Gate 2007-IT
Question 1 |
7/8 | |
1/2 | |
7/16 | |
5/32 |
The probability of obtaining heads
= (1/2)(5/8) + (1/2)(1/4)
= (5/16) + (1/8)
= 7/16
Question 2 |

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Question 3 |
weight (u, v) < 12 | |
weight (u, v) ≤ 12 | |
weight (u, v) > 12 | |
weight (u, v) ≥ 12 |
weight (u,v) ≥ 12
Lets check other options,
A) If the weight (u,v) < 12
Minimum weight of (s,v)
= weight of (s,v) + weight of (u,v)
= 53 + (<12)
= less than 65, which is not possible because shortest path from s to v has weight 65.
Similarly, we can check for option (B) and (C).
Question 4 |
Iteration size | |
Cost | |
Adopted process such as Rational Unified Process or Extreme Programming | |
Risk |
Question 5 |
Expert system | |
DB repository | |
Aircraft flight controller | |
Signal processing |
Question 6 |
1.83 | |
2 | |
3 | |
6 |
For a non-pipelined processor each instruction takes 12 cycles.
So for n instructions total execution time be 12 × n = 12n
For a pipelined processor each instruction takes
max (3, 2, 5, 4, 6, 2) =6
So for n instructions total execution time be,
(1×6 + (n-1) × 1) × 6
= (6 + n - 1) × 6
= (5 + n) × 6
= 30 + 6n

So, if n is very large,

Question 7 |
11, 00 | |
01, 10 | |
10, 01 | |
00, 11 |
So, 00 input cause indeterminate state which may lead to oscillation.
Question 8 |

X1 = b, X2 = 0, X3 = a | |
X1 = b, X2 = 1, X3 = b | |
X1 = a, X2 = b, X3 = 1 | |
X1 = a, X2 = 0, X3 = b
|
If we put
X1 = b
X2 = 0
X3 = a
Then we get,
F = ab
Question 9 |
L (D) ⊂ L (G) | |
L (D) ⊃ L (G) | |
L (D) = L (G) | |
L (D) is empty |
For example, by converting NFA to DFA language will not be changed.
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 P1 and P2 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 |
Question 11 |

16 | |
19 | |
20 | |
37 |

At t = 8

At t = 10

At t = 11

J7 can be finished at t = 19.
Question 12 |
0 | |
4 | |
7 | |
8 |
0200 - Page fault, address till 299 in memory
0430 - Page fault, address till 529 in memory
0499 - No page fault
0510 - No page fault
0530 - Page fault, address till 629 in memory
0560 - No page fault.
0120 - Page fault, address till 219 in memory
0220 - Page fault, address till 319 in memory
0240 - No page fault
0260 - No page fault
0320 - Page fault, address till 419 in memory
0410 - No page fault
So, total page faults = 7.
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 |
The timeout value cannot be fixed for entire duration as it will turn timer to static timer, we need dynamic timer for timeout.
Statement-II: It is True.
Basic algorithm, Jacobson's algorithm, Karl's modification; these three algorithms are to be appropriate to RTT estimation algorithm used to set timeout value dynamically.
Statement-III: It 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 |
Consider a TCP connection in a state where there are no outstanding ACKs. The sender sends two segments back to back. The sequence numbers of the first and second segments are 230 and 290 respectively. The first segment was lost, but the second segment was received correctly by the receiver. Let X be the amount of data carried in the first segment (in bytes), and Y be the ACK number sent by the receiver. The values of X and Y (in that order) are
60 and 290 | |
230 and 291 | |
60 and 231 | |
60 and 230 |
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
|
ii) It uses the P-Box permutation.
Statement-I is false, II is true.
Question 16 |
5 | |
8 | |
12 | |
16 |
ap-1 mod p = 1
Here, p = 17
So, p-1 = 16 is the answer.
Question 17 |
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Recurrence relation T(n) = T(n/2) + O(1)
T(n) = O(log n)
Question 18 |
A firewall is to be configured to allow hosts in a private network to freely open TCP connections and send packets on open connections. However, it will only allow external hosts to send packets on existing open TCP connections or connections that are being opened (by internal hosts) but not allow them to open TCP connections to hosts in the private network. To achieve this the minimum capability of the firewall should be that of
A combinational circuit | |
A finite automaton | |
A pushdown automaton with one stack | |
A pushdown automaton with two stacks
|
Turing machine can do everything as the normal computer can do, so firewall can be created by the TM.
Question 19 |

