Gate 2006-IT

Question 1

In a certain town, the probability that it will rain in the afternoon is known to be 0.6. Moreover, meteorological data indicates that if the temperature at noon is less than or equal to 25°C, the probability that it will rain in the afternoon is 0.4. The temperature at noon is equally likely to be above 25°C, or at/below 25°C. What is the probability that it will rain in the afternoon on a day when the temperature at noon is above 25°C?

A
0.4
B
0.6
C
0.8
D
0.9
       Engineering Mathematics        Probability
Question 1 Explanation: 
Probability rain in afternoon = (0.5×probability of rain when temp≤25) + (0.5×Probability of rain when temp>25)
0.6 = (0.5×0.4) + (0.5×P(rain at temp>25)
0.6 = (2) + (0.5×P(rain at temp>25)
P(rain at temp>25) = 0.8
Question 2
A
1 and 2 only
B
2 and 3 only
C
1 and 3 only
D
None of these
       Engineering Mathematics        Sets and Relations
Question 2 Explanation: 
(1) f(x, y) = x + y -3
y = 3
Here, identity element is 3.
(2) f(x, y) = max(x, y) = x = max(y, x)
⇒ y = 1
Here, identity element = 1
(3) f(x, y) = xny is not equal to f(y, x) = ynx
So, no identity element.
Question 3
 
A
The automaton accepts u and v but not w
B
The automaton accepts each of u, v, and w
C
The automaton rejects each of u, v, and w
D
The automaton accepts u but rejects v and w
       Theory of Computation        Finite Automata
Question 3 Explanation: 
(i) u = abbaba

where t is final state
(ii) v = bab

s is not final state
(iii) w = aabb

s is not final state
Question 4
 
A
aaaa
B
baba
C
abba
D
babaaabab
       Theory of Computation        Context Free Grammar
Question 4 Explanation: 
S → aSa | bSb | a | b | ϵ
Given string accepts all palindromes.
Option B → baba is not palindrome.
So, this is not accpeted by S.
Question 5
 
A
(a + b)* a(a + b)b
B
(abb)*
C
(a + b)* a(a + b)* b(a + b)*
D
(a + b)*
       Theory of Computation        Regular Expression
Question 5 Explanation: 
Question 6
Given a boolean function f (x1, x2, …, xn), which of the following equations is NOT true.
A
f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)
B
f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn)
C
f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)
D
f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn)
       Engineering Mathematics        Sets and Relations
Question 6 Explanation: 
Proceed by taking f(x1) = x,
LHS: f(x1) = 0 where x1 = 0
LHS: f(x1) = 1 when x1 = 1
RHS: f(0) + f(1) = 0 + 1 = always 1
Question 7
The addition of 4-bit, two’s complement, binary numbers 1101 and 0100 results in
A
0001 and an overflow
B
1001 and no overflow
C
0001 and no overflow
D
1001 and an overflow
       Digital Logic Design        Number System
Question 7 Explanation: 
2's complement of 1101 = 0011
2's complement of 1100 = 1100
Add = 1111
Now convert 1111 to normal form.
⇒ 0000 (1's complement)
⇒ 0001 (2's complement) No carry bit.
Question 8
Which of the following DMA transfer modes and interrupt handling mechanisms will enable the highest I/O band-width?
A
Transparent DMA and Polling interrupts
B
Cycle-stealing and Vectored interrupts
C
Block transfer and Vectored interrupts
D
Block transfer and Polling interrupts
       Computer Organization        modes of Transfer
Question 8 Explanation: 
→ CPU has highest bandwidth in transparent DMA and polling. But it asked for I/O bandwidth not CPU bandwidth option. A is wrong.
→ In case of cycle stealing in each cycle time derive send data then wait again few CPU cycle it sends to memory. So, option B is wrong.
→ In case of polling CPU takes the initiative. So I/O bandwidth cannot be high. So, option D is wrong.
→ Consider block transfer in each single block device. Send data so bandwidth must be high. This makes option (C) correct,
Question 9
In a binary tree, the number of internal nodes of degree 1 is 5, and the number of internal nodes of degree 2 is 10. The number of leaf nodes in the binary tree is
A
10
B
11
C
12
D
15
       Data Structures        Binary Tree
Question 9 Explanation: 
A node in a binary tree has degree 0, 1, 2.
No. of 1 degree nodes = 5
No. of 2 degree nodes = 10
Total no. of edges = (1*5) + (2*10) = 5 + 20 = 25
So, Total no. of edges = 25 + 1 = 26 (No. of nodes in a tree is 1 more than no. of edges)
Total no. of leaf nodes (node with 0 degree) = 26 - 5 - 10 = 11
Question 10
A problem in NP is NP-complete if
A
It can be reduced to the 3-SAT problem in polynomial time
B
The 3-SAT problem can be reduced to it in polynomial time
C
It can be reduced to any other problem in NP in polynomial time
D
Some problem in NP can be reduced to it in polynomial time
       Algorithms        P-NP
Question 10 Explanation: 
A problem is in NP becomes NP-complete if all NP problems can be reduced to it in polynomial time. This is same as reducing any of the NPC problem to it. 3SAT being an NPC problem reducing it to a NP problem would mean that NP problem is NPC.
Question 11
If all the edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is a  
A
Hamiltonian cycle
B
grid
C
hypercube
D
tree
       Data Structures        Tree
Question 11 Explanation: 
MST:
If all edge weights of an undirected graph are positive, then any subset of edges that connects all the vertices and has minimum total weight is minimum spanning tree.
Question 12
 
