GATE 2014 [Set2]
Question 1 
Choose the most appropriate phrase from the options given below to complete the following sentence.
India is a postcolonial country because
It was a former British colony  
Indian Information Technology professionals have colonized the world  
India does not follow any colonial practices  
India has helped other countries gain freedom 
India is a postcolonial country because "It was a former British colony" that means if a country is called post colonial then it came into existence after the colonies rules and it can become a independent.
Question 2 
Who ___________ was coming to see us this evening?
you said  
did you say  
did you say that  
had you said 
Question 3 
Match the columns.
Column 1 Column 2 1) eradicate P) misrepresent 2) distort Q) soak completely 3) saturate R) use 4) utilize S) destroy utterly
1 : S, 2 : P, 3 : Q, 4 : R  
1 : P, 2 : Q, 3 : R, 4 : S  
1 : Q, 2 : R, 3 : S, 4 : P  
1 : S, 2 : P, 3 : R, 4 : Q 
Distort = misrepresent
Saturate = Soak completely
Utilize = use
Question 4 
What is the average of all multiples of 10 from 2 to 198?
90  
100  
110  
120 
Average = 10+20+30+...+180+190 / 19 = (19/2 [10+190]) / 19 = 100
Question 5 
The value of is
3.464  
3.932  
4.000  
4.444 
Squaring on both sides,
x^{2} = 12+x
x^{2}  x  12 = 0
(x4) (x+3) = 0
∴ x = 4 (x ≠ 3)
Question 6 
The old city of Koenigsberg, which had a German majority population before World War 2, is now called Kaliningrad. After the events of the war, Kaliningrad is now a Russian territory and has a predominantly Russian population. It is bordered by the Baltic Sea on the north and the countries of Poland to the south and west and Lithuania to the east respectively. Which of the statements below can be inferred from this passage?
Kaliningrad was historically Russian in its ethnic make up  
Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia  
Koenigsberg was renamed Kaliningrad, as that was its original Russian name  
Poland and Lithuania are on the route from Kaliningrad to the rest of Russia 
Question 7 
The number of people diagnosed with dengue fever (contracted from the bite of a mosquito) in north India is twice the number diagnosed last year. Municipal authorities have concluded that measures to control the mosquito population have failed in this region.
Which one of the following statements, if true, does not contradict this conclusion?
A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent  
More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency  
Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test  
The number of people with malarial fever (also contracted from mosquito bites) has increased this year 
Question 8 
If x is real and x^{2}  2x + 3 = 11, then possible values of  x^{3} + x^{2}  x include
2, 4  
2, 14  
4, 52  
14, 52 
x^{2}  2x + 3 = 11, x is real
x^{2}2x+3 = 11
x^{2}2x+8 = 0
(x4)(x+2) = 0
x = 4, 2
x^{2}2x+3 = 11
x^{2}2x+14 = 0
x is not real in this case.
x^{3}+x^{2}x
when x=2
⇒ (2)^{3}+(2)^{2}(2)
= ((8) + (4) + 2 = 14
x=4
⇒ (4)^{3}+(4)^{2}(4)
= 64 + 16 4
= 52
Possible values of x^{3}+x^{2}x include 14, 52.
Question 9 
The ratio of male to female students in a college for five years is plotted in the following line graph. If the number of female students doubled in 2009, by what percent did the number of male students increase in 2009?
140  
141  
142  
143 
Number of female in 2008 = y
Number of female in 2009 = 2y
Given,
x/y = 2.5 ⇒ y = 2x/5
Let, number of male in 2009 = M
Given, M/2y = 3
M/2(2x/5) = 3 ⇒ M = 12x/5
Percentage increase = Mx/x × 100
= 12x/5x/ x × 100
= 7/5 × 100
= 140%
Question 10 
At what time between 6 a.m. and 7 a.m. will the minute hand and hour hand of a clock make an angle closest to 60°?
6:22 am  
6:27 am  
6:38 am  
6:45 am 
And for every minute, the angle between them decreases by 6°  (1/2)° = 5 (1/2)°
∴ For the angle to be closest to 60°, the angle must be reduced by atmost 120°.
1 min  5(1/2)°
x min  120°
x = 2/11 × 120 = 240/11 = 21.81 m ≈ 22 min
i.e. 6.22 a.m. the angle between minute hand and hour hand will be closest to 60°.
Question 11 
The security system at an IT office is composed of 10 computers of which exactly four are working. To check whether the system is functional, the officials inspect four of the computers picked at random (without replacement). The system is deemed functional if at least three of the four computers inspected are working. Let the probability that the system is deemed functional be denoted by p. Then 100p =_____________.
11.90  
11.91  
11.92  
11.93 
Out of which ‘4’ are chosen at random without replaced and created.
If at least three of them are working then system is deemed functional
i.e., there should be only ‘one’ nonworking system in set of ‘4’.
It is possible with combination
W – W – W – N,
W – W – N – W,
W – N – W – W,
N – W – W – W.
