## GATE 2019

Question 1 |

Ten freiends planned to share equally the cost of buying a gift for their teacher. When two of them decided not to contribute, each of the other friends had to pay Rs 150 more. The cost of the gift was Rs. _____.

12000 | |

666 | |

3000 | |

6000 |

Gift cost = 10x

When two people are not going to contribute then remaining friends have to pay Rs 150 more.

Two friends contribution = 150 × 8 = 1200

One friend contribution = 600

Ten friends contribution = 6000

Cost of the Gift = 6000

Question 2 |

The expenditure on the project _____ as follows; equipment Rs.20 lakhs, salaries Rs.12 lakhs, and contingency Rs.3 lakhs.

break down | |

break | |

breaks down | |

breaks |

Question 3 |

The search engine's business model _____ around the fulcrum of trust.

bursts | |

revolves | |

sinks | |

plays |

Question 4 |

Two cars start at the same time from the same location and go in the same direction. The speed of the first car is 50 Km/h and the speed of the second car is 60 Km/h. The number of hours it takes for the distance between the two cars to be 20 Km is _____.

2 | |

3 | |

1 | |

6 |

Speed of the second car = 60 km/h

Let no. of hours = x say

⇒ From the question we can write

(60)x - (50)x = 20

10x = 20

**∴ x = 2 hrs**

Question 5 |

a syllabus | |

a student | |

a punishment | |

a school |

__a school__is to a teacher.

Question 6 |

In a college, there are three student clubs. Sixty students are only in the Drama club, 80 students are only in the Dance club, 30 students are only in the Maths club, 40 students are in both Drama and Dance clubs, 12 students are in both Dance and Maths clubs, 7 students are in both Drama and Maths club, and 2 students are in all the clubs. If 75% of the students in the college are not in any of these clubs, then the total number of students in the college is _____.

900 | |

975 | |

225 | |

1000 |

No. of students present in three student clubs

= 60 + 80 + 30 + 38 + 2 + 10 + 5

= 225 [i.e., 25% of the students in the college]

Total no. of students in the college = 225 × 4 = 900

Question 7 |

The police arrested four criminals - P, Q, R and S. The criminals knew each other. They made the following statements:

P says "Q committed the crime." Q says "S committed the crime." R says "I did not do it." S says "What Q said about me is false."

Assume only one of the arrested four committed the crime and only one of the statements made above is true. Who committed the crime?

P | |

Q | |

R | |

S |

II, III, IV are false.

II is false: 'S' is not criminal.

IV is also false: 'S' committed crime.

There is conflict among I, IV now as P, S both should be criminals.

⇒ II true: 'S' is criminal.

I, III, IV are false.

IV is false. So, 'S' is not criminal.

III is false. So. 'R' is criminal.

There is conflict among II, III as S, R both should be criminal.

III is true: Criminal ≠ R.

I, II, IV are false.

I is false. So, 'Q' is also not criminal.

II is false. So, 'S' is not criminal.

IV is false. So, 'S' is not criminal.

Q, R, S are not criminals. So, 'P' is criminal.

IV is true: I, II, III are false.

II is false. So, 'S' is not criminal.

IV is true. So, 'S' is criminal.

I is false. So, 'Q' is not criminal.

III is false. So, 'R' is criminal.

There is conflict, as R, S both should be criminals.

Question 8 |

"A recent High Court judgement has sought to dispel the idea of begging as a disease — which leads to its stigmatization and criminalization — and to regard it as a symptom. The underlying disease is the failure of the state to protect citizens who fall through the social security net."

Which of the following statements can be inferred from the given passage?

Beggars are created because of the lack of social welfare schemes | |

Beggars are lazy people who beg because they are unwilling to work | |

Begging is an offence that has to be dealt with firmly | |

Begging has to be banned because it adversely affects the welfare of the state |

The court is judged that because of failure of the state to protect citizens through social security schemes beggars are created.

Question 9 |

Three of the five students allocated to a hostel put in special requests to the warden. Given the floor plan of the vacant rooms, select the allocation plan that will accommodate all their requests.

Request by X: Due to pollen allergy, I want to avoid a wing next to the garden. Request by Y: I want to live as far from the washrooms as possible, since I am very sensitive to smell. Request by Z: I believe in Vaastu and so want to stay in the South-west wing.The shaded rooms are already occupied. WR is washroom.

Question 10 |

In the given diagram, teachers are represented in the triangle, researchers in the circle and administrators in the rectngle. Out of the total number of the people, the percentage of administrators shall be in the range of _____.

16 to 30 | |

0 to 15 | |

46 to 60 | |

31 to 45 |

From the given diagram:

Total no. of people = 70 + 10 + 20 + 20 + 40 = 160

No. of Administrators = 50

% of Administrators = 50/160 = 31.25

Question 11 |

If L is a regular language over Σ = {a,b}, which one of the following languages is NOT regular?

Suffix (L) = {y ∈ Σ* such that xy ∈ L} | |

{ww ^{R} │w ∈ L} | |

Prefix (L) = {x ∈ Σ*│∃y ∈ Σ* such that xy ∈ L} | |

L ∙ L ^{R} = {xy │ x ∈ L, y^{R} ∈ L} |

^{R}cannot be recognized without using stack, so it cannot be regular.

