Gate 2004-IT
Question 1 |
In a population of N families, 50% of the families have three children, 30% of the families have two children and the remaining families have one child. What is the probability that a randomly picked child belongs to a family with two children?
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In that 50% of families having 3 childrens
i.e., 50×3 = 150 (No. of children)
→ Likes that 30% have 2 childrens = 30×2 = 60
→ 20% have 1 children = 20×1 = 20
Probability of choosing a children the family have two children
= 60/(150+60+20) = 60/230 = 6/23
Question 2 |
In a class of 200 students, 125 students have taken Programming Language course, 85 students have taken Data Structures course, 65 students have taken Computer Organization course; 50 students have taken both Programming Language and Data Structures, 35 students have taken both Data Structures and Computer Organization; 30 students have taken both Data Structures and Computer Organization, 15 students have taken all the three courses.How many students have not taken any of the three courses?
15 | |
20 | |
25 | |
35 |
PL = 125
DS = 85
CO = 65
PL & DS = 50
DS & CO = 35
CO & PL = 30
PL & DS & CO = 15

⇒ Not taken any course = 200 - (60 +15+15+35+15+15+20)
= 200 - 175
= 25
Question 3 |
(∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)] | |
(∃x)(∀y)[(a(x, y) ∨ b(x, y)) ∧¬ c(x, y)] | |
¬(∀x)(∃y)[(a(x, y) ∧ b(x, y)) → c(x, y)] | |
¬(∀x)(∃y)[(a(x, y) ∨ b(x, y)) → c(x, y)] |

Question 4 |
- An element x in A is related to an element y in B (under R1) if x + y is divisible by 3.
- An element x in B is related to an element y in C (under R2) if x + y is even but not divisible by 3.
R1R2 = {(1, 2), (1, 4), (3, 3), (5, 4), (7, 3)} | |
R1R2 = {(1, 2), (1, 3), (3, 2), (5, 2), (7, 3)} | |
R1R2 = {(1, 2), (3, 2), (3, 4), (5, 4), (7, 2)} | |
R1R2 = {(3, 2), (3, 4), (5, 1), (5, 3), (7, 1)} |
R1 ={(1,2), (1,8), (3,6), (5,4), (7,2), (7,8)}
where x+y is divisible by 3
R2 = {(2,2), (4,4), (6,2), (6,4), (8,2)}
where x+y is not divisible by 3
Then the composition of R1 with R2 denotes R1R2, is the relation from A to C defined by property such as:
(x,z) ∈ R1R2, iff if there is a y ∈ B such that (x,y) ∈ R1 and (y,z) ∈ R2.
Thus, R1R2 = {(1,2), (3,2), (3,4), (5,4), (7,2)}
Question 5 |
n - 1 | |
n | |
n + 1 | |
2n - 1 |
= n - 1
Question 6 |
x = 6, y = 3, z = 2 | |
x = 12, y = 3, z = -4 | |
x = 6, y = 6, z = -4 | |
x = 12, y = -3, z = 0 |

Question 7 |
(a* + b* + c*)* | |
(a*b*c*)* | |
((ab)* + c*)* | |
(a*b* + c*)* |
From option 'c' we cannot be able to create a without b. So option is not equivalent.
Question 8 |
3 | |
4 | |
5 | |
6 |