For W – W – W – N, the probability = (choosing working out of 10) × (choosing working out of 9) × (choosing working out of 8) × (choosing nonworking out of 7)
= 4/10×3/9×2/8×1/7
where 4/10 ⇒ 4 working out of 10
3/9 ⇒ 3 working are remaining out of ‘9’ as ‘1’ is already taken
For ‘4’ Sum combinations
Total probability = 4×[4/10×3/9×2/8×1/7] = 600/5040
We need 100p ⇒ 100×600/5040 = 11.90
Question 12 
Each of the nine words in the sentence "The quick brown fox jumps over the lazy dog" is written on a separate piece of paper. These nine pieces of paper are kept in a box. One of the pieces is drawn at random from the box. The expected length of the word drawn is _____________. (The answer should be rounded to one decimal place.)
3.9  
4.0  
4.1  
4.2 
There are ‘9’ words in this sentence.
No. of characters in each word
The (3)
quick (5)
brown (5)
fox (3)
jumps (5)
over (4)
the (3)
lazy (4)
dog (3)
Each word has equal probability.
So expected length = 3×1/9+5×1/9+5×1/9+3×1/9+5×1/9+ 4×1/9+3×1/9+4×1/9+3×1/9
= 35/9
= 3.9
Question 13 
The maximum number of edges in a bipartite graph on 12 vertices is ______.
36  
37  
38  
39 
Total no. of edges = 6×6 = 36
Question 14 
If the matrix A is such that
then the determinant of A is equal to
0  
1  
2  
3 
Question 15 
A nonzero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?
f(0)f(4) < 0  
f(0)f(4) > 0  
f(0) + f(4) > 0  
f(0) + f(4) < 0 
Polynomial will be
f(x) = (x1)(x2)(x3)
f(0) = 1 × 2 × 3 = 6
f(4) = 3 × 2 × 1 = 6
f(0) ∙ f(4) =  36
f(0) + f(4) = 6  6 = 0
Option (A) is correct.
Question 16 
The dual of a Boolean function F(x_{1}, x_{2}, ..., x_{n}, +, ⋅, '), written as F^{D}, is the same expression as that of F with + and ⋅ swapped. F is said to be selfdual if F = F^{D}. The number of selfdual functions with n Boolean variables is
2^{n}  
2^{(n1)}  
2^{(2n )}  
2^{(2(n1) )} 
Number of mutually exclusive pairs of minterms = 2^{n1}.
There are 2 choices for each pair i.e., we can choose one of the two minterms from each pair of minterms for the function.
Therefore number of functions = 2 x 2 x …. 2^{n1} times.
= 2^{(2(n1)) }
Question 17 
Let k = 2^{n}. A circuit is built by giving the output of an nbit binary counter as input to an nto2^{n} bit decoder. This circuit is equivalent to a
kbit binary up counter.  
kbit binary down counter.  
kbit ring counter.  
kbit Johnson counter. 
A n x 2^{n} decoder is a combinational circuit with only one output line has one and all others (2^{n}1) have zeros.
A nbit binary Counter produces outputs from 0 to 2^{n} i.e 000...00 to 111...11 and repeats.
The n x 2^{n} Decoder gets the input (000..00 to 111...11 ) from the binary counter and only one output line has one and rest have zeros.
This circuit is equivalent to a 2^{n}  bit ring counter.
Question 18 
Consider the equation (123)_{5} = (x8)_{y} with x and y as unknown. The number of possible solutions is __________.
3  
5  
6  
7 
(123)_{5} = (x8)_{y}
In R.H.S. since y is base so y should be greater than x and 8, i.e.,
y > x
y > 8
Now, to solve let's change all the above bases number into base 10 number,
5^{2} × 1 +2 × 5 + 3 = y × x + 8
38 = xy + 8
xy = 30
⇒ yx = 30
So the possible combinations are
(1,30), (2,15), (3,10), (5,6)
But we will reject (5,6) because it violates the condition (y > 8).
So, total solutions possible is 3.
Question 19 
A 4way setassociative cache memory unit with a capacity of 16 KB is built using a block size of 8 words. The word length is 32 bits. The size of the physical address space is 4 GB. The number of bits for the TAG field is __________.
20  
21  
22  
23 
Cache size = 16K bytes = 2^{14} Bytes
block size = 8 words = 8⨯4 Byte = 32 Bytes = 2^{5} Bytes
(where each word = 4 Bytes)
No. of blocks =2^{14}/2^{5}=2^{9}
block offset =9bits
Because it is 4way set associative cache, no. of sets =2^{9}/4=2^{7}
Set of set = 7 bits
TAG = 32 – (7 + 5) = 20 bits
Question 20 
Consider the function func shown below:
int func(int num) { int count = 0; while (num) { count++; num >>= 1; } return (count); }
The value returned by func(435)is __________.
9  
10  
11  
12 
Shift left of 1, which means the number gets doubled.
In program, shift right of 1 is given, means every time we enter the loop the number will get halved, and the value of count will get incremented by 1. And when the value of num will become zero then the while loop will get terminated. So,
num = 435/2 = 217/2 = 108/2 = 54/2 = 27/2= 13/2 = 6/2 = 3/2 = 1/2 = 0
Count = 9
So, the value count that will get returned is 9.