Question 12 |

The chip select logic for a certain DRAM chip in a memory system design is shown below. Assume that the memory system has 16 address lines denoted by A_{15} to A_{0}. What is the range of addresses (in hexadecimal) of the memory system that can get enabled by the chip select (CS) signal?

C800 to C8FF | |

C800 to CFFF | |

DA00 to DFFF | |

CA00 to CAFF |

_{0}- A

_{15}= 2

^{15}

The chip select address for given figure:

A

_{4}- A

_{15}= 2

^{5}

So, total addressable loactions = 2

^{16}/2

^{5}= 2

^{11}

2

^{11}= 2048 or location 0 to 2047

∴ CFFF - C800 = 2047

Answer is C800 to CFFF.

Question 14 |

Which of the following protocol pairs can be used to send and retrieve e-mails (in that order)?

SMTP, MIME | |

SMTP, POP3 | |

IMAP, POP3 | |

IMAP, SMTP |

POP3: Post Office Protocol (Responsible for retrieve email)

SMTP: Simple Mail Transfer Protocol (Responsible for send Email)

IMAP: Internet Message Access protocol (Responsible for store and view)

MIME: Multi purpose Internet Mail Extensions (For media)

Question 15 |

Which one of the following statements is NOT correct about the B^{+} tree data structure used for creating an index of a relational database table?

Each leaf node has a pointer to the next leaf node | |

Non-leaf nodes have pointers to data records | |

B ^{+} Tree is a height-balanced tree | |

Key values in each node are kept in sorted order |

In B+ trees non-leaf nodes do not have pointers to data records.

Question 16 |

#include <stdio.h> int main () { int arr [] = {1,2,3,4,5,6,7,8,9,0,1,2,5}, *ip = arr+4; printf ("%d\n", ip[1]); return 0; }

The number that will be displayed on execution of the program is _____.

5 | |

6 | |

7 | |

8 |

We know that arr is a pointer to arr[ ] & hence arr+4 is pointer to 4

^{th}index of array (starting from 0 to 4).

Now *ip is a pointer of int type pointing to memory location108, which is part of arr.

Hence, when we will print ip[1] it will be equivalent to *(ip+1).

Address of ip will be incremented by 1 & value inside 110 will be printed.

Question 17 |

A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes. Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the *Tag* and the *Index* fields resectively in the addresses generated by the processor?

28 bits and 0 bits | |

24 bits and 4 bits | |

24 bits and 0 bits | |

28 bits and 4 bits |

No index bits is there. So, now for tag bits,

Total bits - Offset bits = 32 - 4 = 28

So, tag bits = 28, Index bits = 0

Question 18 |

Consider the following two statements about database transaction schedules:

I. Strict two-phase locking protocol generates conflict serializable schedules that are also recoverable. II. Timestamp-ordering concurrency control protocol with Thomas Write Rule can generate view serializable schedules that are not conflict serializable.Which of the above statements is/are TRUE?

Both I and II | |

Neither I nor II | |

II only | |

I only |

In strict 2PL, a transaction T does not release any of its exclusive (write) locks until after it commits or aborts.

Hence, no other transaction can read or write an item that is written by T unless T has committed, leading to a strict schedule for recoverability.

(Ref: Fundamentals of Database Systems by Elmasri and Navathe, 7e Pg. No. 789)

By ignoring the write, Thomas write rule allows schedules that are not conflict serializable but are nevertheless correct.

Those non-conflict-serializable schedules allowed satisfy the definition of view serializable schedules.

(Ref: Database System Concepts by Silberschatch, Korth and Sudarshan, 6e Pg No. 686)

Question 19 |

S → Aa A → BD B → b | ε D → d | εLet a, b, d, and $ be indexed as follows:

Compute the FOLLOW set of the non-terminal B and write the index values for the symbols in the FOLLOW set in the descending order. (For example, if the FOLLOW set is {a, b, d, $}, then the answer should be 3210)

30 | |

31 | |

10 | |

21 |

{Follow(B) = Follow(A) when D is epsilon}

Follow(B) = {d} Union {a} = {a,d}

Hence Answer is : 31

Question 20 |

Let G be an arbitrary group. Consider the following relations on G:

R1: ∀a,b ∈ G, aR1b if and only if ∃g ∈ G such that a = g-1bg R2: ∀a,b ∈ G, aR2b if and only if a = b-1Which of the above is/are equivalence relation/relations?

R _{2} only | |

R _{1} and R_{2} | |

Neither R _{1} and R_{2} | |

R _{1} only |

Consider Statement R

_{1}:

Reflexive:

aR

_{1}a

⇒ a = g

^{-1}ag

Left multiply both sides by g

⇒ ga = gg

^{-1}ag

Right multiply both sides by g

^{-1}

⇒ gag

^{-1}= gg

^{-1}agg

^{-1}

⇒ gag

^{-1}= a [∴ The relation is reflexive]

Symmetric:

If aR

_{1}b, then ∃g ∈ G such that gag

^{-1}= b then a = g

^{-1}bg, which is Correct.

⇒ So, given relation is symmetric.

Transitive:

The given relation is Transitive.

So, the given relation R

_{1}is equivalence.

R

_{2}:

The given relation is not reflexive. So, which is not equivalence relation. Such that a ≠ a

^{-1}. So, only R

_{1}is true.

