graduate aptitude test in engineering ( gate )

GRADUATE APTITUDE TEST IN ENGINEERING ( GATE )

The Graduate Aptitude Test in Engineering (GATE) is an All-India Examination conducted by the seven Indian Institutes of Technology and the Indian Institute of Science, Bangalore, on behalf of the National Coordinating Board - GATE, Department of Education, Ministry of Human Resources Development (MHRD), Government of India.

OBJECTIVES OF GATE

Identify, at the National level, meritorious and motivated candidates for admission to Postgraduate Programmes in Engineering, Technology, Architecture, Science and Pharmacy.

Serve as a mandatory qualification for the MHRD scholarship/assistantship. Serve as benchmark of normalization of the Undergraduate Engineering Education in the

country.

GATE - Graduate Aptitude Test in Engineering

Post Engineering, all Engineering students come at crossroads, where they have to make a choice – a choice that defines their career for the rest of their lives: to choose one amongst the options of a job, management or M.Tech. The three most widely used options are management, M. Tech or a job. Given the current situation in India, most of the engineering colleges are not able to give a 100% placement, which results in the candidate making a choice between management and M. Tech. This article is about choosing the second option : M.Tech, and the exam that holds the key to the most important institutes for M.Tech, the GATE.

Why does one choose CAT over GATE or vice versa? To quote the Bible, “A feast is made for laughter, and wine maketh merry: but money answereth all things”. Most of the students choose management because they perceive that a management person makes more money than a research person. This might have been true in the past. But now that companies like Motorola and Honda are beginning to outsource their R&D to India, the demand for Research oriented people is on the rise. "Virtually every single global IT major is rapidly expanding its footprint into India," says Sunil Mehta, vice president, Nasscom. As of now, 100 Fortune 500 companies - including Delphi, Eli Lilly, General Electric, Hewlett Packard, DaimlerChrysler and others - have put up R&D facilities in India over the past five years. Considering the scenario, M.Tech is no longer the underdog. And the number of students giving GATE has been increasing every year.

One needs to consider factors other than monetary as well before he makes the choice. CAT is for management, and management means management of people, finance, affairs of a company, strategizing etc. So people who are good at management, and have interest in pursuing this should take management as a subject. M.Tech, however, entails a research oriented job or a faculty position in a university. Down the line, these people may even take up a management position in a research based enterprise. However, essentially both kinds of careers are for a completely different persona and interests.

Though the number of GATE aspirants is equal to or probably even more than CAT aspirants, it has largely been neglected by the media as well as the coaching institutes. Till date, to my information, there is not a single GATE coaching institute with pan India presence. Check Out the Freshersworld.com Directory for all the Coaching Institutes in India.

After one makes a choice as to what he would take up as a career, he needs to prepare well for it. And to prepare well, he needs to know and understand what GATE is all about....

GATE 2007 - Why Comparative Evaluation

Two friends, Ram and Shyam, are out on a safari in Kenya. En-route they see a vast stretch of land covered with velvette-ish green layer of grass. They get out of their car, relax, put off their shoes and start walking bare footed. Suddenly they see a tiger running towards them. Ram starts running whereas Shyam starts putting on the shoes. Puzzled, Ram asks Shyam, "Do you think you can outrun the tiger just by have the shoes on?" Shyam replies back, "I don't need to outrun the tiger. I just need to outrun YOU!"

India has a huge intellectual capital. With so many intelligent brains competing with each other, percentile/comparative evaluation is the most logical solution for admission to IISc and the IITs, which in turn have a limited number of seats. With this kind of comparative evaluation, the institutions are able to simply take the best of the lot and leave the rest. You should be aware that only 4 out of a thousand students giving GATE get a chance to be interviewed by IITs and the IISc. The numbers getting admission to these institutions are even lesser. So being good is not enough. You should be one of the better ones amongst the best.

The sole purpose of GATE is not only to conduct a test for admission but, which is also the major goal, to identify the suitable engineers/researchers in various areas and make them to choose their area of choice in which they can pursue their master degree. As you must be knowing by now, GATE is only organized by IITs and IISc which always try to innovate and introduce new systems to test the basic knowledge of candidates in various fields.

As a part of that innovation, the previous year GATE papers were fully objective type. This year also the same pattern is being used, and mathematics added! Although this made the job relatively easier for the evaluators, the candidates now have a bound choice to cover all the topics. Being set by the highly qualified faculty, the questions papers are not like a very traditional one. They always try to test the very basic of candidates in respective areas by

bringing the element of innovation in these papers. It is not surprising that sometimes the questions are mere applications of basic concepts in their area of research.

GATE - How should one prepare, specifically GATE 2009

How does one prepare for GATE, specifically GATE 2009

Before the marathon begins, the ‘runner’ puts in months, even years, of effort before he actually runs the race. As the saying goes – get your fundamentals right. GATE is an exam that tests you on your fundamentals. The questions are generally derivations of the fundamentals. Preparation for GATE is an ongoing process, and is supposed to happen in stages. First get your fundamentals right, and then test yourself on those fundamentals. When you have done this, you should pit yourself with the competition, which means a mock test which would give you a percentile to let you know where you stand amongst competitors.

How do you choose which coaching material to go with. This is something that is quite subjective. A coaching material cannot be assessed until you have gone through it. So, you could go by the word of your seniors as to what coaching material they followed to prepare for GATE.

