Question 1
Choose the most appropriate phrase from the options given below to complete the following sentence.
`India is a post-colonial country because`
 A it was a former British colony B Indian Information Technology professionals have colonized the world C India does not follow any colonial practices D India has helped other countries gain freedom
English    GATE-CS-2014-(Set-2)
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Question 1 Explanation:
A country is called postcolonial if it came into existence after the colonies of the British and the Europeans were abolished and the countries then under their rule were declared independent. India was under the British colonial rule till 1947, i.e. it was a former British colony and thus is called a postcolonial country. So, A is the correct option.
 Question 2
Who ___________ was coming to see us this evening?
 A you said B did you say C did you say that D had you
English    GATE-CS-2014-(Set-2)
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Question 2 Explanation:
Only (B) makes sense, others don't fit.
 Question 3
Match the columns.
```   Column 1          Column 2
2) distort         Q) soak completely
3) saturate        R) use
4) utilize         S) destroy utterly ```
 A 1:S, 2:P, 3:Q, 4:R B 1:P, 2:Q, 3:R, 4:S C 1:Q, 2:R, 3:S, 4:P D 1:S, 2:P, 3:R, 4:Q
English    GATE-CS-2014-(Set-2)
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Question 3 Explanation:
"eradicate" means "destroy utterly" "distort" matches with "misrepresent" "saturate" matches with "soak completely" "utilize" matches with "use"
 Question 4
What is the average of all multiples of 10 from 2 to 198?
 A 90 B 100 C 110 D 120
General Aptitude    GATE-CS-2014-(Set-2)
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Question 4 Explanation:
```From 2 to 198 there are 19 multiples of 10.

These are 10,20,30.....180,190.

This is an A.P series, whose sum is (n/2{a1 + aN}),
where a1 and aN are the 1st and last terms respectively.

Putting in the formula,

a1 = 10, aN = 190.

Sum = Sn = (19/2{10 + 190 }) = 1900

Average = Sum / Total number = 1900 / 19 = 100. ```
 Question 5
 A 3.464 B 3.932 C 4 D 4.444
General Aptitude    GATE-CS-2014-(Set-2)
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Question 5 Explanation:
```let x = the given expression.
Now squaring both sides.

(x)^2 = 12 + x
(x)^2 - x - 12 = 0

Now solve this quadratic equation for finding the
roots(value of x, or this expression).
(x-4)(x+3) = 0

hence x = 4.
```
 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?
 A Kaliningrad was historically Russian in its ethnic make up B Kaliningrad is a part of Russia despite it not being contiguous with the rest of Russia C Koenigsberg was renamed Kaliningrad, as that was its original Russian name D Poland and Lithuania are on the route from Kaliningrad to the rest of Russia
English    GATE-CS-2014-(Set-2)
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Question 6 Explanation:
Explanation: A is not true in the light of the given facts
• A is incorrect because First line says that Kaliningrad (Koenigsberg before war) had a majority of the German population before the war. So, it was historically German and not Russian.
• B is correct as although Kaliningrad is not contiguous with the rest of Russia (being surrounded by countries of Poland in the south and west, Lithuania in the east and Baltic sea on the north), it has a predominantly Russian population.
• C cannot be inferred from the passage as it is nowhere in the passage what the original Russian name of Koenigsberg was.
• D is also not true because no data about the route is mentioned in the passage.
So, B is the correct option
 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 A high proportion of the affected population has returned from neighbouring countries where dengue is prevalent B More cases of dengue are now reported because of an increase in the Municipal Office’s administrative efficiency C Many more cases of dengue are being diagnosed this year since the introduction of a new and effective diagnostic test D The number of people with malarial fever (also contracted from mosquito bites) has increased this year
English    GATE-CS-2014-(Set-2)
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Question 7 Explanation:
Explanation:
• A contradicts the conclusion of the municipal authorities. So, A is not the correct choice.
• B is an incorrect choice because there is no data in the passage relating the reporting of cases and the efficiency of the administrative capabilities of the municipal authorities.
• C is also incorrect as it contradicts the conclusion of the municipal authorities.
• D is the correct choice as both malarial fever and dengue fever are caused by mosquito bite. If dengue fever had increased because of some other reason other than that concluded by the municipal authorities, then there would not have been increase in the people with malarial fever. This statement supports the conclusion of the municipal authorities.
 Question 8
If x is real and |x2 - 2x + 3| = 11, then possible values of |- x3 + x2 - x| include
 A 2, 4 B 2, 14 C 4, 52 D 14, 52
General Aptitude    GATE-CS-2014-(Set-2)
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Question 8 Explanation:
```Here we use the modulus property, which says:

|x| = x when x >= 0

|x| = -x when x < 0

i.e. range of a modulus function is always positive.

Now, given that |x^2 – 2x + 3| = 11, we can say that

x^2 – 2x + 3 = +11 ----------------(1)

and

x2 – 2x + 3 = -11------------------(2)

Solving 1st equation, we get real roots as 4 and -2.

Solving 2nd eq, we get imaginary roots, hence we ignore them.

Now, for eq |- x^3 + x^2 – x|, we put 4 and -2 in place of x.

putting x = 4, we get |-4^3 + 4^2-4| = |-64+16-4| =  52

putting x = -2 we get |-(-2)^3 + (-2)^2 - (-2)| = 14

so |- x^3 + x^2 – x| has possible values as 52 and 14 ```
 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?
 A 120 B 140 C 160 D 180
General Aptitude    GATE-CS-2014-(Set-2)
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Question 9 Explanation:
```Let x represent the males and y represent the females.

In 2008.
x / y = 2.5 -----------(1)

In 2009.
M / 2y = 3-------------(2)

where M is the total number of males in 2009.

