Gate IT 2007
Question 1 
7/8  
1/2  
7/16  
5/32 
Discuss it
Probability of 1st coin to be selected = ½
And probability of getting head on it = 5/8
Or
Probability of 2nd coin to be selected = 1/2
And pobability of getting head on 2nd coin = ¼
Probability of getting head (overall) = (Probability of selecting 1st coin * Probability of head on it) +(Probability of selecting 2nd coin * Probability of getting head on it) = (1/2)*(5/8) + (1/2)*(1/4) = 7/16
Question 2 
5  
(5 + √5)/2  
3  
(5  √5)/2 
Discuss it
A(λ*I) = 0
(3λ)(2λ)1 = 0
63λ 2λ + λ^{2}+1=0
λ^{2}5λ+5=0
λ = (5+√5)/2 and (5√5)/2,
λ = (5+√5)/2 is max value another root with negative sign which will not be max value.
For, λ=5+5√2, x^{T}Ax=[18.131 21.231 21.231 34.331]
For, λ=5−5√2, x^{T}Ax=[Tex]\begin{pmatrix} 14.300&−0.700\\ 6.200&1.200 \end{pmatrix}[/Tex]
Hence, for λ=5+5√2 the value of x^{T}Ax is maximum.
Question 3 
weight (u, v) < 12  
weight (u, v) ≤ 12  
weight (u, v) > 12  
weight (u, v) ≥ 12 
Discuss it
Question 4 
Iteration size  
Cost  
Adopted process such as Rational Unified Process or Extreme Programming  
Risk 
Discuss it
Question 5 
Expert system  
DB repository  
Aircraft flight controller  
Signal processing 
Discuss it
Question 6 
1.83  
2  
3  
6 
Discuss it
It takes, 12 cycles to complete 1 instruction
So, for n instructions it will take 12n cycle
For pipelined processor,
Each stage time = max{each stage cycles} = max{3, 2, 5, 4, 6 and 2} = 6 cycles
So, for n instructions it will take = 6*6+ (n1)*6 {6*6 for 1st instruction and for rest of n1 it will take 6}
For a large number of instructions:
Lim_{n>∞} 12n/36 + (n1)*6 = 12/6 =2
Question 7 
11, 00  
01, 10  
10, 01  
00, 11 
Discuss it
Question 8 
What are the values of X_{1}, X_{2}, X_{3}?
X_{1}=b, X_{2}=0, X_{3}=a  
X_{1}=b, X_{2}=1, X_{3}=b  
X_{1}=a, X_{2}=b, X_{3}=1  
X_{1}=a, X_{2}=0, X_{3}=b 
Discuss it
Question 9 
L (D) ⊂ L (G)  
L (D) ⊃ L (G)  
L (D) = L (G)  
L (D) is empty 
Discuss it
Question 10 
get_exclusive_access ( )
{
if (critical _flag == FALSE) {
critical_flag = TRUE ;
critical_region () ;
critical_flag = FALSE;
}
}
Consider the following statements.i. It is possible for both P_{1} and P_{2} to access critical_region concurrently.
ii. This may lead to a deadlock.
Which of the following holds?
(i) is false and (ii) is true  
Both (i) and (ii) are false  
(i) is true and (ii) is false  
Both (i) and (ii) are true 
Discuss it
Say P_{1} starts first and executes statement 1, after that system context switches to P_{2} (before executing statement 2), and it enters inside if statement, since the flag is still false. So now both processes are in critical section!! so (i) is true.. (ii) is false By no way it happens that flag is true and no process' are inside the if clause, if someone enters the critical section, it will definitely make flag = false. So no deadlock.
Question 11 
The time at which the request for J_{7} will be completed will be
16  
19  
20  
37 
Discuss it
Memory Block  Size  Job (t=0)  Job(t=8)  Job(t=10)  Job(t=11) 
1  4k  J3  2 units (1K free left)  
2  8k  J4 – 8 units (2K free left)  J5  14 units  J5 – 14 units  J5 – 14 units 
3  20k  J2 10 units(6K free left)  J2 10 units  J6 – 11 units  J7 – 19 units 
4  2k  J1 4 units 
Question 12 
0100, 0200, 0430, 0499, 0510, 0530, 0560, 0120, 0220, 0240, 0260, 0320, 0410.