0 | |
1 | |
2 | |
3 |
Question 20 |

The Title and Content elements | |
The Content and TOC elements | |
The Title and TOC elements | |
The Title, Content and TOC elements
|
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)
|
RHS = if P(x) holds for all x, then Q(x) holds for all x
LHS ⇒ RHS (✔)
RHS ⇒ LHS (️❌)
Question 22 |

(iv) only | |
(iii) and (iv) only | |
(ii), (iii) and (iv) only | |
(i), (ii), (iii) and (iv) |
Question 23 |
(i) and (iii) | |
(ii) and (iv) | |
(i) and (iv) | |
(iii) and (iv) |
The given condition can be satisfied by
(iii) (145, 265) → 5 ≤ 5, 4 < 6 and 1< 2
(iv) (0, 153) → 0 < 3
Question 24 |
d[u] < d[v] | |
d[u] < f[v] | |
f[u] < f[v] | |
f[u] > f[v] |

Option A:
d[u]
d[u]
f[u]
Question 25 |
1 | |
2 | |
3 | |
n |
→ If we create a different spanning tree by removing one edge at a time.
→ For cycle required 3 minimum edges. So there is at least 3 spanning trees in such a graph.
Question 26 |
Consider n jobs J1, J2,......Jn such that job Ji has execution time ti and a non-negative integer weight wi. The weighted mean completion time of the jobs is defined to be
where Ti is the completion time of job Ji. Assuming that there is only one processor available, in what order must the jobs be executed in order to minimize the weighted mean completion time of the jobs?
Non-decreasing order of ti | |
Non-increasing order of wi | |
Non-increasing order of witi | |
None-increasing order of wi/ti |
So answer is the non-increasing order of wi/ti.
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) |
→ Let n=3, then it is terminated in 2nd iteration.
→ Let n=5, then sequence is 5→14→7→20→10 and it will repeat.
→ Any number with factor 5 and 2 leads to infinite recursion.
So, (iv) is True and (iii) is False.
Question 28 |
5 | |
6 | |
7 | |
10 |
Probability of collision for each entry = 1/20
After inserting X values then probability becomes 1/2
i.e., (1/20)X = 1/2
X = b
Question 29 |
35 | |
64 | |
128 | |
5040 |
Smaller values 90, 80 and 70 are visited in order
i.e., 7!/(4!3!) = 35
Question 30 |
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 |
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 ("%dn", maxsum); }
9 | |
8 | |
7 | |
6 |
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 |
push 2
pop 2
pop 5
5 * 2 = 10
push 10
push 3
push 3
push 2
pop 2
pop 3
3 + 2 = 5
push 5
pop 5
pop 3
3 * 5 = 15
push 15
pop 15
pop 10
10 + 15 = 25
push 25
Finally, pop 25 and print 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 |
'i' is a global variable. Then in main( ) a local variable 'j' as integer declared, i.e., j=60 and global varible 'i' is initialized to 50 by i=50.
Now procedure f called and values of 'i' and 'j' are passed to it, i.e., in f(i, j)→f(x, y).
Content of memory location of 'i' (here 50) is copied to memory location of x (which is different from i) and content of memory location of 'j' (here 60) is copied to memory location of y. Then in f(x, y) i=100 changes the global i to 100, x=10 changes the local x from 50 to 10 and y=y+i means y=60+100=160. Now, when return back to main, i & j will be 100 & 60 respectively.
Call by reference:
Now procedure f called and passed reference of i and j to it, i.e., in f(i,j)→f(x,y). x and y are pointing to same memory location of i and j respectively. So i=100 changes the global i to 100 and x=10 means as well as global i=10 (as the i being passed is the global variable and x and i share the same address).
y=y+i means y=60+10=70 and this changes the value of j also to as j and y have the same addresses. Now when return to main, i and j will be 10 and 70.
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
|
Question 35 |
Early, late, decrease, increase | |
Late, early, increase, decrease | |
Late, early, decrease, increase | |
Early, late, increase, decrease |
Dynamic scoping requires late binding (during execution time).
Late binding decreases efficiency as this binding needs to be done at run time.
Question 36 |
The floating point unit of a processor using a design D takes 2t cycles compared to t cycles taken by the fixed point unit. There are two more design suggestions D1 and D2. D1 uses 30% more cycles for fixed point unit but 30% less cycles for floating point unit as compared to design D. D2 uses 40% less cycles for fixed point unit but 10% more cycles for floating point unit as compared to design D. For a given program which has 80% fixed point operations and 20% floating point operations, which of the following ordering reflects the relative performances of three designs? (Di > Dj denotes that Di is faster than Dj)
D1 > D > D2 | |
D2 > D > D1 | |
D > D2 > D1 | |
D > D1 > D2 |
D = 0.8 × t + 0.2 × 2t = 1.2t
D1 = 0.8 × 1.3t + 0.2 × 0.7 × 2t = 1.32t
D2 = 0.8 × 0.6t + 0.2 × 1.1 × 2t = 0.92t
⇒ D2 > D > D1
Question 37 |
3 | |
18 | |
20 | |
30 |

