gate
Content
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For IIT-JAM, JNU, GATE, NET, NIMCET and Other Entrance Exams
Pattern of questions : MCQs
There are a total of 65 questions carrying 100 marks.
Questions (56-65) belongs to general aptitude (GA).
Questions (56-60) will carry 1-mark each, and
question (61-65) will carry 2-marks each
GATE - COMPUTER SCI ENCE
1-C-8, Sheela Chowdhary Road, Talwandi, Kota (Raj.) Tel No. 0744-2429614
Web Site www.vpmclasses.com [email protected]
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Total marks : 100
Duration of test : 3 Hours
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MOCK TEST PAPER
Questions (1-25) will carry 1-mark each and
questions (26-55) will carry 2-marks each.
For Q.1-25 and Q.56-60 1/3 mark will be deducted
for each wrong answer.For Q.26-51 and Q. 61-65 2/3
mark will be deducted for each wrong answer.
The question pairs (Q.52, Q.53) and (Q.54, Q.55)
are linked questions.For Q.52 &54 2/3 mark will be
deducted. There is no negative marking for Q.53 &Q.55.
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Q.48-51 are common data questions.
If first question is attempted wrongly then answer of
second question will not be evaluated.
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1. Find the minimum number of edges in a connected graph having 11 vertices.
(A) 5
(B) 10
(C) 15
(D) 20
2. If A be (n × n) non-singular matrix then which of the following is true?
(A) ( )
2(n 1)
Adj Adj A A
÷
=
(B) ( )
2
(n 1)
Adj Adj A A
÷
=
(C) ( )
n 1
Adj Adj A A
÷
=
(D) None of these
3. The multiplicative inverse of matrix A =
3 0 0
0 6 2
0 2 6
(
(
(
(
¸ ¸
is given by
(A)
2
1
(A –15A 68 )
96
+ I
(B)
2
1
(A –15A 96 )
96
+ I
(C)
2
1
(A –15A 68 )
68
+ I
(D)
2
1
(A –15A 96 )
68
+ I
4. If Successive approximations are given by x
1
, x
2
, x
3
............, x
n
, x
n+1
, then Newton Raphson formula is
given by
(A)
( )
( )
n 1
n 1 n
n
f X
X X
f ' x
+
+
= +
(B)
( )
( )
n
n 1 n
n
f X
X X
f ' x
+
= +
(C)
( )
( )
n
n 1 n
n
f X
X X
f ' x
+
= ÷
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(D)
( )
( )
n–1
n 1 n
n
f ' X
X X
f ' x
+
= ÷
5. Use Trapezoidal rule to obtain
5.2
4
nxdx
}
using the table given below
x: 4.0 4.2 4.4 4.6 4.8 5.0 5.2
nx: 1.386 1.435 1.482 1.526 1.569 1.609 1.649
(A) 1.8277
(B) 1.9284
(C) 1.6424
(D) 1.9127
6. If,
dx x
tan a b
1 sinx 2
| |
= + +
|
+
\ .
}
then
(A) a =
4
t
, b = 3
(B) a = –
4
t
, b = 3
(C) a =
4
t
, b = arbitrary constant
(D) a = –
4
t
, b = arbitrary constant
7. If a continuous random variable X follows rectangular distribution in the range (2, 7), then find the
probability
p(2.5 s X s 4).
(A) 0.3
(B) 0.4
(C) 0.5
(D) 0.6
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8. The value of C for which P(X = k) = Ck
2
can serve as the probability function of a random variable X
that takes
0, 1, 2, 3, 4 is
(A)
1
30
(B)
1
10
(C)
1
3
(D)
1
15
9. In the above group, the inverse of cos –
2r 2r
i sin is
n n
t + t
(A)
–2 r –2 r
cos i sin
n n
t t | | | |
+
| |
\ . \ .
(B)
2 r 2 r
cos i sin
n n
t t | | | |
+
| |
\ . \ .