Suppose that the memory can store only one page and if x is the address which causes a page fault then the bytes from addresses x to x + 99 are loaded on to the memory.
How many page faults will occur ?
0  
4  
7  
8 
Discuss it
Address Page faults last byte in memory 0100 page fault, 199 0200 page fault, 299 0430 page fault, 529 0499 no page fault 0510 no page fault 0530 page fault, 629 0560 no page fault 0120 page fault, 219 0220 page fault, 319 0240 no page fault 0260 no page fault 0320 page fault, 419 0410 no page fault So, 7 is the answer (C)
Question 13 
(i) is false, but (ii) and (iii) are true  
(i) and (iii) are false, but (ii) is title  
(i) and (ii) are false, but (iii) is true  
(i), (ii) and (iii) are false 
Discuss it
 The timeout value is set to the RTT (Round Trip Time) measured during TCP connection establishment for the entire duration of the connection. FALSE The timeout value can’t be fixed for entire duration as it will turn timer to static timer, we need dynamic timer for timeout.
 Appropriate RTT estimation algorithm is used to set the timeout value of a TCP connection.TRUE Yes, all three algorithm are appropriate RTT estimation algorithm used to set timeout value dynamically.
 Timeout value is set to twice the propagation delay from the sender to the receiver.FALSE This statement is false because, timeout value is set to twice the propagation delay in data link layer where, hop to hop distance is known, not in TCP layer.
Question 14 
60 and 290  
230 and 291  
60 and 231  
60 and 230 
Discuss it
Question 15 
Both are false  
Statement (i) is true and the other is false  
Statement (ii) is true and the other is false  
Both are true 
Discuss it
Injective Function:
A function F(X) is said to be injective if it has onetoone mapping. Statement 1: Hash function is an injective function Statement 2: DES Encryption technique performs a permutation on the elements of its input alphabet
1) Generally, a hash function H(X) is mapping from a larger set to a predefined output set For example, let H(X) = (X)%5 The above function H(X) is not injective because Let X1 = 10, X2 = 15 H(10) = H(15) = 0 As the output of H(X1) = H(X2) where X1!= X2 => H(X) is manytoone function. Statement 1 is false.
2) In DES encryption scheme, it performs PBox permutation. Statement 1 is false, Statement 2 is true.
This solution is contributed by Anil Saikrishna Devarasetty.
Question 16 
5  
8  
12  
16 
Discuss it
Or put the value of p and get the answer
Question 17 
O(logn)  
O(√n)  
O(n/logn)  
O(n) 
Discuss it
Binary_exp(b,n) // Compute bn mod m { if(n == 0) Return 1; Else if(n == 1) Return b mod m; Else { Half = Binary_exp(b,n/2); if(n%2 == 0) // n is even Return (Half*Half) mod m; Else // n is odd Return (((Half*Half) mod m)*n) mod m; } }Recurrence Relation for computing the time complexity of the above given algorithm is T(n) = T(n/2) + constant = (log2n) This solution is contributed by Pranjul Ahuja.
Question 18 
A combinational circuit  
A finite automaton  
A pushdown automaton with one stack  
A pushdown automaton with two stacks 
Discuss it
B) A finite automaton => We need infinite memory, there is no upper limit on Number of TCP ckt so Not this.
C) A pushdown automaton with one stack => Stack is infinite. Suppose we have 2 connections , we have pushed details of those on stack we can not access the details of connection which was pushed first, without popping it off. So Big NO.
D) pushdown automaton with two stacks => This is TM. It can do everything our normal computer can do so Yes. Firewall can be created out of TM.
Question 19 
With reference to the HTML lines given above, consider the following statements.