Here are some parameters on which you can decide whether you should go for a particular coaching institute, use a particular coaching material, or tests:

1. Uniqueness in their study material: Study material provided by the coaching institutes is of little assistance if they are compact copy-paste or rewrite of materials taken from other books. If the reading of such material doesn’t increase interest and enjoyment then they are not worth it. There are plenty of standard books on each subject by good authors, which can make your study enjoyable during preparation.

2. Collection of quality books in their library: During the process of theory conceptualization and building application capabilities, you need good books, which can really put your brain on exercise. Check out their library!

3. Flexibility in the Coaching Model: What happens when your pace of learning is much faster or slower than the average? Is there any mechanism by which the model can identify exactly where you need help and provide the same? Is it possible in that coaching model to minimize the wastage of your time?

4. Quality of questions discussed: Number of questions discussed is not that important. By discussing and solving 10-15 conceptual questions on each topic you can build a good

application capability. On the other hand solving many tricky non-conceptual questions will simply waste your time.

5. Tests and evaluation model: How is the progress of your preparation tested and analyzed? To what extent the feedback helps in identifying the areas for further work? Here I must say that this is the most crucial part of the preparation. This is the area where most of the students fail due to lack of proper test materials which can help them to build in themselves a real-test-like environment and temperament. Once you are able to choose the correct assistance for your GATE journey, it will be an enjoying and thrilling experience.

Here are some things apart from working on your engineering concepts, that you should do for a complete preparation for GATE:

1. Solve previous years’ GATE papers: Solving previous years’ papers gives you a fair idea of what the actual paper would be like. It also brushes up your basics and exposes your ‘areas of improvement’.

2. Solve test papers: Solve as many test papers as possible. This actually is the best way to keep improving as you prepare for GATE.

3. Analyze : Analyzing your test results is a very important part of taking the test. If you do not analyze, the test does not add value. You should minutely analyze and define as to where you could have scored more; analyze your accuracy rates in various topics and maintain a topic wise datasheet which lists your performance topic wise for different test papers.

4. Simulate actual test environment : This is very important. The actual test happens in a classroom, and is timed. When you take up the test, switch off your cell phone, have a timer which times your tests, and avoid taking any breaks. Also, if possible, take up a mock test series which enables you to take the test in a classroom environment.

GATE - Structure of the Examination

The Latest Pattern From 2007 onwards

The GATE 2008 examination consists of a single paper of 3 hours duration and carries a maximum of 150 marks.

The question paper of GATE 2009 will be fully objective type.

Candidates have to mark the correct choice by darkening the appropriate bubble against each question on an Objective Response Sheet (ORS).

There will be negative marking for wrong answers. The deduction will be 25% of the marks allotted. A candidate will have to choose any one of the papers listed below:

Aerospace Engineering AE Information Technology IT

Agricultural Engineering AG Mathematics MA

Architecture AR Mechanical Engineering ME

Civil Engineering CE Mining Engineering MN

Chemical Engineering CH Metallurgical Engineering MT

Computer Science & Engg. CS Physics PH

Chemistry CY Production & Industrial Engg. PI

Electronics & Comm. Engg. EC Pharmaceutical Sciences PY

Electrical Engineering EE Textile Engg.& Fibre Science TF

Geology & Geophysics GG Engineering Sciences XE*

Instrumentation Engineering IN Life Sciences XL*

Papers XE and XL are of general nature and will comprise the following Sections:

Three Sections, one compulsory as indicated below:

ENGINEERING SCIENCES(XE) CODE LIFE SCIENCES(XL) CODE

Engg. Maths (Compulsory) A Food Technology I

Computational Science B Chemistry (Compulsory) J

Electrical Sciences C Biochemistry K

Fluid Mechanics D Biotechnology L

Materials Science E Botany M

Solid Mechanics F Microbiology N

Thermodynamics G Zoology O

Polymer Science and Engineering H

GATE Results

The GATE result will be announced on March 15, 2009 at 10000 hrs. at GATE offices of IITs/ IISc.

GATE scorecard

Scorecard will be sent only to the qualified candidates. No information will be sent to candidates who are not qualified.

The GATE scorecard is a valuable document. Care should be taken to preserve it. Additional scorecards (upto a maximum of two) will be issued on payment basis only once.

The scorecard cannot be treated as a proof of date of birth, category and disability status.

The scorecard will indicate GATE score and rank of the qualified candidates.

GATE score

The GATE score of a candidate is in the range 0 to 1000. It reflects the performance of a candidate, irrespective of the GATE paper or year in which he/she has qualified. Candidates with same GATE score from different GATE papers and/or years can be considered to have the same performance level.

The marks obtained by the candidate is normalized on the basis of the average and standard deviation of marks of all candidates who appeared in the paper mentioned on the scorecard in GATE 2009. Subsequently, this is scaled with respect to the global average and global standard deviation so as to facilitate performance comparison across GATE papers and over a block of years since GATE 2005.

GATE SCORE = , where

= marks obtained by the candidate.

= average of marks of all candidates who appeared in the paper mentioned on this scorecard in GATE 2008

= standard deviation of marks of all candidates who appeared in the paper mentioned on this scorecard in GATE 2008

= average (global) of marks of all candidates who appeared across all paper and years (2005-2008)

= standard deviation (global) of marks of all candidates who appeared across all papers and years (2005-2008) The maximum score can be 1000.