Now x = 2.5y (from (1))
and M = 6y ( from (2))

hence increased number of males in 2009 are
M - x = 6y - 2.5y = 3.5y

Now the increase in % of males =

(change in no of males / initial number)*100

= (3.5y / x )*100 = (3.5 / 2.5)*100 = 140
(given x/y = 2.5)

Hence, B is the right option.```
 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 60o
 A 6:22 am B 6:27 am C 6:38 am D 6:45 am
General Aptitude    GATE-CS-2014-(Set-2)
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Question 10 Explanation:
```Speed of the Hour hand = ( 360 degree ) / ( 12 * 60 minutes )
= 0.5 degree / minute.
Speed of the minute hand = ( 360 degree ) / ( 60 minutes )
= 6 degree / minute.```
At 6 am, hour hand will point vertically downward to 6 in the clock, and minute hand will point vertically upward to 12 in the clock. ( Hence they are opposite with a gap of 180 degree between them). Now after t minutes , The angle between these two hands will be either of the below two:
```
(180 - 6t) + 0.5t, when the minute hand is
behind the hour hand or,

6t - (180 + 0.5t), when the minute hand is

Now the the question asks for angle of 60 degree.

Hence,
(180 - 6t) + 0.5t = 60 , we get t = 21.8 minute
and
6t - ( 180 + 0.5t ) = 60, we get t = 48 minutess```
Hence the closest to answer is 22 minutes, i.e. 6.22 am. Hence answer is A.
 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 = _____________.
 A 11.85
Probability    GATE-CS-2014-(Set-2)
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Question 11 Explanation:
```Total ways to pick 4 computers = 10*9*8*7

Total ways that at least three computers are fine =
Total ways that all 4 are fine + Total ways any 3 are fine

Total ways that all 4 are fine = 4*3*2*1

Total ways three are fine = 1st is Not working and other 3 working +
2nd is Not working and other 3 working +
3rd is Not working and other 3 working +
4th is Not working and other 3 working +
= 6*4*3*2 + 4*6*3*2 + 4*3*6*2 + 4*3*2*6
= 6*4*3*2*4

The probability = Total ways that at least three computers are fine /
Total ways to pick 4 computers
=  (4*3*2*1 + 6*4*3*2*4) / (10*9*8*7)
= (4*3*2*25) / (10*9*8*7)
= 11.9% ```
 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.)
 A 3.8 to 3.9
Probability    GATE-CS-2014-(Set-2)
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Question 12 Explanation:
There are total 9 words in the sentence. To find expected length first we count total length of the sentence i.e 35. Now to find expected length divide total length by total words i.e 35/9 so answer is 3.8 or 3.9 Another Explanation: Expected value = ∑( x * P(x) ) = 3*4/9 + 4*2/9 + 5*3/9 = 35/9 = 3.9
 Question 13
The maximum number of edges in a bipartite graph on 12 vertices is __________________________.
 A 36
Graph Theory    GATE-CS-2014-(Set-2)
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Question 13 Explanation:
Number of edges would be maximum when there are 6 edges on each side and every vertex is connected to all 6 vertices of the other side.
 Question 14
If the matrix A is such that then the determinant of A is equal to
 A 0
Linear Algebra    GATE-CS-2014-(Set-2)
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Question 14 Explanation:
This is a numerical answer question of gate paper, in which no options are provided, and the answer is to given by filling a numeral into a text box provided. In the question, matrix A is given as the product of 2 matrices which are of order 3 x 1 and 1 x 3 respectively. So after multiplication of these matrices, matrix A would be a square matrix of order 3 x 3. So, matrix A is : 2 18 10 -4 -36 -20 7 63 35 Now, we can observe by looking at the matrix that row 2 can be made completely zero by using row 1, this is to be done by using the row operation of matrix which here is : R2 <- R2 + 2R1 After applying above row operation in the matrix, the resultant matrix would be: 2 18 10 0 0 0 7 63 35 i.e. Row 2 has become zero now. And if a square matrix has a row or column with all its elements as 0, then its determinant is 0. ( A property of a square matrix ) Hence answer is 0. Note: Determinant is defined only for square matrices, and it is a number which encodes certain properties of a matrix, for ex: a square matrix with determinant 0 does not has its inverse matrix.
 Question 15
A non-zero polynomial f(x) of degree 3 has roots at x = 1, x = 2 and x = 3. Which one of the following must be TRUE?
 A f(0)f(4) < 0 B f(0)f(4) > 0 C f(0) + f(4) < 0 D f(0) + f(4) > 0
Numerical Methods and Calculus    GATE-CS-2014-(Set-2)
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Question 15 Explanation:
The graph of a degree 3 polynomial f(x) = a0 + a1x + a2(x^2) + a3(x^3), where a3 ≠ 0 is a cubic curve, as can be seen here https://en.wikipedia.org/wiki/... Now as given, the polynomial is zero at x = 1, x = 2 and x = 3, i.e. these are the only 3 real roots of this polynomial. Hence we can write the polynomial as f(x) = K (x-1)(x-2)(x-3) where K is some constant coefficient. Now f(0) = -6K and f(4) = 6K ( by putting x = 0 and x = 4 in the above polynomial ) and f(0)*f(4) = -36(k^2), which is always negative. Hence option A. We can also get the answer by just looking at the graph. At x < 1, the cubic graph (or say f(x) ) is at one side of x-axis, and at x > 3 it should be at other side of x-axis. Hence +ve and -ve values, whose multiplication gives negative.
 Question 16
The dual of a Boolean function f(x1, x2, … , xn, +, ∙ , ′ ), written as FD, is the same expression as that of F with + and . swapped. F is said to be self-dual if F = FD. The number of self-dual functions with n Boolean variables is
 A 2n B 2n-1 C 22n D 22n-1
GATE-CS-2014-(Set-2)
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Question 16 Explanation:
The question asks for no: of self-dual functions for n variables. Principle of Duality: Any theorem or identity in Boolean algebra remains true if 0 and 1 are swapped and . and + are swapped throughout. There are two properties for self-dual functions: 1. It is neutral (no of minterms = no of maxterms) 2. A single function does not contains two mutually exclusive terms. Considering the above properties. If we have n-variable then we have 2^n minterms/maxterms From 2^n minterms/maxterms there are (2^n)/2 mutually exclusive pairs. ie 2^(n-1) So we have 2^(n-1) pairs to use to make self-dual functions. So, by Fundamental principle of counting, because each pair in 2^(n-1) has two choices. No of self-dual functions from n-variables are = 2*2*2...2^(n-1) times = 2^(2^(n-1)) or example if n=3 We have minterms as(000,001,010,...,111) Mutually exclusive pairs are (0,7),(1,6),(2,5),(3,4) The pairs are mutually exclusive since they cannot come in a self-dual function together. So here we have.2*2*2*2 functions ie 16.
 Question 17
Let k = 2n. A circuit is built by giving the output of an n-bit binary counter as input to an n-to-2n bit decoder. This circuit is equivalent to a
 A k-bit binary up counter. B k-bit binary down counter. C k-bit ring counter. D k-bit Johnson counter.
Digital Logic & Number representation    GATE-CS-2014-(Set-2)
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Question 17 Explanation:
For output of a decoder , only single output will be ‘1’ and remaining will be ‘0’ at the same time. So high output  will give the count of the ring counter. Hence Ans is ( C) part.
 Question 18
Consider the equation (123)5 = (x8)y with x and y as unknown. The number of possible solutions is _____ .
 A 1 B 2 C 3 D 4
Digital Logic & Number representation    GATE-CS-2014-(Set-2)
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Question 18 Explanation:
Changing (123) base 5 into base 10= 1*25+2*5+3*1=38 Changing x8 base y in decimal= x*y+8 Equating both we get xy+8=38
• xy=30
• possible combinations =(1,30),(2,15),(3,10)
but we have ‘8’ present in x8 so base y>8 as all three are satisfying the conditions so total solutions =3 hence ans is ( C) part
 Question 19
A 4-way set-associative 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 _____
 A 5 B 15 C 20 D 25
Computer Organization and Architecture    GATE-CS-2014-(Set-2)
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Question 19 Explanation:
In a k-way set associate mapping, cache memory is divided into sets, each of size k blocks. Size of Cache memory = 16 KB As it is 4-way set associative,K = 4 Block size B = 8 words The word length is 32 bits. size of Physical address space = 4 GB. --------------------------------------------------- No of blocks in Cache Memory(N) = (size of cache memory / size of a block) = (16*1024 bytes / 8*4 bytes) = 512 (as 1 word = 4 bytes) No of sets(S) = (No of blocks in cache memory/ no of blocks in a set) = N/K = 512/4 = 128 Now,size of physical address = 4GB = 4*(2^30) Bytes = 2^32 Bytes These physical adresses are divided equally among the sets. Hence, each set can access ((2^32)/128) bytes = 2^25 bytes = 2^23 words = 2^20 blocks So, each set can access total of 2^20 blocks. So to identify these 2^20 blocks, each set needs TAG bits of length 20 bits. Hence option C.
 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 __________.
 A 8 B 9 C 10 D 11
GATE-CS-2014-(Set-2)
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Question 20 Explanation:
The function mainly returns position of Most significant bit in binary representation of n. The MSD in binary representation of 435 is 9th bit. Another explanation: >> in right shift. In other words, it means divide by 2. If keep on dividing by 2, we get: 435, 217, 108, 54, 27, 13, 6, 3, 1. Therefore, the count is 9.
 Question 21
Suppose n and p are unsigned int variables in a C program. We wish to set p to nC3. If n is large, which of the following statements is most likely to set p correctly?
 A p = n * (n-1) * (n-2) / 6; B p = n * (n-1) / 2 * (n-2) / 3; C p = n * (n-1) / 3 * (n-2) / 2; D p = n * (n-1) * (n-2) / 6.0;
Data Types    GATE-CS-2014-(Set-2)
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Question 21 Explanation:
As n is large, the product n*(n-1)*(n-2) will go out of the range(overflow) and it will return a value different from what is expected. So we consider a shorter product n*(n-1). n*(n-1) is always an even number. So the subexpression "n * (n-1) / 2 " in option B would always produce an integer, which means no precision loss in this subexpression. And when we consider `n*(n-1)/2*(n-2)`, it will always give a number which is a multiple of 3. So dividing it with 3 won't have any loss.
 Question 22
A priority queue is implemented as a Max-Heap. Initially, it has 5 elements. The level-order 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 level-order traversal of the heap after the insertion of the elements is:
 A 10, 8, 7, 3, 2, 1, 5 B 10, 8, 7, 2, 3, 1, 5 C 10, 8, 7, 1, 2, 3, 5 D 10, 8, 7, 5, 3, 2, 1
Heap    GATE-CS-2014-(Set-2)
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Question 22 Explanation:
```Initially heap has 10, 8, 5, 3, 2
10
/  \
8    5
/ \
3   2