A
1 only
B
2 only
C
Neither 1 nor 2
D
Both 1 and 2
       Operating Systems        Thrashing
Question 12 Explanation: 
According to the concept of thrashing both statement (1) and (2) are true.
Question 13
 
A
It is a multiprogrammed operating system
B
It uses preemptive scheduling
C
It uses non-preemptive scheduling
D
It is a multi-user operating system
       Operating Systems        Process Scheduling
Question 13 Explanation: 
Option B is wrong. Because given OS is non-prempting scheduling algorithm.
Question 14

Consider the relations r1(P, Q, R) and r2(R, S, T) with primary keys P and R respectively. The relation r1 contains 2000 tuples and r2 contains 2500 tuples. The maximum size of the join r1⋈ r2 is :

A
2000
B
2500
C
4500
D
5000
       Database Management System        Relational Algebra
Question 14 Explanation: 
r1⋈ r2 is a join opertaion done on the common attribute R. So this can have 2000 tuples.
Question 15
 
A
2 and 3 only
B
1 and 2 only
C
1 and 3 only
D
1, 2 and 3
       Database Management System        Relational Query Languages
Question 15 Explanation: 
Given all three languages have same expressive power.
Question 16
The cyclomatic complexity of the flow graph of a program provides
A
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at most once
B
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at most once
C
an upper bound for the number of tests that must be conducted to ensure that all statements have been executed at least once
D
a lower bound for the number of tests that must be conducted to ensure that all statements have been executed at least once
Question 16 Explanation: 
Note: Out of syllabus.
Question 17
 
A
1 or 3
B
2 or 3
C
2 or 4
D
1 or 4
Question 17 Explanation: 
Note: Out of syllabus.
Question 18
HELO and PORT, respectively, are commands from the protocols
A
FTP and HTTP
B
TELNET and POP3
C
HTTP and TELNET
D
SMTP and FTP
       Computer Networks        Network Protocols
Question 18 Explanation: 
Note: Out of syllabus.
Question 19
Which of the following statements is TRUE?
A
Both Ethernet frame and IP packet include checksum fields
B
Ethernet frame includes a checksum field and IP packet includes a CRC field
C
Ethernet frame includes a CRC field and IP packet includes a checksum field
D
Both Ethernet frame and IP packet include CRC fields
       Computer Networks        Ethernet
Question 19 Explanation: 
Ethernet frame:

IP packet:

IP Datagram:
Question 20
 
A
1 only
B
2 and 3 only
C
1 and 3 only
D
2 only
       Computer Networks        Network Security
Question 20 Explanation: 
(1) A hash function takes a message of arbitary length and generates a fixed length code. So this is correct.
(2) Statement-2 is wrong, refer statement-1.
(3) Statement-3 is correct, for example hash function N%10, this will generate same values for 1 as well as 2!
Question 21
 
A
satisfiable and valid
B
satisfiable and so is its negation
C
unsatisfiable but its negation is valid
D
satisfiable but its negation is unsatisfiable
       Engineering Mathematics        Propositional Logic
Question 21 Explanation: 
The given relationis known to be symmetry. We have both symmetric relations possible as well as antisymmetric but neither always holds for all sets. So they both are valid but are satisfiable.
Question 22

When a coin is tossed, the probability of getting a Head is p0<p<1. Let be the random variable denoting the number of tosses till the first Head appears, including the toss where the Head appears. Assuming that successive tosses are independent, the expected value of N is

 
A
1/p
B
1/(1-p)
C
1/p2
D
1/(1-p2)
       Engineering Mathematics        Probability
Question 22 Explanation: 
E = 1 × P + (2 × (1 - p) p) + (3 × ( 1 - p) (1 - p) p) + ......
Multiply both sides with (1 - p) and subtract,
E - (1 - p) E = 1 × p + (1 - p) p + (1 - p) (1 - p) p + ......
E - (1 - p) E = p/(1 - (1 - p))
(1 - 1 + p) E = 1
pE = 1
E = 1/p
Question 23
 
A
1 only
B
2 only
C
Neither 1 nor 2
D
Both 1 and 2
       Engineering Mathematics        Sets and Relations
Question 23 Explanation: 

