GATE-CS-2017 (Set 2)

Q.1

A test has questions worth 100 marks totals. There are two types of questions, multiple choice questions are worth 3 marks
each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have?

Ans .

Explanation :

Let us consider all choices :

Choice A : 12 [Wrong] 
There is no integer x that satisfies below equation
12 * 3  + x * 11 = 100
x = 64/11 [Which is not an integer]

Choice B : 15 [Right] 
15 * 3  + x * 11 = 100
x = 55/11  = 5

Choice C : 18 [Wrong] 
There is no integer x that satisfies below equation
18 * 3  + x * 11 = 100
x = 46/11 [Which is not an integer]

Choice D : 19 [Wrong] 
There is no integer x that satisfies below equation
19 * 3  + x * 11 = 100
x = 43/11 [Which is not an integer]

Q.2

There are 5 buildings called V, W, X, Y, Z in a row (not necessarily in order). V is to the West of W, Z is to the East of X and
the West of V. W is to West of Y. Which building is in the middle?

Ans .

Explanation :

V is to the West of W,
V

W
Z is to the East of X and the West of V.
X

Z

V

W
W is to West of Y.
X

Z

V

W

Y
Therefore, V is in middle. So option A is correct.

Q.3

Saturn is __________ to be seen on a clear night with the naked eye?

enough bright

bright enough

bright as enough

as enough bright

Ans .

Explanation :

With adjectives and adverbs, enough comes after adjectives and adverbs. 
With nouns, enough comes before noun.
In the given question, enough is used with bright which is an adjective, so enough will come after the adjective.
So bright enough is the correct option.

Q.4

There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability
that they are of the same color is

1/5

7/30

1/4

4/15

Ans .

Explanation :

Ways to collect both Red = 3C2
Ways to collect both Green = 4C2
Ways to collect both Blue = 3C2

Total Ways = 10C2

P(Socks of Same Colour) = (3C2 + 4C2+ 3C2 ) /  10C2

= 4/15 
Therefore, Option D is correct

Q.5

Chose the option with words that are not synonyms.

aversion, dislike

luminous, radiant

plunder, loot

yielding, resistant

Ans .

Explanation :

Yield - produce or provide
Resistance - the refusal to accept or comply with something. Rest are all synonyms. Hence, (D) is correct.

Q.6

The numbers of the roots of e^x + 0.5x² -2 = 0 in the range [-5, 5] are

Ans .

Explanation :

The given function is a continuous function in range [-5, 5]. 
The Intermediate Value Theorem states if a continuous function has values of opposite sign inside an interval, then it has a root in that interval. 
The function changes sign twice in range [-5, 5].
It becomes positive to negative in range [-5, 0] when ex + 0.5x2 becomes less than 2.
It becomes positive to negative in range [0, 5] when ex + 0.5x2 becomes more than 2.

Q.7

There are three boxes, one contains apples, another contains oranges and last one contains both apples and oranges.
All three are known to be incorrectly labelled. You are permitted to open just one box and then pull out and inspect only one fruit.
Which box would you open to determine the contents of all three boxes?
Which of the following statements best reflects the author's opinion?

The box labelled apples

The box labelled both apples and oranges

The box labelled oranges

Can't be determined

Ans .

Explanation :

In given question there are three boxes incorrectly labeled:
Box 3 labeled as Apples and Oranges - Original Content can be Apple or Orangle (This box will have only one fruit, so whichever we pick would
be the correct label). Let's say the picked item was Apple, therefore the box labeled Orange will have Apples and Oranges.)
Box 2 - Oranges - This box will definitely not have orange (as its mislabeled) and it can not contain only apple (As it is in box 3) 
So this will have Original Content- both oranges and Apples
Box 1 - Apples - This is the leftover box and it can now have Original Content - Orange
Therefore option B is correct.

Q.8

X is 30 digit number starting with 4 followed 7. Then number X³ will have

Ans .