To create 2-input Exclusive-OR function we require 4 NAND gates.
Question 9 |
There exist context-free languages such that all the context-free grammars generating them are ambiguous | |
An unambiguous context free grammar always has a unique parse tree for each string of the language generated by it | |
Both deterministic and non-deterministic pushdown automata always accept the same set of languages | |
A finite set of string from one alphabet is always a regular language
|
But non-deterministic ones can recognize all context-free languages.
So, option C is false.
Question 10 |
32 K × 16 bits | |
64 K × 16 bits | |
16 K × 32 bits | |
64 K × 32 bits |
Possible combination in ROM = (28 × (28) [size of truth table]
= 216
= 64 KB
= 64 K ×16 bits
Question 11 |
What is the bit rate of a video terminal unit with 80 characters/line, 8 bits/character and horizontal sweep time of lOOµs (including 20 µs of retrace time)?
8 Mbps | |
6.4 Mbps | |
0.8 Mbps | |
0.64 Mbps |
Total number of bits transmitted = 80 * 8 = 640 bits
Bit rate = (640 * 106)/100 = 6.4 Mbps
Question 12 |
13.0 ns | |
12.8 ns | |
12.6 ns | |
12.4 ns |
H1 = 0.8, (1 - H1) = 0.2
H2 = 0.9, (1 - H2) = 0.1
T1 = Access time for level 1 cache = 1ns
T2 = Access time for level 2 cache = 10ns
Hm = Hit rate of main memory = 1
Tm = Access time for main memory = 500ns
Average access time = [(0.8 * 1) + (0.2 * 0.9 * 10) + (0.2)(0.1) * 1 * 500]
= 0.8 + 1.8 + 10
= 12.6ns
Question 13 |
Let P be a singly linked list. Let Q be the pointer to an intermediate node x in the list. What is the worst-case time complexity of the best known algorithm to delete the node x from the list?
O(n) | |
O(log2 n) | |
O(logn) | |
O(1) |
Question 14 |
Stack | |
Address Space | |
File Descriptor Table | |
Message Queue |
Question 15 |
x = 1 + x; | |
x = 1 - x; | |
x = x - 1; | |
x = 1 % x; |
For x = 0, it gives 1.
For x = 1, it gives 0.
Question 16 |
II and III | |
II and IV | |
I and III | |
II only |
Question 17 |
effort and duration based on the size of the software | |
size and duration based on the effort of the software | |
effort and cost based on the duration of the software | |
size, effort and duration based on the cost of the software |
Question 18 |
Entity Relationship Diagram | |
Deployment Diagram | |
Data Flow Diagram | |
Use Case Diagram |
Question 19 |
A software organization has been assessed at SEI CMM Level 4. Which of the following does the organization need to practice beside Process Change Management and Technology Change Management in order to achieve Level 5?
Defect Detection | |
Defect Prevention | |
Defect Isolation | |
Defect Propagation |
Question 20 |
keeping track of the schedule based on the milestones reached | |
maintaining different versions of the configurable items | |
managing manpower distribution by changing the project structure | |
all of the above |
Question 21 |
Page | |
Table | |
Row | |
Page, table and row level locking allow the same degree of concurrency
|
Table locking can be used for concurrency control with DDL operations.
In row share table is less restrictive but it consists of highest degree of concurrency compared to page and table.
Question 22 |
Packet switching leads to better utilization of bandwidth resources than circuit switching. | |
Packet switching results in less variation in delay than circuit switching. | |
Packet switching requires more per packet processing than circuit switching. | |
Packet switching can lead to reordering unlike in circuit switching. |
Question 23 |
TCP guarantees a minimum communication rate | |
TCP ensures in-order delivery | |
TCP reacts to congestion by reducing sender window size | |
TCP employs retransmission to compensate for packet loss
|
Sequence numbers can allow receivers to discard duplicate packets and properly sequence reordered packets.
Option C:
If the congestion is deleted, the transmitter decreases the transmission rate by a multiplicative factor.
Option D:
Acknowledgement allows the sender to determine when to retransmit lost packets.
Question 24 |
HTTP runs over TCP | |
HTTP describes the structure of web pages | |
HTTP allows information to be stored in a URL | |
HTTP can be used to test the validity of a hypertext link |
Question 25 |
Sender encrypts using receiver’s public key | |
Sender encrypts using his own public key | |
Receiver decrypts using sender’s public key | |
Receiver decrypts using his own public key |
Question 26 |
14 | |
30 | |
62 | |
126 |
= 26- 2
= 64 - 2
= 62
Question 27 |
A host is connected to a Department network which is part of a University network. The University network, in turn, is part of the Internet. The largest network in which the Ethernet address of the host is unique is:
the subnet to which the host belongs | |
the Department network | |
the University network | |
the Internet |
Question 28 |
byte | |
word | |
segment | |
message |
Question 29 |
session and request only | |
request and response only | |
response and session only | |
session, request and response |
Question 30 |
l and II only | |
II and III only | |
l and III only | |
I, II and III |
Then there are certain compliance rules which may be used for inheritance. So other statement (I) and (II) are True.
Question 31 |
P and Q only | |
P and R only | |
P and S only | |
P, Q, R and S |
If we somehow get this fallacy (T→F) then an argument is invalid.
For options P and S you don't get any such combinations for T→F, so P and S are valid.
For option Q: If we put p=F, q=T, r=T then we get T→F. So its INVALID.
For option R: If we put p=F, q=F, r=F then we get T→F. So it is INVALID.
So, answer is (C).
Question 32 |

