1. Which of the following data structure is Non-linear type ?
(A) Strings
(B) Lists
(C) Stacks
(D) None of the above
(B) Lists
(C) Stacks
(D) None of the above
Ans:-D
Explanation:- In Linear type data can be traversed sequentially. For example, arrays, linked list. In non linear type data is traversed in a non-sequential,hierarchal manner. Example are trees, Binary search trees etc. So option D will be the correct answer.
2. The total number of comparisons in a bubble sort is
(A) 0(log n)
(B) 0(n log n)
(C) 0(n)
(D) None of the above
(B) 0(n log n)
(C) 0(n)
(D) None of the above
Ans:-D
Explanation:-
The total number of comparisons in a bubble sort is
=(N-1)+(N-2)+…+2+1
=(N-1)*N/2=O(N2)
Since this value is not available in the list of options,we go with D.
The total number of comparisons in a bubble sort is
=(N-1)+(N-2)+…+2+1
=(N-1)*N/2=O(N2)
Since this value is not available in the list of options,we go with D.
3. Which of the following is a bad example of recursion ?
(A) Factorial
(B) Fibonacci numbers
(C) Tower of Hanai
(D) Tree traversal
(B) Fibonacci numbers
(C) Tower of Hanai
(D) Tree traversal
Ans:-B
4. Domain and Range of the function Y=– sqrt(–2x+3)is
(A) x≥3/2,y≥0
(B) x>3/2,y<=0
(C) x≥3/2,y≤0
(D) x≤3/2,y≤0
(B) x>3/2,y<=0
(C) x≥3/2,y≤0
(D) x≤3/2,y≤0
Ans:-D
Explanation:- Since i am not such a mathematical person, i am just giving you the link from where you can get the explanation for this answer. The same problem along with how to arrive at the answer is given in purple math.
5. Maximum number of edges in a n- Node undirected graph without self loop is
(A) n2
(B) n(n-1)
(C) n(n + 1)
(D) n(n-1)/2
(B) n(n-1)
(C) n(n + 1)
(D) n(n-1)/2
Ans:-D
Explanation:-
Theorem.
In an undirected simple graph with N vertices without loops, there are at most N(N-1)/2 edges.
This can be proved by induction.
This can also be verified by drawing some simple graphs of 2,3, 4 and 5 vertices. But No multiple edges. So the answer is option D.
Theorem.
In an undirected simple graph with N vertices without loops, there are at most N(N-1)/2 edges.
This can be proved by induction.
This can also be verified by drawing some simple graphs of 2,3, 4 and 5 vertices. But No multiple edges. So the answer is option D.
6. A hash table has space for 75 records, then the probability of collision before the table is 6% full
(A) .25
(B) .20
(C ).35
(D).30
(B) .20
(C ).35
(D).30
Explanation:
The answer which i am getting out of calculation is not available in any of the options given here. Anyhow i will go ahead and give the explanation for the same. The hash table has space for 75 records. So let us say n=75. We have to calculate the probability of collision before the table is 6% full. If the table is 6% full, then 5 slots are occupied in the table. Since it is given before the table is 6% full, let us take the occupying items as 4. Let us calculate the probability of n collision and then subtract them from 1 to get the final answer.
Prob. of no collision on first insertion = 1
Prob. of no collision on 2nd insertion=(n-1)/n=74/75
Prob. of no collision on 3rd insertion=(n-2)/n=73/75
Prob. of no collision on 4th insertion=(n-3)/n=72/75
So the probability of no collision on any of m insertions is the product of these values which is
=1 * 0.987 * 0.973 * 0.96* 0.947
= 0.873
So the likelihood of at least one collision is just 1 minus this value.
=1-0.873 = 0.123 which seems to be the answer. But there is no option like this at all. So i am not answering this question.
The answer which i am getting out of calculation is not available in any of the options given here. Anyhow i will go ahead and give the explanation for the same. The hash table has space for 75 records. So let us say n=75. We have to calculate the probability of collision before the table is 6% full. If the table is 6% full, then 5 slots are occupied in the table. Since it is given before the table is 6% full, let us take the occupying items as 4. Let us calculate the probability of n collision and then subtract them from 1 to get the final answer.