Question 21 
Suppose n and p are unsigned int variables in a C program. We wish to set p to ^{n}C_{3}. If n is large, which one of the following statements is most likely to set p correctly?
p = n * (n1) * (n2) / 6;  
p = n * (n1) / 2 * (n2) / 3;  
p = n * (n1) / 3 * (n2) / 2;  
p = n * (n1) * (n2) / 6.0; 
From the options n*(n1)*(n2) will go out of range. So eliminate A & D.
n*(n1) is always an even number. So subexpression n(n1)/2 also an even number.
n*(n1)/ 2*(n2), gives a number which is a multiple of 3. So dividing with 3 will not have any loss. Hence B is option.
Question 22 
A priority queue is implemented as a MaxHeap. Initially, it has 5 elements. The levelorder traversal of the heap is: 10, 8, 5, 3, 2. Two new elements 1 and 7 are inserted into the heap in that order. The levelorder traversal of the heap after the insertion of the elements is:
10, 8, 7, 3, 2, 1, 5  
10, 8, 7, 2, 3, 1, 5  
10, 8, 7, 1, 2, 3, 5  
10, 8, 7, 5, 3, 2, 1 
The level order traversal in this max heap final is:
10, 8, 7, 3, 2, 1, 5.
Question 23 
Which one of the following correctly determines the solution of the recurrence relation with T(1) = 1?
T(n) = 2T(n/2) + Logn
Θ(n)  
Θ(nlog n)  
Θ(n^{2})  
Θ(logn) 
Apply Master’s theorem,
a=2, b=2, k=0, p=1
According to Master’s theorem, we can apply CaseI
(I) If a>b^{k}, then T(n) = θ(n^{(logba )}) = θ(n^{(log22)}) = θ (n)
Question 24 
Consider the tree arcs of a BFS traversal from a source node W in an unweighted, connected, undirected graph. The tree T formed by the tree arcs is a data structure for computing
the shortest path between every pair of vertices.  
the shortest path from W to every vertex in the graph.  
the shortest paths from W to only those nodes that are leaves of T.  
the longest path in the graph. 
But in the given question the BFS algorithm starts from the source vertex w and we can find the shortest path from W to every vertex of the graph.
Question 25 
If L_{1} = {a^{n}n≥0} and L_{2} = {b^{n}n≥0}, consider

(I) L_{1}·L_{2} is a regular language
(II) L_{1}·L_{2} = {a^{n}b^{n}n≥0}
Which one of the following is CORRECT?
Only (I)  
Only (II)  
Both (I) and (II)  
Neither (I) nor (II) 
Since L_{1} and L_{2} both are regular languages and regular languages are closed under concatenation. So their concatenation (i.e., L_{1}⋅ L_{2}) must also be a regular language.
So, L_{1}⋅L_{2} = { a^{n}b^{n}  n ≥ 0}
Hence, statement (i) is True but statement (ii) is False.
Question 26 
Let A≤_{m}B denotes that language A is mapping reducible (also known as manytoone reducible) to language B. Which one of the following is FALSE?
If A≤_{m} B and B is recursive then A is recursive.  
If A≤_{m} Band A is undecidable then B is undecidable.  
If A≤_{m} Band B is recursively enumerable then A is recursively enumerable.  
If A≤_{m} B and B is not recursively enumerable then A is not recursively enumerable. 
Rule 1: If B is recursive then A is recursive.
Rule 2: If B is recursively enumerable then A is recursively enumerable.
Rule 3: If A is not recursively enumerable then B is not recursively enumerable.
Rule 4: If A is undecidable then B is undecidable.
Other than these rules, all conclusion are false.
Question 27 
Consider the grammar defined by the following production rules, with two operators ∗ and +
S > T * P T > U  T * U P > Q + P  Q Q > Id U > Id
Which one of the following is TRUE?
+ is left associative, while ∗ is right associative  
+ is right associative, while ∗ is left associative  
Both + and ∗ are right associative  
Both + and ∗ are left associative 
P ⟶ Q + P, here P is right recursive, so + is right associative.
Question 28 
Which one of the following is NOT performed during compilation?
Dynamic memory allocation  
Type checking  
Symbol table management  
Inline expansion 
Question 29 
Which one of the following is TRUE?
The requirements document also describes how the requirements that are listed in the document are implemented efficiently.  
Consistency and completeness of functional requirements are always achieved in practice.  
Prototyping is a method of requirements validation.  
Requirements review is carried out to find the errors in system design. 
Question 30 
A FAT (file allocation table) based file system is being used and the total overhead of each entry in the FAT is 4 bytes in size. Given a 100 × 10^{6} bytes disk on which the file system is stored and data block size is 10^{3} bytes, the maximum size of a file that can be stored on this disk in units of 10^{6} bytes is ________.