Question 21 |

1111 1111 1110 0100 | |

1111 1111 0001 1100 | |

0000 0000 1110 0100 | |

1000 0000 1110 0100 |

1’s complement = 1111 1111 1110 0011

2’s complement = 1’s complement + 1

2’s complement = 1111 1111 1110 0100

= (-28)

Question 22 |

Two numbers are chosen independently and uniformly at random from the set {1, 2, ..., 13}. The probability (rounded off to 3 decimal places) that their 4-bit (unsigned) binary representations have the same most significant bit is ______.

0.502 | |

0.461 | |

0.402 | |

0.561 |

1 - 0001

2 - 0010

3 - 0011

4 - 0100

5 - 0101

6 - 0110

7 - 0111

8 - 1000

9 - 1001

10 - 1010

11 - 1011

12 - 1100

13 - 1101

The probability that their 4-bit binary representations have the same most significant bit is

= P(MSB is 0) + P(MSB is 1)

= (7×7)/(13×13) + (6×6)/(13×13)

= (49+36)/169

= 85/169

= 0.502

Question 23 |

Leftmost in reverse | |

Rightmost in reverse | |

Leftmost | |

Rightmost |

Question 24 |

Consider a sequence of 14 elements: A = [-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0]. The subsequence sum . Determine the maximum of S(i,j), where 0 ≤ i ≤ j < 14. (Divide and conquer approach may be used)

19 | |

39 | |

29 | |

09 |

Ex: {A,B,C,D}

{ A,AB,AC,AD,ABC,ABD,ACD,

B, BC,BD,BCD,

C,CD,

D }

Step-1: Array of elements A = [-5, -10, 6, 3, -1, -2, 13, 4, -9, -1, 4, 12, -3, 0 ]

Step-2: As per the given question, if they want to find maximum subsequence means

{6,3,13,4,4,12}

= 42

Step-3: But according to given recurrence relation, the sequence should be continuous. {6,3,13,4,4,12}.

This is not continuous subsequence.

Step-4: The continuous sequence is {6, 3, -1, -2, 13, 4, -9, -1, 4, 12}

Total is {29}

**Note**: We can't get more than 29 maximum subsequence sum.

Question 25 |

Consider three concurrent processes P1, P2 and P3 as shown below, which access a shared variable D that has been initialized to 100.

The process are executed on a uniprocessor system running a time-shared operating system. If the minimum and maximum possible values of D after the three processes have completed execution are X and Y respectively, then the value of Y–X is _______.

10 | |

40 | |

60 | |

80 |

P2 reads D=100, preempted.

P1 executes D=D+20, D=120.

P3 executes D=D+10, D=130.

Now, P2 has D=100, executes

D = D-50 = 100-50 = 50

P2 writes D=50 final value. This is minimum.

Next,

P2 reads D=100, executes D = D-50, before that assume P1 & P3 has read D=100.

P2 makes D=50 & writes it.

P1 executes (D=100), D=D+20 & P3 executes D=D+10 gives maximum value D=130.

So, Y - X = 130 - 50 =80.

Question 26 |

An array of 25 distinct elements is to be sorted using quicksort. Assume that the pivot element is chosen uniformly at random. The probability that the pivot element gets placed in the worst possible location in the first round of partitioning (rounded off to 2 decimal places) is _____.

0.08 | |

0.01 | |

1 | |

8 |

Step-2: Pivot element = uniformly random.

Step-3: Worst case position in the pivot element is either first (or) last.

Step-4: So total 2 possibilities among 25 distinct elements

= 2/25

= 0.08

Question 27 |

Let X be a square matrix. Consider the following two statements on X.

I. X is invertible. II. Determinant of X is non-zero.Which one of the following is TRUE?

I implies II; II does not imply I. | |

II implies I; I does not imply II. | |

I and II are equivalent statements. | |

I does not imply II; II does not imply I. |

That means we can also say that determinant of X is non-zero.

Question 28 |

For Σ = {a,b}, let us consider the regular language L = {x|x = a^{2+3k} or x = b^{10+12k}, k ≥ 0}. Which one of the following can be a pumping length (the constant guaranteed by the pumping lemma) for L?

3 | |

9 | |

5 | |

24 |

**Pumping Lemma for Regular Languages**:

For any language L, there exists an integer n, such that for all x ∈ L with |x| ≥ n, there exists u,v, w ∈ Σ*, such that x = uvw, and

(1) |uv| ≤ n

(2) |v| ≥ 1

(3) for all i ≥ 0: uv

^{i}w ∈ L

We have to find "n" which satisfies for all the strings in L.

Considering strings derived by b

^{10+12k}.

The minimum string in L = "bbbbbbbbbb" but this string b

^{10}cannot be broken in uvw.

So, pumping length 3, 9 and 5 cannot be the correct answer.

So, the minimum pumping length, such that any string in L can be divided into three parts "uvw" must be greater than 10 .

Question 29 |

Let G be an undirected complete graph on n vertices, where n > 2. Then, the number of different Hamiltonian cycles in G is equal to

n! | |

1 | |

(n-1)! | |

The total number of hamiltonian cycles in a complete graph are

(n-1)!/2, where n is number of vertices.

Question 30 |

Consider Z = X - Y, where X, Y and Z are all in sign-magnitude form. X and Y are each represented in n bits. To avoid overflow, the representation of Z would require a minimum of:

n bits | |

n + 2 bits | |

n - 1 bits | |

n + 1 bits |

To store overflow/carry bit there should be extra space to accommodate it.

Hence, Z should be n+1 bits.