(C)
2(n–r) –2(n–r)
cos i sin
n n
t+ t
(D) None of these
10. The number of prime implicants of the Boolean expression
E = xyz + xyz’ + x’yz’ + x’y’z’ is,
(A) 1
(B) 3
(C) 4
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(D) None of these
11. For the logic circuit shown in the figure below the output y is equal to
(A) ABC
(B) A B C + +
(C) AB+ BC + A +C
(D) AB + BC
12. The Venn diagram representing the Boolean expression A +
( )
A B · is
(A)
A B
(B)
B
(C)
A
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(D)
B
13. The number of full adders and half adders required to add 16 - bit numbers are
(A) 8 half adders 8 full adders
(B) 16 half adders only
(C) 1 half adders 15 full adders
(D) 4 half adders 12 full adders
14. What will be the output?
(A) AB + BC +CA +BC
(B) A © B © C
(C) A © B
(D) ABC ABC ABC ABC + + +
15. For a binary half-subtractor having two inputs A and B, the correct set of logical expression for the
outputs D (= A minimum B) and X ( = borrow) are
(A) D = AB + AB, X AB =
(B) D = AB AB, X AB + =
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(C) D = + = AB AB , X AB
(D) D = AB + AB, X AB =
16. Consider the DFA shown below:
ˆ
o (A, 01) will be:
(A) {D}
(B) {C, D}
(C) {B,C,D}
(D) {A,B,C,D}
17. How many variables does the following grammar have when converted to CNF?
E ÷ E + T
E ÷ T
T ÷ (E)
T ÷ i
(A) 6
(B) 7
(C) 8
(D) 9
18. G
1
= {{(S, A, B}, {a,b}, {S ÷ bA| aB, A ÷ aS|bAA|aBB|b}, S}
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G
2
= {{S}, {a,b}, {S ÷ aSbS|bSaS| e}, S}
Choose the correct alternative?
(A) L (G
1
) = L (G
2
)
(B) L (G
1
) c L (G
2
)
(C) L (G
1
) L (G
2
)
(D) L (G
1
) L (G
2
) = |
19. X: Given a TM M over E and any word e e E
*
does M loop on forever on e?
You any assume that the halting problem of TM is un-decidable but partially decidable. Then which of
the following is true?
(A) X is decidable
(B) X is un-decidable but partially decidable
(C) X is un-decidable and not even partially decidable
(D) X is not a decision problem
20. Consider the following grammar:
S ÷ Aa|b
A ÷ Ac|Sd|e
After removing left recursion the resulting grammar will be,
(A)
S bS'
S' daS' |
A' SdA' |
A' cA
÷
÷ e
÷ e
÷
(B)
S Aa | b
A bdA' |
A' cA' | adA' |
÷
÷ e
÷ e
(C) Both (A) and (B)
(D) None of these
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21. Which of the following is an advantage of virtual memory?
(A) Faster access to memory on an average.
(B) Processes can be given protected address spaces.
(C) Linker can assign addresses independent of where the program will be loaded in physical
memory
(D) None of these
22. The output of a 3-input logic circuit f(x, y, z) is 1 if ax + by + cz < d and 0, otherwise (a, b, c, d) are
constant). For what values of a, b, c, d does this represent an implementation of the AND gate.
(A) a = b = c = 1; d = 5/2
(B) a = b = c = – 1; d = – 5/2
(C) a = b = c = 1; d = 3/2
(D) a = b = c = – 1; d = – 3/2
22. How many gate input are required to realize the following expression?
1
F ABC ABCD EF AD = + + +
(A) 14
(B) 11
(C) 13
(D) 15
24. Which of the following can be recognized by a Deterministic Finite-state Automation?
(A) The numbers, 1, 2, 4, ....z
n
...... written in binary.
(B) The numbers 1, 2, 4, ......, z
n
,.... written in un-binary.
(C) The set of binary string in which the number of zeros is the same as the number of ones.
(D) The set {1, 101, 11011, 1110111, ....}
25. What is the idea behind using insertion sort to sort, sub files generated by shell sort?
(A) Insertion sort works best for small files
(B) Insertion sort works well for nearly sorted files
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(C) There is no such need for insertion sort; the sorts like selection sort that are simple to code may
also be used.
(D) Insertion sort is preferred because comparisons are not expensive.
26. What is the output of the following ‘C’ program?
main()
{
Int x = 10, y = 10, z = 5, i;
i = x < y < z;
printf(“\n%d”, i);
(A) 1
(B) 0
(C) Error
(D) 5
27. T(n) = (n+1) + T(n – 1) Then T(n) is equal to
(A) O(nlogn)
(B) O(logn)
(C) O(n
2
)
(D) None of these
28. What is the minimal expression for
f = tM(2, 3, 8, 12, 13), d(10, 14)?