P = {1, 2, 4, 5}
Q = {2, 3, 5, 6}
R = {4, 5, 6, 7}
(1) P Δ (Q ∩ R) = (P Δ Q) ∩ (P Δ R)
P Δ ({2,3,5,6} ∩ {4,5,6,7}) = ({1,2,4,5} Δ {2,3,5,6} ∩ {1,2,4,5} Δ {4,5,6,7})
P Δ {5,6} = ({1,2,3,4,5,6} - {2,5}) ∩ ({1,2,4,5,6,7} - {4,5})
({1,2,4,5} Δ {5,6}) = {1,3,4,6} ∩ {1,2,6,7}
{1,2,4,5,6} - {5} = {1,6}
{1,2,4,6} ≠ {1,6}
Statement-1 is False.
(2) P ∩ (Q ∩ R) = (P ∩ Q) Δ (P Δ R)
LHS:
{1,2,4,5} ∩ {5,6} = {5}
RHS:
({1,2,4,5} ∩ {2,3,5,6}) Δ ({1,2,4,5} Δ {4,5,6,7})
{2,5} Δ ({1,2,4,5,6,7} - {4,5})
{2,5} Δ {1,2,6,7}
{1,2,5,6,7} - {2}
{1,5,6,7}
LHS ≠ RHS
Statement - 2 is also wrong.
Question 24
 
A
28
B
33
C
37
D
44
       Engineering Mathematics        Sets
Question 24 Explanation: 
n = 123
A = set of numbers divisible by 2
B = set of numbers divisible by 3
C = set of numbers divisible by 5
n(A∪B∪C) = n(A) + n(B) + n(C) - n(A∩B) - n(B∩C) - n(C∩A) + n(A∩B∩C)
= ⌊123/2⌋ + ⌊123/3⌋ + ⌊123/5⌋ - ⌊123/6⌋ - ⌊123/15⌋ - ⌊123/10⌋ + ⌊123/30⌋
= 61 + 41 + 24 - 20 -12 - 8 + 4
= 90
Total no. that are not divisible
= n - n(A∪B∪C)
= 123 - 90
= 33
Question 25

Consider the undirected graph G defined as follows. The vertices of G are bit strings of length n. We have an edge between vertex u and vertex v if and only if u and v differ in exactly one bit position (in other words, v can be obtained from u by flipping a single bit). The ratio of the chromatic number of G to the diameter of G is

A
1/(2n-1)
B
1/n
C
2/n
D
3/n
       Engineering Mathematics        Graph Theory
Question 25 Explanation: 
For the given condition we can simply design a K-map and mark an edge between every two adjacent cells in K-map.
That will give us a bipartite graph, with chromatic number = 2
Also from the same we can conclude that we need for a 'n' bit string, to traverse no more than (n-1) edges or 'n' vertices to get a path between two arbitary points. So the ratio is (2/n).
Question 26
 
A
B
a, a, a
C
0, a, 2a
D
-a, 2a, 2a
       Engineering Mathematics        Linear Algebra
Question 26 Explanation: 
Question 27
 
A
1-R, 2-S, 3-P, 4-Q
B
1-S, 2-R, 3-Q, 4-P
C
1-S, 2-Q, 3-R, 4-P
D
1-S, 2-P, 3-Q, 4-R
       Engineering Mathematics        Match The Following
Question 27 Explanation: 
Note: Out of syllabus.
Question 28
   
A
B
C
D
E
None of the above
       Engineering Mathematics        Calculus
Question 28 Explanation: 
Question 29
 
A
{w ∈ (a + b)* | #a(w) is even) and {w ∈ (a + b)* | #a(w) is odd}
B
{w ∈ (a + b)* | #a(w) is even) and {w ∈ (a + b)* | #b(w) is odd}
C
{w ∈ (a + b)* | #a(w) = #b(w) and {w ∈ (a + b)* | #a(w) ≠ #b(w)}
D
{ϵ}, {wa | w ∈ (a + b)* and {wb | w ∈ (a + b)*}
       Theory of Computation        Regular Grammar
Question 29 Explanation: 

⇒ This results even number, no. of a's.
Question 30
Which of the following statements about regular languages is NOT true?
A
Every language has a regular superset
B
Every language has a regular subset
C
Every subset of a regular language is regular
D
Every subset of a finite language is regular
       Theory of Computation        Regular Languages
Question 30 Explanation: 
Regular languages are not closed under subset.
Question 31
Which of the following languages is accepted by a non-deterministic pushdown automaton (PDA) but NOT by a deterministic PDA?
A
{anbncn ∣ n≥0}
B
{albmcn ∣ l≠m or m≠n}
C
{anbn ∣ n≥0}
D
{ambn∣ m,n≥0}
       Theory of Computation        Push Down automata
Question 31 Explanation: 
At a time, the DPDA can compare 'a' and 'b' or 'b' and 'c' but not both.
To compare both conditions at the same time, we need a NPDA.
Question 32
Let L be a context-free language and M a regular language. Then the language L ∩ M is
A
always regular
B
never regular
C
always a deterministic context-free language
D
always a context-free language
       Theory of Computation        Identify Class Language
Question 32 Explanation: 
CFL is closed under regular intersection.
Question 33
A
{albmcn | l = m = n}
B
{albmcn | l = m}
C
{albmcn | 2l = m+n}
D
{albmcn | m=n}
       Theory of Computation        Push Down Automata
Question 33 Explanation: 