The evaluation of the ORS is carried out by a computerized process using scanning machines, with utmost care. Requests for revaluation of the answer script and re-totaling of marks will not be entertained.

The GATE results of the qualified candidates will be made available to interested organizations (educational institutions, R & D laboratories, industries, etc.) in India and abroad based on written request by the organization and on payment. Details can be obtained from GATE Chairmen of IITs/IISc.

GATE - The Pattern

QUESTION PAPER PATTERN

The question paper will consist of only objective type questions. Candidates have to mark the correct choice by darkening the appropriate bubble against each question on an Objective Response Sheet (ORS). There will be negative marking for wrong answers. The deduction for each wrong answer will be 25% of the allotted marks.

MAIN PAPERS

The question paper will be for a total of 150 marks divided into three groups:

(i) Group I: Question Numbers 1 to 20 (20 questions) will carry one mark each (sub total 20 marks).

(ii) Group II: Question Numbers 21 to 75 (55 questions) will carry two marks each (sub total 110 marks). Out of these, Q.71 to Q.75 may be common data based questions.

(iii) Group III: Question Numbers 76 to 85 (10 questions) will carry two marks each (sub total 20 marks). These questions are called linked answer questions. These 10 questions comprise five pairs of questions (76 & 77, 78 & 79, etc.). The solution to the second question of each pair (e.g. Q.77) will be linked to the correct answer to the first one (e.g. Q.76) in the pair.

Each question will have four choices for the answer. Only one choice is correct. Wrong answers carry 25% negative marks. In Q.1 to Q.20, 0.25 mark will be deducted

for each wrong answer and in Q.21 to Q.76, Q.78, Q.80, Q.82 and Q.84, 0.5 mark will be deducted for each wrong answer. If the first question in the linked pair is wrong, then the second question in the pair will not be evaluated. However, there is no negative marking for the linked answer questions in Q.77, Q.79, Q.81, Q.83 and Q.85.

Papers bearing the code AE, AG, CE, CH, CS, EC, EE, IN, IT, ME, MN, MT, PI, TF will contain questions on Engineering Mathematics to the extent of 20 to 25 marks.

The multiple choice objective test questions can be of the following type:

(i) Each choice containing a single stand-alone statement/phrase/data.

Example Q. The state of an ideal gas is changed from (T1, P1) to (T2, P2) in a constant volume process. To calculate the change in enthalpy, h, ALL of the following properties/variables are required. (A) Cv, P1, P2 (B) Cp, T1, T2

(C) Cp, T1, T2, P1, P2 (D) Cv, P1, P2, T1, T2

(ii) Each choice containing a combination of option codes. The question may be accompanied by four options P, Q, R, S and the choices may be a combination of these options. The candidate has to choose the right combination as the correct answer.

Example Q. The following list of options P, Q, R and S are some of the important considerations in the design of a shell and tube heat exchanger. P: square pitch permits the use of more tubes in a given shell diameter Q: the tube side clearance should not be less than one fourth of the tube diameter R: baffle spacing is not greater than the diameter of the shell or less than one fifth of the shell diameter. S: the pressure drop on the tube side is less than 10 psi (A) P, Q and R (B) Q, R and S (C) R, S and P (D) P, Q, R and S

(iii) Assertion[a]/Reason[r] type with the choices stating if [a]/[r] are True/False and/or stating if [r] is correct/incorrect reasoning of [a]

Example Q. Assertion [a]: Bernoulli’s equation can be applied along the central streamline in a steady laminar fully-developed flow through a straight circular pipe. Reason [r]: The shear stress is zero at the centre-line for the above flow. (A) Both [a] and [r] are true and [r] is the correct reason for [a] (B) Both [a] and [r] are true but [r] is not the correct reason for [a] (C) Both [a] and [r] are false (D) [a] is false but [r] is true

(iv) Match items: Match all items in Group 1 with correct options from those given in Group 2 and choose the correct set of combinations from the choices E, F, G and H.

Example Q. Group 1 contains some CPU scheduling algorithms and Group 2 contains some applications. Match entries in Group 1 to entries in Group 2. Group 1 Group 2 P- Gang Scheduling 1- Guaranteed Scheduling Q- Rate Monotonic Scheduling 2- Real-time Scheduling R- Fair Share Scheduling 3- Thread Scheduling (A) P-3; Q-2; R-1 (B) P-1; Q-2; R-3 (C) P-2; Q-3; R-1 (D) P-1; Q-3; R-2

(v) Common data based questions: Multiple questions may be linked to a common problem data, passage and the like. Two or three questions can be formed from the given common problem data. Each question is independent and its solution obtainable from the above problem data/passage directly. (Answer of the previous question is not required to solve the next question). Each question under this group will carry two marks.

Example

Common Data for Questions 74 and 75:

Let X and Y be jointly distributed random variables such that the conditional distribution of Y, given X=x, is uniform on the interval (x-1,x+1). Suppose E(X)=1 and Var(X)=5/3.

First question using common data

Q.74 The mean of the random variable Y is

(A) 1/2 (B) 1 (C) 3/2 (D) 2

Second question using common data

Q.75 The variance of the random variable Y is

(A) 1/2 (B) 2/3 (C) 1 (D) 2

(vi) Linked answer questions: These questions are of problem solving type. A problem statement is followed by two questions based on the problem statement. The two questions are designed such that the solution to the second question depends upon the answer to the first one. In other words, the first answer is an intermediate step in working out the second answer. Each question in such ‘linked answer questions’ will carry two marks.