So memory block 18 is not in the cache.
Question 38 |
1 | |
a’ + b’ + c’ + d’ | |
a’ + b + c’ + d’ | |
a’ + b’ + c + d’ |
= ((ab)'c)' + ((a'c)'d)' + ((bc)'d)' + (ad)'
= ab + c' + a'c + d' + bc + d' + a' + d'
= ab + c' + a'c + bc + a' + d'
= ab + c' + bc + a' + d'
= b + c' + bc + a' + d'
= a' + b + c' + d'
Question 39 |
Data forwarding techniques can be used to speed up the operation in presence of data dependencies. Consider the following replacements of LHS with RHS.
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) |
(ii) Is false. Because R2 gets the correct data in both LHS and RHS, but Loc is not updated in RHS.
(iii) Is false. Because R2 is writing last in LHS, but not in RHS.
(iv) Is true. The first write to Loc in LHS is useless as it is overwritten by the next write.
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 |
If an instruction is in execution phase then any other instructions cannot be in the execution phase. So, atleast 7 clock cycles will be taken.
Now, it is given that between two instructions latency or delay should be there based on their operation.
= 1100000000010010.00100101 - 0000010111001110.10100000
= 1011101001000011.10000101
= 1011101000011.100001010
= (135103.412)O
Question 43 |
0 | |
1 | |
2 | |
3 |
The no. of bits that can be corrected is

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 |
For Avg. seek time:
Given that, time to move between successive tracks is 1ms.
Time to move from track1 to track1 = 0ms
Time to move from track1 to track2 = 1ms
Time to move from track1 to track3 = 2ms
⋮ Time to move from track1 to track500 = 499ms
∴ Avg. seek time = 0+1+2+...+499/500 = 249.5ms
Avg. rotational delay:
600 rotations in 60sec.
One rotation takes 60/600 s = 100ms ∴ Avg. rotational delay = 100/2 = 50ms
Data transfer time:
In one rotation, we can read data on one complete track
= 100 × 500 = 50,000B data is read in one complete rotation
One complete rotation takes 100ms.
100ms → 50,000B
250B → 100/50000 × 250 = 0.5ms
∴ Avg. time to transfer = 249.5 × 50 + 0.5 = 300ms
Question 45 |

(The line T in the following figure is permanently connected to the ground)
0000 | |
0111 | |
1111 | |
None of these |

Since the problem is in the link T which is connected as input to NOR gate. So to check link T we have to make the output dependent on T by deactivating link M. So to deactivate link M, the output at M should be 0, as link M is input to NOR gate. So, to output at M as 0,
X1 = 1
X2 = 1
X3 = 1
X4 = 0
∴ None of the given option is correct.
Question 46 |