1. Clicking on the point <80, 75> does not have any effect.
2. The web browser can identify the area applicable to the mouseclick within the image and the subsequent action to be taken without additional responses from the web server.
3. The dots in the cgibin URL will be resolved by the web browser before it is sent to the web server
4. The "fd.html" request when sent to the web server will result in a GET request.
Exactly how many of the statements given above are correct?
0  
1  
2  
3 
Discuss it
Question 20 
Consider the XPath expression: *[not (self ) : : TOC]
What would be the result of the given XPath expression when the current node is Book?
The Title and Content elements  
The Content and TOC elements  
The Title and TOC elements  
The Title Content and TOC elements 
Discuss it
Question 21 
∀x(P(x) => Q(x)) => (∀xP(x) => ∀xQ(x))  
∃x(P(x) ∨ Q(x)) => (∃xP(x) => ∃xQ(x))  
∃x(P(x) ∧ Q(x)) <=> (∃xP(x) ∧ ∃xQ(x))  
∀x∃y P(x, y) => ∃y∀x P(x, y) 
Discuss it
LHS: For every x (if P holds then Q holds)
RHS: If P(x) holds for all x, then Q(x) holds for all x.
(B) LHS !>RHS
LHS: An x exist for which either P(x) is true or Q(x) is true.
RHS: If an x exist for which P(x) is true then another x exist for which Q(x) is true.
(C) It is not necessary that on RHS both x are same.
LHS: There exist an x for which both P(x) and Q(x) are true.
RHS: There exist an x for which P(x) is true and there exist an x for which Q(x) is true.
(D) LHS!>RHS
LHS: For every x, there exist a y such that P(x, y) holds.
RHS: There exist a y such that for all x P(x, y) holds.
Question 22 
i. 10^{2}
ii. 10^{3}
iii. 10^{4}
iv. 10^{5}
For which of these values of the step size h, is the computed value guaranteed to be correct to seven decimal places. Assume that there are no roundoff errors in the computation.
(iv) only  
(iii) and (iv) only  
(ii), (iii) and (iv) only  
(i), (ii), (iii) and (iv) 
Discuss it
Question 23 
i. (0, 0) ∊ P.
ii. (a, b) ∊ P if and only if a % 10 ≤ b % 10 and (a/10, b/10) ∊ P.
Consider the following ordered pairs:
i. (101, 22)
ii. (22, 101)
iii. (145, 265)
iv. (0, 153)
Which of these ordered pairs of natural numbers are contained in P?
(i) and (iii)  
(ii) and (iv)  
(i) and (iv)  
(iii) and (iv) 
Discuss it
Question 24 
d[u] < d[v]  
d[u] < f[v]  
f[u] < f[v]  
f[u] > f[v] 
Discuss it
Question 25 
1  
2  
3  
n 
Discuss it
Given that there are n edges, that means a cycle is there in the graph.
The simplex graph with these conditions may be:
Now we can make a different spanning tree by removing one edge from the cycle, one at a time.
Minimum cycle length can be 3, So, there must be atleast 3 spanning trees in any such Graph.
Question 26 
Nondecreasing order of t_{i}  
Nonincreasing order of w_{i}  
Nonincreasing order of w_{i}t_{i}  
Noneincreasing order of w_{i}/t_{i} 
Discuss it
Question 27 
int f (int n) { if (n <= 1) return 1; else if (n % 2 == 0) return f(n/2); else return f(3n  1); }Assuming that arbitrarily large integers can be passed as a parameter to the function, consider the following statements.
1. The function f terminates for finitely many different values of n ≥ 1.
ii. The function f terminates for infinitely many different values of n ≥ 1.
iii. The function f does not terminate for finitely many different values of n ≥ 1.
iv. The function f does not terminate for infinitely many different values of n ≥ 1.
Which one of the following options is true of the above?
(i) and (iii)  
(i) and (iv)  
(ii) and (iii)  
(ii) and (iv) 
Discuss it
So, (i) is false and (ii) is TRUE.
Let n = 3, it will terminate in 2nd iteration.