Example

Statement for Linked Answer Questions 80 and 81:

Consider a machine with a byte addressable main memory of 216 bytes. Assume that a direct mapped data cache consisting of 32 lines of 64 bytes each is used in the system. A 50x50 two dimensional array of bytes is stored in the main memory starting from memory location 1100H. Assume that the data cache is initially empty. The complete array is accessed twice. Assume that the contents of the data cache do not change in between the two accesses.

First question of the pair

Q.80 How many data cache misses will occur in total?

(A) 48 (B) 50 (C) 56 (D) 59

Second question of the pair

Q.81 Which of the following lines of the data cache will be replaced by new blocks in accessing the array for the second time?

(A) line 4 to line 11 (B) line 4 to line 12 (C) line 0 to line 7 (D) line 0 to line 8

XE SECTION PAPERS EXCEPT FOR SECTION A

All sections apart from section A are for 60 marks each. Each of the 60 mark section will be fully objective type and the questions are divided into three groups.

(i) Group I: Question Numbers 1 to 8 (8 questions) will carry one mark each (subtotal 8 marks).

(ii) Group II: Question Numbers 9 to 30 (22 questions) will carry two marks each (subtotal 44 marks). Out of this, Q.29 and Q.30 may be common data based questions.

(iii) Group III: Question Numbers 31 to 34 (4 questions) will carry two marks each. These questions are called linked answer questions. These 4 questions comprise two pairs of questions (31 and 32, 33 and 34). The solution to the second question of each pair (e.g. Q.34) will be linked to the correct answer to the first one (e.g. Q.33) in the pair (subtotal 8 marks).

All questions have four choices with only one being correct. Wrong answers carry 25% negative marks. In Q.1 to Q.8 of each section, 0.25 mark will

be deducted for each wrong answer and in Q.9 to Q.30, 0.5 mark will be deducted for each wrong answer. However, there is no negative marking in Q.32 and Q.34.

The pattern of multiple-choice questions is the same as described for the main papers.

XE - SECTION A PAPER (Engineering Mathematics)

XE section A paper will be for 30 marks. It will consist of fully objective type and the questions are divided into two groups.


(ii) Group II: Question Numbers 7 to 18 (12 questions) will carry two marks each (subtotal 24 marks).


be deducted for each wrong answer and in Q.7 to Q.18, 0.5 mark will be deducted for each wrong answer.


XL SECTION PAPERS

Each Section is of 50 marks. Each section will be fully objective type and the questions are divided into three groups.


(ii) Group II: Question Numbers 7 to 24 (18 questions) will carry two marks each (subtotal 36 marks). Out of this, Q.23 and Q.24 may be common data based questions.

(iii) Group III: Question Numbers 25 to 28 (4 questions) will carry two marks each. These questions are called linked answer questions. These 4 questions comprise two pairs of questions (25 and 26 and 27 and 28). The solution to the second question of each pair (e.g. Q.26) will be linked to the correct answer to the first one (e.g. Q.25) in the pair (subtotal 8 marks).


be deducted for each wrong answer and in Q.7 to Q.25 and Q.27, 0.5 mark will be deducted for each wrong answer. However, there is no negative marking in Q.26 and Q.28.


Questions And Answers

No. Question

1 Consider the following program segment for a hypothetical CPU having three user registers Rl, R2 and R3.

Instruction Operation Instruction Size (in words)

MOV Rl,5000 ; Rl Memory[5000] 2

MOV R2,(R1) ; R2 Memory[(Rl)] 1

ADD R2,R3 ; R2 R2 + R3 1

MOV 6000, R2 ; Memory[6000] R2 2

HALT ; Machine halts 1Let the clock cycles required for various operations be as follows:Register to/from memory transfer : 3 clock cyclesADD with both operands in register : 1 clock cycleInstruction fetch and decode : 2 clock cycles per wordThe total number of clock cycles required to execute the program is

OptionsA) 29 B) 24C) 23 D) 20

Correct Answer

B

2The order of an internal node in a B+ tree index is the maximum number of children it can have. Suppose that a child pointer takes 6 bytes, the search field value takes 14 bytes, and the block size is 512 bytes. What is the order of the internal node?

OptionsA) 24 B) 25C) 26 D) 27

Correct Answer

C

3 The Boolean function x, y, + xy + x, y

OptionsA) x, + y, B) x + yC) x + y, D) x, + y

Correct Answer

D

4In an MxN matrix such that all non-zero entries are covered in a rows and b columns. Then the maximum number of non-zero entries, such that no two are on the same row or column, is

OptionsA) a + b B) max {a, b}C) min {M-a, N-b} D) min {a, b}

Correct Answer

A

5

The relation scheme Student Performance (name, courseNo, rollNo, grade) has the following functional dependencies:name, courseNo graderollNo, courseNo gradename rollNorollNo nameThe highest normal form of this relation scheme is

Correct Answer

A

6The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by

OptionsA) the instruction set architecture B) page sizeC) physical memory size D) number of processes in memory

Correct Answer

D

7

Consider the following program segment for a hypothetical CPU having three user registers Rl, R2 and R3.




