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GATE 2014 – A Brief Analysis (Based on student test experiences in the stream of CS on 1st March, 2014 - Second Session) Section wise analysis of the paper

Engineering Mathematics Data Structures and Algorithms DBMS Theory of Computation Compiler Design Computer Organization Computer Networks Digital Logic Operating systems Software Engineering and Web Technologies Verbal Ability Numerical Ability

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1 Mark 6 5 2 2 1 1 2 2 2 2 2 3 30

2 Marks 9 5 3 2 1 3 3 1 3 0 2 3 35

Total No of Questions 15 10 5 4 2 4 5 3 5 2 4 6 65

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Types of questions asked from each section

Engineering Mathematics

There were questions from Graph, Calculus, Functions, Probability, Matrices and Predicate Logic Questions from C-programs, Time Complexity, NP Complete, Quick sort, Analyzing Algorithms Questions from finding candidate keys, Normal Forms, Minimal Cover, SQL, Conflict Serializability Questions from Regular Expression, Finite Automata, Closure Properties Questions from SR and RR Conflicts, Code Optimization Questions from addressing mode, Cache Organization Questions from Network Security, Routing Protocols, Efficiency of SR Protocol Questions from Multiplexer, SOP, Radix Questions from SRTF, Optimal Page Replacement, Mutli threading Matching question on software models

Data Structures and Algorithms DBMS Theory of Computation Compiler Design Computer Organization Computer Networks Digital Logic Operating Systems SEWT

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Question’s & Solutions for 2nd March ( 2nd Session Paper ) 1.

Consider minterm expansion of the function F F(P,Q,R,S) =   0,2,5, 7, 8,10,13,15 Where minterms 2, 7, 8, 13 are “do not care” terms. The minimal Sum of Product form of F is (A) QS  QS

(B) QS  QS

(C) QRS  QRS  QRS  QRS

(D) PQS  PQS  PQS  PQS

Solution: (B) RS PQ

00

01

11

1

10

X

01

1

X

11

X

1

10

2.

00

X

Minimal SOP form  QS  Q S

1

Let X, Y be finite sets and F: XY be a function. Which of the following statement is TRUE? (A) For any subsets A and B of X, f  A  B   f  A   f B  (B) For any subsets A and B of X, f  A  B   f  A   f B 



(C) For any subsets A and B of X, f  A  B   min f  A  , f B 



(D) For any subsets S and T of Y, f 1  S  T   f 1  S   f 1  T  Solution: (D) 3.

Consider the following four pipeline systems P1: Four pipeline stages with delay 1, 2, 2, 1 P2: Four pipeline stages with delay 1, 1.5, 1.5, 1.5 P3: Five pipeline stages with delay 0.5, 1, 1, 0.6, 1 P4: Five pipeline stages with delay 0.5, 0.5,1, 1, 1.1 Which of the following has peak clock cycle rate? (A) P1 (B) P2 (C) P3 Solution: (A) 4.

(D) P4

Which of the following statement is true about every nxn matrix with only real eigen values? (A) If trace of matrix is positive and the determinant of the matrix is negative, at least one of its eigen values is negative (B) If the trace of the matrix is positive, all its eigen values are positive (C) If the determinant of the matrix is positive, all its eigen values are positive (D) If the product of the trace and determinant of the matrix is positive, all its eigen values are positive

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Solution: (A) If either the trace or determinant is positive, there is at least one positive eigen value. Trace of matrix is positive and the determinant of the matrix is negative, this is possible only when there is odd number of negative eigen values. Hence at least one eigen value is negative. 2

5.

If



x sin x dx  K, then the value of K is _______

0

2

Solution:



x sin x dx  K

0

2







x sin x dx 

0

x sin x dx  K



2







 x sin x dx     x sin x  dx  K 0





2

0



  x cos x  sin x    x cos x  sin x 

 4  K 6.

 K

K 4

Consider the following combinational function block involving four Boolean variable x, y, a, b where x, a, b are inputs and y is the output. F(x, y, a, b) { If (x is 1) y = a; else y = b; } Which of the following digital block is most suitable for implementing the function? (A) Full Adder

(B) Priority Encoder

(C) Multiplexer

(D) Flip flop

Solution: (C)

b

I0 2 1 MUX

a

y

I1

x

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Let G be a group with 15 elements. Let L be a subgroup of G. It is known that LG and that the size of L is at least 4. The size of L is _____

Solution: Order of subgroup divides order of group (Lagrange’s theorem). 3, 5 and 15 can be the order of subgroup. As subgroup has at least 4 elements and it is not equal to the given group, also order of subgroup can’t be 3 and 15. Hence it is 5. 8.

