2010

Question 1
   
A
|S| = 2|T|
B
|S| = |T| - 1
C
|S| = |T|
D
|S| = |T| + 1
Question 1 Explanation: 

id= no. of vertices of degree ‘d’ in ‘G’
Eg:

No. of vertices with degree ‘2’ = 3
ξ(G')=3×2='6' i.e., sum of degrees
By Handshaking Theorem,
The sum of degrees would be equal to twice the no. of edges
|V|=2|E|
It is given that ξ(G)=ξ(S) then
Sum of degrees of vertices in G is equal to sum of degrees of vertices in S
i.e., 2*(no. of edges in G)=2*no. of edges in S no. of edges in G=no. of edges in S
Eg:

ξ(G)=(2×2)+(2×3)=4+6=10

ξ(S)=2×5=10
You can observe that, though no. of vertices are different, but still no. of edges are same.
Question 2
Newton-Raphson method is used to compute a root of the equation x2 - 13 = 0 with 3.5 as the initial value. The approximation after one iteration is
A
3.575
B
3.676
C
3.667
D
3.607
Question 2 Explanation: 
Note: Out of syllabus.
(Numerical methods are not explicitly mentioned in the GATE syllabus after 2017)
dx=dz

Newton Raphson’s Method: (Root finding algorithm)
It is used for finding better approximations to the roots of real valued function.
x1=xn-n+f(xn)/ f'(xn)
xn+1=xn-f(xn)/ f'(xn)
xn=previous root
f(xn)=function of real values
f'(xn)=derivative of f(x)
f(x)=x2-13
f'(x)=2x
Initial value x0=3.5(given)
x1=x0-f(x0)/f'(x0)=3.5-((3.5)2-13/ 2(3.5))=3.607
Question 3
What is the possible number of reflexive relations on a set of 5 elements?
A
210
B
215
C
220
D
225
Question 3 Explanation: 
Let set = ‘A’ with ‘n’ elements,
Definition of Reflexive relation:
A relation ‘R’ is reflexive if it contains xRx ∀ x∈A
A relation with all diagonal elements, it can contain any combination of non-diagonal elements.
Eg:
A={1, 2, 3}


So for a relation to be reflexive, it should contain all diagonal elements. In addition to them, we can have possible combination of (n2-n)non-diagonal elements (i.e., 2n2-n)
Ex:
{(1,1)(2,2)(3,3)} ----- ‘0’ non-diagonal element
{(1,1)(2,2)(3,3)(1,2)} ----- ‘1’ non-diagonal element
{(1,1)(2,2)(3,3)(1,2)(1,3)} “
___________ “
___________ “
{(1,1)(2,2)(3,3)(1,2)(1,3)(2,1)(2,3)(3,1)(3,2)} (n2-n) diagonal elements
____________________
Total: 2n2-n
For the given question n = 5.
The number of reflexive relations =2(25-5)=220
Question 4
Consider the set S = {1, ω, ω2}, where ω and ω2 are cube roots of unity. If * denotes the multiplication operation, the structure (S,  *) forms
A
A group
B
A ring
C
An integral domain
D
A field
Question 4 Explanation: 
A Group is an algebraic structure which satisfies
1) closure
2) Associativity
3) Have Identity element
4) Invertible
Over ‘*’ operation the S = {1, ω, ω2} satisfies the above properties.
The identity element is ‘1’ and inverse of 1 is 1, inverse of ‘w’ is 'w2' and inverse of 'w2' is 'w'.
Question 5
 