Let n=5, it will go like 5  14  7  20  10  5 – and now it will repeat.
And any number with a factor of 5 and 2, there are infinite recursions possible.
So, (iv) is TRUE and (iii) is false.
Question 28 
5  
6  
7  
10 
Discuss it
Say after inserting x values probability becomes ½
(1/20).x = ½
X=10
Question 29 
35  
64  
128  
5040 
Discuss it
= 7!/(4!3!)
= 35
Question 30 
i. isEmpty (Q) — returns true if the queue is empty, false otherwise.
ii. delete (Q) — deletes the element at the front of the queue and returns its value.
iii. insert (Q, i) — inserts the integer i at the rear of the queue.
Consider the following function:
void f (queue Q) { int i ; if (!isEmpty(Q)) { i = delete(Q); f(Q); insert(Q, i); } }What operation is performed by the above function f ?
Leaves the queue Q unchanged  
Reverses the order of the elements in the queue Q  
Deletes the element at the front of the queue Q and inserts it at the rear keeping the other elements in the same order  
Empties the queue Q

Discuss it
So, the queue will be in reverse.
Question 31 
#include <stdio.h> int main () { int sum = 0, maxsum = 0, i, n = 6; int a [] = {2, 2, 1, 3, 4, 2}; for (i = 0; i < n; i++) { if (i == 0  a [i] < 0  a [i] < a [i  1]) { if (sum > maxsum) maxsum = sum; sum = (a [i] > 0) ? a [i] : 0; } else sum += a [i]; } if (sum > maxsum) maxsum = sum ; printf ("%d\n", maxsum); }What is the value printed out when this program is executed?
9  
8  
7  
6 
Discuss it
When i=2 > i==0 is false, a[i]<0 is true and so condition (1) is true.Now if (sum > maxsum) is false, since sum=0 and maxsum=2.Since sum = (a [i] > 0) ? a [i] : 0; , sum=0 since a[i]<0.
When i=3 > i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 3. When i=4 > i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is false so condition (1) is false. Now sum += a [i] = 7. When i=5 > i==0 is false , a[i]<0 is false and a [i] < a [i – 1] is true so condition (1) is true. sum > maxsum is true, since sum=7 and maxsum=2,so maxsum=7.Since sum = (a [i] > 0) ? a [i] : 0, so sum=2 since a[5]>0. This solution is contributed by nirmal Bharadwaj
Question 32 
#include #define EOF 1 void push (int); /* push the argument on the stack */ int pop (void); /* pop the top of the stack */ void flagError (); int main () { int c, m, n, r; while ((c = getchar ()) != EOF) { if (isdigit (c) ) push (c); else if ((c == '+')  (c == '*')) { m = pop (); n = pop (); r = (c == '+') ? n + m : n*m; push (r); } else if (c != ' ') flagError (); } printf("% c", pop ()); }What is the output of the program for the following input ? 5 2 * 3 3 2 + * +
15  
25  
30  
150 
Discuss it
Question 33 
int i ; program main () { int j = 60; i = 50; call f (i, j); print i, j; } procedure f (x, y) { i = 100; x = 10; y = y + i ; }Which one of the following options represents the correct output of the program for the two parameter passing mechanisms?
Call by value : i = 70, j = 10; Call by reference : i = 60, j = 70  
Call by value : i = 50, j = 60; Call by reference : i = 50, j = 70  
Call by value : i = 10, j = 70; Call by reference : i = 100, j = 60  
Call by value : i = 100, j = 60; Call by reference : i = 10, j = 70 
Discuss it
Call by reference: A link to original variables will be passed, by allowing the function to manipulate the original variables.
Question 34 
int i ; program main () { i = 10; call f(); } procedure f() { int i = 20; call g (); } procedure g () { print i; }Let x be the value printed under static scoping and y be the value printed under dynamic scoping. Then, x and y are
x = 10, y = 10  
x = 20, y = 10  
x = 10, y = 20  
x = 20, y = 20 
Discuss it
Question 35 
i. Static scope facilitates w1 bindings.
ii. Dynamic scope requires w2 bindings.
iii. Early bindings w3 execution efficiency.
iv. Late bindings w4 execution efficiency.