Explanation :

X = 4777....(till 30 digits)

or X = 4.7777? * 1029

X29 = (4.777?*1029 )3 
  = (4.777...)3 * 10 87

So, (X3 )3  requires 3 + 87 = 90 digits. 
Therefore, option A.

Q.9

"We lived in culture and denied any merit to literally works, Considering them important only when they were handmaidens
to something seemingly more urgent - namely ideology. This was a country where all gestures even the most private , interpreted as political terms.
" The author's believes that ideology is not as important as literature is revealed by the word :

culture

seemingly

urgent

political

Ans .

Explanation :

This paragraph is taken from the book - Reading Lolita in Tehran by- AZAR NAFISI. 
"Seemingly" means appearing to be something which is Not True. Or, not what its look like.
Author is criticizing the fake ideology of Iranian politics. Author is trying to say that Literature is more important than fake ideology. 
But this is how she is expressing- ..... they were handmaidens to something seemingly more urgent - namely ideology..
means- Ideology is not more important than litrature. So, (B) seemingly is CORRECT.

Q.10

An air pressure contour line join the locations in the region having the same atmospheric pressure.
The following is the air pressure contour of a geographical region. The contour line are shown as 0.05 bar intervals in this plot.
GATE CSE Paper solved
If the probability of thunderstorm is given by how fast air pressure rises or drops over a region. Which of the following region is most likely to have
thunderstorm.

Ans .

Explanation :

Region R has maximum air pressure variation as it  which has maximum number of lines crossing the area. 
Therefore option C is correct 

Alternate Solution
We have to see for the region which has maximum change in air pressure for thunderstorm.
In region P : the air pressure changes only once by 0.05 as only 2 lines are crossing the region.
The first line with the air pressure 0.95 and the next one is with pressure 0.9. 
In region Q : No line crosses the region which implies that the air pressure doesn't change.
In region R : the air pressure changes at 4 places resulting in maximum change in air pressure in the whole region. 
The minimum pressure in R is 0.65 and the maximum is 0.8.
In region S : the air pressure in the region in 0.95 and other contouring line is outside the region thus no change in air pressure within the region.
Thus the probability of thunderstorm is maximum in region R.

Q.11

Let p, q, r denote the statement "It is raining", "It is cold", and "It is pleasant", respectively.
Then the statement "It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold" is represented by:

(¬ p ∧ r) ∧ ((¬ r → (p ∧ q))

(¬ p ∧ r) ∧ ((p ∧ q) → ¬ r)

(¬ p ∧ r) ∨ ((p ∧ q) → ¬ r)

(¬ p ∧ r) ∨ ((r → (p ∧ q))

Ans .

Explanation :

It is not raining (~p) and it is pleasent (r) : (¬p ∧ r)

 It is raining (p) and it is cold (q) : (p∧q) it is not pleasant (¬r) only if it is raining and it is cold : ¬r → (p∧q) 

So the answer should be- (¬ p ∧ r) ∧ ((¬ r → (p ∧ q)) So, option (A) is true.

Q.12

Match the following according to input(from the left column) to the compiler phase(in the right column) that process it:

P -> (ii), Q -> (iii), R -> (iv), S -> (i)

P -> (ii), Q -> (i), R -> (iii), S -> (iv)

P -> (iii), Q -> (iv), R -> (i), S -> (ii)

P -> (i), Q -> (iv), R -> (ii), S -> (iii)

Ans .

Explanation :

P - iii Syntax tree is input to semantic analyser Q - iv Character stream is input to lexical analyser. 

R - i Intermediate code is input to code generator S -ii  Token stream is input to syntax analyser. Therefore option C.

Q.13

In a file allocation system, which of the following allocation scheme(s) can be used if no external fragmentation is allowed?
I. Contiguous
II. Linked
III. Indexed

(I) and (III) only

(II) only

(III) only

(II) and (III) only

Ans .