ADD R2,R3 ; R2 R2 + R3 1

MOV 6000, R2 ; Memory[6000] R2 2

HALT ; Machine halts 1Consider that the memory is byte addressable with size 32 bits, and the program has been loaded starting from memory location 1000 (decimal). If an interrupt occurs while the CPU has been halted after executing the HALT instruction, the return address (in decimal) saved in the stack will be

OptionsA) 1007 B) 1020C) 1024 D) 1028

Correct Answer

A

8Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is

OptionsA) 12 B) 8C) Less than 8 D) More than 12

Correct Answer

A

9

What does the following algorithm approximate? (Assume m > 1, > 0).x = m;y-i;while (x - y > ){ x = (x + y) / 2 ; y = m/x ;}print (x) ;

Options A) log m B)

m2

C) m1/2 D) m1/3

Correct Answer

C

10

Consider the following C programmain (){ int x, y, m, n ;scanf ("%d %d", &x, &y);/ * Assume x > 0 and y > 0 * / m = x; n = y ;while ( m ! = n){ if (m > n)m = m — n; elsen = n - m ; }printf("%d",n); }The program computes

OptionsA) x + y, using repeated subtraction B) x mod y using repeated subtractionC) the greatest common divisor of x and y D) the least common multiple of x and y

Correct Answer

C

11 The best data structure to check whether an arithmetic expression has balanced parentheses is a

OptionsA) queue B) stackC) tree D) list

Correct Answer

B

12

A Priority-Queue is implemented as a Max-Heap. Initially, it has 5 elements. The level-order traversal of the heap is given below:10, 8,5,3,2Two new elements 1 and 7 are inserted in the heap in that order. The level-order traversal of the heap after the insertion of the elements is

OptionsA) 10,8,7,5,3,2,1 B) 10,8,7,2,3,1,5C) 10,8,7,1,2,3,5 D) 10,8,7,3,2,1,5

Correct Answer

D

13 Consider the following C program segment struct CellNode {

struct CellNode *leftChild ; int element;struct CellNode *rightChild ;}; int DoSomething (struct CellNode *ptr) {int value = 0 ; if (ptr ! = NULL){ if (ptr->leftChild ! = NULL) value = 1 + DoSomething (ptr - > leftChild) ;if (ptr - > rightChild ! = NULL)value = max (value, 1 + DoSomething (ptr - > rightChild)) ;}return (value); }

The value returned by the function DoSomething when a pointer to the root of a non-empty tree is passed as argument is

OptionsA) The number of leaf nodes in the tree B) The number of nodes in the tree C) The number of internal nodes in the tree D) The height of the tree

Correct Answer

D

14An organization has a class B network and wishes to form subnets for 64 departments. The subnet mask would be

OptionsA) 255.255.0.0 B) 255.255.64.0C) 255.255.128.0 D) 255.255.252.0

Correct Answer

D

15Suppose the round trip propagation delay for a 10 Mbps Ethernet having 48-bit jamming signal is 46.4 s. The minimum frame size is:

OptionsA) 94 B) 416C) 464 D) 512

Correct Answer

C

16 Consider the following C program segment:

char p [ 20];

char * s = "string" ;

int length = strlen (s) ;

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

p[ i ] = s [length - i] ;

print f ("%s", p) ;The output of the program is

Correct Answer

A

17

Consider the grammarS (S) | aLet the number of states in SLR(1), LR(1) and LALR(1) parsers for the grammar be n1, n2 and n3

respectively. The following relationship holds good

OptionsA) n1< n2 < n3 B) n1= n3 < n2

C) n1= n2 = n3 D) n1 n3 n2

Correct Answer

B

18

Consider the following C function:int f (int n){ static int i = 1;if (n >= 5) return n;n = n + i;i ++;return f (n);}The value returned by f(1) is

OptionsA) 5 B) 6C) 7 D) 8

Correct Answer

C

19

Consider the following code fragment:if (fork ( ) = = 0){a = a + 5; print f (“%d, %d / n”, a, and a); }else {a - 5; print f (“ %d, %d / n”, a,& a); }Let u, v be the values printed by the parent process, and x, y be the values printed by the child process. Which one of the following is TRUE?

OptionsA) u = x + 10 and v = y B) u = x + 10 and v yC) u + 10 =x and v = y D) u + 10 = x and v y

Correct Answer

B

20The following numbers are inserted into an empty binary search tree in the given order: 10, 1, 3, 5, 15, 12, 16. What is the height of the binary search tree (the height is the maximum distance of a leaf node from the root)?


Correct Answer

B

No. Question 1 Consider the following C functionvoid swap (int a, int b) { int temp ;temp = a ;a = b ;b = temp ;}In order to exchange the values of two variables x and y. Options A) call swap (x, y) B) call swap (&x, &y)C) swap (x, y) cannot be used as it does not return any value

D) swap (x, y) cannot be used as the parameters are passed by value

Your Answer (Not Answered) Correct Answer B 2 A 5 stage pipelined CPU has the following sequence of stages:IF - Instruction fetch from instruction memory,RD - Instruction decode and register read,EX - Execute: ALU operation for data and address computation,MA - Data memory access - for write access, the register read at RD stage is used,WB - Register write back.Consider the following sequence of instructions:I1 : L R0, 1oc1; R0 <= M[1oc1]I2 : A R0, R0; R0 <= R0 + R0I3: A R2, R0; R2 <= R2 - R0Let each stage take one clock cycle.What is the number of clock cycles taken to complete the above sequence of instructions starting from the fetch of I1? Options A) 8 B) 10C) 12 D) 15Your Answer (Not Answered) Correct Answer A 3 The address resolution protocol (ARP) is used for Options A) Finding the IP address from the DNS B) Finding the IP address of the default gatewayC) Finding the IP address that corresponds to a MAC address