99.60  
99.61  
99.62  
99.63 
= Total disk size/ disk block size
= 100 × 10^{6} / 10^{3}
= 100 × 10^{3}
∴ Total overhead = 100 × 10^{3} × 4B
= 400 × 10^{3} B
= 0.4 × 10^{6} B
So, Maximum size of file that can be stored in disk is,
Total disk size  Total overhead
= 100 × 10^{6}  0.4 × 10^{6}
= 99.6 × 10^{6} B
Question 31 
The maximum number of superkeys for the relation schema R(E, F, G, H) with E as the key is _____.
8  
9  
10  
11 
Here, n = 4.
So, the possible super keys = 2^{41} = 8
The super keys are: E, EH, EG, EF, EGH, EFH, EFG, EFGH.
Question 32 
Given the STUDENTS relation as shown below.
For (StudentName, StudentAge) to be the key for this instance, the value X should not be equal to
19  
20  
21  
22 
Question 33 
Which one of the following is TRUE about the interior gateway routing protocols – Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)?
RIP uses distance vector routing and OSPF uses link state routing  
OSPF uses distance vector routing and RIP uses link state routing  
Both RIP and OSPF use link state routing  
Both RIP and OSPF use distance vector routing 
Question 34 
Which one of the following socket API functions converts an unconnected active TCP socket into a passive socket?
connect  
bind  
listen  
accept 
(b) The bind function assigns a local protocol address to a socket. With the Internet protocols, the protocol address is the combination of either a 32bit IPv4 address or a 128bit IPv6 address, along with a 16bit TCP or UDP port number.
(c) The listen function converts an unconnected socket into a passive socket, indicating that the kernel should accept incoming connection requests directed to this socket.
(d) The accept function is called by a TCP server to return the next completed connection from the front of the completed connection queue. If the completed connection queue is empty, the process is put to sleep (assuming the default of a blocking socket).
Question 35 
In the diagram shown below L1 is an Ethernet LAN and L2 is a TokenRing LAN. An IP packet originates from sender S and traverses to R, as shown. The link within each ISP, and across two ISPs, are all point to point optical links. The initial value of TTL is 32. The maximum possible value of TTL field when R receives the datagram is
26  
27  
28  
29 
Initially TTL value was 32, so at the Receiver it will become 32 – 6 = 26.
Question 36 
Consider the store and forward packet switched network given below. Assume that the bandwidth of each link is 10^{6} bytes / sec. A user on host A sends a file of size 10^{3} bytes to host B through routers R1 and R2 in three different ways. In the first case a single packet containing the complete file is transmitted from A to B. In the second case, the file is split into 10 equal parts, and these packets are transmitted from A to B. In the third case, the file is split into 20 equal parts and these packets are sent from A to B. Each packet contains 100 bytes of header information along with the user data. Consider only transmission time and ignore processing, queuing and propagation delays. Also assume that there are no errors during transmission. Let T1, T2 and T3 be the times taken to transmit the file in the first, second and third case respectively. Which one of the following is CORRECT?
T1 < T2 < T3  
T1 > T2 > T3  
T2 = T3, T3 < T1  
T1 = T3, T3 > T2 
L = 10^{3} byte
Case: 1
L = 1000 bytes
Header size = 100 bytes
Total Frame size = 1000 + 100 = 1100 bytes
∴ T_{x} = 1100 × 8/ 10^{6}×8 = 1100μs
So, T_{1}=3300μs
Case: 2
L = 100 bytes
Header size = 100 bytes
Total Frame size = 100 + 100 = 200 bytes
∴ T_{x} = 200×8/ 10^{6}×8 = 200μs for 1 packet
For 10 packets ⇒ T_{x} = 2000μs
So, T_{2} = 2000+200+200 = 2400μs
Case: 2
L = 50 bytes
Header size = 100 bytes
Total Frame size = 50 + 100 = 150 bytes
∴ T_{x} = 150×8/ 10^{6}×8 = 150μs for 1 packet
For 20 packets ⇒ T_{x} = 3000μs
So, T_{3} = 3000+150+150 = 3300μs
∴ T_{1} = T_{3}
T_{3} > T_{2}
Question 37 
An IP machine Q has a path to another IP machine H via three IP routers R1, R2, and R3.
Q—R1—R2—R3—H
H acts as an HTTP server, and Q connects to H via HTTP and downloads a file. Session layer encryption is used, with DES as the shared key encryption protocol. Consider the following four pieces of information:
[I1] The URL of the file downloaded by Q [I2] The TCP port numbers at Q and H [I3] The IP addresses of Q and H [I4] The link layer addresses of Q and H
Which of I1, I2, I3, and I4 can an intruder learn through sniffing at R2 alone?
Only I1 and I2  
Only I1  
Only I2 and I3  
Only I3 and I4 
An Intruder can learn [I2] through sniffing at R2 because Port Numbers are encapsulated in the payload field of IP Datagram.
An Intruder can learn [I3] through sniffing at R2 because IP Addresses and Routers are functioned at network layer of OSI Model.
An Intruder can’t learn [I4] through sniffing at R2 because it is related to Data Link Layer of OSI Model.