A
0
B
e-2
C
e-1/2
D
1
Question 5 Explanation: 
Question 6
     
A
m2+m4+m6+m7
B
m0+m1+m3+m5
C
m0+m1+m6+m7
D
m2+m3+m4+m5
Question 6 Explanation: 
Convert PQ + QR' + PR' into canonical form
= PQR + PQR' + PQR' + P'QR' + PQR' + PQ'R'
= PQR + PQR' + P'QR' + PQ'R'
=m7 + m6 + m2 + m4
Question 7
A main memory unit with a capacity of 4 megabytes is built using 1M 1-bit DRAM chips. Each DRAM chip has 1K rows of cells with 1K cells in each row. The time taken for a single refresh operation is 100 nanoseconds. The time required to perform one refresh operation on all the cells in the memory unit is
A
100 nanoseconds
B
100*210 nanoseconds
C
100*220 nanoseconds
D
3200*220 nanoseconds
Question 7 Explanation: 
Each chip capacity = 1M x 1-bit
Required capacity = 4MB
Number of chips needed = 4M*8 bits / 1M x 1-bit = 32 (1M x 1-bit)/(1M x 1-bit) = 32
Irrespective of the number of chips, all chips can be refreshed in parallel.
And all the cells in a row are refreshed in parallel too. So, the total time for refresh will be number of rows times the refresh time of one row.
Here we have 1K rows in a chip and refresh time of single row is 100ns.
So total time required =1K×100
=100 ×210 nanoseconds
Question 8
P is a 16-bit signed integer. The 2's complement representation of P is (F87B)16. The 2's complement representation of 8*P is
A
(C3D8)16
B
(187B)16
C
(F878)16
D
(987B)16
Question 8 Explanation: 
(F87B)16 is 2's complement representation of P.
(F87B)16=(1111 1000 0111 1011)2. (It is a negative number which is in 2's complement form)
P=1111 1000 0111 1011 (2's complement form)
8 * P = 23* P= 1100 0011 1101 1000. ( NOTE: Left shift k times is equivalent to Multiplication by 2k)
Hence, 1100 0011 1101 1000 is 2's complement representation of 8P.
1100 0011 1101 1000 = (C3D8)16.
Question 9
     
A
B
P⊕Q⊕R
C
P+Q+R
D
Question 9 Explanation: 
f= P’Q’ R + P’Q R’ + PQ’ R’ + PQR
= (P’Q’ + PQ)R + (P’Q+PQ’)R’
= (P⊕Q)’ R + (P⊕Q)R’
= (P⊕Q⊕R)
Question 10
In a binary tree with n nodes, every node has an odd number of descendants. Every node is considered to be its own descendant. What is the number of nodes in the tree that have exactly one child?
A
0
B
1
C
(n-1)/2
D
n-1
Question 10 Explanation: 
Given a Binary Tree with n nodes.
Every node has an odd number of descendants.
Also given every node is considered to be its own descendant.

― This is even number of descendants (2), because A is also considered as a descendant.

― This is odd number of descendants (3), A, B, C are descendants here.
Condition satisfied – odd number, but number of nodes in a tree that has exactly one child is 0.
Question 11
 
A
2 2
B
2 1
C
0 1
D
0 2
Question 11 Explanation: 

Both pointers points to j = 1
now *p = 2
where j is updated with value 2.
printf (i, j) i = 0, j = 2
Question 12
Two alternative packages A and B are available for processing a database having 10k records. Package A requires 0.0001n2 time units and package B requires 10nlog10n time units to process n records.What is the smallest value of k for which package B will be preferred over A?
A
12
B
10
C
6
D
5
Question 12 Explanation: 
As per given information Package B 10nlog10n is lesser than or equals to Package A 0.0001n2 0 because n2 is asymptotically larger than nlogn. Finally, 10nlog10n ≤ 0.0001n2
Let n = 10k records. Substitute into 10nlog10n ≤ 0.0001n2
10(10k)log1010k ≤ 0.0001(10k)2
10k+1k ≤ 0.0001 × 102k
k ≤ 102k−k−1−4
k ≤ 10k−5
According to the problem value 6 is suitable for K.
Question 13
Which data structure in a compiler is used for managing information about variables and their attributes?
A
Abstract syntax tree
B
Symbol table
C
Semantic stack
D
Parse Table
Question 13 Explanation: 
Symbol tables are data structures that are used by compilers to hold information about source-program constructs. The information is collected incrementally by the analysis phases of a compiler and used by the synthesis phases to generate the target code. Entries in the symbol table contain information about an identifier such as its character string (or lexeme) , its type, its position in storage, and any other relevant information.
Question 14
Which languages necessarily need heap allocation in the runtime environment?
A
Those that support recursion
B
Those that use dynamic scoping
C
Those that allow dynamic data structures
D
Those that use global variables
Question 14 Explanation: 
Dynamic memory is allocated on the heap by the system. So the languages which allow dynamic data structure require heap allocation at runtime.
Question 15
One of the header fields in an IP datagram is the Time to Live (TTL) field. Which of the following statements best explains the need for this field?
A
It can be used to prioritize packets
B
It can be used to reduce delays
C
It can be used to optimize throughput
D
It can be used to prevent packet looping
Question 15 Explanation: 
Time to Live (TTL) is a limit on the period of time or transmissions in computer and computer network technology that a unit of data (e.g. a packet) can experience before it should be discarded. If the limit is not defined then the packets can go into an indefinite loop. The packet is discarded when the Time to Live field reaches 0 to prevent looping.
Question 16
Which one of the following is not a client server application?
A
Internet chat
B
Web browsing
C
E-mail
D
ping
Question 16 Explanation: 
Ping is used for knowing status of a host by another host. it is not a client server application.
Question 17
Let L1 be a recursive language. Let L2 and L3 be languages that are recursively enumerable but not recursive. Which of the following statements is not necessarily true?
A
L2 – L1 is recursively enumerable
B
L1 – L3 is recursively enumerable
C
L2 ∩ L1 is recursively enumerable
D
L2 ∪ L1 is recursively enumerable
Question 17 Explanation: 
L2 − L1 means L2 ∩ L1c , since L1 is recursive hence L1c must also be recursive, So L2∩L1c is equivalent to (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2− L1 must be recursive enumerable. Hence A is always true.
L1 − L3 means L1 ∩ L3c , since recursive enumerable is not closed under complement, so L3c may or may not be recursive enumerable, hence we cannot say that L1 − L3 will always be recursive enumerable. So B is not necessarily true always.
L2 ∩ L1 means (Recursive enum ∩ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∩ Recursive enum) and recursive enum is closed under intersection, so L2∩ L1 must be recursive enumerable. Hence C is always true.
L2 ∪ L1 means (Recursive enum ∪ Recursive) , as every recursive is recursive enum also, so it is equivalent to (Recursive enum ∪ Recursive enum) and recursive enum is closed under union, so L2 ∪ L1 must be recursive enumerable. Hence D is always true.
Question 18
Consider a B+-tree in which the maximum number of keys in a node is 5. What is the minimum number of keys in any non-root node?
A
1
B
2
C
3
D
4
Question 18 Explanation: 
― In B+ tree, the root contains minimum two block pointers and maximum ‘p’ block pointers.
― Here,
p = order
key = order – 1 = p – 1
― In the non-root node, the minimum no. of keys =⌈p/2⌉-1
― So, key = 5
order = 6
― So, minimum no. of keys in root node =⌈6/2⌉-1=2
Question 19
   