Question 31 |

^{51}mod 5 is ______.

3 | |

5 | |

2 | |

1 |

^{51}mod 5

⇒ 3

^{1}= 3 ⇒ 3 mod 5 = 3

3

^{2}⇒ 9 mod 5 = 4

3

^{3}⇒ 27 mod 5 = 2

3

^{4}⇒ 81 mod 5 = 1

3

^{5}⇒ 243 mod 5 = 3

For every four numbers sequence is repeating.

So, (51 % 4) = 3

⇒ 3

^{3}= 27

⇒ 27 mod 5 = 2

Question 32 |

(x + y) ⊕ z = x ⊕ (y + z) | |

(x ⊕ y) ⊕ z = x ⊕ (y ⊕ z) | |

x ⊕ y = x + y, if xy = 0 | |

x ⊕ y = (xy + x'y')' |

(x+y) ⊕ z = (1+1)⊕ 0 = 1 ⊕ 0 = 1

x ⊕ (y+z) = 1⊕(1+0) = 1 ⊕ 1 = 0

So,

(x+y) ⊕ z ≠ x ⊕ (y+z)

Question 33 |

#include <stdio.h> int jumble (int x, int y) { x = 2 * x + y ; return x ; } int main ( ) { int x=2, y=5 ; y = jumble (y, x) ; x = jumble (y, x) ; printf ("%d \n", x) ; return 0 ; }The value printed by the program is ______.

26 | |

67 | |

25 | |

13 |

#include

int jumble(int x, int y)

{

printf("Inside jumble : 2*%d + %d\n", x,y);

x = 2*x +y;

return x;

}

int main()

{

int x=2, y=5;

printf("Initial x=%d, y=%d\n",x,y);

printf("1st jumble call:jumble(%d,%d)\n",y,x);

y = jumble(y,x);

printf("Value of y after 1st jumble = %d\n", y);

printf("2

^{nd}jumble call: jumble(%d,%d)\n", y,x);

x = jumble(y,x);

printf("Value of x after 2nd jumble = %d\n", x);

printf("Final : %d\n", x);

return 0;

}

////////////////////////////////////OUTPUT

Initial x=2, y=5

1

^{st}jumble call: jumble(5,2)

Inside jumble : 2*5 + 2

Value of y after 1

^{st}jumble = 12

2

^{nd}jumble call: jumble(12,2)

Inside jumble : 2*12 + 2

Value of x after 2nd jumble = 26

Final : 26

Question 34 |

The following C program is executed on a Unix/Linux system:

#include <unistd.h> int main () { int i ; for (i=0; i<10; i++) if (i%2 == 0) fork ( ) ; return 0 ; }The total number of child processes created is _____.

26 | |

33 | |

31 | |

28 |

#include

int main( )

{

int i;

fork( );

fork( );

fork( );

fork( );

fork( );

}

n - fork statements will have 2

^{n}-1 child.

Hence, n = 5 ⇒ We have 31 child processes.

Question 35 |

Let U = {1,2,...,n}. Let A = {(x,X)|x ∈ X, X ⊆ U}. Consider the following two statements on |A|.

Which of the above statements is/are TRUE?

Only II | |

Only I | |

Neither I nor II | |

Both I and II |

and given A = {(x, X), x∈X and X⊆U}

Possible sets for U = {Φ, {1}, {2}, {1, 2}}

if x=1 then no. of possible sets = 2

x=2 then no. of possible sets = 2

⇒ No. of possible sets for A = (no. of sets at x=1) + (no. of sets at x=2) = 2 + 2 = 4

Consider statement (i) & (ii) and put n=2

Statement (i) is true

Statement (i) and (ii) both are true. Answer: (C)

Question 36 |

Consider the following relations P(X,Y,Z), Q(X,Y,T) and R(Y,V).

How many tuples will be returned by the following relational algebra query?

∏x(σ(P.Y=R.Y ∧ R.V=V2)(P × R)) - ∏x(σ(Q.Y=R.Y ∧ Q.T>2)(Q × R))

0 | |

1 | |

2 | |

3 |

_{(P.Y=R.Y ∧ R.V=V2)}(P × R)

∏

_{x}(σ

_{(P.Y=R.Y ∧ R.V=V2)}(P × R))

σ

_{(Q.Y=R.Y ∧ Q.T>2)}(Q × R)

∏

_{x}(σ

_{(Q.Y=R.Y ∧ Q.T>2)}(Q × R))

∏

_{x}(σ

_{(P.Y=R.Y ∧ R.V=V2)}(P × R)) - ∏

_{x}(σ

_{(Q.Y=R.Y ∧ Q.T>2)}(Q × R))

Question 37 |

What is the minimum number of 2-input NOR gates required to implement a 4-variable function function expressed in sum-of-minterms form as f = Σ(0, 2, 5, 7, 8, 10, 13, 15)? Assume that all the inputs and their complements are available.

2 | |

4 | |

7 | |

1 | |

3(Option not given) |

Question 38 |

Consider three machines M, N, and P with IP addresses 100.10.5.2, 100.10.5.5, and 100.10.5.6 respectively. The subnet mask is set to 255.255.255.252 for all the three machines. Which one of the following is true?

M, N, and P all belong to the same subnet | |

Only M and N belong to the same subnet | |

M, N, and P belong to three different subnets | |

Only N and P belong to the same subnet |

Therefore, N and P belong to the same subnet.