After insertion of 1
10
/   \
8     5
/ \   /
3   2 1
No need to heapify as 5 is greater than 1.

After insertion of 7
10
/   \
8     5
/ \   / \
3   2 1   7
Heapify 5 as 7 is greater than 5
10
/   \
8     7
/ \   / \
3   2 1   5
No need to heapify any further as 10 is
greater than 7 ```
 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 `
 A Θ(n) B Θ(nLogn) C Θ(n*n) D Θ(log n)
Analysis of Algorithms (Recurrences)    GATE-CS-2014-(Set-2)
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Question 23 Explanation:
``` T(n) = 2T(n/2) + log n
T(1) = 1
Substitute n = 2^k

T(2^k)  = k + 2T(2^(k-1))
T(2^k)  = k + 2(k-1) + 4T(2^(k-2))
= k + 2(k-1) + 4(K-2) + 8T(2^(k-3))
= k + 2(k-1) + 4(K-2) + 8(k-3) + 16T(2^(k-4))
= k + 2(k-1) + 4(K-2) + 8(k-3) + ...... + 2^kT(2^(k-k))
= k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^kT(1)
= k + 2(k-1) + 4(K-2) + 8(k-3) + .......+ 2^k  --------(1)

2T(2^k) =     2k     + 4(k-1) + 8(K-2) + ...... + 2*2^k + 2^(k+1) --------(2)

Subtracting 1 from 2, we get below
T(2^k) = - k + 2 + 4 ......    2^(k-2) + 2^(k-1) + 2^k + 2^(k+1)
= - k + 2 * (1 + 2 + 4 + ..... 2^k)
= -k + [2*(2^k - 1)] / [2-1]
= -k + [2*(2^k - 1)]

T(n) = -Logn + 2*(n - 1)

T(n)  = 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.
 A the shortest path between every pair of vertices. B the shortest path from W to every vertex in the graph. C the shortest paths from W to only those nodes that are leaves of T. D the longest path in the graph
Graph Traversals    GATE-CS-2014-(Set-2)
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Question 24 Explanation:
BFS always produces shortest path from source to all other vertices in an unweighted graph.
 Question 25
Which one of the following is CORRECT?
 A Only (I) B Only (II) C Both (I) and (II) D Neither (I) nor (II)
Regular languages and finite automata    GATE-CS-2014-(Set-2)
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 Question 26
Let A ≤m B denotes that language A is mapping reducible (also known as many-to-one reducible) to language B. Which one of the following is FALSE?
 A If A ≤m B and B is recursive then A is recursive. B If A ≤m B and A is undecidable then B is undecidable. C If A ≤m B and B is recursively enumerable then A is recursively enumerable. D If A ≤m B and B is not recursively enumerable then A is not recursively enumerable.
Recursively enumerable sets and Turing machines    GATE-CS-2014-(Set-2)
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Question 26 Explanation:
• A ≤m B means language A is mapping reducible to language B.Thus, A cannot be harder than B. Since, A can be reduced to B, instead of deciding A we can now decide B. So, the first three options are correct.
• As B is not recursively enumerable, it doesn't guarantee A is not recursively enumerable.Thus, if A ≤m B and B is not recursively enumerable then A is not recursively enumerable. Therefore, answer is D is correct
Please comment below if you find anything wrong in the above post.
 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?
 A + is left associative, while ∗ is right associative B + is right associative, while ∗ is left associative C Both + and ∗ are right associative D Both + and ∗ are left associative
Parsing and Syntax directed translation    GATE-CS-2014-(Set-2)
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Question 27 Explanation:
From the grammar we can find out associative by looking at grammar.
```Let us consider the 2nd production
T -> T * U
T is generating T*U recursively (left recursive) so * is
left associative.