Question 38 
A graphical HTML browser resident at a network client machine Q accesses a static HTML webpage from a HTTP server S. The static HTML page has exactly one static embedded image which is also at S. Assuming no caching, which one of the following is correct about the HTML webpage loading (including the embedded image)?
Q needs to send at least 2 HTTP requests to S, each necessarily in a separate TCP connection to server S  
Q needs to send at least 2 HTTP requests to S, but a single TCP connection to server S is sufficient  
A single HTTP request from Q to S is sufficient, and a single TCP connection between Q and S is necessary for this  
A single HTTP request from Q to S is sufficient, and this is possible without any TCP connection between Q and S 
Whenever a browser opens a webpage, a separate HTML request must be sent for each image or component in HTML like css file or javascript. But all can be done in the same TCP connection.
Question 39 
Consider the following schedule S of transactions T1, T2, T3, T4:
Which one of the following statements is CORRECT?
S is conflictserializable but not recoverable  
S is not conflictserializable but is recoverable  
S is both conflictserializable and recoverable  
S is neither conflictserializable nor is it recoverable 
In the precedence graph, there are no cycles. So, it is conflict serializable and recoverable also.
Question 40 
Consider a join (relation algebra) between relations r(R) and s(S) using the nested loop method. There are 3 buffers each of size equal to disk block size, out of which one buffer is reserved for intermediate results. Assuming size(r(R)) < size(s(S)), the join will have fewer number of disk block accesses if
relation r(R) is in the outer loop.  
relation s(S) is in the outer loop.  
join selection factor between r(R) and s(S) is more than 0.5.  
join selection factor between r(R) and s(S) is less than 0.5. 
For each tuple r in R do
For each tuple s in S do
If r and s satisfy the join condition then
output the tuple
The above algorithm involves (n_{r}*b_{s}+b_{r})
block transfer and (n_{r}+b_{r}) seeks.
where,
b_{r} → no. of blocks in R
b_{s} → no. of blocks in S
h_{r} → no. of tuples in R
To have less block accesses in n_{r} should be less and in question it is given that size(r(R)) < size(s(S)).
To have fewer no. of disk block accesses the relation r(R) should be in outer loop.
Question 41 
Consider the procedure below for the ProducerConsumer problem which uses semaphores:
semaphore n=0; semaphore s=1; void producer() void consumer() { { while(true) while(true) { { produce(); semWait(s); semWait(s); semWait(n); addToBuffer(); removeFromBuffer(); semSignal(s); semSignal(); semSignal(n); consume(); } } } }
Which one of the following is TRUE?
The producer will be able to add an item to the buffer, but the consumer can never consume it.  
The consumer will remove no more than one item from the buffer.  
Deadlock occurs if the consumer succeeds in acquiring semaphore s when the buffer is empty.  
The starting value for the semaphore n must be 1 and not 0 for deadlockfree operation. 
Question 42 
Three processes A, B and C each execute a loop of 100 iterations. In each iteration of the loop, a process performs a single computation that requires tc CPU milliseconds and then initiates a single I/O operation that lasts for tio milliseconds. It is assumed that the computer where the processes execute has sufficient number of I/O devices and the OS of the computer assigns different I/O devices to each process. Also, the scheduling overhead of the OS is negligible. The processes have the following characteristics:
Process id tc tio A 100 ms 500 ms B 350 ms 500 ms C 200 ms 500 ms
The processes A, B, and C are started at times 0, 5 and 10 milliseconds respectively, in a pure time sharing system (round robin scheduling) that uses a time slice of 50 milliseconds. The time in milliseconds at which process C would complete its first I/O operation is ___________.
1000  
1001  
1002  
1003 
So 'C' completes its I/O at 1000 time units.
Question 43 
A computer has twenty physical page frames which contain pages numbered 101 through 120. Now a program accesses the pages numbered 1, 2, …, 100 in that order, and repeats the access sequence THRICE. Which one of the following page replacement policies experiences the same number of page faults as the optimal page replacement policy for this program?
Leastrecentlyused  
Firstinfirstout  
Lastinfirstout  
Mostrecentlyused 
So, there would be 300 page faults in total (each access 100 page faults). Also it is visible that every time a replacement is done for the page which is most recently referred as it will be least recently referred in future. So, for the given page reference string optimal page replacement policy is working same as most recently used policy and thus number of page faults will be same in both of them.
Question 44 
For a C program accessing X[i][j][k], the following intermediate code is generated by a compiler. Assume that the size of an integer is 32 bits and the size of a character is 8 bits.
t0 = i ∗ 1024 t1 = j ∗ 32 t2 = k ∗ 4 t3 = t1 + t0 t4 = t3 + t2 t5 = X[t4]
Which one of the following statements about the source code for the C program is CORRECT?
X is declared as “int X[32][32][8]”.  
X is declared as “int X[4][1024][32]”.  
X is declared as “char X[4][32][8]”.  
X is declared as “char X[32][16][2]”. 
Arr[0][1][2], this 3D array contains
1 two dimensional array as i value is zero (if i =1 then it has 2, two D array), & the two dimension array has 2 row and 3 column.
So, In a 3D array, i represent number of 2D arrays, j represent number of rows and k represent number of columns.