Question 39 |

Consider three 4-variable functions f_{1}, f_{2} and f_{3}, which are expressed in sum-of-minterms as

f_{1}= Σ(0, 2, 5, 8, 14), f_{2}= Σ(2, 3, 6, 8, 14, 15), f_{3}= Σ(2, 7, 11, 14)

For the following circuit with one AND gate and one XOR gate, the output function f can be expressed as:

Σ (2, 14) | |

Σ (7, 8, 11) | |

Σ (2, 7, 8, 11, 14) | |

Σ (0, 2, 3, 5, 6, 7, 8, 11, 14, 15) |

f3 = ∑(2,7,11,14)

f1*f2 ⊕ f3 = ∑(2,8,14) ⊕ ∑(2,7,11,14)

= ∑(8,7,11) (Note: Choose the terms which are not common)

Question 40 |

S' → S S → 〈L〉 | id L → L,S | S

Let I_{0} = CLOSURE ({[S' → ·S]}). The number of items in the set GOTO (I_{0} , 〈 ) is: _____.

4 | |

5 | |

6 | |

7 |

_{0}= CLOSURE ({[S' → ·S]})

Hence, the set GOTO (I0 , 〈 ) has 5 items.

Question 41 |

I. The smallest element in a max-heap is always at a leaf node II. The second largest element in a max-heap is always a child of the root node III. A max-heap can be constructed from a binary search tree in Θ(n) time IV. A binary search tree can be constructed from a max-heap in Θ(n) timeWhich of the above statements are TRUE?

I, II and III | |

II, III and IV | |

I, III and IV | |

I, II and IV |

(ii) TRUE: The second smallest element in a heap is always a child of root node.

(iii) TRUE: Converting from binary search tree to max heap will take O(n) time as well as O(n) space complexity.

(iv) FALSE: We can’t convert max heap to binary search tree in O(n) time.

Question 42 |

In an RSA cryptosystem, the value of the public modulus parameter n is 3007. If it is also known that Φ(n) = 2880, where Φ() denotes Euler's Totient Function, then the prime factor of n which is greater than 50 is ______.

107 | |

97 | |

45 | |

92 |

n = 3007, fi(n) = 2880 → fi(n) = (p – 1) (q – 1),

where p, q are prime factor of n.

The unit place of n is 7, it is a prime number and factor will be

1.7=7

11*17

21*37

31*47

….

31*97 =>3007

n = 3007 => 31*97

Therefore, 31 & 97 are the two prime numbers, which is satisfying the condition and 97 is greater than 50.

So, 97 is the correct answer.

Other methods:

When ϕ(n) is given when n=pq where p and q are prime numbers, then we have

ϕ(n) = (p−1)(q−1) = pq−(p+q)+1

But pq=n,

therefore, ϕ(n) = n−(p+q)+1 and p+q = n+1−ϕ(n).

Now, p and q are the roots of the equation,

x

^{2}− (p+q)x + pq = (x-p)(x-q)

Substituting for p+q and pq in the above equation

x

^{2}- (n+1-ϕ(n))x + n

Question 43 |

A relational database contains two tables Student and Performance as shown below:

The primary key of the Student table is Roll_no. For the Performance table, the columns Roll_no. and Subject_code together from the primary key. Consider the SQL query given below:

SELECT S.Student_name, sum (P.Marks) FROM Student S, Performance P WHERE P.Marks > 84 GROUP BY S.Student_name;

The number of rows returned by the above SQL query is _____.

0 | |

9 | |

7 | |

5 |

SQL> SELECT S.Student_name,sum(P.Marks)

2 FROM Student S,Performance P

3 WHERE P.Marks>84

4 GROUP BY S.Student_name;

Question 44 |

Assume that in a certain computer, the virtual addresses are 64 bits long and the physical addresses are 48 bits long. The memory is word addressable. The page size is 8 kB and the word size is 4 bytes. The Translation Look-aside Buffer (TLB) in the address translation path has 128 valid entries. At most how many distinct virtual addresses can be translated without any TLB miss?

8×2 ^{20} | |

4×2 ^{20} | |

16×2 ^{10} | |

256×2 ^{10} |

So, it can refer to 2

^{7}pages.

Each page size is 8 kB & word is 4 bytes.

So, the total addresses of virtual address spaces that can be addressed

Question 45 |

Consider the following four processes with arrival times (in milliseconds) and their length of CPU bursts (in milliseconds) as shown below:

These processes are run on a single processor using preemptive Shortest Remaining Time First scheduling algorithm. If the average waiting time of the processes is 1 millisecond, then the value of Z is _____.

2 | |

7 | |

1 | |

4 |

At this point P4 arrives with burst 'Z' & P3 is in queue with burst 3.

P1 & P2 have executed with P1 incurred delay 1unit & P2 0units.

Hence, Avg = 0+1+x/4 =1

⇒ x=3, the next delay should be 3. It would happen if assume Z=2.

It executes and completes at 6.

P3 will wait totally for 3units.

Hence, Avg=1.

Z=2

Question 46 |

Let T be a full binary tree with 8 leaves. (A full binary tree has every level full). Suppose two leaves a and b of T are chosen uniformly and independently at random. The expected value of the distance between a and b in T (i.e., the number of edges in the unique path between a and b) is (rounded off to 2 decimal places) _____.