The right choices of wl, w2, w3 and w4 (in that order) are
Early, late, decrease, increase  
Late, early, increase, decrease  
Late, early, decrease, increase  
Early, late, increase, decrease 
Discuss it
Late binding can be done in Dynamic scoping (during execution time). Late binding decreases efficiency as it needs to be done at runtime. (but it increases flexibility)
Question 36 
(Di > Dj denotes that Di is faster than Dj)
D1 > D > D2  
D2 > D > D1  
D > D2 > D1  
D > D1 > D2 
Discuss it
Say, t=1
D = 0.8(1) + 0.2(2)
=1.2
D1 = 0.8(1.3)+0.2(1.4)
=1.04 +.28
=1.32
D2 = 0.8(10.04) + 0.2(22*0.1)
= 0.8*0.96 + 0.2*1.8
= 0.768 +0.36 = 1.128
D1>D>D2
Question 37 
3  
18  
20  
30 
Discuss it
Memory Block  Blocks 
0  8, 0, 16, 24 
1  9, 17, 25, 63, 17 
2  2, 18, 2, 82 
3  3 
4  20 
5  5, 5 
6  30 
7  63,63 
Question 38 
(a.b).c + (a'.c).d + (b.c).d + a. d
What is the function finally realized ?
1  
a' + b' + c' + d'  
a' + b + c' + d'  
a' + b' + c + d' 
Discuss it
Question 39 
In which of the following options, will the result of executing the RHS be the same as executing the LHS irrespective of the instructions that follow ?
(i) and (iii)  
(i) and (iv)  
(ii) and (iii)  
(ii) and (iv)  
Only (i) 
Discuss it
(iii) and (iv) are the same!! and both are wrong because R2 is writing last, not R1.
(ii) false, because R2 get the correct data, but location has not got updated.
Question 40 
0110  
1011  
1101  
1111 
Discuss it
The linear feedback shift register makes use of XOR gate for tapping and providing feedback. It works by shifting and placing the feedback value. Feedback value is determined by last bit and second bit tapped and fed to XOR gate.
A 
B 
XOR 
0 
0 
0 
0 
1 
1 
1 
0 
1 
1 
1 
0 
Question 41 
Load R1, Loc 1; Load R1 from memory location Loc1 Load R2, Loc 2; Load R2 from memory location Loc 2 Add R1, R2, R1; Add R1 and R2 and save result in R1 Dec R2; Decrement R2 Dec R1; Decrement R1 Mpy R1, R2, R3; Multiply R1 and R2 and save result in R3 Store R3, Loc 3; Store R3 in memory location Loc 3What is the number of cycles needed to execute the above code segment assuming each instruction takes one cycle to execute ?
7  
10  
13  
14 
Discuss it
Question 42 
(135103.412)_{o}  
(564411.412)_{o}  
(564411.205)_{o}  
(135103.205)_{o} 
Discuss it
= 1100 0000 0001 0010. 0010 0101 (convert each digit of hexa to 4 bit binary)
 0000 0101 1100 1110. 1010 0000
= 1011 1010 0100 0011. 1000 0101
= 1 011 101 001 000 011 . 100 001 010 (make pairing of 3 to convert in octal)
= (135103.412)_{o}
Question 43 
00000000, 00001111, 01010101, 10101010, 11110000.
What is the maximum number of bit errors that can be corrected ?
0  
1  
2  
3 
Discuss it
maximum hamming distance between any two of the given code = 8 (between 01010101 and 10101010)
t=3.5, t=3
Question 44 
 Number of tracks = 500
 Number of sectors/track = 100
 Number of bytes /sector = 500
 Time taken by the head to move from one track to adjacent track = 1 ms
 Rotation speed = 600 rpm.