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Find its determinant, Determinant = 3.
Now check options, by putting n=1, I am getting following results,
A) 5
B) 7
C) 3
D) 3
(A), (B) can't be the answer.
Now, check for n=2, Determinant = 9-1 = 8.
Put n=2 in (C), (D)
C) 7
D) 8
So, (D) is the answer.
Question 33 |
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Question 34 |
nHn+1 – (n + 1) | |
(n + 1)Hn – n | |
(n + 1)Hn – n | |
(n+1)Hn+1 – (n+1) |
Question 35 |
44 | |
96 | |
120 | |
3125 |

Question 37 |
10 | |
11 | |
18 | |
19 |
By Handshaking Lemma,
6 * 2 + 3 * 4 + (x - 9) * 3 = 27 * 2
24 + (x - 9) * 3 = 54
x = 19
Question 38 |
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Question 39 |
Q-II, R-IV, S-II, T-I | |
Q-III, R-II, S-I, T-IV | |
Q-II, R-I, S-IV, T-III | |
Q-I, R-IV, S-II, T-III |
Question 40 |
aaa | |
aabab | |
baaba | |
bab |

Now, here transition ((s,a,t), (s,a)) implies reading input symbol 'a' in the state 's' we have to move 's' having any symbol on the top of stack ... epsilon here implies "anything on of Top of stack".
Now, observe the PDA carefully, it is saying that in the starting you have to push one 'a' for each of 'a' and 'b'. And in the end you have to pop one 'a' by one 'a' by one 'a' or one 'b'. Thus the count of a's and b's in first half of the string should be equal to second half of string. Now to move from first half to second half we are required one 'a', i.e., moving from s to f.
So, all odd strings in which 'a' is the middle element will be accpeted.
Thus in our question, option (B) is aabab having 'b' in the middle and thus can't be accepted.
Question 41 |
1 only | |
2 only | |
3 only | |
1 and 3 only |
1) Sign bit of two input numbers is 0, and the result has sign bit 1.
2) Sign bit of two input numbers is 1, and the result has sign bit 0.
So, only (2) causes overflow.
Question 42 |
(A72E)16 and (22130232)4 | |
(A72E)16 and (22131122)4 | |
(A73E)16 and (22130232)4 | |
(A62E)16 and (22120232)4
|
= (00 1010 0111 0010 1110)2
= (A72E)16
Also,
(001 010 011 100 101 110)2
= (00 10 10 01 11 00 10 11 10)2
= (22130232)4
Question 43 |
AC’+ AB + A’C | |
AB’+ AC’+ A’C | |
A’B+ AC’+ AB’ | |
A’B + AC + AB’ |

⇒ A'C + AC' + AB'
Question 44 |
A serial transmission T1 uses 8 information bits, 2 start bits, 1 stop bit and 1 parity bit for each character. A synchronous transmission T2 uses 3 eight bit sync characters followed by 30 eight bit information characters. If the bit rate is 1200 bits/second in both cases, what are the transfer rates of T1 and T2?
100 characters/sec, 153 characters/sec | |
80 characters/sec, 136 characters/sec | |
100 characters/sec, 136 characters/sec | |
80 characters/sec, 153 characters/sec |
Transfer rate = 1200/12 = 100 char/sec
T2: Transfer character in bits = 24 + 240 = 264 bits
In 264 = 30 characters
Then in 1200 = ? 264/30 = 1200/x
x = 136.3 char/sec
So, correct option is (C).
Question 45 |

R1 → R3 R2 → M[100] | |
M[100] → R2 R1 → R2 R1 → R3 | |
R1 → M[100] R2 → R3 | |
R1 → R2 R1 → R3 R1 → M[100] |
Question 46 |