(A) (A D) (A B C) (B C A) + + + + +
(B) (A D) (A B C) (A B C) + + + + +
(C) (A B) (B C A) (A B C) + + + + +
(D) (A B) (A C B) (A B C) + + + + +
29. Dividing (11101010)
2
by (1001)
2
will be
(A) 1100
(B) 1001
(C) 1101
(D) None of these
30. The control memory is
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(A) RAM
(B) ROM
(C) Both A and B
(D) None of these
31. Consider a computer with 4 floating point processors. Suppose that each processor uses a cycle time
of 40nsec. How long will it take to perform 400 floating point operations? What is the time if the
operations are carried out on a single processor with cycle time of 10nsec?
(A) 400 n sec, 4000 n sec
(B) 4000 n sec, 4000 n sec
(C) 4000 n sec, 400 n sec
(D) None of these
32. In sign magnitude notation take an n-bit integer and store it in m bit, then
(A) m < n
(B) m = n
(C) m > n
(D) None of these
33. What would be the output of the following program?
int i = 4; printf (“%d%d%d%, i ++,++ i, i) ;
(A) 5 5 4
(B) 4 garbage value garbage value
(C) 4 8 10
(D) 5 5 5
34. main ( ) {
unsigned i = 5;
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clrscr( );
if (i > – 1)
printf (“Hell”);
else
printf(“Heaven”);
}
The output will be:
(A) Heaven
(B) Hell
(C) Compiler error
(D) Unexpected results
35. Which of the following will represent correct expression for the given tree?
2
4
3
5 6
+
*
+
*
+
(A) 654321 + + * * +
(B) 1 + 2 * 3 + 4 * 5 + 6
(C) 12 + 34 + 56 + * *
(D) * + 12 * + 34 + 56
36. Which of the following statements is/are true ?
S1: Semaphores and monitors can be used in distributed system
S2: For distributed system one must use message passing
(A) Only S1
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(B) Only S2
(C) Both S1 and S2
(D) None of these
37. Consider the page sequence 4,3,2,1,4,3,5,4,3,2,1,5.
If FIFO page replacement algorithm is used and frame size is 3, then the percentage of page fault is,
(A) 75%
(B) 85%
(C) 83%
(D) 77%
38. Consider relation schema R with set functional dependencies (F) as:
R(ABCDE) and F = {A ÷ BCDE, B ÷ C,D ÷ E }
The above relation is in
(A) 1NF
(B) 2NF
(C) 3NF
(D) BCNF
39. ( ) ( )
s _name, title card_no
title F53
supp book borrow book
=
| |
t ÷ t o
|
\ .
(A) Name of suppliers who have supplied at least one copy of all the titles which are issued to card no
F53
(B) Name of suppliers who have supplied all copies of at least one title issued to card no F53
(C) Name of suppliers who have supplied all copies of all titles which are issued under card no F53
(D) None of these
40. With regard to the expressive power of the formula relational Query languages which of the following
statements is TRUE?
(A) Relational algebra is more powerful than relational calculus
(B) Relational algebra has the same power as relational calculus
(C) Relational algebra has the same power as safe relational calculus
(D) None of these
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41. Which of the following queries finds the client of banker Agassi and the city they live in?
(A) ( )
client _customer _name.customer _city client _customer name customer customer _name Banker_name " Agassi "
( ( Client Customer
÷ = =
t o o ×
(B) ( )
customer _name.customer _ city Bankeer _name " Aggasi
( Client Customer
=
t o ×
(C)
Client.Customer _name.Customer _city Banker_name " Aggasi " client.Customer _name Customer.Customer _name
( ( (Client Customer)))
= =
t o o ×
(D)
Customer _name,Customer _city Banker_names " Aggasi "
( (Client Customer)
=
t o ×
42. Match the following
A. Secondary index I. functional dependency
B. Non-procedural II. B-Tree
Query language
C. Closure of a set of III. domain calculus
Attributes
D. Natural-join IV. Relational algebraic operations
Codes:
(A) A–II, B–III, C–I, D–IV
(B) A–III, B-II, C–IV, D–I
(C) A–III, B-I, C–II, D–IV
(D) A–II, B-I, D–IV, C–III
43. In B
+
tree index file if there are K search keys and each leaf node contains n records then number of
nodes to be accessed
(A) log
2
n
k
(B) log
n
k
(C)
k
log n
(
(
(D)
n/ 2
log k
(
(
44. A decision table
(A) Documents rules that select one or more action based on one or more conditions from a set of
possible conditions
(B) Represents an information flow
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(C) Show the decision path
(D) Gets an accurate picture of the system
45. The basic COCOMO model predicts an effort of 45PM for a development project. The effort
multipliers for the intermediate COCOMO model have nominal values except for following:
RELY = 1.15, STOR = 1.21 TIME = 1.11, TOL = 1.10
Estimate the adjusted effort?