The smallest length of the string not generated by the given regular expression is _____ a*b*(ba)*a*

Solution: Smallest length is 3 and string is bab. 9.

Consider the following statements: P: Good mobile phones are not cheap Q: Cheap mobile phones are not good L: P implies Q M: Q implies P N: P is equivalent to Q (A) Only L is true

(B) Only M is true

(C) Only N is true

(D) L, M, N are true

Solution: (D) g : mobile is good

c : mobile is cheap

P : Good mobile phones are not cheap  g   c    g   c 

Q a  b   a  b

Q : Cheap mobile phones are not good  c   g    c   g  Both P and Q are equivalent which means P and Q imply each other

10.

If V1 and V2 are four dimensional subspace of a six dimensional vector space V then the smallest possible dimension of V1V2 is _____

11.

Consider  = {a,b} and * denotes all possible string that can be generated by {a,b}, and 2 (A) 2 (B) 2 (C) 2 (D) 2

* * * *

*

is the power set of *. Which of the following is true?

is countable * is uncountable is countable * is countable is uncountable * is countable is uncountable * is uncountable

Solution: (C) With diagonalization we can prove that 2

*

is uncountable and

*

 is countable as one to one mapping is possible with natural numbers.

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Consider the set of all functions, f: {0, 1, …., 2014}{0, 1, …., 2014} such that f(f(i))= i  0i2014. Consider the following statements: P: For each such function it must be the case that for every i, f(i) = i Q: For each such function it must be the case that for some i, f(i) = i R: Each such function must be onto. Which of the following is CORRECT? (A) P, Q and R are true

(B) Only Q and R are true

(C) Only P and Q are true

(D) Only R is true

Solution: (B) P : For example, let f  0   1 and f 1  0  f  f  0    f 1  0 and f  f 1   f  0   1  f  f i   i but f i  i P is not true.



For some i, f(i) = i since there are odd number of elements in the set on which function is defined. 13.

The value of the integral given below is  2

x

cos xdx

0

(A) -2

(B) 

(C) -

(D) 2

Solution: (A) 

 2

x

2

cos x dx  x

 sin x   2x   cos x   2   sin x 

O

o





 2 sin   2 cos   2 sin    0  0  0   2.0  2  1  0

14.

 2

Consider a basic block a = b+c c = a+d d = b+c e = d-b a = e+b Maximum number of nodes and edges in DAG generated for the above block (A) 6, 6

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(B) 9, 12

(C) 8, 10

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Solution: (C)

d

e 

a



c 

a 

d

15.

c

b

Consider a decision problem 2CNFSAT = { | is satisfiable propositional formula in CNF with at most 2 literals per clauses} For example  = (x1  x2)  (x1  x3 )  (x2  x4) a Boolean formula 2 CNFSAT The decision problem 2CNFSAT is (A) NP- complete (B) Solvable in P time by reduction to directed graph reachability problem (C) Solves in constant time if every input sequence is satisfiable (D) NP hard but not NP complete

Solution: (B) 16.

With respect to the numerical evaluation of the definite integral, b

K

2

 x dx , where a and b are given, which of the following is/are TRUE? a

I. The value of K obtained using Trapezoidal rule is always greater than or equal to the exact value of the definite integral. II. The value of K obtained using Simpson’s rule is always equal to the exact value of the definite integral. (A) I only

(B) II only

(C) Both I and II

(D) Neither I nor II

Solution: (C) Let us consider 1

x3 x dx  o 3

1

2

 0

1  0.33333 up to 5decimals  3

Let n = 4

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x

0

yx

2

0 y0

1 2

x

0.25

0.75

1

0.0625 0.25 0.5625 1 y1 y2 y3 y4

h  y 0  y 4   2  y1  y2  y3   (By trapezoidal rule) 2

dx 

0



0.5

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0.25  0  1  2  0.0625  0.25  0.5625   2 

 0.34375

Clearly the value is greater than 0.33333, also for greater values of ‘n’ there will be slight difference, hence given integral is always greater than 0.33333 Using Simpson’s 1

h

2

 x dx  3  y

0

1 rd rule 3

 y 4   2  y2   4  y1  y3  

o



0.25  0  1  2  0.25   4  0.0625  0.5625   3 

 0.33333 up to 5  decimals  , which is equal to exact value.

17.

J

J

Q2

Q1

C

C

K

J

Q0

K

Q0

C

K

Q2

Q1

The above synchronous sequential circuit built using JK flip flops is initialized with Q2Q1Q0 = 000. The state sequence for this circuit for the next three clock cycles is (A) 001, 010, 011

(B) 111, 110, 101

(C) 100, 110, 111

(D) 100, 011, 001

Solution: (C) P.S. Q2

Q1

FF inputs Q0

J2

K2

J1

N.S.