11 ns | |
12 ns | |
13 ns | |
28 ns |

So, total time would be 13 ns.
Question 47 |
4 | |
5 | |
6 | |
7 |
4, 3, 25, 8, 19, 6, 16, 35
CPU array (first element being least recently used)
[4, 3, 19, 6, 25, 8, 16, 35]
So, 45 replaces 4.
45, 3, 25, 8, 19, 6, 16, 35 [3, 19, 6, 25, 8, 16, 35, 45]
Similarly, 22 replaces 3 to give,
45, 22, 25, 8, 19, 6, 16, 35 [19, 6, 25, 8, 16, 35, 45, 22]
8 hits in cache.
45, 22, 25, 8, 19, 6, 16, 35 [19, 6, 25, 16, 35, 45, 22, 8]
3 replaces 19,
45, 22, 25, 8, 3, 6, 16, 35 [6, 25, 16, 35, 45, 22, 8, 3]
16 and 25 hits in cache,
45, 22, 25, 8, 3, 6, 16, 35 [6, 35, 45, 22, 8, 3, 16, 25]
Finally, 7 replaces 6, which is in block 5.
Question 48 |
T1.I1 + T2.I3 + T4.I3 + T3 | |
(T1 + T2 + T3).I3 + T1.I1 | |
(T1 + T2 ).I1 + (T2 + T4).I3 + T3 | |
(T1 + T2 ).I2 + (T1 + T3).I1 + T3 |
So, Ain is 1 in T3 of I1, I2 and I3. Also during T1, and T2 and T4 of I3. So, answer will be
T1.I1 + T2.I3 + T4.I3 + T3.I1 + T3.I2 + T3.I3
Since, CPU is having only 3 instructions, T3.I1 + T3.I2 + T3.I3 can be replaced by T3.
So, T1.I1 + T2.I3 + T4.I3 + T3
Question 49 |
1.155 | |
1.185 | |
1.255 | |
1.285 |

Question 50 |
The storage area of a disk has innermost diameter of 10 cm and outermost diameter of 20 cm. The maximum storage density of the disk is 1400 bits/cm. The disk rotates at a speed of 4200 RPM. The main memory of a computer has 64-bit word length and 1µs cycle time. If cycle stealing is used for data transfer from the disk, the percentage of memory cycles stolen for transferring one word is
0.5% | |
1% | |
5% | |
10% |
x μs is data transfer time.
% of time CPU idle is,
y/x × 100
Maximum storage density is given, so consider innermost track to get the capacity
= 2 × 3.14 × 5 × 1700 bits
= 3.14 × 14000 bits
Rotational latency = 60/4200 s = 1/70 s
Therefore, to read 64 bits, time required
(106 × 64)/(70 × 3.14 × 17000) μs = 20.8 μs
As memory cycle time given is 1μs,
Therefore, % CPU cycle stolen = (1/20.8) × 100 ≈ 5%
Question 51 |
A program attempts to generate as many permutations as possible of the string, 'abcd' by pushing the characters a, b, c, d in the same order onto a stack, but it may pop off the top character at any time. Which one of the following strings CANNOT be generated using this program?
abcd | |
dcba | |
abad | |
cabd |
B) First push abcd, and after that pop one by one. Sequence of popped elements will come to dcba.
C) push abc, and after that pop one by one. Sequence of popped elements will come to cba. Now push 'd' and pop 'd', final sequence comes to cbad.
D) This sequence is not possible because 'a' cannot be popped before 'b' anyhow.
Question 52 |
An array of integers of size n can be converted into a heap by adjusting the heaps rooted at each internal node of the complete binary tree starting at the node ⌊(n - 1) /2⌋, and doing this adjustment up to the root node (root node is at index 0) in the order ⌊(n - 1)/2⌋, ⌊(n - 3)/2⌋, ....., 0. The time required to construct a heap in this manner is
O(log n) | |
O(n) | |
O(n log log n) | |
O(n log n) |
And we know that time complexity for building the heap is O(n)
Question 53 |
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Left root right.
Preorder traversal is,
Root left right.
Question 54 |
f(n) + g(n) = O(h(n)) + h(n)) | |
f(n) = O(h(n)) | |
fh(n) ≠ O(f(n)) | |
f(n)h(n) ≠ O(g(n)h(n))
|
So, f(n)g(n) = O(g(n)h(n)) is True.
Hence, (D) is false.
Question 55 |