Prob. of no collision on first insertion = 1
Prob. of no collision on 2nd insertion=(n-1)/n=74/75
Prob. of no collision on 3rd insertion=(n-2)/n=73/75
Prob. of no collision on 4th insertion=(n-3)/n=72/75
So the probability of no collision on any of m insertions is the product of these values which is
=1 * 0.987 * 0.973 * 0.96* 0.947
= 0.873
So the likelihood of at least one collision is just 1 minus this value.
=1-0.873 = 0.123 which seems to be the answer. But there is no option like this at all. So i am not answering this question.
7. BCC in the internet refers to
(A) Black carbon copy
(B) Blind carbon copy
(C) Blank carbon copy
(D) Beautiful carbon copy
Ans:- B
(B) Blind carbon copy
(C) Blank carbon copy
(D) Beautiful carbon copy
Ans:- B
8.Hub is a term used with (A) A Star Networks
(B) A Ring Networks
(C) A Router
(D) A Bridge
(B) A Ring Networks
(C) A Router
(D) A Bridge
Ans:-A
Explanation:-
In network topology, a star network is one where each host is connected to a central hub via a point-to—point link. It is considered a s the easiest topology to design and implement. It is also very easy to add nodes. But the disadvantage is that the hub represents a single point of failure. If the hub is down then the entire network becomes down.
Explanation:-
In network topology, a star network is one where each host is connected to a central hub via a point-to—point link. It is considered a s the easiest topology to design and implement. It is also very easy to add nodes. But the disadvantage is that the hub represents a single point of failure. If the hub is down then the entire network becomes down.
9.The amount of uncertainty in a system of symbol is called
(A) Bandwidth
(B) Entropy
(C ) Loss
(D) Quantum
(B) Entropy
(C ) Loss
(D) Quantum
Ans:- B
10. Which of the following network access standard disassembler is used for connection station to a packet switched network ?
(A) X.3
(B) X.21
(C) X.25
(D) X.75
Ans:- C
Explanation:-
X.25 is used in packed switched network. The idea is to connect a dumb terminal to a packet-switched network.
(B) X.21
(C) X.25
(D) X.75
Ans:- C
Explanation:-
X.25 is used in packed switched network. The idea is to connect a dumb terminal to a packet-switched network.
11. A station in a network forward incoming packets by placing them on its shortest output queue. What routing algorithm is being used ?
(A) Hot potato routing
(B) Flooding
(C) Static routing
(D) Delta routing
(B) Flooding
(C) Static routing
(D) Delta routing
Ans:-A
12. Start and stop bits are used in serial communications for
(A) Error detection
(B) Error correction
(C) Synchronization
(D) Slowing down the communication
(B) Error correction
(C) Synchronization
(D) Slowing down the communication
Ans:-C
13. For a data entry project for office staff who have never used computers before (user interface and user- friendliness are extremely important), one will use
(A) Spiral model
(B) Component based model
(C) Prototyping
(D) Waterfall model
(B) Component based model
(C) Prototyping
(D) Waterfall model
Ans:-C
14. An SRS
(A) establishes the basis for agreement between client and the supplier.
(B) provides a reference for validation of the final product.
(C) is a prerequisite to high quality software.
(D) all of the above.
(B) provides a reference for validation of the final product.
(C) is a prerequisite to high quality software.
(D) all of the above.
Ans:-D
15. McCabe’ s cyclomatic metric V(G) of a graph G with n vertices, e edges and p connected component is
(A) e
(B) n
(C) e– n +p
(D) e – n + 2p
(B) n
(C) e– n +p
(D) e – n + 2p
Ans:-C
Explanation:-
The cyclomatic number V(G) of a graph with n vertices, e edges and p connected component is
V(G) = e – n + p
The cyclomatic number V(G) of a graph with n vertices, e edges and p connected component is
V(G) = e – n + p
16. Emergency fixes known as patches are result of
(A) adaptive maintenance
(B) perfective maintenance
(C) corrective maintenance
(D) none of the above
(B) perfective maintenance
(C) corrective maintenance
(D) none of the above
Ans:-C
17. Design recovery from source code is done during
(A) reverse engineering
(B) re-engineering
(C) reuse
(D) all of the above
(B) re-engineering
(C) reuse
(D) all of the above
Ans:-A
18. Following is used to demonstrate that the new release of software still performs the old one did by rerunning the old tests :