For every 'a' we put two X in stacks [at state S].
After that for every 'b' we pop out one X [reach to state t].
After that for every 'c' we pop out one X [reach to state u].
If all X are popped out then reached to final state f, means for every 'b' and 'c' there is 'a'. 'a' is followed by 'b' and 'b' is followed by 'c'.
Means,
Sum of no. of b's and no. of c's = twice of no. of a's
i.e., {albmcn | 2l = m+n}
Question 34
A
((a + b)* b)*
B
{ambn | m ≤ n}
C
{ambn | m = n}
D
a* b*
       Theory of Computation        Context afree Grammar
Question 34 Explanation: 
Option A:
((a + b)* b)* → It accepts string aa but given grammar does not accepts.
Option C&D:
→ abb accepted by given grammar but option C & D are not accepting.
Question 35
 
A
QRS
B
PQS
C
PQ'S'
D
Q'S'
       Digital Logic Design        K Map
Question 35 Explanation: 
Essential prime implicants which are grouped only by only one method or way. So, in the given question cornor's ones are grouped by only one method.
Question 36
A
XOR, AND
B
XOR, XOR
C
OR, OR
D
OR, AND
       Digital Logic Design        Boolean Function
Question 36 Explanation: 

Thus we have OR and AND which gives different outputs on (0,0) and (1,1).
The encodes can be hence select from the two and decide output of the function according to x.
Question 37
A
S+ = S’ . y’ + S . x
B
S+ =S. x . y’ + S’ . y . x’
C
S+ =x . y’
D
S+ =S’ . y + S . x’col
       Theory of Computation        Finite Automata
Question 37 Explanation: 

From the table:
S' = S'y' + Sx
Question 38
 
A
2Y and Y
B
-2Y and 2Y
C
-2Y and 0
D
0 and Y
       Digital Logic Design        Number System
Question 38 Explanation: 

⇒ -2Y and 0
Question 39
Which of the following statements about relative addressing mode is FALSE?
A
It enables reduced instruction size
B
It allows indexing of array elements with same instruction
C
It enables easy relocation of data
D
It enables faster address calculations than absolute addressing
       Computer Organization        Addressing Modes
Question 39 Explanation: 
As relative address are calculated from the absolute address. So relative addressing cannot be faster than absolute addressing.
Question 40
 
A
Memory location 1000 has value 20
B
Memory location 1020 has value 20
C
Memory location 1021 has value 20
D
Memory location 1001 has value 20
       Computer Organization        Machine Instructions
Question 40 Explanation: 
Rs ← 1
Rd ← 1
Rd ← 1001
Store in address 1001 ← 20.
Question 41
 
A
32, 5, 010
B
5, 32, 010
C
5, 31, 011
D
5, 31, 010
       Computer Organization        Micro Instructions
Question 41 Explanation: 
→ If a number is to be even LSB = 0.
→ So, there may be only 31, is for an unsigned even integer.
→ And 31 left shifts are needed to determine the number of 1's.
Question 42

A cache line is 64 bytes. The main memory has latency 32 ns and bandwidth 1 GBytes/s. The time required to fetch the entire cache line from the main memory is

A
32 ns
B
64 ns
C
96 ns
D
128 ns
       Computer Organization        Cache
Question 42 Explanation: 
For 1GBytes/s bandwidth → it takes 1 sec to load 109 bytes on line.
→ So, for 64 bytes it will take 64*1/109 = 64ns.
Main memory latency = 32
Total time required to place cache line is
64+32 = 96 ns
Question 43
 
A
1K × 18-bit, 1K × 19-bit, 4K × 16-bit
B
1K × 16-bit, 1K × 19-bit, 4K × 18-bit
C
1K × 16-bit, 512 × 18-bit, 1K × 16-bit
D
1K × 18-bit, 512 × 18-bit, 1K × 18-bit
       Computer Organization        Cache
Question 43 Explanation: 
No. of blocks in cache = Capacity/Block size = 2m
Bits to represent blocks = m
No. of words in a block = 2n
Bits to represent a word = n
Tag bits = (length of physical address of a word) - (bits to represent blocks) - (bits to represent a word)
⇒ Each block will have its own tag bits.
So, total tag bits = no. of blocks × tag bits.
Question 44
Which of the following sequences of array elements forms a heap?
A
{23, 17, 14, 6, 13, 10, 1, 12, 7, 5}
B
{23, 17, 14, 6, 13, 10, 1, 5, 7, 12}
C
{23, 17, 14, 7, 13, 10, 1, 5, 6, 12}
D
{23, 17, 14, 7, 13, 10, 1, 12, 5, 7}
       Data Structures        Heap tree
Question 44 Explanation: 

In this every children and parent satisfies Max heap properties.
Question 45

Suppose that we have numbers between 1 and 100 in a binary search tree and want to search for the number 55. Which of the following sequences CANNOT be the sequence of nodes examined?