Similarly
P -> Q + P
Right recursion so + is right associative.
So option B is correct. ```
NOTE: Above is the shortcut trick that can be observed after drawing few parse trees. One can also find out correct answer by drawing the parse tree.
 Question 28
Which one of the following is NOT performed during compilation?
 A Dynamic memory allocation B Type checking C Symbol table management D Inline expansion
Principles of Programming Languages    GATE-CS-2014-(Set-2)
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Question 28 Explanation:
Dynamic memory allocation happens at run time only. Compiler only compiles instructions for dynamic memory allocation like malloc(), calloc().
 Question 29
Which one of the following is TRUE?
 A The requirements document also describes how the requirements that are listed in the document are implemented efficiently. B Consistency and completeness of functional requirements are always achieved in practice. C Prototyping is a method of requirements validation. D Requirements review is carried out to find the errors in system design
Software Engineering    GATE-CS-2014-(Set-2)
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 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 x 106 bytes disk on which the file system is stored and data block size is 103 bytes, the maximum size of a file that can be stored on this disk in units of 106 bytes is ____________.
 A 99.55 to 99.65
File structures (sequential files, indexing, B and B+ trees)    GATE-CS-2014-(Set-2)
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Question 30 Explanation:
```Here block size is 10^3 B.
No. of entries in the FAT = Disk capacity/ Block size
= 10^8/10^3
= 10^5
Total space consumed by FAT = 10^5 *4B
= 0.4*10^6B
Max. size of file that can be stored = 100*10^6-0.4*10^6
= 99.6*10^6B.
 Question 31
The maximum number of superkeys for the relation schema R(E,F,G,H) with E as the key is
 A 5 B 6 C 7 D 8
Database Design(Normal Forms)    GATE-CS-2014-(Set-2)
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Question 31 Explanation:
Maximum no. of possible superkeys for a table with n attributes = 2^(n-1) Here, n = 4. So, the possible superkeys = 24-1 = 8 The possible superkeys 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
 A 18 B 19
Database Design(Normal Forms)    GATE-CS-2014-(Set-2)
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Question 32 Explanation:
There is already an entry with same name and age as 19. So the age of this entry must be something other than 19.
 Question 33
Which one of the following is TRUE interior Gateway routing protocols - Routing Information Protocol (RIP) and Open Shortest Path First (OSPF)
 A RIP uses distance vector routing and OSPF uses link state routing B OSPF uses distance vector routing and RIP uses link state routing C Both RIP and OSPF use link state routing D Both RIP and OSPF use distance vector routing
Network Layer    GATE-CS-2014-(Set-2)
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Question 33 Explanation:
Both Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) are Interior Gateway Protocol, i.e., they both are used within an autonomous system. RIP is an old protocol (not used anymore) based on distance vector routing. OSPF is based Link State Routing.
 Question 34
Which one of the following socket API functions converts an unconnected active TCP socket into a passive socket.
 A connect B bind C listen D accept
Transport Layer    GATE-CS-2014-(Set-2)
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Question 34 Explanation:
listen() marks the socket referred to by sockfd as a passive socket, that is, as a socket that will be used to accept incoming connection requests using accept(). Source: http://linux.die.net/man/2/listen
 Question 35
In the diagram shown below L1 is an Ethernet LAN and L2 is a Token-Ring 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
 A 25 B 24 C 26 D 28
Discuss it

Question 35 Explanation:
TTL(Time to Live) - It is a mechanism that limits the lifespan of a packet in a computer network. It is implemented with the help of a counter or timestamp which is set by the sender of the packet and embedded in the packet itself. It represents the maximum lifetime of a packet in the network. When a packet routes through a network, each router checks the current value of its TTL, if the TTL value is not zero then only the router accepts the packet, and decrements its value by 1. This process takes place at every router. If some router founds the TTL value of the incoming packet to be 0, then it simply discards/destroys the packet ( because the lifetime of the packet is over, and hence the packet is not eligible to be in the network ). One of the main purposes of setting the TTL value and doing all this process is to ensure that there is no undelivered packet in the network which is circulating indefinitely, and to avoid the problem of duplicate delivery of the same packet, which may arise in the case of network congestion. Now, routing decisions are taken place at the network layer. hence we have to see in the above question that when the packet is going through the network layer. LANs work at Data Link Layer only, hence the packet doesn't reach network layer in LANs. So in the question above, Except LANs, at all other points routing decisions have to be taken. Hence TTL value will be checked and manipulated at those points/routers. At Receiver end also, the packet has to go through the network layer so as to reach to application layer, hence Receiver will also check and decrement the TTL value. So there are 6 routers in the above diagram. 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 106 bytes / sec. A user on host A sends a file of size 103 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?
 A T1 < T2 < T3 B T1 > T2 > T3 C T2 = T3, T3 < T1 D T1 = T3, T3 > T2
Discuss it

Question 36 Explanation:
The important thing to note here is in first case, the whole packet is being transmitted, so no piplelining of packet happens. In second and third case, we have advantage of pipelining (While packet 'i' is being transmitted from R1 to R2, packet 'i-1' is being transmitted from A to R1 at the same time). Following are complete calculations.
```File Size = 1000 bytes
Transmission Speed of all  links = 10^6 bytes/sec

Ist Case:
= packetsize/bandwidth
= (1000 + 100)/10^6
= 1100 micros
Total time = 3*1100
= 3300 microsec.