A
1, 0
B
1, 2
C
1, 3
D
1, 5
Question 19 Explanation: 
Passenger:

― 1, 3 Pids are returned
Question 20
 
A
I only
B
II only
C
Both I and II
D
Neither I nor II
Question 20 Explanation: 
― Two-phase locking protocol (2PLP) ensures the conflict serializable schedule but it may not free from deadlock.
― Timestamp ordering protocol ensures conflict serializability and free from deadlock.
Question 21
     
A
19
B
21
C
20
D
10
Question 21 Explanation: 
Note: Out of syllabus.
Question 22
 
A
P-3, Q-2, R-4, S-1
B
P-2, Q-3, R-1, S-4
C
P-3, Q-2, R-1, S-4
D
P-2, Q-3, R-4, S-1
Question 22 Explanation: 
Note: Out of syllabus.
Question 23
   
A
Mutual exclusion but not progress
B
Progress but not mutual exclusion
C
Neither mutual exclusion nor progress
D
Both mutual exclusion and progress
Question 23 Explanation: 
In this mutual exclusion is saisfied because at any point of time either S1 = S2 or S1 ≠ S2, but not both. But here progress is not satisfied because suppose S1 = 1 and S2 = 0 and P1 is not interested to enter into critical section but P2 wants to enter into critical section, and P2 will not be able to enter, because until P1 will not enter critical section, S1 will not become equal to S2. So if one process do not interested in entering critical section, will not allow other process to enter critical section which is interested. So progress is not satisfied.
Question 24
A system uses FIFO policy for page replacement. It has 4 page frames with no pages loaded to begin with. The system first accesses 100 distinct pages in some order and then accesses the same 100 pages but now in the reverse order. How many page faults will occur?  
A
196
B
192
C
197
D
195
Question 24 Explanation: 
The first 100 accesses causes 100 page faults.
Page 1 …….. Page 100 causes 100 faults.
Now, when they are accesses in reverse order page 100, 99, 98, 97 are already present. So we get 96 page faults. So, total of 196 page faults.
Question 25
 