300.5 ms  
255.5 ms  
255.0 ms  
300.0 ms 
Discuss it
 Avg Seek Time  time taken to move from 1st track to 1sr track : 0ms, 1st to 2nd : 1ms, 2ms, 3ms,....499ms Avg Seek time =( ∑0+1+2+3+...+499)/500 = 249.5 ms
 Avg Rotational Delay  RMP : 600 , 600 rotations in 60 sec (one Rotation = 60/600 sec = 0.1 sec) So, Avg Rotational Delay = 0.1/2= 50 ms
 Data Transfer Time: In One 1 Rotation we can read data on one track = 100 * 500 = 50,000 B data is read in one rotation. 250 bytes > 0.1 * 250 / 50,000 = 0.5 ms
Question 45 
0000  
0111  
1111  
None of these 
Discuss it
We consider circuit in two different configurations to check the inputs. And then apply the three available inputs to get output values. On doing so, we observe that all inputs give same outputs in both the configurations. Thus None of these is the answer.
This solution is contributed by Kriti Kushwaha.
Question 46 
Which one of the following choices describes the properties satisfied by the strings in these languages?
G1 : No y appears before any x G2 : Every x is followed by at least one y  
G1 : No y appears before any x G2 : No x appears before any y  
G1 : No y appears after any x G2 : Every x is followed by at least one y  
G1 : No y appears after any x G2 : Every y is followed by at least one x 
Discuss it
Question 47 
What language does this DFA recognize ?
All strings of x and y  
All strings of x and y which have either even number of x and even number of y or odd number or x and odd number of y  
All strings of x and y which have equal number of x and y  
All strings of x and y with either even number of x and odd number of y or odd number of x and even number of y 
Discuss it
Question 48 
S → x B  y A
A → x  x S  y A A
B → y  y S  y B B
Consider the following strings.
(i) xxyyx
(ii) xxyyxy
(iii) xyxy
(iv) yxxy
(v) yxx
(vi) xyx
Which of the above strings are generated by the grammar ?
(i), (ii), and (iii)  
(ii), (v), and (vi)  
(ii), (iii), and (iv)  
(i), (iii), and (iv) 
Discuss it
Question 49 
Which one of the following statements is true?
G_{1} is contextfree but not regular and G_{2} is regular  
G_{2} is contextfree but not regular and G_{1} is regular  
Both G_{1} and G_{2} are regular  
Both G_{1} and G_{2} are contextfree but neither of them is regular 
Discuss it
Say, while = w, expr =E, stmt = S, other = o
Here, we can write a right linear grammar for G_{1} as
S > w(E)S
S > o
E>ID
So, L(G_{1}) is regular.
Now for G_{2} also we can write a right linear grammar:
S > w(E)S
E > E+E
E > E*E
S > o
So, L(G_{2}) is regular
Question 50 
The automation which recognizes the language L(P) ∩ L(Q) is :
a  
b  
c  
d 
Discuss it
In P(L) after c both a and b could be accepted, but in Q(L) only a could be accepted. Option C drops out here,
In P(L) aa could be accepted, but in Q(L) aa could not be accepted. Option B drops out here,
Question 51 
The Branch coverage is
3/4  
2/3  
1/2  
3/8 
Discuss it
Question 52 
The set of activities that lie on the critical path are
T1, T2, T8, T10  
T1, T3, T8, T10  
T1, T2, T3, T4, T5, T6, T7, T8, T9, T10  
T1, T4, T5, T7, T8, T10 
Discuss it
Question 53 
IF ((A > B) AND (C > D)) THEN A = A + 1 B = B + 1 ENDIFThe cyclomatic complexity of the pseudocode is
2  
3  
4  
5 
Discuss it
where π is the number of decision points in the program, and s is the number of exit points.
