amcat computer science questions answer and syllabus
Crack amcat computer science with GetWays Solution
here you will find all the details about amcat like amcat computer science questions with answers and amcat computer science syllabus.
Number of Questions: 20
Module Duration: 20 minutes
Detailed Syllabus:
Operating System and Computer Architecture
Basics of OS
Computer Architecture
I/O and File Management
Memory Management
Process Management and Synchronization
Process Management
DBMS
The basic concept, Data model, Views, Operation, TRC, DRC, Architecture
Normalization, Generalization, ERD, Key, Database, SQL, Joins, Indexing
Computer Networks
Basics of Computer Networks & Communication
Routing
Reference Network Model and Protocols
Sample Question with answer
Q1. Rahul was studying about the network topology of his company. He asked one of his fellow mates to guess the right topology on the following clues:
The topology used is a point to point configured.
The topology provides fault isolation property.
If one link becomes unusable, then the whole system is not incapacitated.
a. Bus
b. Mesh
c. Star
d. Tree
Ans: c
GetWays Solution Explanation:
Topology
A Network Topology is the arrangement with which laptop systems or network gadgets are related to every other. Topologies can also outline each physical and logical factor of the network. Both logical and physical topologies may be identical or exclusive in an equal community.
Statement
- The topology used is a point to point configured.
- The topology provides fault isolation property.
- If one link becomes unusable, then the whole system is not incapacitated
| tree | Bus | Mesh | Star |
|---|---|---|---|
YES
|
No
|
Yes
|
Yes
|
No
|
No
|
Yes
|
Yes |
No
|
No
|
No
|
Yes
|
Q2. Champu is designing a database of motor vehicles. It has one base entity ‘Vehicles’ which is classified into two sub-entities, two-wheeler and four-wheeler. He further breaks them down into more entities. What is the process being used here?
a. Generalization.
b. Specialization.
c. Aggregation.
d. Segregation
Ans: b
GetWays Solution Explanation
Generalization
Generalization is the process of extracting common properties from a set of entities and create a generalized entity from it. It is a bottom-up approach.
Specialization
In specialization, an entity is divided into sub-entities based on their characteristics. It is a top-down approach where the higher-level entity is specialized into two or more lower-level entities
Aggregation
An ER diagram is not capable of representing a relationship between an entity and a relationship which may be required in some scenarios. In those cases, a relationship with its corresponding entities is aggregated into a higher-level entity.
Segregation
data separation
Q3. Sunil is working on a database management system. He wants to program transactions such that a transaction is completed only if all the database operations are completed and committed, otherwise, the transaction is aborted and rolled back. Which of the following database characteristics is he trying to implement?
a. Atomicity.
b. Consistency.
c. Isolation.
d. Durability.
Ans: a
GetWays Solution Explanation
(ACID)
A database guarantees the following four properties to ensure database reliability, as follows:
Atomicity: A database follows the all or nothing rule, i.e., the database considers all transaction operations as one whole unit or atom. Thus, when a database processes a transaction, it is either fully completed or not executed at all.
Consistency: Ensures that only valid data following all rules and constraints are written in the database. When a transaction results in invalid data, the database reverts to its previous state, which abides by all customary rules and constraints.
Isolation: Ensures that transactions are securely and independently processed at the same time without interference, but it does not ensure the order of transactions. For example, user Amit withdraws $100 and user Ballu withdraws $250 from user Zinnu’s account, which has a balance of $1000. Since both Amit and Ballu draw from Zinnu’s account, one of the users is required to wait until the other user transaction is completed, avoiding inconsistent data. If Ballu is required to wait, then Ballu must wait until Amit’s transaction is completed, and Zinnu’s account balance changes to $900. Now, Ballu can withdraw $250 from this $900 balance.
Durability: In the above example, user Ballu may withdraw $100 only after user Amit’s transaction is completed and is updated in the database. If the system fails before Amit’s transaction is logged in the database, Amit cannot withdraw any money, and Zinnu’s account returns to its previous consistent state.
Q4. There are 7 processes P1, P2, P3, P4, P5, P6 and P7 given. Their respective priorities, Arrival Times and Burst times are given in the table below. Calculate the average waiting time, if the lower number in the priority column in the given image denotes the Lower priority.?
a. 10 unit.
b. 8 unit.
c. 9 unit.
d. 8.5 unit
Ans:c
GetWays Solution Explanation
Ans:c
GetWays Solution Explanation
GANTT chart Preparation
At time 0, P1 arrives with the burst time of 1 units and priority 2. Since no other process is available hence this will be scheduled till next job arrives or its completion (whichever is lesser).
At time 1, P2 arrives. P1 has completed its execution and no other process is available at this time hence the Operating system has to schedule it regardless of the priority assigned to it.
The Next process P3 arrives at time unit 2, the priority of P3 is higher to P2. Hence the execution of P2 will be stopped and P3 will be scheduled on the CPU.
During the execution of P3, three more processes P4, P5 and P6 become available. Since all these three have the priority lower to the process in execution so PS can't preempt the process. P3 will complete its execution and then P5 will be scheduled with the priority highest among the available processes.
Meanwhile the execution of P5, all the processes got available in the ready queue. At this point, the algorithm will start behaving as Non Preemptive Priority Scheduling. Hence now, once all the processes get available in the ready queue, the OS just took the process with the highest priority and execute that process till completion. In this case, P4 will be scheduled and will be executed until the completion.
Since P4 is completed, the other process with the highest priority available in the ready queue is P2. Hence P2 will be scheduled next.
P2 is given the CPU till the completion. Since its remaining burst time is 6 units hence P7 will be scheduled after this.
The only remaining process is P6 with the least priority, the Operating System has no choice unless of executing it. This will be executed at the last.
The Completion Time of each process is determined with the help of GANTT chart. The turnaround time and the waiting time can be calculated by the following formula.
Avg Waiting Time = (0+14+0+7+1+25+16)/7 = 63/7 = 9 units
Preemptive Priority Scheduling: If the new process arrived at the ready queue has a higher priority than the currently running process, the CPU is preempted, which means the processing of the current process is stopped and the incoming new process with higher priority gets the CPU for its execution.
Q5. Any program residing in the memory contains a set of instructions that need to be executed by the computer in a sequential manner. This cycle for every instruction is known as the instruction cycle.
The cycle consists of the following steps which may or may not be the proper sequence of the execution.
1. Fetch the operand from the memory
2. Execution
3. Fetch instruction
4. Decode the instruction
5. Result
In which of the following steps, is the Program Counter(PC) loaded with the address of the next instruction?
a. Execution
b. Fetch instruction
c. Decode the instruction
d. Result
Ans:b
GetWays Solution Explanation
After the fetch cycle, the next instruction will be updated...
Therefore after fetch comes to the decode instruction step and therefore according to the question in this step itself the address of next instruction will be loaded in the program counter.
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