Explanation :

In contiguous allocation there is always possibility of external fragmentation. Both indexed and linked allocation are free from external fragmentation.Therefore, option D

Q.14

GATE CSE Paper solved

Ans .

Explanation :

Q.15

The representation of the value of a 16-bit unsigned integer X in a hexadecimal number system is BCA9.
The representation of the value of X in octal number system is:

571244

736251

571247

136251

Ans .

Explanation :

Q.16

An ER model of a database consists of entity types A and B. These are connected by a relationship R which does not have its own attribute.
Under which of the following conditions, can the relational table for R be merged with that of A?
(A) Relation R is one-to-many and the participation of A in R is total.
(B) Relation R is one-to-many and the participation of A in R is partial.
(C) Relation R is many-to-one and the participation of A in R is total.
(D) Relation R is many-to-one and the participation of A in R is partial.

Ans .

Explanation :

In one to many or many to one participation, relation is merged in table which is on many side and if there is total participation then relation is merged in that side. 

Therefore, option C is correct

Q.17

Match the algorithms with their time complexities:
GATE CSE Paper solved

P-> (iii), Q -> (iv), R -> (i), S -> (ii)

P-> (iv), Q -> (iii), R -> (i), S -> (ii)

P-> (iii), Q -> (iv), R -> (ii), S -> (i)

P-> (iv), Q -> (iii), R -> (ii), S -> (i)

Ans .

Explanation :


Tower of Hanoi - Ɵ(2n)
Heap sort worst case - Ɵ(n log n)
Binary Search - Ɵ(log n)
Addition of two nxn matrices - Ɵ (n2)

Therefore Option C is correct

Q.18

Consider the following statements about the routing protocols, Routing Information Protocol(RIP) and Open Shortest Path First(OSPF) in an IPv4 network.
I. RIP uses distance vector routing
II. RIP packets are sent using UDP
III. OSPF packets are sent using TCP
IV. OSPF operation is based on link-state routing
Which of the following above are CORRECT?

I and IV only

I, II and III only

I, II and IV only

II, III and IV only

Ans .

Explanation :

RIP uses distance vector routing (DVR) protocol which employ the hop count as a routing metric. Also, RIP uses the UDP as its

 transport protocol with port no 520. OSPF uses link state routing (LSR) protocol works within a single Autonomous System. OSPF encapsulates its data

 directly into IP Packets and does not use either TCP or UDP. Therefore, statement (III) is false and only statements I, II and IV are correct. Option (C) is true.

Q.19

GATE CSE Paper solved

Ans .

Explanation :

Q.20

The maximum number of IPv4 router address addresses that can be listed in the record route (RR) option field of an IPv4 header is ____.

Ans .

Explanation :

In IPV4 frame format there are 40 bytes of option field, out of which only 38 bytes can be used, 2 bytes are reserved . Also one IPV4 adress is of 4 bytes 
Thus maxmimum IPV4 address that can be hold <= 38/4 = 9

Q.21

Consider the following tables T1 and T2:
GATE CSE Paper solved
In table T1, P is the primary key, Q is the foreign key referencing R in table T2 with on-delete cascade and on-update cascade.
In table T2, R is the primary key and S is the foreign key referencing P in the table T1 with on-delete set NULL and on-update cascade.
In order to delete record (3,8) from table, numbers of additional record that need to be deleted from table T1 is ______.

Ans .

Explanation :

Given,
Q -> R(Primary Key)
S -> P (Primary Key)
Entry to be deleted - P (3) and Q(8)
Q can be deleted directly
Now, S - > P but the relationship given is on delete set NULL, Therefore when  we delete 3 from  T1 ,the  entry in T2 having 3 will be NULL.
Therfore, Option A  - Answer is 0 entries

Q.22

Breath First Search(BFS) has been implemented using queue data structure.
GATE CSE Paper solved
Which one of the following is a possible order of visiting the nodes in the graph above.