Question 54 
wait (wrt) writing is performed signal (wrt) wait (mutex) readcount = readcount + 1 if readcount = 1 then S1 S2 reading is performed S3 readcount = readcount  1 if readcount = 0 then S4 signal (mutex)The values of S1, S2, S3, S4, (in that order) are
signal (mutex), wait (wrt), signal (wrt), wait (mutex)  
signal (wrt), signal (mutex), wait (mutex), wait (wrt)  
wait (wrt), signal (mutex), wait (mutex), signal (wrt)  
signal (mutex), wait (mutex), signal (mutex), wait (mutex) 
Discuss it
Question 55 
6.9 × 10^{6} × e^{20}  
1.02 × 10^{6} × e^{20}  
6.9 × 10^{3} × e^{20}  
1.02 × 10^{3} × e^{20} 
Discuss it
So in 45 min requests received = 15
So, lemda = 15
So, using poission distribution formula: P(x; μ) = (e^{μ}) (μ^{x}) / x!
p(one request) + p(3 request) + p(5 request) = p(1; 15) + p(3; 15) + p(5; 15) = 6.9 * 10^{3} * e^{15} = 1.02 × 10^{6} × e^{20}
Question 56 
0.194  
0.233  
0.514  
0.981 
Discuss it
page fault service time =300 (if there is dirty page)
probability of page fault =p
probability of being dirty is =p
so total page fault service time(ps) = p(300)+ (1p)(100) = 200p+100
now given EAT =3 so
EAT = p(page fault service time +memory access time)+(1p) memory access time
3= p*page fault service time + memory access time
3 =p(200p+100)+1= 200p^{2}+100p+1
p=0.0194
Question 57 
a1 b1
a2 b2
a3 b2
a4 b1
The contents of the text file t2 txt containing five lines are as follows :
a1 c1
a2 c2
a3 c3
a4 c3
a5 c4
Consider the following Bourne shell script :
awk  F ' ' ' {Print $1, $2} ' t1.txt  while read a b ; do awk v aV = $ a  v by = $b  F ' ' aV = = $1 (print aV, bV, $2 ) ' t2.txt doneWhich one of the following strings will NOT be present in the output generated when the above script in run?
(Note that the given strings may be substrings of a printed line.)
"b1 c1"  
"b2 c3"  
"b1 c2"  
"b1 c3" 
Discuss it
Question 58 
A.
B.
C.
D.
A  
B  
C  
D 
Discuss it
Using DV routing protocol, F > D > B yields distance as 20 that drops option B and D.
Question 59 
1000010111 and Differential Manchester respectively  
0111101000 and Differential Manchester respectively  
1000010111 and Integral Manchester respectively  
0111101000 and Integral Manchester respectively 
Discuss it
Question 60 
5  
4.45  
3.45  
0 
Discuss it
Number of packets (+backlog from previous)  Backlog 
6  1 
9+1  5 
3+5  3 
7+3  5 
2+5  2 
2+2  0 
2  0 
3  0 
4  0 
6  1 
1+1  0 
10  5 
7+5  7 
5+7  7 
8+7  10 
3+10  8 
6+8  9 
2+9  6 
9+6  10 
5+10  10 
Total Backlog  89 
Question 61 
3  
4.26  
4.53  
5.26 
Discuss it
{Same as above}
So, Total Hops = 2 * 2.267 = 4.53
Question 62 
1 Mbps  
100/11 Mbps  
10 Mbps  
100 Mbps 
Discuss it
T_{t} =1000 bytes/10Mbps =800μs.
Polling delay is = 80 μs
Efficiency=800/(800+80)= 10/11
Maximum throughput is =(10/11) * 10 Mbps= 100/11 Mbps
Question 63 
50  
100  
150  
200 
Discuss it
Here σ_{A<=100} signifies a selection query which selects all the tuples which has attribute A’s value less than or equal to 100. Attribute A’s values in the range [0,500] are uniformaly distributed among 1000 tuples. So first when we arrange all 1000 tuples in the ascending order of attribute A’s value, then first 200 tuples will have attribute A’s values in the range [0,99] and next 200 tuples will have attribute A’s values in the range [100,199] and so on .So number of tuples returned by the given selection query is 200.
This solution is contributed by Nirmal Bharadwaj.