A
I only
B
I and III only
C
II and III only
D
I, II and III
Question 25 Explanation: 
- In SRTF longer bursts will suffer from starvation.
- Pre-emptive scheduling causes starvation (for example lower priority jobs are waiting).
- Best Response time is given by RR.
Question 26
Consider a company that assembles computers. The probability of a faulty assembly of any computer is p. The company therefore subjects each computer to a testing process. This testing process gives the correct result for any computer with a probability of q. What is the probability of a computer being declared faulty?
A
pq + (1 - p)(1 - q)
B
(1 - q)p
C
(1 - p)q
D
pq
Question 26 Explanation: 
The product can be ( correctly assembled or faulty assembled ) & Tested ( correctly & incorrectly ) We have ( AT, A’T + AT’ + A’T’ ) choices where A= correctly assembled, and T for correct test results.
A faulty product can be correctly assembled &resulted correctly with (AT) case, where the probability is p*q.
A faulty product can be assembled faulty & resulted wrongly which is A’T’ case, where probability = assembly fail probability* computer testing fail probability = (1-p)*(1-q)
Total probability = pq +(1-p)*(1-q).
Question 27
What is the probability that divisor of 1099 is a multiple of 1096?
A
1/625
B
4/625
C
12/625
D
16/625
Question 27 Explanation: 
Probability that divisor of 1099 is a multiple of 1096
We can write 1099 as 1096×103
So, (1099)/(1096) to be a whole number, [1096×103/1096]➝ (1)
We can observe that every divisor of 103 is a multiple of 1096
So number of divisor of 103 to be found first
⇒ 103=(5×2)3=23×53
No. of divisors = (3 + 1) (3 + 1) = 16
Total number of divisor of 1099 are 1099=299×599=100×100=10000
Probability that divisor of 1099 is a multiple of 1096 is
⇒16/10,000
Question 28
 
A
I and II
B
III and IV
C
IV only
D
II and IV
Question 28 Explanation: 
Havel Hakimi theorem:
⇾ Arrange the degree of vertices in descending order
eg. d1,d2, d3...dn
⇾ Discard d1, subtrack ‘1’ from the next 'd1'degrees
eg:
⇒ 1 1 0 1
⇾ We should not get any negative value if its negative, this is not valid sequence
⇾ Repeat it till we get ‘0’ sequence
I. 7, 6, 5, 4, 4, 3, 2, 1
➡️5, 4, 3, 3, 2, 1, 0
➡️3, 2, 2, 1, 0, 0
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
II. 6, 6, 6, 6, 3, 3, 2, 2
➡️5, 5, 5, 2, 2, 1, 2
put them in descending order
➡️5, 5, 5, 2, 2, 2, 1
➡️4, 4, 1, 1, 1, 1
➡️3, 0, 0, 0, 1 (descending order)
➡️3, 1, 0, 0, 0
➡️0, -1, -1, 0
[This is not valid]
III. 7, 6, 6, 4, 4, 3, 2, 2
➡️5, 5, 3, 3, 2, 1, 1
➡️4, 2, 2, 1, 0, 1
➡️4, 2, 2, 1, 1, 0 (descending order)
➡️1, 1, 0, 0, 0
➡️0, 0, 0, 0
[valid]
IV. 8, 7, 7, 6, 4, 2, 1, 1
There is a degree ‘8’, but there are only ‘8’ vertices.
A vertex cannot have edge to itself in a simple graph. This is not valid sequence.
Question 29
 
A
x=4, y=10
B
x=5, y=8
C
x=-3, y=9
D
x=-4, y=10
Question 29 Explanation: 

Trace = {Sum of diagonal elements of matrix}

Here given that eigen values are 4, 8
Sum = 8 + 4 = 12
Trace = 2 + y
⇒ 2 + y = 12
y = 10

Determinant = |2y - 3x|
Product of eigen values = 8 × 4 = 32
2y - 3x = 32
(y = 10)
20 - 3x = 32
-12 = 3x
x = -4
∴ x = -4, y = 10
Question 30
Suppose the predicate F(x, y, t) is used to represent the statement that person x can fool person y at time t. Which one of the statements below expresses best the meaning of the formula ∀x∃y∃t(¬F(x, y, t))?
A
Everyone can fool some person at some time
B
No one can fool everyone all the time
C
Everyone cannot fool some person all the time
D
No one can fool some person at some time
Question 30 Explanation: 
F(x,y,t) ⇒ Person 'x' can fool person 'y' at time 't'.
For better understanding propagate negation sign outward by applying Demorgan's law.
∀x∃y∃t(¬F(x, y, t)) ≡ ¬∃x∀y∀t(F(x,y,t))
Now converting ¬∃x∀y∀t(F(x,y,t)) to English is simple.
¬∃x∀y∀t(F(x,y,t)) ⇒ There does not exist a person who can fool everyone all the time.
Which means "No one can fool everyone all the time".
Hence, Option (B) is correct.
Question 31
 
A
B
C
D
Question 31 Explanation: 
f = ((P’Q’ + Q’R’)’ + ( P’R’ + Q’R’)’ )’
= (P’Q’ + Q’R’)( P’R’ + Q’R’)
= (P’Q’P’R’ + P’Q’Q’R’ + Q’R’P’R’ + Q’R’Q’R’)
= (P’Q’R’ + P’Q’R’ + P’Q’R’ + Q’R’)
= (P’Q’R’ + Q’R’)
= (Q’R’)
= (Q+R)’
Question 32
   