(A) 70 PM
(B) 76 PM
(C) 80 PM
(D) 86 PM
46. Compute cyclometric complexity.
(A) 4
(B) 2
(C) 3
(D) 1
47. Link tag is used
(A) To inform the location of the based document location
(B) To inform the browser the banner
(C) To inform the banner
(D) None of these
COMMON DATA QUESTION 48 & 49:
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Let’s look about the algorithm
int temp, i, j;
for (i = 1; i< n; i + + )
{
temp = A[i];
for (j = i – 1; j > i && (A [j] > temp); j --)
A[j + 1] = A[j];
A[j] = temp;
}
}
48. If the Array is sorted then the complexity is
(A) O(n)
(B) O(n
2
)
(C) O(nlog
2
n)
(D) O(log
2
n)
49. If the array is in reverse sorted order then time complexities will be
(A) O(n)
(B) O (nlog
2
n)
(C) O(n
2
)
(D) O(log
2
)
COMMON DATA QUESTION 50- 51
void x (int A [ ], int n)
{
int i, j;
for (i = 0; i < n ; i + + )
{
j = n – 1;
while (j > i)
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{
if (A [j] < A[j –1] )
{
swap (A[j], A [j – 1]);
{
j --;
}
}
}
50. What will be time complexity of the above algorithm?
(A) O(n)
(B) O(n
2
)
(C) O(nlog
2
n)
(D) O(n
3
)
51. If the array is in sorted order then time complexity will be
(A) O (n)
(B) O(n
2
)
(C) O(nlog
2
n)
(D) O(log
2
n)
Statement for Linked Answer Questions: 52 & 53
Consider the following 5 processes with the length of CPU burst time given in milliseconds (assuming
their arrival time be 0).
Process Burst time
P
1
10
P
2
29
P
3
3
P
4
7
P
5
12
52. Which scheduling algorithm gives the minimal average waiting time?
(A) FCFS
(B) SJF
(C) RR (q = 10) (q : time quantum)
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(D) Both (A) & (B)
53. The proper sequence of the execution of jobs is,
(A) P
1
, P
2
, P
3
, P
4
, P
5
(B) P
3
, P
4
, P
1
, P
5
, P
2
(C) P
1
, P
2
, P
3
,P
4
, P
5
, P
1
, P
2
(D) None of these
Statement for Linked Answer Questions: 54 & 55
Consider an 8085-microprocessor system.
The following program starts at location 0100H.
LXI SP, 00FF
LXI H, 0701
MVI A, 20H
SUB M
54. The content of accumulator when the program counter reaches 0109H is
(A) 20H
(B) 02H
(C) 00H
(D) FFH
55. If in addition following code exists from 0109H onwards.
OR 40H
ADD M
What will be the result in the accumulator after the last instruction is executed?
(A) 40H
(B) 20H
(C) 60H
(D) 42H
56. No doubt, it was our own government but it was being run on borrowed ideas, using
_________solutions.
(A) worn out
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(B) second hand
(C) impractical
(D) appropriate
The question below consists of pair of related words followed by four pairs of words. Select the pair that
best expresses the relation in the original pair:
57. Ratify: Approval::
(A) mutate: change
(B) pacify : conquest
(C) duel: combat
(D) appeal : authority
58. A car goes 35 km in 1 hour, next 270 km in 3 hrs. and next 80 km in
1
2
2
hrs.Find the average speed
of the car
(A) 59.23 km/h.
(B) 61.5 km/h
(C) 80 km/h
(D)52.3km/h
59. Some critics believe that Satyajit Ray never quite came back to the great beginning he made in this
path breaking film Pather Panchali. ______have endured decades of well-traveled bad prints to
become a signpost in cinematic history.
(A) The bizarre history of its misty origins
(B) Its haunting images
(C) Its compelling munificence
(D) The breathtaking awe it inspires
Choose the most appropriate word from the options given below that is most nearly opposite in
meaning to the given word:
60. Valedictory
(A) sad
(B) collegiate
(C) derivative
(D) generosity
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61. If the area of a given square ABCD is 3 find the total area of the entire figure?
(A) 45
2
(B) 45
(C) 48
(D) 31
62. The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the amino
terminal portions of both the H and L chains.