A
{10, 75, 64, 43, 60, 57, 55}
B
{90, 12, 68, 34, 62, 45, 55}
C
{9, 85, 47, 68, 43, 57, 55}
D
{79, 14, 72, 56, 16, 53, 55}
       Data Structures        Binary tree
Question 45 Explanation: 
In the binary search tree on right side of parent number, the number should be greater than it. But in option C, after 47, 43 appears. So, this is False.
Question 46
A
{P, Q, R, S}, {T}, {U}, {V}
B
{P, Q, R, S, T, V}, {U}
C
{P, Q, S, T, V}, {R}, {U}
D
{P, Q, R, S, T, U, V}
       Data Structures        Graphs
Question 46 Explanation: 
In a strongly connected component every two vertices must be reachable from one to other and itis maximal component.
From given graph {P, Q, R, S, T, V} and {U} are strongly connected components.
Question 47
 
A
There is only one connected component
B
There are two connected components, and P and R are connected
C
There are two connected components, and Q and R are connected
D
There are two connected components, and P and Q are connected
       Data Structures        Graphs
Question 47 Explanation: 
Since, d(q) = d(p) + 1 and f(q) < f(p) which means p and q are connected and r is separate, so (D) is the answer.
Question 48
A
fdheg
B
ecgdf
C
dchfg
D
fehdg
       Algorithms        Greedy Method
Question 48 Explanation: 
Huffman's tree is as follows. The two least frequently characters are taken as the children of a newly made node and the frequency of the newly made node is made equal to the sum of those two child nodes. Then the same procedure is repeated till all nodes are finished.

∴ 110111100111010 = fdheg
Question 49
   
A
jungle, n, 8, ncestor
B
etter, u, 6, ungle
C
cetter, k, 6, jungle
D
etter, u, 8, ncestor
       Programming       Program
Question 49 Explanation: 
Lets take the part of program,
Line 1 - main ( )
Line 2 - {
Line 3 - struct test *p = st;
Line 4 - p += 1;
Line 5 - ++p → c;
Line 6 - printf("%s", p++→ c);
Line 7 - printf("%c", +++p → c);
Line 8 - printf("%d", p[0].i);
Line 9 - printf("%s\n", p → c);
Line 10 - }
Now,
Line 3: Initially p is pointing to st, i.e., first element of st which is {5, "become"}
Line 4: Now p is pointing to {4, "better"}
Line 5: ++(p → c), since → has higher precedence, so p → c points to 'e' of "better".
Line 6: prints 'enter' and p now points to {6, "jungle"}
Line 7: ***(p → c), since → has higher precedence. So, prints 'u'.
Line 8: p → i, which is 6 so prints '6'.
Line 9: prints 'ungle' since p is pointing to 'u'.
So, output is "enter, u, 6, ungle".
Question 50
 
A
a = 0, b = 3
a = 0, b = 3
B
a = 3, b = 0
a = 12, b = 9
C
a = 3, b = 6
a = 3, b = 6
D
a = 6, b = 3
a = 15, b = 12
       Programming       Program
Question 50 Explanation: 
First of all, the swap function just swaps the pointers inside the function and has no effect on the variable being passed.
Inside print 'a' and 'b' are added to odd integers from 1 to 5, i.e., 1+3+5=9. So, in first call to print ab,
a = -3+9 = 6
b = -6+9 = 3
Static variable have one memory throughout the program run (initialized during program start) and they keep their values across function calls. So during second call to print ab,
a = 6+9 = 15
b = 3+9 = 12
Question 51
 
A
8, -12, 7, 23, 8
B
8, 8, 7, 23, 7
C
-12, -12, 27, -31, 23
D
-12, -12, 27, -31, 56
       Programming       Program
Question 51 Explanation: 
1) a[0][2] = *(*(a+0)+2)
It returns the value of 3rd element in a1.
First printf print 8.
2) *a[2] = *(*(a+2))
It returns the value of 1st element in a3.
Second printf print -12.
3) *++a[0] = *(++(*(a+0)))
a[0] is pointing to 1st element in a1.
++a[0] - after preincrement performed, now a[0] is pointing to 2nd element in a1.
*++a[0] return the value of 2nd element in a1.
Third printf print 7.
4) *(++a)[0]
++a - after preincrement is performed 'a' is pointing to a2.
(++a)[0] is pointing to 1st element in a2.
*(++a)[0] returns the value of 1st element in a2.
Fourth printf print 23.
5) a[-1][+1] = *(*(a-1)+1)
(a-1) is pointing to a1.
*(a-1) is pointing to the 2nd element in a1, because in 3rd printf already a1 was incremented by 1.
*(a-1)+1 is pointing 3rd element in a1.
*(*(a-1)+1) returns the value of 3rd element in a1, i.e., 8.
Question 52
   
A
(n == 0) || (m == 1)
B
(n == 0) && (m == 1)
C
(n == 0) || (m == n)
D
(n == 0) && (m == n)
       Programming       Program
Question 52 Explanation: 
We know that,
mC0 = 1
nCn = 1
Question 53
 
A
(i) - B, (ii) - D, (iii) - E, (iv) - F, (v) - G, (vi) - A
B
(i) - C, (ii) - A, (iii) - E, (iv) - D, (v) - H, (vi) - F
C
(i) - C, (ii) - F, (iii) - H, (iv) - A, (v) - G, (vi) - D
D
(i) - B, (ii) - E, (iii) - C, (iv) - F, (v) - G, (vi) - H
       Programming       Match The Following
Question 53 Explanation: 
Note: Out of syllabus.
Question 54

A
11, 15, 9
B
10, 15, 9
C
11, 16, 10
D
12, 17, 11
       Operating Systems        Process Scheduling
Question 54 Explanation: 
Gantt-chart:

Hence, finish times of process.
P1 - 10
P2 - 11
P3 - 9
Question 55
 
A
1 only
B
2 only
C
Neither 1 nor 2
D
Both 1 and 2
       Operating Systems        Process Synchronization
Question 55 Explanation: 
Suppose Wait(F) and Wait(S) are interchanged. And let the slots are full → F=0.
Now if Wait(S) in producer succeeds, then producer will wait for Wait(F) which is never going to succeed as consumer would be waiting for Wait(S). So deadlock, can happen.
If Signal(S) and Signal(F) are interchanged in consumer, deadlock won't happen. It will just give priority to a producer compared to the next consumer waiting.
Question 56
 
A
P < S < T
B
S < P < T
C
S < T < P
D
T < S < P
       Operating Systems        Virtual memory
Question 56 Explanation: 
Case-1 (Two level paging):
For P1,
Page size is 1KB. So, no. of pages required for P1=195. An entry in page table is of size 4 bytes and assuming an inner level page table takes the size of a page, we can have upto 256 entries in second level page table and we require only 195 for P1. Thus only 1 second level page table is enough. So, memory overhead = 1KB (for first level) + 1KB for second level = 2KB.
For P2 and P3 also, we get 2KB each and for P4 we get 1+2=3KB as it requires 1 first level page table and 2 second level page tables (364 > 256). So, total overhead for their concurrent execution = 2×3+3 = 9KB. Thus P = 9KB.
Case-2 (For segmentation method):
P1 uses 4 segments → 4 entries in segment table = 4×8 = 32Bytes
Similarly, for P2, P3 and P4 we get 5×8, 3×8 and 8×8 Bytes respectively and the total overhead will be
32+40+24+64 = 160 Bytes
So, S = 160 Bytes
Case-3 (For segmentation with paging):
Here, we segment first and then page. So, we need the page table size. We are given maximum size of a segment is 256KB and page size is 1KB and thus we require 256 entries in the page table. So, total size of page table = 256 × 4 = 1024 Bytes (exactly one page size).
So, now for P1 we require 1 segment table of size 32 Bytes plus 1 page table of size 1KB.
Similarly,
P2 - 40 Bytes and 1 KB
P3 - 24 Bytes and 1 KB
P4 - 64 Bytes and 1KB
Thus, total overhead = 160 Bytes + 4KB = 4256 Bytes
So, T = 4256 Bytes
So, answer should be S < T < P.
Question 57
 
A
P(x_sem), V(next)
B
V(next), P(x_sem)
C
P(next), V(x_sem)
D
P(x_sem), V(x_sem)
       Operating Systems        Process Synchronization
Question 57 Explanation: 
x_count is the no. of processes waiting on semaphore x_sem, initially 0.
x_count is incremented and decremented in x_wait, which shows that in between them wait(x_sem) must happen which is P(x_sem). Correspondingly V(x_sem) must happen in x_signal. So, D choice.
Question 58

A software program consists of two modules M1 and M2 that can fail independently, but never simultaneously. The program is considered to have failed if any of these modules fails. Both the modules are ‘repairable’ and so the program starts working again as soon as the repair is done. Assume that the mean time to failure (MTTF) of M1is T1 with a mean time to repair (MTTR) of R1. The MTTF of M2 is T2 with an MTTR of R2. What is the availability of the overall program given that the failure and repair times are all exponentially distributed random variables?

A
B
C
D
Question 58 Explanation: 
Note: Out of syllabus.
Question 59
 
A
III and IV
B
I and IV
C
II and IV
D
I, II and IV
Question 59 Explanation: 
Note: Out of syllabus.
Question 60
 
A
VXZ
B
VXY
C
VWXY
D
VWXYZ
       Database Management System        Functional Dependencies
Question 60 Explanation: 
As we can see that attribute XY do not appear in RHS of an FD, they need to be a part of key.
Candidate keys are
VXY, WXY, ZXY
Question 61

In a database file structure, the search key field is 9 bytes long, the block size is 512 bytes, a record pointer is 7 bytes and a block pointer is 6 bytes. The largest possible order of a non-leaf node in a B+ tree implementing this file structure is

A
23
B
24
C
34
D
44
       Database Management System        B+ Tree
Question 61 Explanation: 
From the structure of B+ tree we can get this question:
n × p + (n - 1) × k ≤ B (for non-leaf node)
Here, n = order, p = tree/block/index pointer, B = size of block
So,
n × p + (n - 1) × k ≤ B
n × 6 + (n - 1) × 9 ≤ 512
n ≤ 34.77
∴ n = 34
Question 62
A
The professor with the lowest course evaluation
B
Professors who have all their course evaluations above the university average
C
The course with the lowest evaluation
D
Courses with all evaluations above the university average
Question 62 Explanation: 
Note: Out of syllabus.
Question 63
 
A
eth0
B
eth1
C
eth2
D
eth3
       Computer Networks        IP Address
Question 63 Explanation: 
Firstly start with longest mask.
144.16.68.117 = 144.16.68.01110101 AND 255.255.255.224 = 255.255.255.11100000
= 144.16.68.96(Not matching with destination)
Now, take 255.255.255.0
144.16.68.117 AND 255.255.255.0
= 144.16.68.0 (matched)
Hence, option (C) is correct.
Question 64