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 |
(x + z)* + (x + z)* y(y + z)*
Regular expression for G2:
(y + z + xy)*
Question 47 |

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 |
Question 48 |
(i), (ii), and (iii) | |
(ii), (v), and (vi) | |
(ii), (iii), and (iv) | |
(i), (iii), and (iv) |
S → xB
S → xxBB
S → xxyB
S → xxyyS
S → xxyyxB
S → xxyyxy
(iii) xyxy
S → xB
S → xyS
S → xyxB
S → xyxy
(iv) yxxy
S → yA
S → yxS
S → yxxB
S → yxxy
Question 49 |

G1 is context-free but not regular and G2 is regular | |
G2 is context-free but not regular and G1 is regular | |
Both G1 and G2 are regular | |
Both G1 and G2 are context-free but neither of them is regular |
Similarly, in right linear grammar, non-terminal appears at the right most position.
Here we can write a right linear grammar for G1 as
S → w(E
E → id)S
S → o
(w-while, o-other)
So, L(G1) is regular.
Now for G2 also we can write a right linear grammar:
S → w(E
E → id)S
E → id+E
E → id*E
S → o
making its language regular.
So, both G1 and G2 have an equivalent regular grammar. But given in the question both these grammars are neither right linear nor left linear and hence not a regular grammar. So, (D) must be the answer.
Question 50 |

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Question 51 |

0.45 | |
0.63 | |
0.84 | |
0.95 |

The probability or reliability that the product will work for a defined period of time without failure is given by

Question 53 |

3/4 | |
2/3 | |
1/2 | |
3/8 |
Question 54 |
Consider the CPM activity chart where an arc connecting two milestones is labeled with a task identifier and the time taken in days. For example in order to go from A to B, task T1 takes 180 days. A dashed line depicts an additional dependency that is equivalent to a zero time task.
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 |
Question 55 |
IF ((A > B) AND (C > D)) THEN A = A + 1 B = B + 1 ENDIFThe cyclomatic complexity of the pseudo-code is
2 | |
3 | |
4 | |
5 |
Question 56 |
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)
|
S2: After readcount has been updated, UP on mutex.
S3: DOWN on mutex to update readcount.
S4: If readcount is zero, i.e., no reader is reading, UP on wrt to allow some writer to write.
Question 57 |
In a multi-user operating system on an average, 20 requests are made to use a particular resource per hour. The arrival of requests follows a Poisson distribution. The probability that either one, three or five requests are made in 45 minutes is given by :
6.9 × 106 × e-20 | |
1.02 × 106 × e-20 | |
6.9 × 103 × e-20 | |
1.02 × 103 × e-20 |
So, λ=15

Question 58 |
A demand paging system takes 100 time units to service a page fault and 300 time units to replace a dirty page. Memory access time is 1 time unit. The probability of a page fault is p. In case of a page fault, the probability of page being dirty is also p. It is observed that the average access time is 3 time units. Then the value of p is
0.194 | |
0.233 | |
0.514 | |
0.981 | |
0.0194 |
(if the page is not dirty)
Page fault service time = 300
(if there is dirty page)
Probability of page fault = P
Probability pf page being dirty = P
So, total page fault service time
= P(300) + (1 - P)100
= 300P + 100 - 100P
= 300P + 100
Now given,
Effective average memory access time = 3
So,
3 = m + P × total page fault service time
= 1 + P(200P + 100)
= 1+ 200P2 + 100P
⇒ 200P2 + 100P - 2 = 0
⇒ P = 0.0194
Question 59 |
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" |
Question 60 |


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Using distance vector routing protocol, F→D→B yeilds distance as 20 which eliminates option (B) and (D).
Question 61 |

1000010111 and Differential Manchester respectively | |
0111101000 and Differential Manchester respectively | |
1000010111 and Integral Manchester respectively | |
0111101000 and Integral Manchester respectively
|
As per IEEE (B) is correct.
As per GE Thomas (A) is correct.
As we usually follow IEEE thus (B) is correct.
Question 62 |
Let us consider a statistical time division multiplexing of packets. The number of sources is 10. In a time unit, a source transmits a packet of 1000 bits. The number of sources sending data for the first 20 time units is 6, 9, 3, 7, 2, 2, 2, 3, 4, 6, 1, 10, 7, 5, 8, 3, 6, 2, 9, 5 respectively. The output capacity of multiplexer is 5000 bits per time unit. Then the average number of backlogged of packets per time unit during the given period is
5 | |
4.45 | |
3.45 | |
0 |
If the no. of packets transmitted is larger than 5 then the extra packets are backlogged. This means gets added to the next number and further backlog is calculated.