D) Finding the MAC address that corresponds to an IP address

Your Answer (Not Answered) Correct Answer D 4 Consider the following relation schema pertaining to a students database:

Student (rollno, name, address)Enroll (rollno, courseno, coursename)where the primary keys are shown underlined. The number of tuples in the Student and Enroll tables are 120 and 8 respectively. What are the maximum and minimum number of tuples that can be present in (Student * Enroll), where ‘*’denotes natural join? Options A) 8, 8 B) 120, 8C) 960, 8 D) 960, 120Your Answer (Not Answered) Correct Answer C 5 Consider a direct mapped cache of size 32 KB with block size 32 bytes. The CPU generates 32 bit addresses. The number of bits needed for cache indexing and the number of tag bits are respectively Options A) 10, 17 B) 10, 22C) 15, 17 D) 5, 17Your Answer (Not Answered) Correct Answer A 6 Which one of the following is a key factor for preferring B+ -trees to binary search trees for indexing database relations? Options A) Database relations have a large number of records

B) Database relations are sorted on the primary key

C) B+ -trees require less memory than binary search trees

D) Data transfer from disks is in blocks

Your Answer (Not Answered) Correct Answer D 7 The goal of structured programming is to Options

A) have well indented programsB) be able to infer the flow of control from the compiled code

C) be able to infer the flow of control from the program text

D) avoid the use of GOTO statements

Your Answer (Not Answered) Correct Answer C 8 The tightest lower bound on the number of comparisons, in the worst ease, for comparison-based sorting is of the order of Options A) n B) n2

C) n log n D) n log2 nYour Answer (Not Answered) Correct Answer B 9 Consider the following two problems on undirected graphsa: Given G(V, E), does G have an independent set of size |V| -4?b: Given G(V, E), does G have an independent set of size 5?Which one of the following is TRUE? Options A) is in P and is NP-complete B) is NP-complete and is in PC) Both and are NP-complete D) Both and are in PYour Answer (Not Answered) Correct Answer C 10 Consider the languagesL1 = {wwR | w {0, 1}*}L2 = {w# wR | w {0, 1}*}, where # is a special symbolL3 = {ww | w {0, 1}*}Which one of the following is TRUE? Options A) L1 is a deterministic CFL B) L2 is a deterministic CFLC) L3 is a CFL, but not a deterministic CFL D) L3 is a deterministic CFLYour Answer (Not Answered) Correct Answer B 11 Let G be a simple graph with 20 vertices and 100 edges. The size of the minimum vertex cover of G is 8. Then, the size of the maximum independent set of G is Options

A) 12 B) 8C) Less than 8 D) More than 12Your Answer (Not Answered) Correct Answer A 12 A and B are the only two stations on an Ethernet. Each has a steady queue of frames to send. Both A and B attempt to transmit a frame, collide, and A wins the first backoff race. At the end of this successful transmission by A, both A and B attempt to transmit and collide. The probability that A wins the second backoff race is Options A) 0.5 B) 0.625C) 0.75 D) 1.0Your Answer (Not Answered) Correct Answer A 13 Consider the grammarE E + n | E x n | nFor a sentence n + n x n, the handles in the right-sentential form of the reduction are Options A) n, E + n and E + n x n B) n, E + n and E + E x nC) n, n + n and n + n x n D) n, E + n and E x nYour Answer (Not Answered) Correct Answer D 14 Consider the grammar with the following translation rules and E as the start symbol.E E1 # T {E.value = E1.value * T.value } | T {E.value = T.value }T T1 & F { T.value = T1.value + F.value } | F {T.value = F.value}F num {F.value = num.value }Compute E. value for the root of the parse tree for the expression: 2 # 3 & 5 # 6 & 4. Options A) 200 B) 180C) 160 D) 40Your Answer (Not Answered) Correct Answer C 15 Let A be a sequence of 8 distinct integers sorted in ascending order. How many distinct pairs of

sequences, B and C are there such that (i) each is sorted in ascending order, (ii) B has 5 and C has 3 elements, and (iii) the result of merging B and C gives A?

Options A) 2 B) 30C) 56 D) 256Your Answer (Not Answered) Correct Answer D 16 Which one of the following is true for a CPU having a single interrupt request line and a single interrrupt grant line? Options A) Neither vectored interrupt nor multiple interrupting devices are possible.

B) Vectored interrupts are not possible but multiple interrupting devices are possible.

C) Vectored interrupts and multiple interrupting devices are both possible.

D) Vectored interrupt is possible but multiple interrupting devices are not possible.

Your Answer (Not Answered) Correct Answer B 17 Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold : (A B, BC D, E C, D A). What are the candidate keys of R? Options A) AE, BE B) AE, BE, DEC) AEH, BEH, BCH D) AEH, BEH, DEHYour Answer (Not Answered) Correct Answer D 18

In a network of LANs connected by bridges, packets are sent from one LAN to another through intermediate bridges. Since more than one path may exist between two LANs, packets may have to be routed through multiple bridges.

Why is the spanning tree algorithm used for bridge-routing?