(E, G), (C, F), (F, G), (A, D), (A, B), (A, C) | |
(A, D), (A, B), (A, C), (C, F), (G, E), (F, G) | |
(A, B), (A, D), (D, F), (F, G), (G, E), (F, C) | |
(A, D), (A, B), (D, F), (F, C), (F, G), (G, E)
|
(C) produces connected component every instead a new edge is added but when first vertex is chosen (first vertex is chosen randomly) first edge must be minimum weight edge that is chosen. Therefore, (A, D) must be chosen before (A, B). Therefore (C) is false.
Question 56 |
RECURSIVE ALGORITHM | RECURRENCE RELATION | ||
---|---|---|---|
P. | Binary search | I. | T(n) = T(n-k) + T(k) + cn |
Q. | Merge sort | II. | T(n) = 2T(n-1) + 1 |
R. | Quick sort | III. | T(n) = 2T(n/2) + cn |
S. | Tower of Hanoi | IV. | T(n) = T(n/2) + 1 |
P-II, Q-III, R-IV, S-I | |
P-IV, Q-III, R-I, S-II | |
P-III, Q-II, R-IV, S-I | |
P-IV, Q-II, R-I, S-III |
Binary search - T(n/2) + 1
Merge sort - T(n) = 2T(n/2)+ cn
Tower of hanoi - T(n) = 2T(n-1) +1
Question 57 |
#include<stdio.h> #define ROW 4 #define COL 4 int M[ROW][COL] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16}; main() { int i, j, t; for (i = 0; i < 4; ++i) { X } for (1 = 0; i < 4; ++i) for (j = 0; j < 4; ++j) printf ("%d", M[i][j]); }
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In (D) , given statements is wrong as temporary variable needs to be assigned some value and not vice-versa.
Question 58 |
#include <stdio.h> int funcf (int x); int funcg (int y); main() { int x = 5, y = 10, count; for (count = 1; count <= 2; ++count) { y += funcf(x) + funcg(x); printf ("%d ", y); } } funcf(int x) { int y; y = funcg(x); return (y); } funcg(int x) { static int y = 10; y += 1; return (y+x); }
43 80 | |
42 74 | |
33 37 | |
32 32 |
In first case of funcf, which in turn calls funcg, y becomes 11 and it returns 5+11=16.
In second call of funcg, y becomes 12 and it returns 5+12=17.
So, in main y is incremented by 16+17=33 to become 10+33 =43.
In second iteration:
y will be incremented by 18+19=37 to give 43+37=80.
Question 59 |
#include <stdio.h> void wrt_it (void); int main (void) { printf("Enter Text"); printf ("n"); wrt_ it(); printf ("n"); return 0; } void wrt_it (void) { int c; if (?1) wrt_it(); ?2 }
?1 is getchar() ! = ‘\n’ ?2 is getchar(c); | |
?1 is (c = getchar()); ! = ‘\n’ ?2 is getchar(c); | |
?1 is c! = ‘\n’ ?2 is putchar(c); | |
?1 is (c = getchar()) ! = ‘\n’ ?2 is putchar(c); |
putchar( ) = writes a character specified by the argument to stdout.
As getchar( ) and putchar( ), both are needed to read the string and prints its reverse and only option (D) contains both the function. (D) is the answer.
Now coming to the code, wrt_id(void) is calling itself recursively. When \n is encountered, putchat( ) gets executed and prints the last character and then the function returns to its previous call and prints last 2nd character and so on.
Question 60 |
#include <stdio.h> typedef struct { char *a; char *b; } t; void f1(t s); void f2(t *p); main() { static t s = {"A", "B"}; printf ("%s %sn", s.a, s.b); f1(s); printf ("%s %sn", s.a, s.b); f2(&s); } void f1(t s) { s.a = "U"; s.b = "V"; printf ("%s %sn", s.a, s.b); return; } void f2(t *p) { p -> a = "V"; p -> b = "W"; printf("%s %sn", p -> a, p -> b); return; }What is the output generated by the program ?
AB UV VW VW | |
AB UV AB VW | |
AB UV UV VW | |
AB UV VW UV |
→ f1 is call by value. The changes applicable only for local from f1. UV is printed.
→ Back in main( ), AB is printed.
→ Then in f2, VW is printed.
Hence, answer is (B).
Question 61 |
2 and 3 | |
3 and 3 | |
3 and 4 | |
4 and 4
|
Direction changes at 120, 110, 130.
FCFS: (90) 120 30 70 115 130 110 80 20 25
Direction changes at 120, 30, 130, 20.
Question 62 |
if T(Pr) < T(Ph) then kill Pr else waitWhich one of the following is TRUE?
The scheme is deadlock-free, but not starvation-free | |
The scheme is not deadlock-free, but starvation-free | |
The scheme is neither deadlock-free nor starvation-free | |
The scheme is both deadlock-free and starvation-free |
So, starvation is possible, but deadlock is not possible.
Question 63 |
for(i = 1; i < = n; i++) fork ();The number of new processes created is