5.54 | |

1.34 | |

4.25 | |

3.82 |

__uniformly & independently__chosen leaf nodes.

A node can be chosen twice and the path from that node to itself will be zero.

∴ Path 1 = 0

Similarly,

Path 2 = 2

Path 3 = 4

Path 4 = 4

Path 5 = 6

Path 6 = 6

Path 7 = 6

Path 8 = 6

∴ Expected value = Σ Path length × Probability of selecting path

= 2×1/8 + 4×2/8 + 6×4/8 + 0×1/8

= 1/4 + 1/1 + 3/1 + 0

= 4 + 1/4

= 17/4

= 4.25

Question 47 |

Consider that 15 machines need to be connected in a LAN using 8-port Ethernet switches. Assume that these switches do not have any separate uplink ports. The minimum number of switches needed is _____.

3 | |

7 | |

1 | |

5 |

Therefore the total required number of the switches = Ceil (15 /7) = 3

Question 48 |

There are n unsorted arrays: A_{1}, A_{2}, ..., A_{n}. Assume that n is odd. Each of A_{1}, A_{2}, ..., A_{n} contains n distinct elements. There are no common elements between any two arrays. The worst-case time complexity of computing the median of the medians of A_{1}, A_{2}, ..., A_{n} is

O(n) | |

O(n log n) | |

Ω(n ^{2} log n) | |

O(n ^{2}) |

But it is similar to quicksort but in quicksort, partitioning will take extra time.

→ Find the median will be (i+j)/2

1. If n is odd, the value is Ceil((i+j)/2)

2. If n is even, the value is floor((i+j)/2)

-> Here, total number of arrays are

⇒ O(n)*O(n)

⇒ O(n

^{2})

**Note:**

They are clearly saying that all are distinct elements.

There is no common elements between any two arrays.

Question 49 |

Suppose Y is distributed uniformly in the open interval (1,6). The probability that the polynomial 3x^{2} + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _____.

0.3 | |

0.9 | |

0.1 | |

0.8 |

3x

^{2}+ 6xY + 3Y + 6

= 3x

^{2}+ (6Y)x + (3Y + 6)

whch is in the form: ax

^{2}+ bx + c

For real roots: b

^{2}- 4ac ≥ 0

⇒ (6Y)

^{2}- 4(3)(3Y + 6) ≥ 0

⇒ 36Y

^{2}- 36Y - 72 ≥ 0

⇒ Y

^{2}- Y - 2 ≥ 0

⇒ (Y+1)(Y-2) ≥ 0

Y = -1 (or) 2

The given interval is (1,6).

So, we need to consider the range (2,6).

The probability = (1/(6-1)) * (6-2) = 1/5 * 4 = 0.8

Question 50 |

#include <stdio.h> int main() { int a[] = {2, 4, 6, 8, 10} ; int i, sum = 0, *b = a + 4 ; for (i = 0; i < 5; i++) sum = sum + (*b - i) - *(b - i) ; printf ("%d\n", sum) ; return 0 ; }The output of the above C program is _____.

3 | |

7 | |

11 | |

10 |

#include

int main()

{

int a[] = {2,4,6,8,10};

int i, sum = 0, *b = a+4;

for(i=0; i<5; i++)

{ printf("*b, (*b-i): %d , %d\n",*b, (*b-i) );

printf("*(b-i): %d\n",*(b-i) );

printf("sum = %d + %d - %d\n",sum, (*b-i),*(b-i));

sum = sum + (*b-i) - *(b-i);

printf("sum = %d\n", sum);

}

printf("%d\n", sum);

return 0;

}

//////////////////////////////OUTPUT

*b, (*b-i): 10 , 10

*(b-i): 10

sum = 0 + 10 - 10

sum = 0

*b, (*b-i): 10 , 9

*(b-i): 8

sum = 0 + 9 - 8

sum = 1

*b, (*b-i): 10 , 8

*(b-i): 6

sum = 1 + 8 - 6

sum = 3

*b, (*b-i): 10 , 7

*(b-i): 4

sum = 3 + 7 - 4

sum = 6

*b, (*b-i): 10 , 6

*(b-i): 2

sum = 6 + 6 - 2

sum = 10

10

Question 51 |

Let the set of functional dependencies F = {QR → S, R → P, S → Q} hold on a relation schema X = (PQRS). X is not in BCNF. Suppose X is decomposed into two schemas Y and Z, where Y = (PR) and Z = (QRS).

Consider the two statements given below.I. Both Y and Z are in BCNF II. Decomposition of X into Y and Z is dependency preserving and losslessWhich of the above statements is/are correct?

I only | |

Neither I nor II | |

II only | |

Both I and II |

R → P

R

^{+}= RP

* In R → P, 'R' is a super key. So, Y is in BCNF.

Z = (QRS)

QR → S

S → Q

CK's = QR, RS

* In, S → Q, 'S' is not a super key. So, Z is not in BCNF.

* Y is in BCNF and Z is not in BCNF.

* 'R' is common attribute in the relations Y and Z and R is candidate key for Y. So, the decomposition is lossless.

* The FD, R → P is applicable on Y and QR → S, S → Q are applicablein 2.

So, the decomposition is dependency preserving.

* Hence, Statement II is correct.