(A) Functional testing
(B) Path testing
(C) Stress testing
(D) Regression testing
(B) Path testing
(C) Stress testing
(D) Regression testing
Ans:-D
19. The post order traversal of a binary tree is DEBFCA. Find out the pre- order traversal.
(A) ABFCDE
(B) ADBFEC
(C) ABDECF
(D) ABDCEF
(B) ADBFEC
(C) ABDECF
(D) ABDCEF
Ans:-C
Explanation:- The same question is given in Paper II of June 2012 and i have explained the same in this blog. Please refer to paper ii – question no 2 of June 2012 paper.
20. B + tree are preferred to binary tree in database because
(A) Disk capacities are greater than memory capacities
(B) Disk access much slower than memory access
(C) Disk data transfer rates are much less than memory data transfer rate
(D) Disk are more reliable than memory
(B) Disk access much slower than memory access
(C) Disk data transfer rates are much less than memory data transfer rate
(D) Disk are more reliable than memory
Ans:-B
Explanation:- Again the same question is repeated in June 2012 paper and the answer is B. Explanation for the same is provided there. In Question no 11 of paper ii – june 2012.
21. What deletes the entire file except the file structure ?
(A) ERASE
(B) DELETE
(C) ZAP
(D) PACK
(B) DELETE
(C) ZAP
(D) PACK
Ans:-C
Explanation:- Again this question is given in June 2012 paper-ii. Since i have solved june 2012-paper II before the answer for the same is available there also. Question No 9 of June 2012-Paper II.
22. Which command classes text file, which has been created using “SET ALTERNA TIVE”“Command” ?
(A) SET ALTERNATE OFF
(B) CLOSE DATABASE
(C) CLOSE ALTERNATE
(D) CLEAR ALL
(B) CLOSE DATABASE
(C) CLOSE ALTERNATE
(D) CLEAR ALL
Ans:-A
Explanation:-
Actually the question should be which command is used for closing the text file, which has been created using SET ALTERNATE TO . The answer is SET ALTERNATE OFF. SET ALTERNATE ON directs the output to text file. OFF disables the output to text file.
Actually the question should be which command is used for closing the text file, which has been created using SET ALTERNATE TO . The answer is SET ALTERNATE OFF. SET ALTERNATE ON directs the output to text file. OFF disables the output to text file.
23. Data security threats include
(A) privacy invasion
(B) hardware failure
(C) fraudulent manipulation of data
(D) encryption and decryption
(B) hardware failure
(C) fraudulent manipulation of data
(D) encryption and decryption
Ans:-C
Explanation:-
Option B and D are completely ruled out. Since the question talks about data security threat choosing option C is correct than option A.
Option B and D are completely ruled out. Since the question talks about data security threat choosing option C is correct than option A.
24. Which of the following statements is true, when structure of database file with 20 records is modified ?
A) ? EOF ( ) Prints. T
(B) ?BOF()Prints F
(C) ?BOF()Prints T
(D) ?EOF()Prints F
(B) ?BOF()Prints F
(C) ?BOF()Prints T
(D) ?EOF()Prints F
Ans:-A
25. The SQL Expression Select distinct T. branch name from branch T, branch S where T. assets > S. assets and S. branch-city = DELHI, finds the name of
(A) all branches that have greater asset than any branch located in DELHI.
(B) all branches that have greater assets than allocated in DELHI.
(C) the branch that has the greatest asset in DELHI.
(D) any branch that has greater asset than any branch located in DELHI.
(B) all branches that have greater assets than allocated in DELHI.
(C) the branch that has the greatest asset in DELHI.
(D) any branch that has greater asset than any branch located in DELHI.