Options A) For shortest path routing between LANs B) For avoiding loops in the routing pathsC) For fault tolerance D) For minimizing collisionsYour Answer (Not Answered) Correct Answer B 19 Suppose T(n) = 2T (n/2) + n, T(0) = T(1) = 1Which one of the following is FALSE? Options A) T(n) = O(n2 ) B) T(n) = (n log n)C) T(n) = (n2) D) T(n) = O(n log n)Your Answer (Not Answered) Correct Answer B 20

Consider the following C program segment struct CellNode {struct CellNode *leftChild ; int element;struct CellNode *rightChild ;}; int DoSomething (struct CellNode *ptr) {int value = 0 ; if (ptr ! = NULL){ if (ptr->leftChild ! = NULL) value = 1 + DoSomething (ptr - > leftChild) ;if (ptr - > rightChild ! = NULL)value = max (value, 1 + DoSomething (ptr - > rightChild)) ;}return (value); }

The value returned by the function DoSomething when a pointer to the root of a non-empty tree is passed as argument is

Options A) The number of leaf nodes in the tree B) The number of nodes in the tree C) The number of internal nodes in the tree D) The height of the tree

Your Answer (Not Answered) Correct Answer D

Questions And Answers

No. Question

1

What does the following algorithm approximate? (Assume m > 1, > 0).x = m;y-i;while (x - y > ){ x = (x + y) / 2 ; y = m/x ;}print (x) ;

OptionsA) log m

B)

m2

C) m1/2 D) m1/3

Your Answer

(Not Answered)

Correct Answer

C

2 The problems 3-SAT and 2-SAT are

OptionsA) both in P B) both NP-completeC) NP-complete and in P respectively D) undecidable and NP-complete respectively

Your Answer

(Not Answered)

Correct Answer

C

3Consider a relation scheme R = (A, B, C, D, E, H) on which the following functional dependencies hold : (A B, BC D, E C, D A). What are the candidate keys of R?

OptionsA) AE, BE B) AE, BE, DEC) AEH, BEH, BCH D) AEH, BEH, DEH

Your Answer

(Not Answered)

Correct D

Answer

4

A circuit outputs a digit in the form of 4 bits. 0 is represented by 0000,1 by 0001,..., 9 by 1001. A combinational circuit is to be designed which takes these 4 bits as input and outputs 1 if the digit ³ 5, and 0 otherwise. If only AND, OR and NOT gates may be used, what is the minimum number of gates required?


Your Answer

(Not Answered)

Correct Answer

C

5If 73x (in base-x number system) is equal to 54y (in base-y number system), the possible values of x and y are

OptionsA) 8, 16 B) 10, 12C) 9, 13 D) 8, 11

Your Answer

(Not Answered)

Correct Answer

D

6In a packet switching network, packets are routed from source to destination along a single path having two intermediate nodes. If the message size is 24 bytes and each packet contains a header of 3 bytes, then the optimum packet size is


Your Answer

(Not Answered)

Correct Answer

D

7

Consider the following program fragment for reversing the digits in a given integer to obtain a new integer. Let n = d1d2... dm.int n, rev;rev = 0;while (n > 0) {rev = rev * 10 + n % 10 ;n = n/10;}The loop invariant condition at the end of the ith iteration is:

OptionsA) n = d1d2... dm-i and rev = dmdm-1….dm-i+1 B) n = dm-i+1...dm-1dm (or) rev = dm-i...d2d1

C) n rev D) n = d1d2...dm (or) rev = dm...d2d1

Your Answer

(Not Answered)

Correct Answer

A

8

Consider the following program segment for a hypothetical CPU having three user registers Rl, R2 and R3.




ADD R2,R3 ; R2 R2 + R3 1

MOV 6000, R2 ; Memory[6000] R2 2

HALT ; Machine halts 1Let the clock cycles required for various operations be as follows:Register to/from memory transfer : 3 clock cyclesADD with both operands in register : 1 clock cycleInstruction fetch and decode : 2 clock cycles per wordThe total number of clock cycles required to execute the program is

OptionsA) 29 B) 24C) 23 D) 20

Your Answer

(Not Answered)

Correct Answer

B

9

Consider an operating system capable of loading and executing a singlesequential user process at a time. The disk head scheduling algorithm used is First Come First Served (FCFS). If FCFS is replaced by Shortest Seek Time First (SSTF), claimed by the vendor to give 50% better benchmark results, what is the expected improvement in the I/O performance of user programs ?

OptionsA) 50% B) 40%C) 25% D) 0%

Your Answer

(Not Answered)

Correct Answer

D

10 Consider the following statements with respect to user-level threads andkernel-supported threads(i) Context switch is faster with kernel-supported threads

(ii) For user-level threads, a system call can block the entire process(iii) Kernel-supported threads can be scheduled independently(iv) User-level threads are transparent to the kernelWhich of the above statements are true?

OptionsA) (ii), (iii) and (iv) only B) (ii) and (iii) onlyC) (i), and (iii) only D) (i) and (ii) only

Your Answer

(Not Answered)

Correct Answer

A

11Consider three decision problems P1, P2 and P3. It is known that P1 is decidable and P2 is undecidable. Which one of the following is TRUE?

OptionsA) P3 is decidable if P1 is reducible to P3 B) P3 is undecidable if P3 is reducible to P2

C) P3 is undecidable if P2 is reducible to P3

D) P3 is decidable if P3 is reducible to P2s complement

Your Answer

(Not Answered)

Correct Answer

A

12

Which one of the following are essential features of an object-orientedprogramming language?