MNOPQR

NQMPOR

QMNROP

POQNMR

Ans .

Explanation :

In BFS, we print a starting node, then its adjacent, then adjacent of adjacent, and so on..
Option A : MNOPQR Wrong
We cannot visit "O" before "R" as we start from "M". 
Note that "O" is adjacent of adjacent for "M" and
"R" is adjacent of "M".

Option B : NQMPOR Wrong
We cannot visit "P" before "O" as we start from "N". 
Note that "P" is adjacent of adjacent for "N" and 
"O" is adjacent.

Option C : QMNROP Wrong
We cannot visit "R" before "O" as we start from "Q". 
Note that "R" is adjacent of adjacent for "Q" and
"O" is adjacent of "Q"

Option D : POQNMR Right
We visit "P", then its adjacent "O", "Q" and "N"
Finally we visit adjacent of adjacent "M" and "R"

Q.23

Consider a socket API on Linux machine that supports UDP socket. A connected UDP socket is a UDP socket on which connect function has already been called.
Which of the following statements is/are correct ?
I. A connected UDP socket can be used to communicate with multiple peers simultaneously.
II. A process can successfully call connect function again for an already connected UDP socket.

I only

II only

Both I and II only

Neither I nor II

Ans .

Explanation :

I. A connected UDP socket can be used to communicate with multiple peers simultaneously is WRONG : 

Even if UDP is connectionless, every UDP socket has a single specified IP number and port number to connect to. 

Therefore we can not use same socket to connect to multiple peers simultaneously. 

II. A process can successfully call connect function again for an already connected UDP socket is RIGHT : We can call connect again to connect to a new 

peer and disconnect previous peer if implementation allows.

Q.24

Let L1 and L2 be any context-free language and R be any regular language. Then, which of the following is correct ?
I. L1 ∪ L2 is context-free.
II. L1' is context-free.
III. L1-R is context-free.
IV. L1 ∩ L2

I, II and IV only

I and III only

II and IV only

I only

Ans .

Explanation :

 Context free language are closed under union and difference with regular language.
It is not closed under complementation and intersection . Complement of CFL is recursive language .

Q.25

Match the following:
GATE CSE Paper solved

P-->(ii), Q-->(iv), R-->(i), S-->(iii)

P-->(ii), Q-->(i), R-->(iv), S-->(iii)

P-->(ii), Q-->(iv), R-->(iii), S-->(i)

P-->(iii), Q-->(iv), R-->(i), S-->(ii)

Ans .

Explanation :


static char var; -> A variable located in data section of memory as it is static in nature
m = malloc(10); m = null; ->This is a lost memory which cannot be freed as m=NULL
Char *Ptr[10]; -> 10 memory locations of char type are allocated to store addresses
register int var1;-> Request to allocate a CPU register to store data

Therefore, Option is A

Q.26

Identity the language generated by following grammar where S is the start variable.
S --> XY
X --> aX | a
Y --> aYb | ∈

{a^m bⁿ| m>=n, n>0 }

{a^m bⁿ| m>=n, n>=0 }

{a^m bⁿ| m>n, n>=0 }

{a^m bⁿ| m>n, n>0 }

Ans .

Explanation :

S --> XY
X --> aX | a  // This produces only "a"
Y --> aYb | ∈  // This produces and "a" for every "b"                     
Option (A) and (B) are wrong because n can be zero also due to epsilon in Y Option (D) is wrong because Y-->aYb produces equal number of a's and b's.
Since there is one variable X which produces at least one a.
Therefore numbers of a's are always greater than numbers of b's.

Q.27

G is undirected graph with n vertices and 25 edges such that each vertex has degree at least 3. Then the maximum possible value of n is ________

Ans .

Explanation :

TAs per handshaking lemma,
Sum of degree of all vertices < = 2 * no. of edges.