Suppose that it takes 1 unit of time to transmit a packet (of fixed size) on a communication link. The link layer uses a window flow control protocol with a window size of N packets. Each packet causes an ack or a nak to be generated by the receiver, and ack/nak transmission times are negligible. Further, the round trip time on the link is equal to N units. Consider time i > N. If only acks have been received till time i(no naks), then the goodput evaluated at the transmitter at time i(in packets per unit time) is

A
1 – N/i
B
i/(N + i)
C
1
D
1 – e(i/N)
       Computer Networks        Sliding Window Protocol
Question 64 Explanation: 
Goodput is the application level throughtout, i.e., the no. of useful information bits delivered by the network to a certain destination per unit of time.
So, successful delivery of packet can be assured if ack has been received for it.
So till time 'i' we would have transmitted 'i' packets but only (i - N) can be acknowledged as minimum time for a packet to get acknowledged is N (since RTT is N which is equal to the window size, there is no waiting for the sender).
So, successfully delivered packets = (i - N)
Time for transmission = i
Goodput = Successfully delivered data/Time
= (i - N)/i
= 1 - N/i
Question 65

In the 4B/5B encoding scheme, every 4 bits of data are encoded in a 5-bit codeword. It is required that the codewords have at most 1 leading and at most 1 trailing zero. How many such codewords are possible?

A
14
B
16
C
18
D
20
       Computer Networks        Encoding
Question 65 Explanation: 
It says we have 5 bit codeword such that "it can't have two consecutive zeros in first and second bit" and also "can't have two consecutive zeros in last two bits".
Codeword with first two bits '0'
= 0 0 x x x
= 23
= 8
Codeword with last two bits '0'
= x x x 0 0
= 23 = 8
Codeword with first two and last two bits '0'
= 0 0 x 0 0
= 2
Codeword with first or last two bits '0'
= 8 + 8 - 2
= 14
Therefore possible codewords
= 32 - 14
= 18
Question 66

A router has two full-duplex Ethernet interfaces each operating at 100 Mb/s. Ethernet frames are at least 84 bytes long (including the Preamble and the Inter-Packet-Gap). The maximum packet processing time at the router for wirespeed forwarding to be possible is (in micro­seconds)

A
0.01
B
3.36
C
6.72
D
8
       Computer Networks        Ethernet
Question 66 Explanation: 
Let's first calculate transmission time Tt,
Tt = 84×8/10×106 = 6.72μs
But since a router has two full-duplex ethernet interfaces, so the maximum processing time should be,
6.72/2 μs = 3.36μs
Question 67

A link of capacity 100 Mbps is carrying traffic from a number of sources. Each source generates an on-off traffic stream; when the source is on, the rate of traffic is 10 Mbps, and when the source is off, the rate of traffic is zero. The duty cycle, which is the ratio of on-time to off-time, is 1:2. When there is no buffer at the link, the minimum number of sources that can be multiplexed on the link so that link capacity is not wasted and no data loss occurs is S1. Assuming that all sources are synchronized and that the link is provided with a large buffer, the maximum number of sources that can be multiplexed so that no data loss occurs is S2. The values of S1 and S2 are, respectively,

A
10 and 30
B
12 and 25
C
5 and 33
D
15 and 22
       Computer Networks        Network Flow
Question 67 Explanation: 
For S1,
Since there is no buffer and constraint given is there should not be any data lost, and no wastage of capacity as well.
Since data should not be lost, we calculate for the extreme case when all sources are ontime (that is transmitting).
10Mbps × n-station ≤ 100Mbps
n = 10 = S1
In the next part of question, it is given that the link is provided with large buffer and we are asked to find out large no. of stations.
For that we calculate expected value of bandwidth usage,
E = 1/3 × 10 + 1/3 × 10 + .......+ ....... n-station times ≤ 100Mbps
⇒ 1/3 × 10 × n-station ≤ 100Mbps ⇒ n-station = 30 = S2
So option (A) is answer.
Question 68

On a wireless link, the probability of packet error is 0.2. A stop-and-wait protocol is used to transfer data across the link. The channel condition is assumed to be independent from transmission to transmission. What is the average number of transmission attempts required to transfer 100 packets?

A
100
B
125
C
150
D
200
       Computer Networks        Stop and Wait Protocol
Question 68 Explanation: 
Total no. of re-transmissions for one frame, in general, is 1/(1-P) where P is probability of error.
So here it would be for one frame = 1/(1-0.2) = 1/0.8
So for 100 frames = 100/0.8 = 125
Question 69

A program on machine X attempts to open a UDP connection to port 5376 on a machine Y, and a TCP connection to port 8632 on machine Z. However, there are no applications listening at the corresponding ports on Y and Z. An ICMP Port Unreachable error will be generated by