K1

J0

K0

Q2



Q1

Q0 

 Q  Q  Q  Q  Q  Q  1

0

2

2

1

0

0 1

0 0

0 0

1 1

0 0

0 1

1 0

0 0

1 1

1 1

0 1

0 0

1

1

0

0

0

1

0

1

1

1

1

1

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Which of the following decision problem is undecidable? (A) Ambiguity problem of context free grammar (B)If the string generated by the context free grammar (C) If context free grammar generates empty language (D) If context free grammar generates finite language

Solution: (A) There is no algorithm for determining the ambiguity of a given CFG, hence it is undecidable. 19.

A cache memory has memory access time for read operation as 1ns if cache hit occurs and 5ns for read operation if cache miss occurs. Memory access time for write operation is 2ns if cache hit occurs and 10ns for write operation if cache miss occurs. Suppose 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations are being performed. The cache hit rate is 0.9, what is the average execution time for the instruction?

Solution: For fetch operation 90 × 1ns + 10× 5ns = 140ns Memory operand read operation 54ns + 30ns = 84 ns Memory operand write operation 72ns+40ns = 112ns Total time for 100 instruction is 140 +84+112 = 336ns Average time is 3.36ns. 20.

Consider the five stage pipeline system with Instruction Fetch (IF) , Instruction Decode and Register Fetch (ID/RF), Execute (EX), Memory access( MEM) and Write Back (WB) operations. The respective stages are taking time 1ns, 2.2ns, 2ns, 1ns, 0.75ns. A modification is proposed on this pipeline system and ID/RF stage is further divided in ID, RF1 and RF2 taking 2.2/3ns each, further EX stage is also divided in EX1 and EX2 each taking equal time. If the instruction is a branch instruction, next instruction pointer will be available at the end of EX stage of first pipeline and after EX2 for the second pipeline. All instruction other than branch instruction have average CPI as 1. 20% instructions are branch instruction. If time average access time for first pipeline is P and for second is Q, the ratio of P & Q is______

Solution: Tavg = (1+ stall frequency × Stall cycle) × Tclock TPavg = (1+ .2 × 2) × 2.2ns = 3.08ns IF

TQavg = (1+ .2 × 5) × 1ns = 2ns

P  1.54 Q

21.

IF

ID

RF1

RF2

EX1

stall

stall

stall

stall

ID/RF

EX

stall

stall

EX2 stall

IF

For first pipeline 2 stall cycles IF

For second pipeline 5 stall cycles

If G is a forest with n vertices and k connected components, how may edges does G have? (A) n   k

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(B) n   k

(C) n  k

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(D) n  k  1

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Solution: (C) Forest is collection of trees and in a forest each tree will form one connected component. Assume that there are k connected components where each is having n1, n2, n3….. nk vertices respectively. Total number of edges in a forest with n vertices and k connected components will be as follows

n1  1  n2  1  n3  1  ........  nk  1  n1  n2  n3  ...  nk   1  1  1  ......k times  nk

22.

Let  denotes the minimum degree of vertex in a graph. For all planar graphs on n vertices with 3, which of the following is TRUE? (A) In any planar embedding, the number of faces is at least

n 2 2

(B) In any planar embedding, the number of faces is less than

n 2 2

(C) There is a planar embedding in which the number of faces is less than (D) There is a planar embedding in which the number of faces is at most

23.

n 2 2

n  1

The correct formula for the sentence “not all rainy days are cold” is (A) d Rainy  d  ~ Cold  d 

(B) d ~ Rainy  d  Cold  d  

(C) d ~ Rainy  d  Cold  d 

(D) d Rainy  d  ~ Cold  d 

Solution: (D)



not all rainy days are cold:~ d Rainy  d  Cold  d 





 ~ d ~ Rainy d   Cold  d 



 d Rainy  d  ~ Cold  d  

24.

Consider the following language defined over  = {0,1,c} L1 = {0n1n|n ≥ 0} L2 = {wcwr| w  {0,1}*} L3 = {wwr| w  {0,1}*} Which of the above language can be recognized by deterministic pushdown automata? (A) None of the language (C) L1 & L2

(B) Only L1 (D) All the languages

Solution: (C) L1 and L2 are deterministic CFLs and so those can be recognized by deterministic PDA where as L3 is not deterministic.

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Let S be a sample space and two mutually exclusive events A and B be such that AB = S. If P(.) denotes the probability of the event, the maximum value of P(A)P(B) is _____

Solution: A  B  S  P  A  B   P S   1  P  A   P B   P  A  B   1  P  A   P B   1 [Q A & B are M.E.] Let P  A   x

 P B   1  x

y  P  A  P B   x 1  x   d2 y  dy 1 d2 y  1  2x  0  x  ;   2  0;  2  0  2 dx 2 dx2  dx x  1 2

 y has maximum at x   ymax 

26.