A
11, 10, 01, 00
B
10, 11, 01, 00
C
10, 00, 01, 11
D
11, 10, 00, 01
Question 32 Explanation: 

The next four values of Q1Q0 are 11, 10, 01, 00.
Question 33
 
A
13
B
15
C
17
D
19
Question 33 Explanation: 
It is given that there is operand forwarding. In the case of operand forwarding the updated value from previous instruction’s PO stage is forwarded to the present instruction’s PO stage. Here there’s RAW dependency between I1-I2 for R5 and between I2-I3 for R2. These dependencies are resolved by using operand forwarding as shown in the below timeline diagram. The total number of clock cycles needed is 15.
Question 34
The weight of a sequence a0, a1, ..., an-1 of real numbers is defined as a0+a1/2+...+ an-1/2n-1.  A subsequence of a sequence is obtained by deleting some elements from the sequence, keeping the order of the remaining elements the same. Let X denote the maximum possible weight of a subsequence of a0, a1, ...,an-1 and Y the maximum possible weight of a subsequence of a1, a2, ..., an-1. Then X is equal to
A
max(Y, a0+Y)
B
max(Y, a0+Y/2)
C
max(Y, a0+2Y)
D
a0+Y/2
Question 34 Explanation: 
Using concepts of Dynamic Programming, to find the maximum possible weight of a subsequence of X, we will have two alternatives:
1. Do not include a0 in the subsequence: then the maximum possible weight will be equal to maximum possible weight of a subsequence of {a1, a2,….an} which is represented by Y
2. Include a0: then maximum possible weight will be equal to : a0 + (each number picked in Y will get divided by 2) = a0 + Y/2.
Key point here is Y will itself pick optimal subsequence to maximize the weight. The value of a0 can be anything (negative or i∈R it is possible that Y>a0+Y/2.
Note: Y/2 meaning: Each number picked in Y will get divided by 2
Question 35
A
-9
B
5
C
15
D
19
Question 35 Explanation: 
int a[ ] = {12, 7, 13, 4, 11, 6}
if (n <= 0)
return 0;
else if (*a % 2 = = 0)
return *a + f(a+1, n-1);
else
return *a – f(a+1, n-1);


⇒12+(7-(13-(4+(11-(6)))))
⇒12+(7-(13-(4+5)))
⇒12+7-(4)
⇒12+3
⇒15
Question 36
   
A
q = NULL; p→next = head; head = p;
B
q→next = NULL; head = p; p→next = head;
C
head = p; p→next = q; q→next = NULL;
D
q→next = NULL; p→next = head; head = p;
Question 36 Explanation: 
C function takes a simple linked list as input argument.
The function modifies the list by moving the last element to the front of the list.
Let the list be 1 → 2 → 3 → 4 & the modified list must be 4 → 1 → 2 → 3.
Algorithm:
Traverse the list till last node. Use two pointers. One to store the address of last node & other for the address of second last node.
After completion of loop. Do these.
(i) Make second last as last
(ii) Set next of last as head
(iii) Make last as head

while (p → !=NULL)
{
q = p;
p = p → next;
}
― p is travelling till the end of list and assigning q to whatever p had visited & p takes next new node, so travels through the entire list.
Now the list looks like

According to the Algorithm new lines must be
q → next = NULL; p → next = head; head = p
Question 37
A
2
B
3
C
4
D
6
Question 37 Explanation: 
Here a, b, and c all have 3 different values so we need atleast 3 registers r1, r2 and r3.
Assume 'a' is mapped to r1, 'b' to r2 and 'c' to r3.
d = a + b, after this line if u notice 'a' is never present on right hand side, so we can map 'd' to r1.
e = c + d, after this line 'd' is never present on rhs, so we can map 'e' to r1.
at this time mapping is
r1 --- e
r2 --- b
r3 --- c
We have 3 registers for e, b and c.
f = c + e
b = c + e
These two are essentially doing same thing, after these two line 'b' and 'f' are same so we can skip computing 'f' or need not give any new register for 'f'. And wherever 'f' is present we can replace it with 'b', because neither of 'f' and 'b' are changing after these two lines, so value of these will be 'c+e' till the end of the program.
At second last line "d = 5 + e"
here 'd' is introduced, we can map it to any of the register r1 or r3, because after this line neither of 'e' or 'c' is required. Value of 'b' is required because we need to return 'd+f', and 'f' is essentially equal to 'b'
finally code becomes
r1 = 1
r2 = 10
r3 = 20
r1 = r1 + r2
r1 = r3 + r1
r2 = r3 + r1
r2 = r3 + r1
r1 = r2 + r2
r3 = 5 + r1
return r3 + r2
Therefore minimum 3 registers needed.
Question 38
The grammar S → aSa|bS|c is
A
LL(1) but not LR(1)
B
LR(1) but not LR(1)
C
Both LL(1) and LR(1)
D
Neither LL(1) nor LR(1)
Question 38 Explanation: 
The LL(1) parsing table for the given grammar is:

As there is no conflict in LL(1) parsing table, hence the given grammar is LL(1) and since every LL(1) is LR(1) also, so the given grammar is LL(1) as well as LR(1).
Question 39
Let L = {w ∈ (0 + 1)| w has even number of 1s}, i.e. L is the set of all bit strings with even number of 1s. Which one of the regular expression below represents L?  
A
(0 * 10 * 1)*
B
0 * (10 * 10*)*
C
0*(10 * 1*)*0*
D
0 * 1(10 * 1)*10*
Question 39 Explanation: 
The best way to find correct answer is option elimination method. We will guess strings which has even number of 1’s and that is not generated by wrong options OR which generate strings which doesn’t have even number of 1’s.
Option A: (reg expr: (0*10*1)* ) doesn’t generate string such as { 110, 1100,....}
Option C: (reg expr: 0*(10*1*)*0* generate string such as {1, 111,....} which have odd number of 1’s.
Option D: (reg expr: 0*1(10*1)*10* doesn’t generate strings such as { 11101, 1111101, ….}.
Question 40
A
Only L2 is context free
B
Only L2 and L3 are context free
C
Only L1 and L2 are context free
D
All are context free
Question 40 Explanation: 
All languages viz L1, L2, L3 and L4 has only one comparison and it can be accepted by PDA (single stack), hence all are Context Free Languages.
Question 41
Let w be any string of length n is {0, 1}*. Let L be the set of all substrings of w. What is the minimum number of states in a non-deterministic finite automaton that accepts L?
A
n-1
B
n
C
n+1
D
2n-1
Question 41 Explanation: 
In order to accept any string of length “n” with alphabet {0,1}, we require an NFA with “n+1” states. For example, consider a strings of length “3” such as “101”, the NFA with 4 states is given below:

Since L is set of all substrings of “w” (Substring of a string is obtained by deleting any prefix or any suffix from string), so if we consider “w” as “101” , then the substrings of w are { ϵ, 0, 1, 10, 01, 101}.
Since the string “101” is also its substring, so we require 4 states (i.e. for n length string, n+1 states are required) and the NFA would be:
Question 42
 
A
T1 → T3 → T2
B
T2 → T1 → T3
C
T2 → T3 → T1
D
T3 → T1 → T2
Question 42 Explanation: 
The given schedule is

― Precedence graph for the above schedule is,

― From the precedence graph the correct serialization is,
Question 43
 
A
100
B
200
C
300
D
2000
Question 43 Explanation: 
Given
R(A, B, C) – 200 tuples
S(B, D, E) – 100 tuples
FD’s:
B → A
A → C
― ‘B’ is primary key for R and foreign key of S from the given FDs.
― Maximum tuples in natural join of R and S is min(200, 100) = 100.
Question 44
A
T1, T2, T3
B
T2, T4
C
T3, T4
D
T1, T2, T4
Question 44 Explanation: 
T1 covers S1
T2 covers S3
T4 covers S2, S4.
Question 45
A
At least twice
B
Exactly twice
C
Exactly thrice
D
Exactly once
Question 45 Explanation: 
S0=1
S1=0
S2=0
P0 enters the critical section first,
prints (‘0’)
releases S1,S2(i.e., S1=1 & S2=1)
Now P1 & P2 both can enter critical section releases S0 & prints (‘0’)
This process continues, hence the number of zero’s printed ≥2.
Question 46
 