(A) H chain
(B) L chain
(C) Both (a) and (b)
(D) None of these
63. In a certain code Language
134 means “good and tasty”
478 means “see good picture”
729 means “picture are faint”
Which number has been used here for faint?
(A) 9
(B) 2
(C) data are inadequate
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(D) 253
64. A bag contains an equal number of one rupee, 50 paisa and 25 paisa coins. If the value of money in
the bag is Rs. 35, find the total number of coins of each type?
(A) 7
(B) 40
(C) 30
(D) 20
Profit to sale-Table for three companies A, B and C for 1996-97
Companies 1996 1997
Total units 300000 400000
Shares A 5% 25%
B 60% 40%
C 35% 35%
Price A 10% 8%
(per unit) B 7% 14%
(in rupees) C 9% 10%
Profit A 3% 1%
(per unit) B 0.5 5%
(in rupees) C 2% 2.5
65. What is the increase in the total profits of company B in 1997?
(A) 800%
(B) 900%
(C) 750%
(D) 789%
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Answer key
Que Ans. Que Ans. Que Ans. Que Ans. Que Ans.
1 B 16 B 31 B 46 C 61 B
2 B 17 B 32 C 47 A 62 C
3 A 18 B 33 A 48 A 63 C
4 C 19 C 34 C 49 C 64 D
5 A 20 C 35 C 50 B 65 D
6 D 21 B 36 B 51 B
7 A 22 D 37 A 52 B
8 A 23 A 38 C 53 B
9 A 24 A 39 A 54 A
10 B 25 B 40 C 55 B
11 A 26 A 41 A 56 B
12 A 27 C 42 A 57 C
13 C 28 B 43 D 58 A
14 D 29 C 44 A 59 B
15 C 30 C 45 B 60 D
HINTS AND SOLUTIONS
1.(B) The minimum number of edges in a connected graph having 11 vertices is 10 because in a
connected graph the minimum number of edges in a connected graph is n-1 where n is no. of
vertices.
So n = 11
No. of edges = n – 1
= 11 – 1
= 10
2.(B) If A be n × n non-singular matrix then.
|adj (adj(A)| =
2
(n - 1)
|A|
or
det (adj A) =
n – 1
(det(A))
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3.(A) A =
3 0 0
0 6 2
0 2 6
(
(
(
(
¸ ¸
A – ìI = 0
3 0 0 0 0
0 6 2 – 0 0 0
0 2 6 0 0
ì ( (
( (
ì =
( (
( ( ì
¸ ¸ ¸ ¸
¬
3 – 0 0
0 6 – 2 0
0 2 6 –
ì (
(
ì =
(
( ì
¸ ¸
(3 – ì) [(6 – ì)
2
– 4] = 0
(3 – ì) (36 + ì
2
– 12 ì – 4) = 0
(3 – ì) (ì
2
– 12 + 32) = 0
3ì
2
– 36 ì + 96 – ì
2
+ 12 ì
2
– 32 ì = 0
15ì
2
– 68 – ì
3
+ 96 = 0
Characteristics equation of ì –
15A
2
– 68A – A
3
+ 96 = 0
A(A
2
– 15A + 68I) = 96I
1 2
1
A (A – 15A 68 )
96
÷
= + I
4.(C)
n
n 1 n
n
f(X )
X X
f '(X )
+
= ÷
Hint: If we have the successive approximations i.e. X
1
, X
2
, X
3
…..X
n
, X
n + 1
, then we get the recursion
formula (iterative formula) i.e.
n
n 1 n
n
f(X )
X X ,
f '(X )
+
= ÷ Provided f’(X
n
) = 0.
This is known as Newton-Raphson formula or simply Newton’s formula
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5.(A)
5.2
4
n x dx
}
b a 5.2 4
n 0.17
n 7
÷ ÷
= = =
5.2
0 n 1 2 n 1
4
h
n x dx 1(y y ) 2(y y ......y )
2
÷
( = + + + +
¸ ¸ }
5.2
4
0.2
n x dx [(1.386 1.649) 2(1.435 1.482 1.526 1.569 1.609)]
2
= + + + + + +
}
=
0.17
[3.035 2(7.621)]
2
+
=
0.2
18.277
2
×
= 1.8277
6.(D)
dx x
tan a b
1 sin x 2
| |
= + +
|
+
\ .