1 2

1 1 1    0.25 2 2 4

Let  denotes the exclusive XOR operator. Let “1”, “0” denote binary constants. Consider following Boolean expression for F over two variables P and Q:

 1  P   P  Q     P  Q    Q  0 

F P, Q  

The equivalent expression for F is (B) P  Q

(A) P  Q

(C) P  Q

(D) P  Q

Solution: (D) F P, Q  

 1  P   P  Q     P  Q   Q  0  



    PQ  PQ   Q  P PQ  P Q   P PQ  PQ    PQ  PQ  Q  PQ  PQ  Q       PQ  PQ   PQ  PQ  

 P  PQ  PQ

 QP  PQ  PQ  P  Q

27.

There are two elements x, y in a group (G,*) such that every element in the group can be written as a product of some number of x’s and y’s in some order. It is known that x*x = y*y=x*y*x*y=y*x*y*x=e where e is the identity element. The maximum number of elements in such a group is ______

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The table below has question wise data on the performance of students in an examination. The marks for each question are also listed. There is no negative or partial marking in the examination. Question

Marks

Number

Answered

Answered

Correctly

Wrongly

Not attempted

1

2

21

17

6

2

3

15

27

2

3

2

23

18

3

What is the average of the marks obtained by the class in the examination? (A) 1.34

(B) 1.74

(C) 3.02

(D) 3.91

Solution: (C) Average marks :

29.

2  21  3  15  2  23  133  3.02 44

44

While trying to collect from under the table an envelop  I II

,

Mr. X fell down and  III

was losing consciousness .  IV 

Which of the above underlined parts of the sentences is NOT appropriate? (A) I

(B) II

(C) III

(D) IV

Solution: (D) 30.

If she _________ how to calibrate the instrument, she _________ done the experiment. (A) knows, will have

(B) knew, had

(C) had known, could have

(D) should have known, would have

Solution: (C) 31.

Choose the word that is opposite in meaning to the word “coherent”. (A) sticky

(B) well-connected (C) rambling

(D) friendly

Solution: (C) 32.

Which number does not belong to the series below? 2, 5, 10, 17, 26, 37, 50, 64 (A) 17

(B) 37

(C) 64

(D) 26

Solution: (C)

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33.

Consider the equation: (7526)8 - (Y)8=(4364)8, where (X)N stands for X to the base N. Find Y. (A) 1634 (B) 1737 (C) 3142 (D) 3162 Solution: (C) 7 5 2 68  3 1 4 28 4 3 6 48

By the beginning of 20th century, several hypothesis were being proposed, suggesting a paradigm shift in our understanding of the universe. However the clinching evidence was provided by experimental measurements of the positions of a star which was directly behind our sun. Which of the following inference(s) may be drawn from the above paragraph? i. Our understanding of the universe changes based on the position of stars ii. Paradigm shift usually occur at the beginning of centuries iii. Stars are important objects in the universe iv. Experimental evidence was important in confirming this paradigm shift (A) i, ii & iv (B) iii only (C) i & iv (D) iv only Solution: (C) 34.

35.

The Gross Domestic Product (GDP) in Rupees grew at 7% during 2012-2013. For international comparison, the GDP is compared in US dollars (USD) after conversion based on the market exchange rate. During the period 2012-2013 the exchange rate for the USD increased from Rs. 50/USD to Rs. 60/USD. India’s GDP in USD during the period 2012-2013 (A) increased by 5% (B) decreased by 13% (C) decreased by 20% (D) decreased by 11% Solution: (B) The ratio of Male to Female students in a college for 5 years is plotted in the following line graph. If the number of female students in 2011 and 2012 is equal, what is the ratio of male students in 2012 to male students in 2011? Ratio of Male to Female students

36.

(A) 1:1 Solution: (C)

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3.5 3 2.5

2 1.5

1 0.5 0

2008

(B) 2:1

2009

2010

2011

(C) 1.5:1

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2012

(D) 2.5:1

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A dance program is scheduled for 10.00 am. Some students are participating in the program and they need to come 1 hour earlier than the start of the event. These students should be accompanied by a parent. Other students and parents should come in time for the program. The instruction you think that is appropriate for this is (A) Students should come at 9.00 am and parents should come at 10.00 am (B) Participating students should come at 9.00 am accompanied by a parent and other parents and students should come by 10.00 am (C) Students who are not participation should come by 10.00 am and they should not bring their parents. Participating students should come at 9.00 am. (D) Participating students should come before 9.00 am. Parents who accompany them should come at 9.00 am. All other should come at 10.00 am

Solution: (B)

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