Question 52 |

- ∀x[(∀z z|x ⇒ ((z = x) ∨ (z = 1))) ⇒ ∃w (w > x) ∧ (∀z z|w ⇒ ((w = z) ∨ (z = 1)))]

Here 'a|b' denotes that 'a divides b', where a and b are integers. Consider the following sets:

S1. {1, 2, 3, ..., 100} S2. Set of all positive integers S3. Set of all integersWhich of the above sets satisfy φ?

S1 and S3 | |

S1, S2 and S3 | |

S2 and S3 | |

S1 and S2 |

One of the case: If -7 is a number which is prime (either divided by -7 or 1 only). then there exists some number like -3 which is larger than -7 also satisfy the property ( either divided by -3 or 1 only).

So, S3 is correct

It's true for all integers too

Question 53 |

A certain processor deploys a single-level cache. The cache block size is 8 words and the word size is 4 bytes. The memory system uses a 60-MHz clock. To service a cache miss, the memory controller first takes 1 cycle to accept the starting address of the block, it then takes 3 cycles to fetch all the eight words of the block, and finally transmits the words of the requested block at the rate of 1 word per cycle. The maximum bandwidth for the memory system when the program running on the processor issues a series of read operations is ______ × 10^{6} bytes/sec.

160 | |

145 | |

172 | |

124 |

Cache block = 8 words

Word size = 4 bytes

Cache block size = 32 bytes

Clock = 60 MHz

⇒ T = 1/clock = 1/60×10

^{6}seconds

Cache miss

= 1 cycle(Address) + 3 cycles (8 words) + 1word/cycle ×8 (transfer)

= 12 cycles

= 12/60×10

^{6}

Total bandwidth = total data/total time = 32 bytes/(12/60×10

^{6}) = 160 × 10

^{6}bytes/second

Answer: 160

Question 54 |

S1. Set of all recursively enumerable languages over the alphabet {0,1} S2. Set of all syntactically valid C programs S3. Set of all languages over the alphabet {0,1} S4. Set of all non-regular languages over the alphabet {0,1}Which of the above sets are uncountable?

S2 and S3 | |

S3 and S4 | |

S1 and S4 | |

S1 and S2 |

S2 is countable, since a valid C program represents a valid algorithm and every algorithm corresponds to a Turing Machine, so S2 is equivalent to set of all Turing Machines.

S3 is is uncountable, it is proved by diagonalization method.

S4 is uncountable, as set of non-regular languages will have languages which is set of all languages over alphabet {0,1} i.e., S3.

Question 55 |

Consider the following grammar and the semantic actions to support the inheriteatd type declaration attributes. Let X_{1}, X_{2}, X_{3}, X_{4}, X_{5} and X_{6} be the placeholders for the non-terminals D, T, L or L_{1} in the following table:

Which one of the following are the appropriate choices for X_{1}, X_{2}, X_{3} and X_{4}?

X _{1} = L, X_{2} = L, X_{3} = L_{1}, X_{4} = T | |

X _{1} = L, X_{2} = T, X_{3} = L_{1}, X_{4} = L | |

X _{1} = T, X_{2} = L, X_{3} = L_{1}, X_{4} = T | |

X _{1} = T, X_{2} = L, X_{3} = T, X_{4} = L_{1} |

L → L

_{1}, id {X

_{3}.type = X

_{4}.type } , this production has L and L

_{1}, hence X

_{3}and X

_{4}cannot be T.

So option 1, 3 and 4 cannot be correct.

Hence, 2 is correct answer.

Question 56 |

Which one of the following languages over Σ = {a,b} is NOT context-free?

{ww ^{R} |w ∈ {a,b}*} | |

{wa ^{n}w^{R}b^{n} |w ∈ {a,b}*, n ≥ 0} | |

{a ^{n}b^{i} | i ∈ {n, 3n, 5n}, n ≥ 0} | |

{wa ^{n}b^{n}w^{R} |w ∈ {a,b}*, n ≥ 0} |

^{n}w

^{R}b

^{n}|w ∈ {a,b}*, n ≥ 0} cannot be CFL.

This is similar to language

L = {a

^{n}b

^{m}c

^{n}d

^{m}| n, m > 0}

Suppose we push “w” then a

^{n}and then w

^{R}, now we cannot match b

^{n}with a

^{n}, because in top of stack we have w

^{R}.

Question 57 |

#include <stdio.h> int r() { static int num=7 ; return num-- ; } int main () { for (r(); r (); r()) printf ("%d", r()); return 0 ; }

Which one of the following values will be displayed on execution of the programs?

41 | |

63 | |

52 | |

630 |

#include

int r()

{

int x;

static int num=7;

x =num--;

printf("num--: %d\n",x);

return x;

}

int main()

{

for(r(); r(); r())

{

printf("%d\n", r());

}

return 0;

}

//////////////////////////////OUTPUT

num--: 7

num--: 6

num--: 5

5

num--: 4

num--: 3

num--: 2

2

num--: 1

num--: 0

Question 58 |

Suppose that in an IP-over-Ethernet network, a machine X wishes to find the MAC address of another machine Y in its subnet. Which one of the following techniques can be used for this?

X sends an ARP request packet to the local gateway's IP address which then finds the MAC address of Y and sends to X | |

X sends an ARP request packet with broadcast IP address in its local subnet | |

X sends an ARP request packet to the local gateway's MAC address which then finds the MAC address of Y and sends to X | |

X sends an ARP request packet with broadcast MAC address in its local subnet |

Since both are present in the same subnet thus an ARP request packet can be sent as broadcast MAC address, all will see but the only destination will reply as a unicast reply.