A
n = 40, k = 26
B
n = 21, k = 12
C
n = 20, k = 10
D
n = 41, k = 19
Question 46 Explanation: 
Consider the case where i = 10 & i = 11, n = 21 & k = 12
P10 requests R10 & R11
P11 requests R10 & R8
Hence P10 & P11 inorder in deadlock.
Question 47
Suppose computers A and B have IP addresses 10.105.1.113 and 10.105.1.91 respectively and they both use the same net mask N. Which of the values of N given below should not be used if A and B should belong to the same network?
A
255.255.255.0
B
255.255.255.128
C
255.255.255.192
D
255.255.255.224
Question 47 Explanation: 
When we perform bitwise AND operation between IP Address and Subnet Mask, it gives Network ID. If for both IP results is same Network ID. It means, both IP are belong to the same network else it's on different network.
When we perform AND operation between IP address 10.105.1.113 and 255.255.255.224 result is 10.105.1.96 and when we perform AND operation between IP address 10.105.1.91 and 255.255.255.224 result is 10.105.1.64.
Therefore, 10.105.1.96 and 10.105.1.64 are different network, so D is correct answer.
Question 48
 
A
2 nanoseconds
B
20 nanoseconds
C
22 nanoseconds
D
88 nanoseconds
Question 48 Explanation: 
Between L1 cache block is of size 4 words, and the bus width between L1 and L2 is also of size 4 words. So only one time the data needs to be transferred from L2 to L1.
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
Question 49
A
222 nanoseconds
B
888 nanoseconds
C
902 nanoseconds
D
968 nanoseconds
Question 49 Explanation: 
Between L1 cache block is of size 4 words, and the bus width between L1 and L2 is also of size 4 words. So only one time the data needs to be transferred from L2 to L1.
And it requires one access of L1 cache and one access of L2 cache. So time taken = 20+2 = 22ns
L2 cache block is of size 16 words and nothing is mentioned about the main memory block, so we assume the main memory block is also of size 16 words. But the bus between L2 and main memory is only 4 words..so when the data has to be transferred from main memory to L2 cache it has to send 4 times through the data bus.
When a data request comes to L1, if there is a cache miss in L1 then it will be searched in L2 if there is a hit in L2 then the required data is transferred from L2 to L1 in a single transfer through the bus. If there is a miss in L2 then it will be searched in main memory. Then the 16 words data from main memory will be transferred to L2 in 4 transfers through the data bus. Time taken for this = 4*(200+20) = 4*220 = 880 ns
Then from L2 to L1 only 4 words of data will be transferred through the data bus in a single transfer. We know time taken for this is 22ns.
So total time taken = 880 + 22 = 902 ns
Question 50
   
A
7
B
8
C
9
D
10
Question 50 Explanation: 
Minimum Spanning Tree and Shortest Path algorithms computation looks similar they focus on 2 different requirements. In MST, requirement is to reach each vertex once (create graph tree) and total (collective) cost of reaching each vertex is required to be minimum among all possible combinations. In Shortest Path, requirement is to reach destination vertex from source vertex with lowest possible cost (shortest weight). So here we do not worry about reaching each vertex instead only focus on source and destination vertices and that's where lies the difference.

MST with total weight of 10. So cost of reaching A to C in MST is 10.
Question 51
A
7
B
8
C
9
D
10
Question 51 Explanation: 
The possible minimum path is,
B → A, A → E, E → C.
Question 52
A
46, 42, 34, 52, 23, 33
B
34, 42, 23, 52, 33, 46
C
46, 34, 42, 23, 52, 33
D
42, 46, 33, 23, 34, 52
Question 52 Explanation: 
Hash Table consists of 10 slots, uses Open Addressing with hash function and linear probing.
After inserting 6 values, the table looks like

The possible order in which the keys are inserted are:
34, 42, 23, 46 are at their respective slots 4, 2, 3, 6.
52, 33 are at different positions.
― 52 has come after 42, 23, 34 because, it has the collision with 42, because of linear probing, it should have occupy the next empty slot. So 52 is after 42, 23, 34.
― 33 is after 46, because it has the clash with 23. So it got placed in next empty slot 7, which means 2, 3, 4, 5, 6 are filled.
42, 23, 34 may occur in any order but before 52 & 46 may come anywhere but before 33.

So option (C)
Question 53
A
10
B
20
C
30
D
40
Question 53 Explanation: 
Total 6 insertions
― 33 must be inserted at last (only one possibility)
― 46 can be inserted in any of the 5 places remaining. So 5 ways.
― 52 must be inserted only after inserting 42, 23, 34. So only one choice for 52.
〈42,23,34〉can be sequenced in 3! ways.
Hence, 5×3! = 30 ways
Question 54
A
4
B
3
C
2
D
1
Question 54 Explanation: 
Link R1- R2 will not be used because its cost is 6 and link R1-R3-R2 costs 5, which is lesser than R1-R2 link.
Similarly, link R4-R6 will not be used, instead this link we can use R4-R5-R6 link which costs only 5 unit.
Question 55
 