}
2 2
dx 1 sin x 1 sin x
dx
1 sinx 1 – sin x cos x
÷ ÷
= =
+
} } }
¬
2
[sec x sec x tanx] dx ÷
}
¬
2
sec x dx sec x tanx dx ÷
} }
¬ tanx – sec x + c
Or
¬
2
2 2
2
2
2tanxy (1 tan xy )
(1 tan xy )
÷ +
÷
¬
2
2
2
2
(tan xy 1)
(1 tan xy )
÷
÷
¬
2
2
2 2
(1 tan xy )
(1 tanxy ) (1 tan xy )
÷
+ ÷
¬
2
2
1 – tan xy
1 tan xy +
x
tan – c
2 4
t | |
= +
|
\ .
–
a , b constant
4
t
= =
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7.(A) Rectangular function =
1
y x ÷
Rang (2, 7) {y = 7, x = 2}
1 1
R.F
7.2 5
= =
Probability distribution
P(X = 2.5 s X s 4) =
4
2.5
1
dx
5
}
=
1
[4 – 2.5]
5
=
1.5
0.3
5
=
8.(A) P(X = K) = CK
2
4
2
0
CK 1 =
¿
C + 4C + 9C + 16C = 1
30C = 1
C =
1
30
9.(A) Given
2 r 2 r
cos – i sin
n n
t t | | | |
+
| |
\ . \ .
¬
2 r 2 r
cos i sin ( cos(-x) = cosx)}
n n
t t | |
+
|
\ .
–1
2 r 2 r
cos i sin
n n
t t (
+
(
¸ ¸
¬
–2 r 2 r
cos i sin –
n n
t t | | | |
+
| |
\ . \ .
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10.(B) E = xyz + xyz’ + x’yz’ + x’y’z
f = m(0, 2, 6, 7)
¿
Index Mint erm Binary disignation Pair ABCD
0 0 0 0 0 0 (0, 1) 0 0 – 0
1 2 0 0 10 (2, 6) 0 – – 0
2 6 0 110 (6, 7) 0 – – –
3 7 0 111
So there are three prime implicant
(0, 1), (2, 6), (6, 7), = xy x z yz + +
11.(A) ABC
Hint: We have the following circuit i.e.
y = (F + G + H)
= (D E) (A A) (C * C) + + +
= AB BC A C + + +
= A B B C A C + + + + +
= A B C ABC + + ¬
12.(A) Hint: We have = A (A B) + ·
= (A A) (A B) + +
= A + B
Hence the venn diagram will be
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B
13.(C) Hint: We know that half adder is adding two single bit binary values, X, Y produces a sum S bit and a
carry out C out bit.
Also the full adder is adding two single bit binary values, X, Y with a carry input bit C-in produces a
sum bit S and a carry out C out bit.
So to add 16-bit number we need 1 half adder and 15 full adders. The one half adder can add the
least significant bit of the two numbers. Full adders are required to add the remaining 15 bits as they
all involve adding carries.
14.(D) In the figure we have the multiplexer 2 inputs say A and B as they vary with respect to carry at the
ports
I
0
+ I
3
.
So the output will be ABC ABC ABC ABC + + +
15.(C) D AB AB, X AB = + =
Hint: To prove this we have to make the truth table
A B X D
0 0 0 0
0 1 1 1
1 0 0 1
1 1 0 0
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B
A
0 1
0
1
0 0
1 0
B
A
0 1
0
1
0 1
1 0
f = AB
–
f = AB + AB
– –
So the answer will be AB AB + and
AB
16.(B) {C,D}
Hint:
ˆ
(A, 0) o = e –closure (o(e –closure(A), 0))
= e –closure (o({A, B, C}0))
= e –closure {A, B}
= {A, B, C}
ˆ
(A, 01) o = e –closure (o({A, B, C}, 1))
= e –closure {C, D}
= {C, D}
17.(B) Hint: After conversion
E ÷ ET’
T’ ÷ PT
P ÷ t
E ÷ OE’
O ÷ (
E’ ÷ EC
C ÷)
T ÷ OE’
T ÷ i
Number of variables = 7.
18.(B) L(G
2
) = L(G
1
) {e}
L(G
1
) c L(G
2
)
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Both essentially accept all strings having equal number’s a’s and b’s.
19.(C) Hint: Suppose a TM, M’, exists that answer YES when input (M, e) is such that M loops forever on e.
Let Mu be the TM corresponding to the universal language Lu, then the following construction allows
Lu to be recursive , since either M’ or Mu will come up with an answer, M’ answers yes, If M does not
accept e and Mu answers YES, if M accepts e.