Number of 2 D array (M)=1
Number of rows (R)=2
Number of columns (C)=3
As arrays are stored in row major order, so this 2 dimension array will be stored as:
Assume base address of Arr is 1000. The address of required position is calculated as:
Arr[i][j][k]= Arr+ [i*(R*C)+(j*C)+k]*4 // multiplication to 4 is due to int takes 4 Bytes.
Arr[0][1][1] = 1000+[0*(2*3)+(1*3)+1]*4
= 1000+[ 0+3+1 ]*4
= 1000+4*4
= 1016
As you can see that in the given example of row order storing of array also has address of Arr[0][1][1] is 1016.
Now coming to the question:
X [ i ][ j ][ k ] is calculated by 3 address code X[t_{4}]
X [ i ][ j ][ k ] = X [ t_{4} ] // by substituting in reverse
= X [ t_{3} + t_{2}]
= X [ t_{1} + t_{0} + k*4]
= X [ t_{0} + t_{1} + k*4] // t_{0} and t_{1} swapped as swapping doesn't have any impact
= X [ i*1024 + j*32 + k*4]
= X [ i*256 + j*8 +k] *4 // taking 4 (common) outside
= X [ i*(32*8)+ (j*8) +k] *4
By comparing the above line with ....... Arr[i][j][k] = Arr+ [i*(R*C)+(j*C)+k]*4
We get R=32, C=8
It means the the declared array has 32 rows and 8 columns and since multiplication by 4 (common outside) so it was declared as INT.
But how many number of 2D arrays in this 3D array, we don't know.
Since option A is the only option with configuration: INT arr[32] [32] [8]. So it is right option.
Question 45 
Let <M> be the encoding of a Turing machine as a string over Σ = {0, 1}. Let L = {<M> M is a Turing machine that accepts a string of length 2014}. Then, L is
decidable and recursively enumerable  
undecidable but recursively enumerable  
undecidable and not recursively enumerable  
decidable but not recursively enumerable 
Now, since total number of strings of length 2014 is finite, so M1 will run the encoding of M for the string of length 2014 and if the M accept the string then M1 will halt in ACCEPT state. But if M goes for infinte loop for every string of length 2014, then M1 also will go into infinite loop. Hence language L is recursively enumerable but not recursive, as in case of rejectance halting is not guaranteed.
Question 46 
Let L_{1} = {w ∈ {0,1}* w has at least as many occurrences of (110)’s as (011)’s}. Let L_{2} = {w ∈ {0,1}*w has at least as many occurrences of (000)’s as (111)’s}. Which one of the following is TRUE?
L_{1} is regular but not L_{2}  
L_{2} is regular but not L_{1}  
Both L_{1} and L_{2} are regular  
Neither nor L_{1} are L_{2} regular 
{Number of occurrences of (110)} ≥ {Number of occurrences of (011)}
Lets analyse the language, consider a string in which occurrence of (110) is more than one.
The following possibilities are: {1100110, 1101110, 110110, ….}
Please observe whenever strings start with “11” then in every situation whatever comes after “11” the string will never violate the condition. So strings of the form 11(0+1)* will always satisfy the condition.
Consider a string in which occurrence of (011) is more than one.
The following possibilities are: {011011, 0111011, 0110011, ….}
In the following possibilities please observe that number of occurrence “011” is two but number of occurrence of (110) is one, which violate the conditions.
If we add “0” in every string mentioned above, i.e. {0110110, 01110110, 01100110, ….} Now, observe that number of occurrence “011” and the number of occurrence of (110) both are equal, which satisfies the conditions.
With these analysis, we can make the DFA , which is mentioned below.
But language L_{2} requires infinite comparison to count the occurrences of (000’s) and (111’s), hence it is not regular.
Question 47 
Consider two strings A = "qpqrr" and B = "pqprqrp". Let x be the length of the longest common subsequence (not necessarily contiguous) between A and B and let y be the number of such longest common subsequences between A and B. Then x+10y = _________.
34  
35  
36  
37 
A = “qpqrr” B = “pqprqrp”
The longest common subsequence (not necessarily contiguous) between A and B is having 4 as the length, so x=4 and such common subsequences are as follows:
(1) qpqr
(2) pqrr
(3) qprr
So y = 3 (the number of longest common subsequences) hence x+10y = 4+10*3 = 34
Question 48 
Suppose P, Q, R, S, T are sorted sequences having lengths 20, 24, 30, 35, 50 respectively. They are to be merged into a single sequence by merging together two sequences at a time. The number of comparisons that will be needed in the worst case by the optimal algorithm for doing this is ______.
358  
359  
360  
361 
In the above implementation, total number of comparisons is
(441) + (941) + (651) + (1591) = 358
Hint: The number of comparisons for merging two sorted sequences of length m and n is m+n1.
Question 49 
Consider the expression tree shown. Each leaf represents a numerical value, which can either be 0 or 1. Over all possible choices of the values at the leaves, the maximum possible value of the expression represented by the tree is _______.