A
0
B
1
C
2
D
4
Question 55 Explanation: 
Now Graph will look like

And only link that will be removed is R5-R6 link.
Question 56
A
masked
B
belied
C
betrayed
D
suppressed
Question 56 Explanation: 
Masked - opposite to given
Belied - Not appropriate to given
Betrayed - Most appropriate to given (Reveal intentionally)
Suppressed - Irrevalent to given
Answer is option C.
Question 57
Which of the following options is closest in meaning to the word Circuitous.
A
cyclic
B
indirect
C
confusing
D
crooked
Question 57 Explanation: 
Synonyms for Circuitous are indirect, oblique, winding, meandering etc.
Question 58
A
uphold
B
restrain
C
cherish
D
conserve
Question 58 Explanation: 
Uphold - cause to remain
Restrain - keep under control
Cherish - befond of
Conserve - protect from harm i.e., keeping safety, loss of destruction
Answer is option d.
Question 59
25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is:
A
2
B
17
C
13
D
3
Question 59 Explanation: 

Total number of persons = a+b+c = 25 → (1)
Number of persons who play hockey = a+b = 15 → (2)
Number of persons who play football = b+c = 17 → (3)
Number of persons who play hockey and football = b = 10 → (4)
From (2) ⇒ a=5
From (3) ⇒ c=7
From (1) ⇒ d = 3
Number of persons who play neither hockey nor football = d = 3
Question 60
 
A
fallow: land
B
unaware: sleeper
C
wit: jester
D
renovated: house
Question 60 Explanation: 
→ A worker who is working is called unemployed.
→ Same as a land which is not used called fallow.
Question 61
If 137+276 = 435 how much is 731+672?
A
534
B
1403
C
1623
D
1513
Question 61 Explanation: 
Let, base = x
(137)x + (276)x = (435)x
x2 + 3x + 7 + 2x2 + 7x + 6 = 4x2 + 3x + 5
x2 - 7x - 8 = 0
x = 8 (or) -1
∴ x = 8 (-1 cannot be a base)
(731)x + (672)x = (731)8 +( 672)8
= 7 × 82 + 3× 8 + 1×1 + 6 × 82 + 7 × 8 + 2 × 1
= 915
From the options, 915 can be written as 1623 in base 8.
Question 62
   
A
HSIG
B
SGHI
C
IGSH
D
IHSG
Question 62 Explanation: 
Let us solve using option elimination approach .
Option A: HSIG
from (ii) , we can say that Gita and Saira are successive siblings.
Hence, option A is not true.
Option C: IGSH

In any of the above 4 cases, (i) is not true.
Hence, option C is not true.
Option D: IHSG

In any of the above 4 cases, (i) is not true.
Hence, option D is not true.
Option B: SGHI

In last two cases, all the facts are true.
∴ The order is SGHI.
Question 63
 
A
Modern warfare has resulted in civil strife.
B
Chemical agents are useful in modern warfare.
C
Use of chemical agents in warfare would be undesirable.
D
People in military establishments like to use chemical agents in war.
Question 63 Explanation: 
→ People in military establishments like to use chemical agents in war.
Question 64
5 skilled workers can build a wall in 20 days: 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall?
A
20
B
10
C
16
D
15
Question 64 Explanation: 
5 skilled workers can build the wall in 20 days ⇒ 1 skilled worker can build the wall in 100 days
Capacity = 1/100
8 semi-skilled workers can build the wall in 25 days
⇒ 1 semi-skilled worker can build the wall in 200 days
Capacity = 1/200
10 un-skilled workers can build the wall in 30 days
⇒ 1 un-skilled worker can build the wall in 300 days
Capacity = 1/300
1 day work (2 skilled + 6 semi-skilled + 5 unskilled) = 2(1/100) + 6(1/200) + 5(1/300) = 1/15
∴ They can complete the work in 15 days.
Question 65
Given digits 2, 2, 3, 3, 3, 4, 4, 4, 4 how many distinct 4 digit numbers greater than 3000 can be formed?
A
50
B
51
C
52
D
54
Question 65 Explanation: 
Greater than 3000
⇒ First digit: 3 or 4
(i) First digit - 3:
We have to choose 3 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222, 333) is not possible as we have only two 2's and two 3's.
Total = 3 × 3 × 3 - 2 = 25
(ii) First digit - 4:
We have to choose 4 digits from (2, 2, 3, 3, 4, 4, 4, 4).
Any place can have either 2 or 3 or 4, but (222) is not possible we have only two 2's.
Total = 3 × 3 × 3 - 1 = 26
∴ Total number possible = 25 + 26 = 51
There are 65 questions to complete.