Ans:-A
26. Dijkestra banking algorithm in an operating system, solves the problem of
(A) deadlock avoidance
(B) deadlock recovery
(C) mutual exclusion
(D) context switching
(B) deadlock recovery
(C) mutual exclusion
(D) context switching
Ans:-A
27. The multiuser operating system, 20 requests are made to use a particular resource per hour, on an average the probability that no request are made in 45 minutes is
(A) e-15
(B) e-5
(C) 1-e-5
(D) 1-e-10
(B) e-5
(C) 1-e-5
(D) 1-e-10
Ans:-A
28. On receiving an interrupt from an I/O device, the CPU
(A) halts for predetermined time
(B) branches off to the interrupt service routine after completion of the current instruction
(C) branches off to the interrupt service routine immediately.
(D) hands over control of address bus and data bus to the interrupting device.
Ans:-B
(B) branches off to the interrupt service routine after completion of the current instruction
(C) branches off to the interrupt service routine immediately.
(D) hands over control of address bus and data bus to the interrupting device.
Ans:-B
29. The maximum amount of information that is available in one portion of the disk access arm for a removal disk pack (without further movement of the arm with multiple heads)
(A) a plate of data
(B) a cylinder of data
(C) a track of data
(D) a block of data
Ans:-B
(B) a cylinder of data
(C) a track of data
(D) a block of data
Ans:-B
30. Consider a logical address space of 8 pages of 1024 words mapped with memory of 32 frames. How many bits are there in the physical address ?
(A) 9 bits
(B) 11 bits
(C) 13 bits
(D) 15 bits
Ans:-D
(B) 11 bits
(C) 13 bits
(D) 15 bits
Ans:-D
Explanation:-
The question asks about the number of bits in the physical address. The word size is 1024. So, the number of bits required to calculate the offset into each page = 210=1024=10 bits.
So for 32 frames(25) of 1024 words each,we would require a total of 5+10 bits = 15 bits. Therefore the option is D.
The question asks about the number of bits in the physical address. The word size is 1024. So, the number of bits required to calculate the offset into each page = 210=1024=10 bits.
So for 32 frames(25) of 1024 words each,we would require a total of 5+10 bits = 15 bits. Therefore the option is D.
31. CPU does not perform the operation
(A) data transfer
(B) logic operation
(C) arithmetic operation
(D) all of the above
Ans:-D
(B) logic operation
(C) arithmetic operation
(D) all of the above
Ans:-D
32. A chip having 150 gates will be classified as
(A) SSI
(B) MSI
(C) LSI
(D) VLSI
Ans:-B
(B) MSI
(C) LSI
(D) VLSI
Ans:-B
Explanation:- The number of gates for the different classification are as follows.
Small Scale Integration(SSI) – 1000
Very Large Scale Integration(VLSI) – >100000
So the answer is option B.
Small Scale Integration(SSI) – 1000
Very Large Scale Integration(VLSI) – >100000
So the answer is option B.
33.If an integer needs two bytes of storage, then the maximum value of unsigned integer is
(A) 216 –1
(B) 215 –1
(C) 216
(D) 215
Ans:-A
(B) 215 –1
(C) 216
(D) 215
Ans:-A
Explanation:- The same question is asked in June 2012,paper -II but there they have asked for the maximum value of signed integer.The link to that answer is JUNE 2012 – PAPER II. Here it is the maximum value of unsigned integer and the answer is option A.
34. Negative numbers cannot be represented in
(A) signed magnitude form
(B) ’1’ s complement form
(C) ’2’ s complement form
(D) none of the above
Ans:-D
(B) ’1’ s complement form
(C) ’2’ s complement form
(D) none of the above
Ans:-D
Explanation:-
Negative numbers can be represented in signed magnitude form, or ’1′ complement form or ’2”s complement form. So the option is D.
Negative numbers can be represented in signed magnitude form, or ’1′ complement form or ’2”s complement form. So the option is D.
35. The cellular frequency reuse factor for the cluster size N is
(A) N
(B) N2
(C) 1/N
(D) 1/N2
Ans:-C
(B) N2
(C) 1/N
(D) 1/N2
Ans:-C
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