6  
7  
8  
9 
Question 50 
Consider the following function
double f(double x){ if (abs(x*x  3) < 0.01) return x; else return f(x/2 + 1.5/x); }
Give a value q (to 2 decimals) such that f(q) will return q:_____.
1.73  
1.74  
1.75  
1.76 
x^{2}  3 < 0.01 will become True.
So, x^{2}  3 < 0.01
x^{2}  3 < 0.01
x^{2} < 3.01
x < 1.732
Hence, x = 1.73.
Question 51 
Suppose a stack implementation supports an instruction REVERSE, which reverses the order of elements on the stack, in addition to the PUSH and POP instructions. Which one of the following statements is TRUE with respect to this modified stack?
A queue cannot be implemented using this stack.  
A queue can be implemented where ENQUEUE takes a single instruction and DEQUEUE takes a sequence of two instructions.  
A queue can be implemented where ENQUEUE takes a sequence of three instructions and DEQUEUE takes a single instruction.  
A queue can be implemented where both ENQUEUE and DEQUEUE take a single instruction each. 
Suppose:
Dequeue:
If we want to delete an element, that first we need to delete 1.
Enqueue:
Question 52 
Consider the C function given below.
int f(int j) { static int i = 50; int k; if (i == j) { printf(“something”); k = f(i); return 0; } else return 0; }
Which one of the following is TRUE?
The function returns 0 for all values of j.  
The function prints the string something for all values of j.  
The function returns 0 when j = 50.  
The function will exhaust the runtime stack or run into an infinite loop when j = 50. 
int f(int j)
{
static int i = 50;
int k;
if (i == j) // This will be True.
{
printf ("Something");
k = f(i); // Again called f(i) with value of i as 50. So, the function will run into infinite loop.
return 0;
}
else return 0;
}
Question 53 
In designing a computer’s cache system, the cache block (or cache line) size is an important parameter. Which one of the following statements is correct in this context?
A smaller block size implies better spatial locality  
A smaller block size implies a smaller cache tag and hence lower cache tag overhead  
A smaller block size implies a larger cache tag and hence lower cache hit time  
A smaller block size incurs a lower cache miss penalty 
Question 54 
If the associativity of a processor cache is doubled while keeping the capacity and block size unchanged, which one of the following is guaranteed to be NOT affected?
Width of tag comparator  
Width of set index decoder  
Width of way selection multiplexor  
Width of processor to main memory data bus 
Width of set index decoder also will be affected when set offset is changed.
A kway set associative cache needs kto1 way selection multiplexer. If the associativity is doubled the width of way selection multiplexer will also be doubled.
With of processor to main memory data bus is guaranteed to be NOT affected as this is not dependent on the cache associativity.
Question 55 
The value of a float type variable is represented using the singleprecision 32bit floating point format of IEEE754 standard that uses 1 bit for sign, 8 bits for biased exponent and 23 bits for mantissa. A float type variable X is assigned the decimal value of −14.25. The representation of X in hexadecimal notation is
C1640000H  
416C0000H  
41640000H  
C16C0000H 
(14.25)_{10} = 1110.01000
= 1.11001000 x 2^{3}
23 bit Mantissa = 11001000000000000000000
Biased Exponent = exponent + bias
= 3 + 127 = 130 = 1000 0010
(14.25) in 32bit IEEE754 floating point representation is
1 10000010 11001000000000000000000
= 1100 0001 0110 0100 0000 0000 000 0000
= (C 1 6 4 0 0 0 0)_{16}
Question 56 
In the NewtonRaphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function
0.75x^{3} – 2x^{2} – 2x + 4 = 0
Consider the statements
(I) x_{3} = 0. (II) The method converges to a solution in a finite number of iterations.
Which of the following is TRUE?
Only I  
Only II  
Both I and II  
Neither I nor II 
Question 57 
The product of the nonzero eigenvalues of the matrix
1 0 0 0 1 0 1 1 1 0 0 1 1 1 0 0 1 1 1 0 1 0 0 0 1
is ______.
6  
7  
8  
9 
AX = λX
x_{1} + x_{5} = λx_{1}  (1)
x_{1} + x_{5} = λx_{5}  (2)
(1) + (2) ⇒ 2(x_{1} + x_{5}) = λ(x_{1} + x_{5}) ⇒ λ_{1} = 2
x_{2} + x_{3} + x_{4} = λ∙x_{2}  (4)
x_{2} + x_{3} + x_{4} = λ∙x_{3}  (5)
x_{2} + x_{3} + x_{4} = λ∙x_{4}  (6)
(4)+(5)+(6) = 3(x_{2} + x_{3} + x_{4}) = λ(x_{2} + x_{3} + x_{4} ) ⇒ λ_{2} = 3
Product = λ_{1} × λ_{2} = 2×3 = 6
Question 58 
The probability that a given positive integer lying between 1 and 100 (both inclusive) is NOT divisible by 2, 3 or 5 is ______.