This is a contradiction and hence is not possible and the language
Le = {(M, e)}/M loops forever on e} Is not r.e
20.(C) S ÷ Aa|b
A ÷ Ac|Sd|e
S ÷ Aa|Aad|bd|e
S ÷ Aa
A ÷ bd A’|A’
A ÷ cA’|adA’ |e
21.(B) Hint: Virtual memory – separation of users logical memory from physical memory. Also the processes
can be given protected address spaces. Hence logical address space can therefore be much larger
than physical address space.
22.(D)
1
F ABC ABCD EF AD = + + +
AND input = 3 + 4 + 2 + 2 = 11
AND gates feeding the OR gate = 1 + 1 + 1 + 1 = 4
Total gate = 11 + 4 = 15
23.(A) A has function distribute record one by one in space, that can hold x number of records records 1,
2,….m
m 1
(x – 1) (x – 2)........(x – (m – 2)) (m – 1)
x
÷
If m
+h
record is first record then (m – 1) record are full the space so m
th
record will collide with (m – 1)
record.
24.(A) A deterministic finite automata can be recognized the binary string. Such as –
The number 1, 2, 4 …..z
n
written in binary.
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25.(B) Insertion sort is a good middle of the road choice for sorting list of few thousand item it is simpler than
shell sort, with only a small trade off in efficiency.
Insertion sort insert each element into its proper position in the previously sorted sub list.
Shell sort comparing list element that are separated by specific distance.
26.(A) x = 10, y = 10, z = 5
i = 10 < 10 < 5
i = 0 < 5 {10 < 0 is fals so output ÷ 0
i = 1 { 0 < 5 is true so output ÷ 1
27.(C) O(n
2
)
Hint: T(n) = (n + 1) + T(n – 1)
= (n + 1) + n + T(n – 2)
= (n + 1) + n + (n – 1) + T(n – 3)
= (n + 1) + n + (n – 1) + (n – 2)+…….T(1)
=
(n 1) (n 2)
2
+ +
= O(n
2
)
28.(B) Table 1
Column 1
Index Minters
Column 2
Pairs
Column 3
Quads
Index 1 2 \
8 \
3 \
Index 2 10 \
12 \
Index 3 13 \
14 \
2,3(1)S
2,10(8)R
8,10(2) \
8,12(4) \
10,14(4)
12,13(1)Q
12,14(2) \
8, 10, 12, 14(2, 4)P
There are four prime implicants P, Q, R and S. Draw the prime implicant chart as shown in table 2
Table 2 .Prime implicant chart
\ \ \ \ \
Ps/Minters 2 3 8 12 13
*P ÷ 8, 10, 12, 14(2, 4)
*Q ÷ 12, 13(1)
R ÷ 2, 10(8)
*S ÷ 2, 3(1)
×
×
×
× ×
×
×
In table 2 M
8
is covered by P only. Further, M
13
is covered by Q only and M
3
is covered by S only.
Therefore, P, Q, and S are the essential prime implicants. Check them out and also check the
Maxterms covered by them. They cover all the Maxterms. So, the final expression is given by
min
f PQS (A D) (A B C) (A B C) (11gate inputs) = = + + + + +
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29.(C) When multiplying
1 0 0 1 × 1 1 0 1
1 1 0 1
0 0 0 0 ×
0 0 0 0 ×
1 1 0 1 ×
1 1 1 0 1 0 1
This is correct answer.
30.(C) Control signal selection and sequencing information is stored in RAM or ROM called a control
memory.
31.(B) Dividing the 400 operation into each of the four processors
Processing time =
400
40 4000 n sec.
4
× =
Using a single processor
Processing time = 400 × 10 = 4000 n sec.
32.(C) In sign magnitude notation, this is easily accomplished simply move the sign bit to the new leftmost
position and fill in with zeros
33.(A) As c try to execute from right side so
1
st
i = 4
+ + I = 5
Then i + + = 5(as i + + is post increment operator)
Hence the output will be 554.
34(C) Hint: Comparison to be done with both numbers as unsigned. Unsigned representation of –1 is
32769.
35.(C) Hint: From the above picture the correct expression is
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(1 + 2) * ((3 + 4) * (5 + 6)),
Change it in the postfix, we get
12 + 34 + 56 + **
36.(B) Hint: S1 is wrong statement because semaphores and monitors can not be used in a distributed
system together. Only semaphores are used in distributed system.