**Video Reference :**

http://eclassesbyravindra.com/mod/page/view.php?id=147

Question 60 |

void convert (int n) { if (n < 0) printf ("%d", n); else { convert (n/2); printf ("%d", n%2); } }

Which one of the following will happen when the function convert is called with any positive integer n as argument?

It will print the binary representation of n in the reverse order and terminate | |

It will not print anything and will not terminate | |

It will print the binary representation of n and terminate | |

It will print the binary representation of n but will not terminate |

Sequence of function calls

Convert(6)

Convert(3)

Convert(1)

Convert(0)

:

Convert(0)

:

:

It will not terminate and never produce any output.

Note:

There is no instruction which stops the loop.

Question 61 |

The index node (inode) of a Unix-like file system has 12 direct, one single-indirect and one double-indirect pointers. The disk block size is 4 kB, and the disk block address is 32-bits long. The maximum possible file size is (rounded off to 1 decimal place) ______ GB.

7.0 | |

9.0 | |

2.0 | |

4.0 |

Max. file size

= (12 × 1k + 1k × 1k) × 4kB ≈ (1024 × 12 + 1024 × 1024) × 4 × 1024 bytes

≈ 4GB

Question 62 |

Consider the following snapshot of a system running n concurrent processes. Process i is holding X_{i} instances of a resource R, 1 ≤ i ≤ n. Assume that all instances of R are currently in use. Further, for all i, process i can place a request for at most Y_{i} additional instances of R while holding the X_{i} instances it already has. Of the n processes, there are exactly two processes p and q such that Y_{p} = Y_{q} = 0. Which one of the following conditions guarantees that no other process apart from p and q can complete execution?

Min (X _{p}, X_{q}) ≥ Min {Y_{k} | 1 ≤ k ≤ n, k ≠ p, k ≠ q} | |

X _{p} + X_{q} < Max {Y_{k} | 1 ≤ k ≤ n, k ≠ p, k ≠ q} | |

Min (X _{p}, X_{q}) ≤ Max {Y_{k} | 1 ≤ k ≤ n, k ≠ p, k ≠ q} | |

X _{p} + X_{q} < Min {Y_{k} | 1 ≤ k ≤ n, k ≠ p, k ≠ q} |

_{1}, P

_{2}, ..., P

_{n}}

P

_{i}holds X

_{i}instances.

P

_{i}can request additional Y

_{i}instances.

Given two process p & q such that their additional requests are zero.

Y

_{p}= Y

_{q}= 0

{Y

_{k}| 1 ≤ k ≤ n, k ≠ p, k ≠ q} means that out of 'n' processes, we are left with (n-2) process (except p&q), i.e., Y

_{k}indicates additional request of all the processes (n-2) except p & q.

For p & q to complete first, accordingly

X

_{p}+ X

_{q}< Min {Y

_{k}}

Option D is correct.

There are exactly two process p and q which do not need any additional instances of resources.

So, p and q will complete their execution and will release X

_{p}and X

_{q}instances of resources.

Now to guarantee that no other process apart from p and q can complete execution, the no. of instances of resources available must be less than the minimum no. of instances of resources required by any other process, i.e.,

X

_{p}+ X

_{q}< Min {Y

_{k}| 1 ≤ k ≤ n, k ≠ p, k ≠ q}

Question 63 |

Let Σ be the set of all bijections from {1, ..., 5} to {1, ..., 5}, where id denotes the identity function i.e. id(j) = j,∀j. Let º denote composition on functions. For a string x = x_{1} x_{2} ... x_{n} ∈ Σ^{n}, n ≥ 0. Let π(x) = x_{1} º x_{2} º ... º x_{n}.

Consider the language L = {x ∈ Σ* | π(x) = id}. The minimum number of states in any DFA accepting L is ______.

120 | |

136 | |

125 | |

132 |

Question 64 |

#include <stdio.h> int main () { float sum = 0.0, j = 1.0, i = 2.0; while (i/j > 0.0625) { j = j + j; sum = sum + i/j; printf ("%f \n", sum); } return 0; }

The number of times the variable sum will be printed, when the above program is executed , is ______.

5 | |

2 | |

7 | |

10 |

#include

int main()

{

float sum= 0.0, j=1.0, i=2.0;

while(i/j > 0.0625)

{

j = j+j;

sum = sum+i/j;

printf("%f\n",sum);

}

return 0;

}

//////////////////////////////////OUTPUT

1.000000

1.500000

1.750000

1.875000

1.937500

Question 65 |

I. G has a unique minimum spanning tree, if no two edges of G have the same weight. II. G has a unique minimum spanning tree, if, for every cut of G, there is a unique minimum-weight edge crossing the cut.Which of the above statements is/are TRUE?

I only | |

II only | |

Both I and II | |

Neither I nor II |

I. TRUE: G Graph is unique, no two edges of the graph is same.

Step-1: Using Kruskal's algorithm, arrange each weights in ascending order.

17, 18, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30

Step-2:

Step-3: 17 + 18 + 20 + 21 + 22 + 23 + 26 = 147

Step-4: Here, all the elements are distinct. So, the possible MCST is 1.

II. TRUE: As per the above graph, if we are cut the edge, that should the be the minimum edge.

Because we are already given, all minimum edge weights if graph is distinct.