A
Y but not Z
B
Z but not Y
C
Neither Y nor Z
D
Both Y and Z
       Computer Networks        TCP and UDP
Question 69 Explanation: 
When an IP packet is lost or discarded ICMP packet will be generated by receiver or the router. It doesn't matter whether its containing TCP or UDP inside it.
Question 70
A subnetted Class B network has the following broadcase address : 144.16.95.255. Its subnet mask  
A
is necessarily 255.255.224.0
B
is necessarily 255.255.240.0
C
is necessarily 255.255.248.0
D
could be any one of 255.255.224.0, 255.255.240.0, 255.255.248.0
       Computer Networks        IP Adrress
Question 70 Explanation: 
In the broadcast address for a subnet, all the host bits are set to 1. So as long as all the bits to the right are 1, bits left to it can be taken as possible subnet.
Broadcast address for subnet is
.95.255 or .01011111.11111111
(as in class B, 16 bits each are used for network and host)
So, we can take minimum 3 bits (from left) as subnet and make rest as host bits (as they are 1)
.224.0 → 11100000.00000000 (leftmost 3 bits for subnet)
.240.0 → 11110000.00000000 (leftmost 4 bits for subnet)
.248.0 → 11111000.00000000 (leftmost 5 bits for subnet)
Question 71
 
A
⌊i/2⌋
B
⌈(i-1)/2⌉
C
⌈i/2⌉
D
⌈i/2⌉ - 1
       Data Structures        Binary tree
Question 71 Explanation: 
If index of array start with 1 then directly divide the ith value by 2 and take floor. If index start with '0' then ⌈i/2⌉ - 1.
Question 72
 
A
O(n)
B
O(log n)
C
O(n log n)
D
O(n log log n)
       Data Structures        Binary tree
Question 72 Explanation: 
It takes O(log n) to heapify an element of heap.
Question 73
 
A
⌊log2 i⌋
B
⌈log2 (i + 1)⌉
C
⌊log2 (i + 1)⌋
D
⌈log2 i⌉
       Data Structures        Binary tree
Question 73 Explanation: 
If the root node is at level 0 then the level of element X[i] is ⌊log2 (i + 1)⌋.
Question 74
 
A
60
B
63
C
66
D
69
Question 74 Explanation: 
Note: Out of syllabus.
Question 75
 
A
315
B
330
C
393
D
403
Question 75 Explanation: 
Note: Out of syllabus.
Question 76
 
A
0.5
B
0.75
C
1.5
D
2.0
       Engineering Mathematics        Linear algebra
Question 76 Explanation: 
Question 77
 
A
it will converge
B
it will diverse
C
it will neither converge nor diverse
D
It is not applicable
       Engineering Mathematics        Linear Algebra
Question 77 Explanation: 
As,
|1| + |1/2| <= |9|
and |3| + |1| <= |10|
Question 78

A
2, 2, 4
B
3, 2, 3
C
4, 2, 2
D
3, 3, 2
       Computer Organization        Pipelining
Question 78 Explanation: 
RAW:
I1 - I2 (R5)
I2 - I3 (R6)
I3 - I4 (R5)
I4 - I5 (R6)
WAR:
I2 - I3 (R5)
I3 - I4 (R6)
WAW:
I1 - I3 (R5)
I3 - I4 (R6)
Question 79
 
A
10
B
12
C
14
D
16
       Computer Organization        Pipelining
Question 79 Explanation: 
Question 80
   
A
Both I and IV
B
Only I
C
Only IV
D
Both II and III
       Theory of Computation        Regular Language
Question 80 Explanation: 
Repeat(L) = {ww|w ∈ L} is non-regular language.
Half (L), Suffix (L) and Prefix (L) are regular languages.
Question 81
 
A
(a + b)*
B
{ϵ, a, ab, bab}
C
(ab)*
D
{anbn | n ≥ 0}
       Theory of Computation        Regular Language
Question 81 Explanation: 
A counter example which proves all the conclusions of the last question in one go should have the following properties:
1) L should be regular due to demand of question.
2) L should be an infinite set of strings.
3) L should have more than one alphabet in its grammar, otherwise repeat(L) would be regular.
∴ (a + b)* is the perfect example to support the conclusions of last questions.
Question 82
 
A
P1-P2-P4, 1 day
B
P1-P3-P4, 1 day
C
P1-P3-P4, 2 days
D
P1-P2-P4, 2 days
Question 82 Explanation: 
Note: Out of syllabus.
Question 83
 
A
100 and 1000
B
100 and 1200
C
150 and 1200
D
200 and 2000
Question 83 Explanation: 
Note: Out of syllabus.
Question 84
 
A
Karthikeyan, Boris
B
Sachin, Salman
C
Karthikeyan, Boris, Sachin
D
Schumacher, Senna
       Database Management System        SQL
Question 84 Explanation: 
For colour = "Red"
did = {22, 22, 31, 31, 64}
For colour = "Green"
did = {22, 31, 74}
Intersection of Red and Green will be = {22, 31}, which is Karthikeyan and Boris.
Question 85
A
36 - 40
B
44 - 48
C
60 - 64
D
100 - 104
       Database Management System        SQL
Question 85 Explanation: 
(4) for taking red cars with (20) comparisions for did and (4) for finding green cars with (10) for did.
red did : 22, 31, 64
green did : 22, 31, 74
(6) for intersection
(1) for searching 22 in driver relation, and (3) for searching 31.
Total: 38 + 6 + 4 = 48
There are 85 questions to complete.