Hence for distributed system one must use message parsing either semaphore or monitor.
Also semaphores and monitor can be used in shared memory situation not in distributed system.
37.(A) String
4 3 2 1 4 3 5 4 3 2 1 5
4 4 4 1 1 1 5 5 5
3 3 3 4 4 4 2 2
2 2 2 3 3 3 1
Number of page fault = 9
so
page fault 9
100% 100 75%
total no. of page 12
× = × =
38.(C) Relation R is in 3NF, because there is no transitive dependencies.
39.(A) The SQL query give the name of supplies who have supplied at least one copy of all the title which
are issued to card no. f53.
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Page 33
40.(C) Relational algebra is a procedural language. Tuple relation calculus restricted to safe expression is
equivalent in expressive power to relation algebra.
41.(A) t
client_customer_name.customer_city
{o
client_customer_name = customer.customer_name(client × customer)
}
This query return the name of customer is the client of bank agassi and city of customer from
customer relation ‘X’ product operation product or mixed two table and check the validity of
client.cust_name = customer.cust_name
42.(A) B-tree ÷ secondary index
Domain calculus ÷ non procedural language
Functional dependency ÷ closure of set of prime or nonprime attribute
Natural join ÷ join to relation algebra expression.
43.(D) Each internal mode has at most P tree pointer, or at least p/2
Within each leaf node
K
1
< K
2
………….< K
q – 1
, q s p
Each leaf node has at least p/2 value. The record pointed py pointer.
The number of node to be accessed is log
n/2
K, {where: k ÷ search field value}
44.(A) Define rule by indicating what action occur for a set of condition, eliminating impossible combination
of condition, develop every possible permutation of condition.
45.(B) E(effort) = 45 Pm
Adjusted effort = 1.15 × 1.21 × 1.11 × 1.10 × 45 = 76 pm
46.(C) V(G) = e –N + 2
E 7
N 6
= ¦
´
=
¹
= 7 – 6 + 2 = 3
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Page 34
47.(A) To inform the location of the based document location
48.(A) When array is sorted then complexity will be 0(n) to sort the n items.
49.(C) If the array is in reverse sorted order then time complexity will be O(n
2
) because it will require to two
array of size n.
50.(B) Complexity of algorithm is O(n
2
) because the algorithm has array A[ ] of size n and loop is process n
times so O(n
2
) is the complexity.
51.(B) If the array is in sorted order the time complexity will be O(n
2
)
52.(B) In SJF (Shortest Job scheduling)
In this shortest job means whose burst time is lower than other come first
3 4 1 5 2
P P P P P
0 3 10 20 32 61
Average waiting time =
0 3 10 20 32 65
13
5 5
+ + + +
= =
In FCFS (First Come First Serve)
1 2 3 4 5
P P P P P
0 61 10 39 42 49
Average waiting time
0 10 39 42 49
28
5
+ + + +
=
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Page 35
In round robin time quantum is must greater than the time of shortest job (3).
53.(B) The proper sequence of the execution of job is
P
3
,P
4
, P
1
, P
5
, P
2
in SJF scheduling
54.(A)
L × , SP, 00FF
L × , H, 0701
00 FF
01 07
I
I
FF 01
00 07
S B M ¬ S B 20H00
= 20H
55.(B) Accumulator content will not changed.
56.(B) No doubt, it was our own government but it was being run on borrowed ideas, using second hand
solutions.
57.(C) Ratify: Approval:: duel: combat
58.(A) Total distance covered
= 35 + 270 + 80 = 385 km.
= 1 + 3 +
1
2
2
hrs.
= 13/2 hrs
Average speed =
D 385 2
T 13
×
=
= 59.23 km/hr.
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Page 36
59.(B) Some critics believe that Satyajit Ray never quite came back to the great beginning he made in this
path breaking film Pather Panchali. Its haunting images have endured decades of well-travelled bad
prints to become a signpost in cinematic history.
60(D) Generosity is nearly opposite to Valedictory
61.(B) Count the number of squares in the figure and multiply it by 3.
62(C) The portion of the immunoglobulin molecule which binds the Specific antigen is formed by the Amino
terminal portions of both the H and L chains.
63.(C) 4 = good 7 = picture and 2 and 9 = are and faint respectively.so data is inadequate.
64.(D)
X X X
35 or
1 2 4
+ + =
7X
4
= 35 or
7X = 35 × 4 or X = 20
65(D) The increase in the total profits of company B in 1997 is 789%