Let v be number of vertices in graph.
==> 3*v <= 2*25
==> Minimum value of v = 16

Q.28

Given the following binary number in 32 bit (single precision) IEEE-754 format:

00111110011011010000000000000000

The decimal value closest to this floating-point number is:

1.45 X 10¹

1.45 X 10^-1

2.27 X 10^-1

2.27 X 10¹

Ans .

Explanation :

In 32-bit IEEE-754 format
1st bit represent sign
2-9th bit represent exponent
and 10-32 represent Mantissa (Fraction part)
Sign = 0, so positive 
2-9 bits — 01111100 when subtracted by 01111111 i.e., 126 decimal value gives -> 0000 0011 
Which is -3.(negative as the value is less than 126) 
As number is less than 126 it is subtracted otherwise 126 would have been subtracted from it in 32 bit representation. 
(https://www3.ntu.edu.sg/home/ehchua/programming/java/datarepresentation.html) 
Mantissa is normal ,hence, 1.M can be used .Which is 1.1101101. 
Thus, 
Data + 1.1101101 * 2^-3 (±M * B^(±e) ) 
Mantissa shift right 3 times -> 
+0.0011101101 
= 0.228 
= 2.28 * 10^-1 
Thus, option c is correct.

Q.29

A Circular queue has been implemented using singly linked list where each node consists of a value and
a pointer to next node. We maintain exactly two pointers FRONT and REAR pointing to the front node and rear node of queue. Which of the following
statements is/are correct for circular queue so that insertion and deletion operations can be performed in O(1) i.e. constant time.
I. Next pointer of front node points to the rear node.
II. Next pointer of rear node points to the front node.

I only

II only

Both I and II

Neither I and II

Ans .

Explanation :

Since, Circular queue deletes an item using Front pointer and insert an element using Rear pointer. If we want to insert an element into circular queue then we have to 

increment Rear pointer to next node then insert element. Then after update the next pointer of Rear node to the Front node. This method will have O(1) time for Insertion 

and Deletion. Only statement (ii) is true. So, option (B).

Q.30

The minimum possible number of states of a deterministic finite automaton that accepts a regular language

 L = {w₁aw₂ | w₁, w₂ ∈{a,b}* , |w₁| = 2, w₂ >=3}

is_______

Ans .

Explanation :

Q.31

Which of the following statements about the parser is/are correct?
I. Canonical LR is more powerful than SLR.
II. SLR is more powerful than LALR.
III. SLR is more powerful than canonical LR.

I only

II only

III only

II and III only

Ans .

Explanation :

LR parsers in term of power CLR > LALR > SLR > LR(0) Therefore, Option A is only correct option

Q.32

Consider a quadratic equation x² - 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b = ______.

Ans .

Explanation :

x² – 13x + 36 = 0
==> x² – (b + 3)x + (3*b + 6) = 0 …..
{ expanding numbers in base b )
putting the values of x we get b = 8.

Q.33

The minimum number of ordered pairs that need to be added to R to make (X, R) a lattice is _____.
GATE CSE Paper solved

Ans .

Explanation :

Here in the given lattice a should be related as 
(a,a) , (a,b), (a,c), (a,d), (a,e)

b: (b,c), (b,e), (b,b).

c: (c,c), (c,e)

d: (d,d), (d,e)

e: (e,e)

all pairs are already present here . 
Thus answer will be zero.

Q.34

Which of the following is/are shared by all the threads in a process?
I. Program Counter
II. Stack
III. Address space
IV. Registers

I and II only

III only

IV only

III and IV only

Ans .

Explanation :

Every thread have its own stack , register, and PC, so only address space that is shared by all thread for a single process. Therefore,  option B is correct.

Q.35

Consider the following function implemented in C:

void printxy(int x, int y)
{
    int *ptr;
    x = 0;
    ptr = &x;
    y = *ptr;
    *ptr = 1;
    printf("%d,%d", x, y);
}

The output of the printxy(1,1) is

0,0

0,1

1,0

1,1

Ans .