Question 64 
A  
B  
C  
D 
Discuss it
Question 65 
i. If A ↠ B and A ↠ C then A → BC
ii. If A → B and A → C then A ↠ BC
iii. If A ↠ BC and A → B then A → C
iv. If A → BC and A → B then A ↠ C
Exactly how many of the above implications are valid?
0  
1  
2  
3 
Discuss it
Question 66 
 bSchema = (bname, bcity, assets)
 aSchema = (anum, bname, bal)
 dSchema = (cname, anumber)
П_{cname} (σ_{bal < 0} (σ_{bcity = "Agra"} branch ⋈ account) ⋈ depositor)  
П_{cname} (σ_{bcity = "Agra"}branch ⋈ (σ_{bal < 0} account ⋈ depositor))  
П_{cname} (σ_{bcity = "Agra"} branch ⋈ σ_{bcity = "Agra" ⋀ bal < 0} account) ⋈ depositor)  
П_{cname} (σ_{bcity = "Agra" ⋀ bal < 0} account ⋈ depositor)) 
Discuss it
Question 67 
IMAP(i), FTP(ii), HTTP(iii), DNS(iv), POP3(v)  
FTP(i), POP3(ii), SMTP(iii), HTTP(iv), IMAP(v)  
POP3(i), SMTP(ii), DNS(iii), IMAP(iv), HTTP(v)  
SMTP(i), HTTP(ii), IMAP(iii), DNS(iv), FTP(v) 
Discuss it
Question 68 
zdp  
fpq  
qwA  
oze 
Discuss it
Question 69 
A  
B  
C  
D 
Discuss it
Question 70 
A  
B  
C  
D 
Discuss it
Question 71 
(a(ba)* + b(ab)*)(a + b)^{+}  
(a(ba)* + b(ab)*)*(a + b)*  
(a(ba)* (a + bb) + b(ab)*(b + aa))(a + b)*  
(a(ba)* (a + bb) + b(ab)*(b + aa))(a + b)^{+} 
Discuss it
B empty may be accepted, not in given
Question 72 
The maximum utilization of the token ring when t_{t} =3 ms, t_{p} = 5 ms, N = 10 is
0.545  
0.6  
0.857  
0.961 
Discuss it
t_{t}= 3 ms
t_{p} =5 ms and n=10,
efficiency = 3/(3+.5)= .857
Question 73 
The maximum utilization of the token ring when t_{t} = 5 ms, t_{p} = 3 ms, N = 15 is :
0.545  
0.655  
0.9375  
0.961 
Discuss it
Question 74 
(1 + (1 + 8c)^{1/2})/2c  
(1  (1 + 8c)^{1/2})/2c  
2  
2/(2c1) 
Discuss it
Question 75 
ad' + (ac)' + bc'd
Which of the following Karnaugh Maps correctly represents the expression?
A  
B  
C  
D 
Discuss it
Question 76 
ad' + (ac)' + bc'd
Which of the following expressions does not correspond to the Karnaugh Map obtained for the above expression??
c'd'+ ad' + abc' + (ac)'d  
(ac)' + c'd' + ad' + abc'd  
(ac)' + ad' + abc' + c'd  
b'c'd' + acd' + (ac)' + abc' 
Discuss it
Question 77 
P_{a} and P_{b} are adjacent to each other with respect to their xcoordinate  
Either P_{a} or P_{b} has the largest or the smallest ycoordinate among all the points  
The difference between xcoordinatef P_{a} and P_{b} is minimum  
None of the above 
Discuss it
Question 78 
Θ(n)  
Θ(nlogn)  
Θ(nlogsup>2n)  
Θ(n^{2}) 
Discuss it
Question 79 
11, 139, 170, 178, 181, 184, 201, 265  
10, 138, 170, 178, 181, 185, 201, 265  
10, 139, 169, 178, 181, 184, 201, 265  
10, 138, 170, 178, 181, 185, 200, 265 
Discuss it
Question 80 
9  
10  
11  
12 
Discuss it