Second case:
Transmission time for one link and one part
= (100 + 100)/10^6
=  200 microsec

[Note the pipe-lining in packets.  While
packet 'i' is being transmitted from R1 to R2,
packet 'i-1' is being transmitted from A to R1
at the same time]
Total time = 3*200 + 9*200
= 2400 micro sec

Third Case:
Transmission time for one link and one part
= (50+100)/10^6
= 150microsec
Total time = 3*150+19*150
= 3300 microsec ```
 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
Which of I1, I2, I3, and I4 can an intruder learn through sniffing at R2 alone?
 A Only I1 and I2 B Only I1 C Only I2 and I3 D Only I3 and I4
Network Layer    GATE-CS-2014-(Set-2)
Discuss it

Question 37 Explanation:
```An Intruder can’t learn [I1] through sniffing at R2 because

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)?
 A Q needs to send at least 2 HTTP requests to S, each necessarily in a separate TCP connection to server S B Q needs to send at least 2 HTTP requests to S, but a single TCP connection to server S is sufficient C A single HTTP request from Q to S is sufficient, and a single TCP connection between Q and S is necessary for this D A single HTTP request from Q to S is sufficient, and this is possible without any TCP connection between Q and S
Application Layer    GATE-CS-2014-(Set-2)
Discuss it

Question 38 Explanation:
Whenever a browser opens a webpage, it makes a separate request for each object of page like image, css, javascript, etc. However if multiple resources are served from same server, then one TCP connect is sufficient.
 Question 39
Consider the following schedule S of transactions T1, T2, T3, T4: Which one of the following statements is CORRECT?
 A S is conflict-serializable but not recoverable B S is not conflict-serializable but is recoverable C S is both conflict-serializable and recoverable D S is neither conflict-serializable nor is it recoverable
Transactions and concurrency control    GATE-CS-2014-(Set-2)
Discuss it

Question 39 Explanation:
To check for conflict-serializable, we need to make a precedence graph, if the graph contains a cycle, then it's not conflict serializable, else it is. Here, for the precedence graph there will be only two directed edges, one from T2 -> T3 ( Read- Write Conflict), and another from T2 -> T1( Read- Write Conflict), hence no cycle, so the schedule is conflict serializable. Now to check for Recoverable, we need to check for a dirty-read operation( Write by Transaction Ti, followed by Read by Transaction Tj but before Ti commits) between any pair of operations. If no dirty-read then recoverable schedule, if a dirty read is there then we need to check for commit operations. Here no dirty read operation ( as T3 and T1 commits before T4 reads the Write(X) of T3 and T1 , and T2 commits before T4 reads the Write(Y) of T2 ). Therefore the schedule is recoverable. Hence, Option C.
 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
 A relation r(R) is in the outer loop. B relation s(S) is in the outer loop. C join selection factor between r(R) and s(S) is more than 0.5. D join selection factor between r(R) and s(S) is less than 0.5.
ER and Relational Models    GATE-CS-2014-(Set-2)
Discuss it

Question 40 Explanation:
Nested loop join is one of the methods to implement database in memory. A nested loop join is an  algorithm that joins two sets by using two nested loops.
According to nested join,given relation R and S   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 <r,s> Cost estimations for the above loop: – b(R) and  b(S) number of blocks in R and in S – Each block of outer relation is read once – Inner relation is read once for each block of outer relation Summing up : IO= b(R)+b(R)*b(S) total  IO operations Lets assume |R|>|S|  i.e b(R) =10  and b(s) =3 Now,   if R is outer relation then, IO= 10+10*3=40 if S is outer relation then IO=3+10*3=33 As it can be observed , that total IO is lesser if the value of outer variable is less and as it is already given that |R|<|S|.Therefore,  Relation r(R) should be in the outer loop to have fewer number of disk block accesses. References:
 Question 41
Consider the procedure below for the Producer-Consumer problem which uses semaphores: Which one of the following is TRUE?
 A The producer will be able to add an item to the buffer, but the consumer can never consume it. B The consumer will remove no more than one item from the buffer. C Deadlock occurs if the consumer succeeds in acquiring semaphore s when the buffer is empty. D The starting value for the semaphore n must be 1 and not 0 for deadlock-free operation.
Process Management    GATE-CS-2014-(Set-2)
Discuss it

Question 41 Explanation:
Initially, there is no element in the buffer.
Semaphore s = 1 and semaphore n = 0.
We assume that initially control goes to the consumer when buffer is empty.
semWait(s) decrements the value of semaphore ‘s’ . Now, s = 0 and semWait(n) decrements the value of semaphore ‘n’. Since, the value of semaphore ‘n’ becomes less than 0 , the control stucks in while loop of function semWait() and a deadlock arises.

Thus, deadlock occurs if the consumer succeeds in acquiring semaphore s when the buffer is empty.

Please comment below if you find anything wrong in the above post.
 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 ___________.
 A 500 B 1000 C 2000 D 10000
GATE-CS-2014-(Set-2)    CPU Scheduling
Discuss it

Question 42 Explanation:
```There are three processes A, B and C that run in
round robin manner with time slice of 50 ms.

Processes atart at 0, 5 and 10 miliseconds.

The processes are executed in below order
A, B, C, A
50 + 50 + 50 + 50 (200 ms passed)

Now A has completed 100 ms of computations and
goes for I/O now

B, C, B, C, B, C
50 + 50 + 50 + 50 + 50 + 50 (300 ms passed)

C goes for i/o at 500ms and it needs 500ms to
finish the IO.