Explanation :

#include
void main()
{
 int x = 1, y = 1;
 printxy(x,y);
}
void printxy(int x, int y) 
{  
    int *ptr;     
    x = 0;     
    
    // ptr point to variable of value 0 
    ptr = &x;      
    
    // y has value pointed by ptr -> x= 0;
    y = *ptr;   
    
    // value pointed by ptr is set to 1 -> x= 1;
    *ptr = 1;       
    
    //x be changed to 1 and y will remain 0
    printf("%d,%d", x, y);  
}

Q.36

A system shares 9 tape drives. The current allocation and maximum requirement of tape drives for 3 processes are shown below: GATE CSE Paper solved
Which of the following best describes the current state of the system?

Not Safe, Not Deadlocked

Not Safe, Deadlocked

Safe, Deadlocked

Safe, Not Deadlocked

Ans .

Explanation :
```
Therefore, option D is correct.
```

Q.37

The read access times and the hit ratios for different caches in a memory hierarchy are as given below:
gate papers solved
The read access time of main memory in 90 nanoseconds. Assume that the caches use the referred-word-first read policy and the writeback policy. Assume that all the caches are direct mapped caches.
Assume that the dirty bit is always 0 for all the blocks in the caches. In execution of a program, 60% od memory reads are for instruction fetch and 40% are for memory operand fetch.
The average read access time in nanoseconds (up to 2 decimal places) is _________.

4.72

2.74

3.10

2.67

Ans .

Explanation :

Since, L2 cache is shared between Instruction and Data. Average Instruction fetch Time = L1 access time + L1 miss rate * L2 access time + L1 miss rate * L2 miss rate * Memory access time = 2 + 0.2 * 8 + 0.2 * 0.1 * 90 = 5.4 ns 

Average Data fetch Time = L1 access time + L1 miss rate * L2 access time + L1 miss rate * L2 miss rate * Memory access time = 2 + 0.1 * 8 + 0.1 * 0.1 * 90 = 3.7 ns

So, average read (access) time = Fraction of Instruction Fetch * Average Instruction fetch time + Fraction of Data Fetch * Average Data Fetch Time = 0.6×5.4+0.4×3.7 = 4.72ns 

Option (A) is correct.

Q.38

Consider the C program fragment below which is meant to divide x by y using repeated subtractions. The variable x, y, q and r are all unsigned int.

while(r >= y)
{
r = r - y;
q = q + 1;
}

Which of the following conditions on the variables x, y, q and r before the execution of the fragment will ensure that the loop terminates in a state satisfying the condition x == (y*q + r)?

( q == 0 ) && ( r == x ) && ( y > 0 )

( x > 0 ) && ( r == x ) && ( y > 0 )

( q == r ) && ( r == 0)

( q == 0 ) && ( y > 0 )

Ans .

Explanation :

x == ( y * q + r ) x= product, y= multiplicand, q = quotient , r =  remainder

For loop to be terminated, the quotient has to be 0, so only options C and D are left
If q=0 -> r=x should apply

Therefore, option A is most appropriate.

Q.39

Let δ denote the transition function and α denoted the extended transition function of the ε-NFA whose transition table is given below:

Then, α (q2,aba) is

{q1, q2, q3}

{q0, q1, q2}

{q0, q2, q3}

Ans .

Explanation :

Extended transition function describes what happens when we start in any state and follow any sequence of inputs. 
Since this is an epsilon NFA so we have to consider the epsilon moves also and see which states we can reach after the input string is over.
The starting state is q2, from q2 transition with input a is dead so we have to look for epsilon transition. With epsilon transition we reach to q0, at q0 we have a transition with input symbol a so we reach to state q1. 
From q1 we can take transition with symbol b and reach state q3 but from q3 we have no further transition with symbol a as input so we have to take another transition from state q1 that is the epsilon transition which goes to state q2. 
From q2 we reach to state q0 and read input b and then read input a and reach state q1. So q1 is one of the state of extended transition function.
From q1 we can reach q2 with epsilon as input (mans with no input) and from q2 we can reach q0 with epsilon move so state q2 and q0 are also part of extended transition function.
So state q1, q2, q0.