n | |
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2n - 1 | |
3n - 1 |
2n - 1.
Question 64 |
P(full), V(empty), P(full), V(empty) | |
P(full), V(empty), P(empty), V(full) | |
P(empty), V(full), P(empty), V(full) | |
P(empty), V(full), P(full), V(empty) |
In process P1, wait on semaphore 'empty' signifies that if there is no space in buffer then P2 cannot produce more items. Signal on semaphore 'full' is to signify that one item has been added to the buffer.
In process P2, wait on semaphore 'full' signifies that if the buffer is empty then consumer cannot consume any item. Signal on semaphore 'empty' increments a space in the buffer after consumption of an item.
Question 65 |
In a virtual memory system, size of virtual address is 32-bit, size of physical address is 30-bit, page size is 4 Kbyte and size of each page table entry is 32-bit. The main memory is byte addressable. Which one of the following is the maximum number of bits that can be used for storing protection and other information in each page table entry?
2 | |
10 | |
12 | |
14 |
No. of frames = Physical memory size/Page size
= (230)/(212)
= 218
18+4 = 32 (PT entry size = 32 bits)
k = 14 bits
Question 66 |
In a particular Unix OS, each data block is of size 1024 bytes, each node has 10 direct data block addresses and three additional addresses: one for single indirect block, one for double indirect block and one for triple indirect block. Also, each block can contain addresses for 128 blocks. Which one of the following is approximately the maximum size of a file in the file system?
512 MB | |
2 GB | |
8 GB | |
16 GB |
= (10 + 27 + 27 × 27 + 7 × 7 × 7) × 210
≈ 231
≈ 2 GB
Question 67 |
A software project involves execution of 5 tasks T1, T2, T3, T4 and T5 of duration 10, 15, 18, 30 and 40 days, respectively. T2 and T4 can start only after T1 completes. T3 can start after T2 completes. T5 can start only after both T3 and T4 complete. What is the slack time of the task T3 in days?
0 | |
3 | |
18 | |
30 |
Question 68 |
int module1 (int x, int y) { while (x! = y) { if (x > y) x = x - y, else y = y - x; } return x; }What is Cyclomatic complexity of the above module?
1 | |
2 | |
3 | |
4 |
Question 69 |
Assume that the delivered lines of code L of a software is related to the effort E in person months and duration t in calendar months by the relation L = P* (E/B)1/3 * t4/3, where P and B are two constants for the software process and skills factor. For a software project, the effort was estimated to be 20 person months and the duration was estimated to be 8 months. However, the customer asked the project team to complete the software project in 4 months. What would be the required effort in person months?
10 | |
40 | |
160 | |
320 |
Question 70 |
A software was tested using the error seeding strategy in which 20 errors were seeded in the code. When the code was tested using the complete test suite, 16 of the seeded errors were detected. The same test suite also detected 200 non-seeded errors. What is the estimated number of undetected errors in the code after this testing?
4 | |
50 | |
200 | |
250 |
Question 71 |
1% | |
24% | |
99% | |
99.009% |
Question 72 |
Consider the following entity relationship diagram (ERD), where two entities E1 and E2 have a relation R of cardinality 1 : m.
The attributes of E1 are A11, A12 and A13 where A11 is the key attribute. The attributes of E2 are A21, A22 and A23 where A21 is the key attribute and A23 is a multi-valued attribute. Relation R does not have any attribute. A relational database containing minimum number of tables with each table satisfying the requirements of the third normal form (3NF) is designed from the above ERD. The number of tables in the database is
2 | |
3 | |
5 | |
4 |
Then, we get
T1: {A11, A12, A13} - key is A11
T2: {A21, A22, A11} - key is A21
T3: {A21, A23} - key is {A21, A23}
Question 73 |
Insert into department values (1, 'Mathematics') Insert into department values (2, 'Physics') Insert into student values (l, 'Navin', 1) Insert into student values (2, 'Mukesh', 2) Insert into student values (3, 'Gita', 1)How many rows and columns will be retrieved by the following SQL statement?