0.259 to 0.261  
0.260 to 0.262  
0.261 to 0.263  
0.262 to 0.264 
n(A)=50, n(B)=33, n(C)=20
n(A∩B)=16, n(B∩C)=6, n(A∩C)=10
n(A∩B∩C)=3
P(A∪B∪C) = P(A)+P(B)+P(C)P(A∩B)P(B∩C) P(A∩C)+P(A∩B∩C) = 74/100
∴ Required probability is P(A∩B∩C) = 1P(A∪B∪C) = 0.26
Question 59 
The number of distinct positive integral factors of 2014 is _________.
0.26  
0.27  
8  
0.29 
= 2' × 19' × 53'
Now number of distinct integral factors of 2014 will be,
(1+1)×(1+1)×(1+1) = 2×2×2 = 8
Question 60 
Consider the following relation on subsets of the set S of integers between 1 and 2014. For two distinct subsets U and V of S we say U < V if the minimum element in the symmetric difference of the two sets is in U. Consider the following two statements:
S1: There is a subset of S that is larger than every other subset. S2: There is a subset of S that is smaller than every other subset.
Which one of the following is CORRECT?
Both S1 and S2 are true  
S1 is true and S2 is false  
S2 is true and S1 is false  
Neither S1 nor S2 is true 
U⊂S, V⊂S
Let U = {1, 2, 3}
V = {2, 3, 4}
Symmetric difference:
(U – V) ∪ (V – U) = {1} ∪ {4} = {1, 4}
The minimum element in the symmetric difference is 1 and 1∈U.
S1: Let S = Universal set = {1, 2, … 2014}
This universal set is larger than every other subset.
S2: Null set is smaller than every other set.
Let U = { }, V = {1}
Symmetric difference = ({ } – {1}) ∪ ({1} – { }) = { } ∪ {1} = {1}
So, U < V because { } ∈ U.
Question 61 
A cycle on n vertices is isomorphic to its complement. The value of n is _____.
5  
6  
7  
8 
In a cycle of n vertices, each vertex is connected to other two vertices. So each vertex degree is 2.
When we complement it, each vertex will be connected to remaining n3 vertices ( one is self and two other vertices in actual graph).
As per given question,
n3 =2
n=5
Cycle of 5 vertices is
Complement of the above graph1 is
Graph1 and Graph2 are complement each other.
So, the value of n is 5.
Question 62 
The number of distinct minimum spanning trees for the weighted graph below is _______.
6  
7  
8  
9 
Minimum Spanning Tree:
From the diagram, CFDA gives the minimum weight so will not disturb them, but in order to reach BE=1 we have 3 different ways ABE/ DBE/ DEB and we have HI=1, the shortest weight, we can reach HI=1 through GHI/ GIH.
So 3*2=6 ways of forming Minimum Spanning Tree with sum as 11.
Question 63 
Which one of the following Boolean expressions is NOT a tautology?
((a ⟶ b) ∧ (b ⟶ c)) ⟶ (a ⟶ c)  
(a ⟷ c) ⟶ (∽ b ⟶ (a ∧ c))  
(a ∧ b ∧ c) ⟶ (c ∨ a)  
a ⟶ (b ⟶ a) 
((a → b) ∧ (b → c)) → (a → c)
If (a → b) is false with a = T, b = F,
then (F ∧ (b → c)) → (a → c)
F → (a → c)
which is True for any (a → c)
This is tautology.
B:
(a ⟷ c) ⟶ (∽b ⟶ (a ∧ c))
For (a ⟷ c) be True and
∽b → (a ∧ c) should be False
Let a = c = F
(F → F) → (∽b (F ∩ F))
T → (∽b → F)
This is False for b = F
So, this is not True.
C:
(a ∧ b ∧ c) ⟶ (c ∨ a)
(c ∨ a) is False only for a = c = F
if (c ∨ a) is False
(F ∧ b ∧ F) → F
F → F which is Tautology
True always.
D:
a ⟶ (b ⟶ a)
a ⟶ (~b ∨ a)
(~a ∨ a) ∨ ~b = T ∨ ~b = T which is tautology
Question 64 
SQL allows tuples in relations, and correspondingly defines the multiplicity of tuples in the result of joins. Which one of the following queries always gives the same answer as the nested query shown below:
select * from R where a in (select S.a from S)
select R.* from R,S where R.a=S.a  
select distinct R.* from R,S where R.a=S.a  
select R.* from R,(select distinct a from S) as S1 where R.a=S1.a  
select R.* from R,S where R.a=S.a and is unique R 
Question 65 
Consider a main memory system that consists of 8 memory modules attached to the system bus, which is one word wide. When a write request is made, the bus is occupied for 100 nanoseconds (ns) by the data, address, and control signals. During the same 100 ns, and for 500 ns thereafter, the addressed memory module executes one cycle accepting and storing the data. The (internal) operation of different memory modules may overlap in time, but only one request can be on the bus at any time. The maximum number of stores (of one word each) that can be initiated in 1 millisecond is ____________.
10000  
10001  
10002  
10003 
Storing of data requires 100 n.s.
In 100 n.s – 1 store
100/10^{6}m.s = 1 store
1 m.s = 10^{6}/100stores = 10000 stores