So C would complete its first IO at 1000 ms
```
 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?
 A Least-recently-used B First-in-first-out C Last-in-first-out D Most-recently-used
Memory Management    GATE-CS-2014-(Set-2)
Discuss it

Question 43 Explanation:
The optimal page replacement algorithm swaps out the page whose next use will occur farthest in the future. In the given question, the computer has 20 page frames and initially page frames are filled with pages numbered from 101 to 120. Then program accesses the pages numbered 1, 2, …, 100 in that order, and repeats the access sequence THRICE. The first 20 accesses to pages from 1 to 20 would definitely cause page fault. When 21st is accessed, there is another page fault. The page swapped out would be 20 because 20 is going to be accessed farthest in future. When 22nd is accessed, 21st is going to go out as it is going to be the farthest in future. The above optimal page replacement algorithm actually works as most recently used in this case. As a side note, the first 100 would cause 100 page faults, next 100 would cause 81 page faults (1 to 19 would never be removed), the last 100 would also cause 81 page faults.
 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?
 A X is declared as “int X[32][32][8]”. B X is declared as “int X[4][1024][32]”. C X is declared as “char X[4][32][8]”. D X is declared as “char X[32][16][2]”.
Arrays    GATE-CS-2014-(Set-2)
Discuss it

Question 44 Explanation:
The final expression can be simplified in form ofi, j and k by following the intermediate code steps in reverse order
```t5 = X[t4]
= X[t3 + t2]
= X[t1 + t0 + t2]
= X[i*1024 + j*32 + k*4]
= X + i*1024 + j*32 + k*4 ```
Since k is multiplied by 4, the array must be an int array. We are left with 2 choices (A and B) among the 4 given choices. X[i][j][k]'th element in one dimensional array is equivalent to X[i*M*L + j*L + k]'th element in one dimensional array (Note that multi-dimensional arrays are stored in row major order in C). So we get following equations
```j*L*4 = j*32, we get L = 8 (4 is the sizeof(int))
i*1024 = i*M*L*4, we get M = 1024/32 = 32```
Therefore option A is the only correct option as M and L are 32 and 8 respectively only in option A.
 Question 45
```Let L1 = {w ∈ {0,1}∗ | w has at least as many occurrences
of (110)’s as (011)’s}.

Let L2 = { ∈ {0,1}∗ | w has at least as many occurrences
of (000)’s as (111)’s}. ```
Which one of the following is TRUE?
 A L1 is regular but not L2 B L2 is regular but not L! C Both L2 and L1 are regular D Neither L1 nor L2 are regular
Regular languages and finite automata    GATE-CS-2014-(Set-2)
Discuss it

Question 45 Explanation:
L1 is regular let us consider the string 011011011011 In this string, number of occurrences of 011 are 4 but when we see here 110 is also occurred and the number of occurrence of 110 is 3. Note that if i add a 0 at the last of string we can have same number of occurrences of 011 and 110 so this string is accepted. We can say if the string is ending with 011 so by appending a 0 we can make 110 also. Now string2: 110110110110 in this number of occurrences of 110 is 4 and 011 is 3 which already satisfy the condition So we can observe here that whenever 110 will be there string will be accepted So with this idea we can build an automata for this. Therefore, it is regular.
 Question 46
Let < M > be the encoding of a Turing machine as a string over {0, 1}. Let L = { < M > |M is a Turing machine that accepts a string of length 2014 }. Then, L is
 A decidable and recursively enumerable B undecidable but recursively enumerable C undecidable and not recursively enumerable D decidable but not recursively enumerable
Undecidability    GATE-CS-2014-(Set-2)
Discuss it

Question 46 Explanation:
There are finite number of strings of length ‘2014’. So, a turing machine will take the input string of length ‘2014’ and test it.
If, input string is present in the language then turing machine will halt in final state . But, if turing machine is unable to accept the input string then it will halt in non-final state or go in an infinite loop and never halt.

Thus, ‘L’ is undecidable and recursively enumerable .

Please comment below if you find anything wrong in the above post.
 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 = ___.
 A 33 B 23 C 43 D 34
Dynamic Programming    GATE-CS-2014-(Set-2)
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Question 47 Explanation:
//The LCS is of length 4. There are 3 LCS of length 4 "qprr", "pqrr" and qpqr   A subsequence is a sequence that can be derived from another sequence by selecting zero or more elements from it, without changing the order of the remaining elements. Subsequence need not be contiguous. Since the length of given strings A = “qpqrr” and B = “pqprqrp” are very small, we don’t need to build a 5x7 matrix and solve it using dynamic programming. Rather we can solve it manually just by brute force. We will first check whether there exist a subsequence  of length 5 since min_length(A,B) = 5. Since there is no subsequence , we will now check for length 4. “qprr”, “pqrr” and “qpqr” are common in both strings. X = 4 and Y = 3 X + 10Y = 34   This solution is contributed by Pranjul Ahuja
 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 ____.
 A 358 B 438 C 568 D 664
GATE-CS-2014-(Set-2)
Discuss it

Question 48 Explanation:
To merge two lists of size m and n, we need to do m+n-1 comparisons in worst case. Since we need to merge 2 at a time, the optimal strategy would be to take smallest size lists first. The reason for picking smallest two items is to carry minimum items for repetition in merging. We first merge 20 and 24 and get a list of 44 using 43 worst case comparisons. Then we merge 30 and 35 into a list of 65 using 64 worst case comparisons. Then we merge 50 and 44 into a list of 94 using 93 comparisons. Finally we merge 94 and 65 using 158 comparisons. So total number of comparisons is 43 + 64 + 93 + 158 which is 358. Source: http://www.geeksforgeeks.org/data-structures-algorithms-set-34-2/
 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 ___.
 A 4 B 6 C 8 D 10
Misc    GATE-CS-2014-(Set-2)
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Question 49 Explanation:
An Expression Tree is a binary tree in which each internal node corresponds to operator and each leaf node corresponds to operand so for example expression tree for 3 + ((5+9)*2) would be:. Below diagram shows values to pick to get the maximum value in expresison tree
 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:_____.
 A 1.73 B 2.24 C 4.22 D 3.42
Recursion    GATE-CS-2014-(Set-2)
Discuss it

Question 50 Explanation:

This solution is contributed by Anil Saikrishna Devarasetty
 Question 51
Suppose 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 A queue cannot be implemented using this stack. B A queue can be implemented where ENQUEUE takes a single instruction and DEQUEUE takes a sequence of two instructions. C A queue can be implemented where ENQUEUE takes a sequence of three instructions and DEQUEUE takes a single instruction. D A queue can be implemented where both ENQUEUE and DEQUEUE take a single instruction each.
Queue    GATE-CS-2014-(Set-2)
Discuss it

Question 51 Explanation:
To DEQUEUE an item, simply POP. To ENQUEUE an item, we can do following 3 operations 1) REVERSE 2) PUSH 3) REVERSE
 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?
 A The function returns 0 for all values of j. B The function prints the string something for all values of j. C The function returns 0 when j = 50. D The function will exhaust the runtime stack or run into an infinite loop when j = 50
Recursion    GATE-CS-2014-(Set-2)
Discuss it

Question 52 Explanation:
When j is 50, the function would call itself again and again as neither i nor j is changed inside the recursion.
 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 A smaller block size implies better spatial locality B A smaller block size implies a smaller cache tag and hence lower cache tag overhead C A smaller block size implies a larger cache tag and hence lower cache hit time D A smaller block size incurs a lower cache miss penalty
Computer Organization and Architecture    GATE-CS-2014-(Set-2)
Discuss it

Question 53 Explanation:
Block : The memory is divided into equal size segments. Each segment is called a block. Data in cache is retrieved in form of blocks. The idea is to use Spatial Locality (Once a location is retrieved, it is highly probable that the nearby locations would be retrieved in near future). TAG bits : Each cache block is given a set of TAG bits to identify which main memory block is present in that cache block. Option A : If the block size is small, there would be less number of near-by address for future references by CPU to be present into that block. Hence this is not better spatial locality. Option B : If the block size is smaller, no of blocks would be more in cache, hence more cache tag bits would be needed, not less. Option C : Cache tag bits are more ( because more no of blocks due to smaller block size ), but more cache tag bits can't lower the hit time ( even it will increase ). Option D : If there is a miss at cache memory ( i.e. the needed block by the CPU is not present in the cache memory ), then that block has to be moved from next lower level of memory ( lets say main memory ) in the memory hierarchy, and if the block size is lower, then it takes less time to be placed into cache memory, hence less miss penalty. Hence option D.
 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?
 A Width of tag comparator B Width of set index decoder C Width of way selection multiplexor D Width of processor to main memory data bus
Computer Organization and Architecture    GATE-CS-2014-(Set-2)
Discuss it

Question 54 Explanation:
If associativity is doubled, keeping the capacity and block size constant, then the number of sets gets halved. So, width of set index decoder can surely decrease - (B) is false. Width of way-selection multiplexer must be increased as we have to double the ways to choose from- (C) is false As the number of sets gets decreased, the number of possible cache block entries that a set maps to gets increased. So, we need more tag bits to identify the correct entry. So, (A) is also false. (D) is the correct answer- main memory data bus has nothing to do with cache associativity- this can be answered without even looking at other options.
 Question 55
The value of a float type variable is represented using the single-precision 32-bit floating point format IEEE-754 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
 A C1640000H B 416C0000H C 41640000H D C16C0000H
GATE-CS-2014-(Set-2)    Number Representation
Discuss it

Question 55 Explanation:
```Since No is negative S bit will be 1
Convert 14.25 into binary 1110.01
Normalize it : 1.11001 X 2 ^ 3
Biased Exponent (Add 127) : 3 + 127 = 130 (In binary 10000010)
Mantissa : 110010.....0 (Total 23 bits)

Num represented in IEEE 754 single precision format :

1 10000010 11001000000000000000000

In Hex (Group of Four bits) -

1100 0001 0110 0100 0000 0000 0000 0000

Num becomes : C1640000
```
 Question 56
In the Newton-Raphson method, an initial guess of x0 = 2 is made and the sequence x0, x1, x2 … is obtained for the function
```
0.75x3 – 2x2 – 2x + 4 = 0

Consider the statements
(I) x3 = 0.
(II) The method converges to a solution in a finite number of iterations. ```
Which of the following is TRUE?
 A Only I B Only II C Both I and II D Neither I nor II
Numerical Methods and Calculus    GATE-CS-2014-(Set-2)
Discuss it

Question 56 Explanation:
In Newton's method, we apply below formula to get next value.
```f'(x) = 2.25x2 – 4x - 2

x1 = 2 - (0.75*23 – 2*22 – 2*2 + 4)/
(2.25*22 – 4*2 - 2)
= 2 - (-2/-1)
= 0

x2 = 0 - (4/-2) = 2

x3 = 0
```
We get x = 0 and x = 2 repeatedly and it never converges.
 Question 57
The product of the non-zero 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 ______
 A 4 B 5 C 6 D 7
Linear Algebra    GATE-CS-2014-(Set-2)
Discuss it

Question 57 Explanation:
The characteristic equation is : | A - zI | = 0 , where I is an identity matrix of order 5. i.e. determinant of the below shown matrix to be 0. 1-z 0 0 0 1 0 1-z 1 1 0 0 1 1-z 1 0 0 1 1 1-z 0 1 0 0 0 1-z Now solve this equation to find values of z. Steps to solve : 1) Expand the matrix by 1st row. (1-z) [ ( 1-z , 1 , 1, 0 ) (1 , 1-z, 1, 0) (1, 1, 1-z, 0) (0, 0, 0, 1-z ) ] + 1. [ ( 0, 1-z, 1, 1 ) ( 0, 1, 1-z, 1) ( 0, 1, 1, 1-z )( 1, 0, 0, 0) ] Note: ( matrix is represented in brackets, row wise ) 2) Expand both of the above 4x4 matrices along the last row. (1-z)(1-z) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] + 1.(-1) [ (1-z, 1, 1) (1, 1-z,1) (1 , 1, 1-z ) ] 3) Apply row transformations to simplify above matrices ( both the matrices are same). C1 <- C1 + C2 + C3 R2 <- R2- R1 R3 <- R3 - R1 result is : (1-z)(1-z) [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] - 1. [ ( 3-z, 1, 1) (0, -z, 0) (0, 0, -z ) ] 4) Solve the matrix by expanding 1st column. result is : (1-z)(1-z)(z)(z) - (3-z)(z)(z) = 0 solve further to get : z^3 ( 3-z ) ( z-2 ) = 0 hence z = 0 , 0 , 0 , 3 , 2 Therefore product of non zero eigenvales is 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 ______ .
 A 0.259 B 0.459 C 0.325 D 0.225
Probability    GATE-CS-2014-(Set-2)
Discuss it

Question 58 Explanation:
```There are total 100 numbers, out of which