Average no. of backlogged packets = 89/20 = 4.45
Question 63 |
A group of 15 routers are interconnected in a centralized complete binary tree with a router at each tree node. Router j communicates with router j by sending a message to the root of the tree. The root then sends the message back down to router j. The mean number of hops per message, assuming all possible router pairs are equally likely is
3 | |
4.26 | |
4.53 | |
5.26 |
Message goes up from sender to root,
Average hops message goes to root
= (3×8) + (2×4) + (1×2) + (0×1)/15
= 2.267
Here, (3×8) represents 3 hops and 8 routers for bottom most level and so on.
Step 2:
Similarly, average hops when message comes down
= (3×8) + (2×4) + (1×2) + (0×1)/15
= 2.267
So, Total hops = 2×2.267 = 4.53
Question 64 |
1 Mbps | |
100/11 Mbps | |
10 Mbps | |
100 Mbps |

Question 65 |
Consider a selection of the form σA≤100(r), where r is a relation with 1000 tuples. Assume that the attribute values for A among the tuples are uniformly distributed in the interval [0, 500]. Which one of the following options is the best estimate of the number of tuples returned by the given selection query ?
50 | |
100 | |
150 | |
200 |
Values for A among the tuples are uniformly distributed in the interval [0, 500]. This can be split to 5 mutually exclusive and exhaustive intervals of same width of 100 ([0-100], [101-200], [201-300], [301-400], [401-500], 0 makes the first interval larger - this must be typ0 in this question) and we can assume all of them have same number of values due to uniform distribution. So no. of tuples with A value in first interval should be,
Total no. of tuples/5 = 1000/5 = 200
Question 66 |
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i) Exclusive locks should be released after the commit.
ii) No locking can be done after the first unlock.
(A) Wrong because to write x you need Exclusive Lock.
(B) Wrong because violating the 1st requirement.
(C) Correct.
(D) Wrong because violating the 1st requirement.
Question 67 |
0 | |
1 | |
2 | |
3 |
if A→B then A→→B holds but reverseis not true.
So,
(i) False.
(ii) True.
(iii) False.
(iv) True.
Question 68 |
-
- b-Schema = (b-name, b-city, assets)
- a-Schema = (a-num, b-name, bal)
- d-Schema = (c-name, a-number)
Пc-name (σbal < 0 (σb-city = “Agra” branch ⋈ account) ⋈ depositor) | |
Пc-name (σb-city = “Agra”branch ⋈ (σbal < 0 account ⋈ depositor)) | |
Пc-name (σb-city = “Agra” branch ⋈ σb-city = “Agra” ⋀ bal < 0 account) ⋈ depositor) | |
Пc-name (σb-city = “Agra” ⋀ bal < 0 account ⋈ depositor)) |
Options (C) and (D) are invalid as there is no b-city column in a-schema.
Question 69 |
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) |
(v) FTP needs two ports, 20 for data and 21 for control.
Hence, only (D) matches.
Question 70 |
zdp | |
fpq | |
qwA | |
oze |
10100011 00110111 11101001 10101011
So, in total we have 32 bits. And for base 64we need 6 digits of binary no. to represent one digit of base 64 no.
So lets padd 4 bits on RHS, so that total digits will become 36 and we can separate then as group of 6 digits each.

