrect:
- b. Tail: The tail pointer (if maintained) only needs adjustment if the list was previously empty; otherwise, it remains pointing to the last node.
- c. Middle: Inserting at the beginning does not require modifying any nodes in the middle of the list.
- d. Top: 'Top' is a term typically used for Stacks, not as a general pointer for linked list traversal, although a stack can be implemented using a linked list where 'top' acts as the 'head'. In the context of general linked list terminology, 'Head' is the standard term.
In Vertical redundancy check (parity check)
A)
Only 1 bit is used for error detection
B)
2 bits are used for error detection
C)
3 bits are used for error detection
D)
None of these
Vertical Redundancy Check (VRC), commonly known as a Parity Check, is a simple error-detection method used in data communications. In VRC, a single redundant bit (the parity bit) is added to every data unit (usually a character or a byte).
1. How it works: Depending on whether 'Even Parity' or 'Odd Parity' is chosen, the parity bit is set to 0 or 1 to make the total number of 1s in the unit either even or odd.
2. Bit count: Regardless of the data length (e.g., 7 bits or 8 bits), only one additional bit is appended to the end of the block for the purpose of error detection.
3. Limitation: It can only detect single-bit errors (or an odd number of bit errors). If two bits are flipped simultaneously, the parity remains the same, and the error goes undetected.
Comparing this to the options:
- Option (a) is correct because VRC uses exactly 1 bit per unit.
- Options (b) and (c) are incorrect as they describe multi-bit redundancy schemes (like Hamming codes or CRC), which are not VRC.
Considering data transmission in terms of baud rate and bit rate, which of the following is true?
A)
Bit rate is less than or equal to baud rate.
B)
Bit rate is always equal to baud rate.
C)
Baud rate is less than or equal to bit rate.
D)
Baud rate is always greater than bit rate.
To understand the relationship between bit rate and baud rate, we must define both terms:
1. **Bit Rate**: This is the number of bits (0s and 1s) transmitted per second. It represents the actual data throughput.
2. **Baud Rate**: This is the number of signal units (symbols) transmitted per second. A signal unit is the smallest change in the signal (e.g., a change in voltage or phase).
**The Relationship Formula:**
Bit Rate = Baud Rate × log₂(L), where 'L' is the number of distinct signal levels (symbols) used to represent data.
**Analysis:**
- If L = 2 (Binary signaling, where one symbol represents one bit), then log₂(2) = 1. In this case, Bit Rate = Baud Rate.
- If L > 2 (Multi-level signaling, such as QAM or PSK, where one symbol represents multiple bits), then log₂(L) > 1. In this case, Bit Rate > Baud Rate.
- It is physically impossible for a single symbol to carry less than one bit of information in standard data transmission; therefore, the bit rate cannot be less than the baud rate.
**Conclusion:**
Since the bit rate is either equal to the baud rate (when 1 symbol = 1 bit) or greater than the baud rate (when 1 symbol = multiple bits), it follows that the Baud rate is always less than or equal to the bit rate.
Consider the following program fragment: main() { int a,b,c; b=2; a=2*(b++); c=2*(++b); } Which one of the given answers is correct?
A)
a=4,c=6
B)
b=3,c=8
C)
b=3,c=6
D)
a=4,c=8
Let's trace the program step-by-step:
1. Initial state: `b = 2`.
2. Statement `a = 2 * (b++);`:
- This uses the **post-increment** operator (`b++`). In post-increment, the current value of the variable is used in the expression first, and then the variable is incremented.
- Expression: `a = 2 * 2` (since current `b` is 2).
- Result: `a = 4`.
- After the expression is evaluated, `b` is incremented: `b` becomes `3`.
3. Statement `c = 2 * (++b);`:
- This uses the **pre-increment** operator (`++b`). In pre-increment, the variable is incremented first, and then the new value is used in the expression.
- Current `b` is 3. Incrementing it first: `b` becomes `4`.
- Expression: `c = 2 * 4`.
- Result: `c = 8`.
Final values: `a = 4`, `b = 4`, `c = 8`.
Comparing this to the options:
- Option a: a=4, c=6 (Incorrect)
- Option b: b=3, c=8 (Incorrect, b is 4)
- Option c: b=3, c=6 (Incorrect)
- Option d: a=4, c=8 (Correct)
In _________ ARQ, only the specific damaged or lost frame is retransmitted.
A)
Selective-Reject
B)
Go-Back-n
C)
Piggy bagging
D)
All of these
The question asks about a specific Automatic Repeat Request (ARQ) mechanism where only the damaged or lost frame is retransmitted.
1. **Selective-Reject (Selective Repeat) ARQ**: In this protocol, the receiver accepts out-of-order frames and buffers them. If a frame is lost or damaged, the receiver sends a Negative Acknowledgement (NAK) for that specific frame. The sender then retransmits ONLY that specific frame. This is the most efficient method in terms of bandwidth usage.
2. **Go-Back-N ARQ**: In this protocol, if a frame is lost or damaged, the receiver discards all subsequent frames (even if they are correct). The sender must retransmit the lost frame AND all subsequent frames that were sent after it, regardless of whether they were received correctly. This is less efficient than Selective-Reject.
3. **Piggybacking**: This is not an ARQ mechanism but a technique used in bidirectional data transmission where the acknowledgement for a received frame is 'hitched' or attached to the next outgoing data frame to save bandwidth.
Therefore, the only mechanism that retransmits only the specific damaged or lost frame is Selective-Reject.
Identify the most appropriate sentence to describe unions
A)
Unions are like structures
B)
Unions contain members of different data types which share the same storage area in memory
C)
Unions are less frequently used in programs
D)
Unions are used for set operations
In C programming, a 'union' is a special data type that allows storing different data types in the same memory location.
1. **Analysis of Option (b):** This is the technical definition of a union. Unlike a structure, where each member has its own unique memory address, all members of a union share the same starting memory address. The size of the union is determined by the size of its largest member. Only one member can hold a value at any given time.
2. **Analysis of Option (a):** While unions are syntactically similar to structures (both use the 'struct' or 'union' keyword and curly braces), they are fundamentally different in memory allocation. Calling them 'like structures' is too vague to be the 'most appropriate' description.
3. **Analysis of Option (c):** Whether unions are 'less frequently used' is a subjective observation about coding habits, not a technical description of what a union is.
4. **Analysis of Option (d):** Unions in C have nothing to do with mathematical 'set operations' (like union, intersection, or difference in set theory). They are a memory management tool.
Therefore, option (b) is the only technically accurate and descriptive definition.
In case of ordinary int variables
A)
The leftmost bit is reserved for sign
B)
The rightmost bit is reserved for sign
C)
No bit is reserved for sign
D)
None of these
In C programming, an 'ordinary int' is typically a signed integer by default. Signed integers are represented in memory using a system called Two's Complement. In this representation, the Most Significant Bit (MSB), which is the leftmost bit, is used as the sign bit. If the leftmost bit is 0, the number is positive; if the leftmost bit is 1, the number is negative.
- Option (b) is incorrect because the rightmost bit (Least Significant Bit) determines whether a number is even or odd, not its sign.
- Option (c) is incorrect because 'ordinary int' refers to signed integers; only 'unsigned int' variables have no bit reserved for the sign.
- Option (d) is incorrect as option (a) accurately describes the standard memory representation of signed integers in C.
8088 microprocessor has
A)
16 bit data bus
B)
4 byte pre-fetch queue
C)
6 byte pre-fetch queue
D)
16 bit address bus
The 8088 has a 6-byte instruction prefetch queue.