Select * from student, department
0 row and 4 columns | |
3 rows and 4 columns | |
3 rows and 5 columns | |
6 rows and 5 columns |
rows = 3 * 2 = 6
Columns = 3 + 2 = 5
Question 74 |
A relation Empdtl is defined with attributes empcode (unique), name, street, city, state and pincode. For any pincode, there is only one city and state. Also, for any given street, city and state, there is just one pincode. In normalization terms, Empdtl is a relation in
1NF only | |
2NF and hence also in 1NF | |
3NF and hence also in 2NF and 1NF | |
BCNF and hence also in 3NF, 2NF an 1NF |
Here key is empcode and contains only one attribute, hence no partial dependency. But there is transitive dependency in this.
Pincode → city, state, so it is not in 3NF.
Question 75 |
roll no. | marks |
1 | 10 |
2 | 20 |
3 | 30 |
4 | Null |
Update T1 set marks = marks + 5 Select avg(marks) from T1What is the output of the select statement?
18.75 | |
20 | |
25 | |
NULL |
(15+25+35)/3 = 25
Question 76 |
T1 | T2 |
---|---|
Read(A)A = A - 10 | Read (A)Temp = 0.2*A Write(A) Read(B) |
Write(A)Read(B) B = B + 10 Write(B) | B = B + TempWrite(B) |
S is serializable only as T1, T2 | |
S is serializable only as T2, T1 | |
S is serializable both as T1, T2 and T2, T1 | |
S is serializable either as T1 or as T2 | |
None of these |
Question 77 |
ROLL_NO | NAME | DEPT_ID |
---|---|---|
1 | ABC | 1 |
2 | DEF | 1 |
3 | GHI | 2 |
4 | JKL | 3 |
DEPT_ID | DEPT_NAME |
---|---|
1 | A |
2 | B |
3 | C |
- update Student set Dept_id = Null where Roll_on = 1
- update Department set Dept_id = Null where Dept_id = 1
Both (i) and (ii) will fail | |
(i) will fail but (ii) will succeed | |
(i) will succeed but (ii) will fail | |
Both (i) and (ii) will succeed |
But in (ii) if we set in Department table, Dept_id = Null, then it will produce inconsistency because in Student table we will still have the tuples containing the Dept_id = 1.
Question 78 |
Consider a table T in a relational database with a key field K. A B-tree of order p is used as an access structure on K, where p denotes the maximum number of tree pointers in a B-tree index node. Assume that K is 10 bytes long; disk block size is 512 bytes; each data pointer PD is 8 bytes long and each block pointer PB is 5 bytes long. In order for each B-tree node to fit in a single disk block, the maximum value of p is
20 | |
22 | |
23 | |
32 |
r = record_ptr_size
b = block_ptr_size
(p - 1) (k + r) + p × b ≤ 512
(p - 1) (10 + 8) + p × 5 ≤ 512
23p ≤ 530
p ≤ 23.04
So, maximum value of p possible will be 23.
Question 79 |
In a data link protocol, the frame delimiter flag is given by 0111. Assuming that bit stuffing is employed, the transmitter sends the data sequence 01110110 as
01101011 | |
011010110 | |
011101100 | |
0110101100 |
Thus using the above logic,
Delimiter flag: 0111
Data sequence: 01110110
So, for a flag of 4 bits we will compare data sequence with a pattern of 3 bits, i.e., 011.
0 1 1 0 1 0 1 1 0 0
In the above pattern the underlined bits are found matched. Hence, 0 in italics is stuffed. Thus resulting in the data sequence as 0110101100 which is option (D).
Question 80 |
In a sliding window ARQ scheme, the transmitter's window size is N and the receiver's window size is M. The minimum number of distinct sequence numbers required to ensure correct operation of the ARQ scheme is
min (M,N) | |
max (M,N) | |
M + N | |
MN |
Question 81 |
Consider a 10 Mbps token ring LAN with a ring latency of 400 µs. A host that needs to transmit seizes the token. Then it sends a frame of 1000 bytes, removes the frame after it has circulated all around the ring, and finally releases the token. This process is repeated for every frame. Assuming that only a single host wishes to transmit, the effective data rate is
1Mbps | |
2Mbps | |
5Mbps | |
6Mbps |
Question 82 |
A 20 Kbps satellite link has a propagation delay of 400 ms. The transmitter employs the "go back n ARQ" scheme with n set to 10. Assuming that each frame is 100 bytes long, what is the maximum data rate possible?
5 Kbps | |
10 Kbps | |
15 Kbps | |
20 Kbps |