50 numbers are divisible by 2,
33 numbers are divisible by 3,
20 numbers are divisible by 5

Following are counted twice above
16 numbers are divisible by both 2 and 3
10 numbers are divisible by both 2 and 5
6 numbers are divisible by both 3 and 5

Following is counted thrice above
3 numbers are divisible by all 2, 3 and 5
```
So total numbers divisible by 2, 3 and 5 are = = 50 + 33 + 20 - 16 - 10 - 6 + 3 = 103 - 29 = 74 So probability that a number is number is not divisible by 2, 3 and 5 = (100 - 74)/100 = 0.26
 Question 59
The number of distinct positive integral factors of 2014 is _________________________
 A 7 B 8 C 9 D 10
General Aptitude    GATE-CS-2014-(Set-2)
Discuss it

Question 59 Explanation:
the factors are 1, 2, 19, 38, 53, 106, 107 and 2014. So the total makes it 8 number of integral factors.
 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?
 A Both S1 and S2 are true B S1 is true and S2 is false C S2 is true and S1 is false D Neither S1 nor S2 is true
Set Theory & Algebra    GATE-CS-2014-(Set-2)
Discuss it

Question 60 Explanation:
According to the given information :
S1 is true because NULL set is smaller than every other set.
S2 is true because the UNIVERSAL set {1, 2, ..., 2014} is larger than every other set.

Thus, both S1 and S2 are true.

Please comment below if you find anything wrong in the above post.
 Question 61
A cycle on n vertices is isomorphic to its complement. The value of n is _____.
 A 2 B 4 C 6 D 5
Graph Theory    GATE-CS-2014-(Set-2)
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Question 61 Explanation:
Below is a cyclic graph with 5 vertices and its complement graph. The complement graph is also isomorphic (same number of vertices connected in same way) to given graph.
 Question 62
The number of distinct minimum spanning trees for the weighted graph below is ____
 A 4 B 5 C 6 D 7
Graph Minimum Spanning Tree    GATE-CS-2014-(Set-2)
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Question 62 Explanation:
Below diagram shows a minimum spanning tree. Highlighted (in green) are the edges picked to make the MST. In the right side of MST, we could either pick edge ‘a’ or ‘b’. In the left side, we could either pick ‘c’ or ‘d’ or ‘e’ in MST. There are 2 options for one edge to be picked and 3 options for another edge to be picked. Therefore, total 2*3 possible MSTs.
 Question 63
Which one of the following Boolean expressions is NOT a tautology?
 A A B B C C D D
Propositional and First Order Logic.    GATE-CS-2014-(Set-2)
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Question 63 Explanation:

(A) a -> b means if ‘a’ is true then ‘b’ is also true. Now, ‘b’ is true. Therefore, ‘c’ is also true. => Using transitivity rule , a -> c
(B) a <-> c is true if both ‘a’ and ‘c’ have same values. Let ‘a’, ‘b’ and c’ be false. Expression 'a and b' is false and expression 'not b' is true. RHS of the given equation should be true. But it evaluates to be false. Therefore, contradiction is there.
Option (B) is not a tautology.
(C) Let ‘a’ and ‘c’ be false. ’b’ be true a and b and c is false c or a is false
(D) Let ‘b’ be true. b -> a means if ‘b’ is true then ‘a’ is also true. Therefore, ‘a’ and ‘b’ both evaluates to be true.

Thus, option (B) is correct.

Please comment below if you find anything wrong in the above post.
 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) `
 A select R.* from R, S where R.a=S.a (D) B select distinct R.* from R,S where R.a=S.a C select R.* from R,(select distinct a from S) as S1 where R.a=S1.a D select R.* from R,S where R.a=S.a and is unique R
SQL    GATE-CS-2014-(Set-2)
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Question 64 Explanation:
The solution of this question lies in the data set(tuples) of Relations R and S we define. If we miss some case then we may get wrong answer. Let's say, Relation R(BCA) with attributes B, C and A contains the following tuples.
```B C A
---------
7 2 1
7 2 1
8 9 5
8 9 5```
And Relation S(AMN) with attributes A, M, and N contains the following tuples.
```A M N
---------
1 6 7
2 8 4
5 9 6
5 5 3```
----------------------------------------------------------------------------------------------------------- Now ,the original Query will give result as: "select * from R where a in (select S.a from S) " - The query asks to display every tuple of Relation R where R.a is present in the complete set S.a.
```B C A
---------
7 2 1
7 2 1
8 9 5
8 9 5```
----------------------------------------------------------------------------------------------------------- Option A query will result in : "select R.* from R, S where R.a=S.a"
```B C A
---------
7 2 1
7 2 1
8 9 5
8 9 5
8 9 5
8 9 5```
----------------------------------------------------------------------------------------------------------- Option B query will result in : " select distinct R.* from R,S where R.a=S.a"
```B C A
---------
7 2 1
8 9 5```
----------------------------------------------------------------------------------------------------------- Option C query will result in : "select R.* from R,(select distinct a from S) as S1 where R.a=S1.a" B C A --------- 7 2 1 7 2 1 8 9 5 8 9 5 ----------------------------------------------------------------------------------------------------------- Option D query will result in : NULL set "select R.* from R,S where R.a=S.a and is unique R" ---------------------------------------------------------------------------------------------------------- Hence option C query matches the original result set. Note : As mentioned earlier, we should take those data sets which can show us the difference in different queries. Suppose in R if you don't put identical tuples then you will get wrong answers. (Try this yourself, this is left as an exercise for you ).
 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 ____________
 A 1000 B 10000 C 100000 D 100
Computer Organization and Architecture    GATE-CS-2014-(Set-2)
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Question 65 Explanation:
One request initiation takes 100 ns. As the operations of memory module may overlap in time another, request can be initiated before it completes its remaining 500 ns. Thus total requests that can be initiated is 1000000 ns/100 ns =10000.
There are 65 questions to complete.

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