Now, the longest substring will be from checking option is 'fpq'.
Question 71 |
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But option (B), (C), (D) accepts aba, which do not contain aa or bb as substring.
Hence, (A) is correct.
Question 72 |
Which one of the regular expressions given below defines the same language as defined by the regular expression R?
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(C) It is not accepting 'abb' which is in language.
(D) It is not accepting 'aa' and 'bb' which is in language.
Question 73 |
0.545 | |
0.6 | |
0.857 | |
0.961 |
Question 74 |
0.545 | |
0.655 | |
0.9375 | |
0.961 |
Question 75 |
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Then,
x1 = c ⋅ (x0)2 - 2 = 1 ⋅ (1)2 - 2 = -1
x2 = c ⋅ (x1)2 - 2 = 1 ⋅ (-1)2 - 2 = -1
So, the value converges to -1, which is equal to

Exactly, only (B) is answer. As all the term of x converges to -1.
Question 76 |
(i) 0nly | |
(i) and (ii) only | |
(i), (ii) and (iii) only | |
(i), (ii), (iii) and (iv) |
From the recurrence we should have c(xn)2 - xn - 2 < 0
For all the above values of c we have above equation as negative.
Question 77 |
Hence, option (A) matches.
Consider the following expression
ad' + (ac)' + bc'd
Which of the following expressions does not correspond to the Karnaugh Map obtained for the above expression??
Question 78 Explanation: Let's check for option (C): a'c' + ad' + abc' + c'd ![]() Not equivalent to the K-map, we get in previous question.
Let P1, P2,..... , Pn be n points in the xy-plane such that no three of them are collinear. For every pair of points Pi and Pj, let Lij be the line passing through them. Let Lab be the line with the steepest gradient amongst all n(n -1)/2 lines. Which one of the following properties should necessarily be satisfied ?
Question 79 Explanation: Draw the line from the first to third point, and look at whether the middle point is above, at or below it. If the middle point is on the line, all slopes are equal. If the middle point is above the line, then the slope from the first to the second point is the highest. If the middle point is below the line, then the slope from the middle point to the second is highest. In any case, you can find a greater slope betweem two adjacent points.
Let P1,P2,…,Pn be n points in the xy-plane such that no three of them are collinear. For every pair of points Pi and Pj, let Lij be the line passing through them. Let Lab be the line with the steepest gradient among all n(n−1)/2 lines. The time complexity of the best algorithm for finding Pa and Pb is
Question 80 Explanation: ![]() For gradient to be maximum x2-x1 should be minimum. So, sort the points (in Θ(n logn) time) according to n coordinate and find the minimum difference between them (in Θ(n) time). ∴ Complexity = Θ(n logn + n) = Θ(n logn)
The head of a hard disk serves requests following the shortest seek time first (SSTF) policy. The head is initially positioned at truck number 180. Which of the request sets will cause the head to change its direction after servicing every request assuming that the head does not change direction if there is a tie in SSTF and all the requests arrive before the servicing starts?
Question 81 Explanation: ![]() Hence, correct option is (B).
The head of a hard disk serves requests following the shortest seek time first (SSTF) policy. The head is initially positioned at track number 180. What is the maximum cardinality of the request set, so that the head changes its direction after servicing every request if the total number of tracks are 2048 and the head can start from any track?
Question 82 Explanation: We need two conditions to satisfy: 1) The alternating direction with SSTF policy. 2) Maximize the no. of requests. The first condition can be satisfied by not having two requests in the equal distance from the current location. As shown below, we must not have request located in the circle marked positions. Now to maximize the no. of request we need the requests to be located as compact as possible, which can be done by just placing the request in the next position after the circle marked position in particular direction (the direction in which the head need to move). Now to satisfy the 1st criteria: ![]() Seek length sequence for maximum cardinality and alternating head movements: → 1, 3, 7, 15, ... → or, 21-1, 22-1, 23-1, 24-1, ... → We have 2048 tracks so, maximum swing (seek length) can be 2047. → Which corresponds to a seek length of 211-1 in the 11th service.
Question 83 Explanation: It is B+ Tree. After inserting K15 we get, ![]() Now, we insert K25, which gives ![]() So, we see in the final tree only (K20, K25) is present. Hence, 1 (Answer).
Question 84 Explanation: The B+ tree in previous question we get was, ![]() Now after deleting 50 we get B+ tree as ![]() Hence, correct statement is only (i) & (ii).
There are 84 questions to complete.
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