Question 83 |
I and III | |
I, II and III | |
II and IV | |
I, II, III and IV
|
Given transmitted codewords are

By inspection we can find the rule for generating each of the parity bits,

Now from above we can see that (I) and (III) are only codewords.
Question 84 |
Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability p = 0.2 in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support, if each host has to be provided a minimum through put of 0.16 frames per time slot?
1 | |
2 | |
3 | |
4 |
0.16 = 0.2 × 0.8N-1
⇒ 0.8 = 0.8N-1
⇒ N = 2
Question 85 |
Fragment Offset | |
Source IP address | |
Destination IP address | |
Destination port number |
Question 86 |
A TCP message consisting of 2100 bytes is passed to IP for delivery across two networks. The first network can carry a maximum payload of 1200 bytes per frame and the second network can carry a maximum payload of 400 bytes per frame, excluding network overhead. Assume that IP overhead per packet is 20 bytes. What is the total IP overhead in the second network for this transmission?
40 bytes | |
80 bytes | |
120 bytes | |
160 bytes |
2120B reach R1's network layer. It removes original IP header, fragments data part at IP and then appends IP header to all fragments and forwards . So, it divides 2100 Bytes into two fragments of size 1200 and 900. And both fragments are sent to R2.
At Router-2:
Both fragments that reach R2 exceed MTU at R2. So, both are fragmented. First packet of 1200B is fragmented into 3 packets of 400 Bytes each. And second packet of 900B is fragmented into 3 fragments of 400, 400 and 100 Bytes respectively.
So, totally 6 packets reach destinations.
So, total IP overhead = 6 × 20 = 120 Bytes
Question 87 |
Suppose that the maximum transmit window size for a TCP connection is 12000 bytes. Each packet consists of 2000 bytes. At some point of time, the connection is in slow-start phase with a current transmit window of 4000 bytes. Subsequently, the transmitter receives two acknowledgements. Assume that no packets are lost and there are no time-outs. What is the maximum possible value of the current transmit window?
4000 bytes | |
8000 bytes | |
10000 bytes | |
12000 bytes |
and packet size =2000 B (or MSS)
Receiver window size = 6 MSS and
Current sender window size = 2 MSS
Slow start threshold = receiver window/2 = 3 MSS
Now current sender window size = 2 MSS <3 MSS,
which implies transmission is in slow start phase.
After receiving first Ack: Current sender window should increase exponentially to 4 MSS but since threshold = 3 MSS, current sender window size goes to threshold which is 3 MSS, then after receiving second Ack: Since now it is in congestion avoidance phase, sender window size increases linearly which makes current sender window
= 4 MSS
= 4 × 2000 B
= 8000 B
Question 88 |
<?xml version = "1.0"?> <!DOCTYPE myMessage SYSTEM "intro.dtd"› <myMessage> <message>Welcome to XML</message> </myMessage> intro.dtd <! ELEMENT myMessage (message)> <! ELEMENT message (#PCDATA)>A validating parser will classify intro.xml as
Well-formed and validated | |
Well-formed but not validated | |
Validated but not well-formed | |
Neither validated nor well-formed |
Question 89 |
-
<A HREF = "http://www.gate.ac.in/HTML/BASIC/testpage.html">Test Me</A>
-
<A HREF = "/BASIC/testpage.html">Test Me</A>
-
<A HREF = "testpage.html">Test Me</A>
-
<A HREF = "testpage.html#test">Test Me</A>
1 and 2 only | |
1 and 3 only | |
1, 2 and 3 only | |
1, 2, 3 and 4 |