Q.40

Consider the following expression grammar G.

E -> E - T | T
T -> T + F | F
F -> (E) | id

Which of the following grammars are not left recursive, but equivalent to G.

A)
E -> E - T | T
T -> T + F | F
F -> (E) | id

B)
E -> TE'
E' -> -TE' | ε
T -> T + F | F
F -> (E) | id

C)
E -> TX
X -> -TX | ε
T -> FY
Y -> +FY | ε
F -> (E) | id

D)
E -> TX | (TX)
X -> -TX | +TX | ε
T -> id

Ans .

Explanation :

We know for left recursion : A -> Aα/β
After removing left recursion it can be written as A->βA’
A’->αA’/ε
Thus for : E->E- T/T
α= -T , β= T . thus new production after removing left recursion
is E->TE’ and E’->- TE’/ ε
T->FT’ and T’->+FT’/ ε
F->(E)/id .

Q.41

In a two-level cache system, the access times of L1 and L2 1 and 8 clock cycles, respectively. The miss penalty from the L2 cache to main memory is 18 clock cycles.
The miss rate of L1 cache is twice that of L2. The average memory access time(AMAT) of this cache system is 2 cycles. The miss rates of L1 and L2 respectively are:
Note: This questions appeared as Numerical Answer Type.

0.111 and 0.056

0.056 and 0.111

0.0892 and 0.1784

0.1784 and 0.0892

Ans .

Explanation :


Access time of L1 = 1
Access Time of L2=8
miss penalty  L1 cache (2*L2)  = 18*2 = 2*a
miss penalty  L2 cache say a = 18
AMAT (average memory access time) =2

AMAT = Access time of L1 + (MissRate L1 * miss penalty  L1)  
where miss penalty L1 = Access time of L2 + (MissRate L2 * miss penalty L2) 2 = 1+ 2*a *(8 + a* 18) 
Solving the equation, a=0.111

Q.42

Consider the following C program:

#include 
int main()
{
    int m = 10;
    int n, n1;
    n = ++m;
    n1 = m++;
    n--;
    --n1;
    n -= n1;
    printf("%d",n);
    return 0;
}

The output of the program is ______.

Ans .

Explanation :
```
 
```

Q.43

Consider two hosts X and Y, connected by a single direct link of rate 10⁶ bits/sec. The distance between the two hosts is 10,000 km
and the propagation speed along the link is 2 x 10⁸ m/s. Hosts X send a file of 50,000 bytes as one large message to hosts Y continuously. Let the transmission and propagation delays
be p milliseconds and q milliseconds, respectively. Then the vales of p and q are:

p = 50 and q = 100

p = 50 and q = 400

p = 100 and q = 50

p = 400 and q = 50

Ans .

Explanation :

Propagation delay = link length/ signal speed = (10^7)/(2*10^8 )sec = 5 =.005 sec =50 msec 
Transmission delay = data length/signal bandwith = 50000*8/10^6 sec =.04 sec =400 msec

Q.44

In a RSA cryptosystem a particular A uses two prime numbers p = 13 and q =17 to generate her public and private keys. If the public key of Ais 35. Then the private key of A is ____________.
Note: This questions appeared as Numerical Answer Type.

Ans .

Explanation :

In an RSA cryptosystem, for public key: 
GCD( ϕ(n) , e) = 1 

And, for private key:
(e * d) mod ϕ(n) = 1 

Where, 
ϕ(n) = (p -1)*(q - 1) = (13 - 1)(17 - 1) =12*16 = 192 
Such that 1 < e, d < ϕ(n)

Therefore, the private key is:
(35 * d) mod ϕ(n) = 1
d = 11

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GATE-CS-2017 (Set 2)