(i) Abstraction and encapsulation(ii) Strictly-typedness(iii) Type-safe property coupled with sub-type rule(iv) Polymorphism in the presence of inheritance

OptionsA) (i) and (ii) only B) (i) and (iv) onlyC) (i), (ii) and (iv) only D) (i), (iii) and (iv) only

Your Answer

(Not Answered)

Correct Answer

B

13

Consider the following relation schema pertaining to a students database:Student (rollno, name, address)Enroll (rollno, courseno, coursename)where the primary keys are shown underlined. The number of tuples in the Student and Enroll tables are 120 and 8 respectively. What are the maximum and minimum number of tuples that can be present in (Student * Enroll), where ‘*’denotes natural join?

Options A) 8, 8 B) 120, 8

C) 960, 8 D) 960, 120

Your Answer

(Not Answered)

Correct Answer

C

14

Assume the following C variable declaration int *A [10], B [10][10];Of the following expressions

I A[2] II A [2] [3] III B[l] IV B[2][3]which will not give compile-time errors if used as left hand sides of assignment statements in a C program ?

OptionsA) I, II, and IV only B) II, III, and IV only

C) II and IV only D) IV only

Your Answer

(Not Answered)

Correct Answer

D

15 How many distinct binary search trees can be created out of 4 distinct keys?

OptionsA) 5 B) 14C) 24 D) 42

Your Answer

(Not Answered)

Correct Answer

B

16

Given the following input (4322, 1334, 1471, 9679, 1989, 6171, 6173, 4199) and the hash function x mod 10, which of the following statements are true ?(i) 9679, 1989, 4199 hash to the same value(ii) 1471, 6171 hash to the same value(iii) All elements hash to the same value(iv) Each element hashes to a different value

OptionsA) (i) only B) (ii) onlyC) (i) and (ii) only D) (iii) or (iv)

Your Answer

(Not Answered)

Correct Answer

C

17 Packets of the same session may be routed through different paths in

OptionsA) TCP, but not UDP B) TCP and UDPC) UDP, but not TCP D) Neither TCP, nor UDP

Your Answer

(Not Answered)

Correct Answer

B

18The following numbers are inserted into an empty binary search tree in the given order: 10, 1, 3, 5, 15, 12, 16. What is the height of the binary search tree (the height is the maximum distance of a leaf node from the root)?


Your Answer

(Not Answered)

Correct Answer

B

19

Postorder traversal of a given binary search tree, T produces the following sequence of keys10, 9, 23, 22, 27, 25, 15, 50, 95, 60, 40, 29Which one of the following sequences of keys can be the result of an inorder traversal of the tree T?

OptionsA) 9, 10, 15, 22, 23, 25, 27, 29, 40, 50, 60, 95 B) 9, 10, 15, 22, 40, 50, 60, 95, 23, 25, 27, 29C) 29, 15, 9, 10, 25, 22, 23, 27, 40, 60, 50, 95 D) 95, 50, 60, 40, 27, 23, 22, 25, 10, 9, 15, 29

Your Answer

(Not Answered)

Correct Answer

A

20Let f: B C and g: A B be two functions and let h = f o g. Given that h is an onto function. Which one of the following is TRUE?

OptionsA) f and g should both be onto functions. B) f should be onto but g need not be ontoC) g should be onto but f need not be onto D) both f and g need not be onto

Your Answer

(Not Answered)

Correct Answer

B

GATE - Preparation Tips

1.Material Collection

Syllabus All the relevant books based on the subject(Divide the books in two groups - (1)

Fundamental and basic concepts (2) Problem oriented Some books helpful for pre-requisite knowledge on the subject Some good guide books for GATE Previous questions papers

2.Keep contact with some expert and GATE experienced persons

3.Study - Syllabus and Previous questions papers

4.Start from the first chapter

read at least 5 books, it will widen your knowledge(if necessary consult with the books for pre-requisite knowledge or with some expert)

Note down the probable concepts(definitions, unit, dimension etc.) Note down necessary theories, formulae etc Solve problems as maximum as possible(from text books, Guide books etc) Think about various tricks in solving problems(if necessary, note it) Go for series of self tests based on this chapter(take other's help to conduct tests) Continue the self tests until getting a very good score

5.Solve more and more problems, discover more and more new tricks…

6.Follow the same procedure for the rest chapters

7.Finally, go for self tests based on whole syllabus(take other's help to conduct these tests)

8.On the exam day…you will be at the Pick, who can stop you?

The pattern of GATE examination has been CHANGED from 2005.

Main Papers

The question paper will be fully objective type for a total of 150 marks divided into three groups:

i. Group I: Question Numbers 1 to 30 (30 questions) will carry one mark each.ii. Group II: Question numbers 31 to 80 (50 questions) will carry two marks each.iii. Group III: Question Numbers 81a to 85b (10 questions) will carry two marks

each. Each number in this series (81,82,83,84,85) will have two sub-questions

(a & b). The answer to part 'b' will be linked to the correct answer to part 'a', as described below in Section (e)(vi).

a. Each question will have four choices for the answer. Only one choice is correct.b. Wrong answers carry 25% negative marks in Q1 to Q80 and Q81a, 82a, 83a, 84a and

85a. Marks for correct answers to Q81b, 82b, 83b, 84b and 85b will be given only if the answer to the corresponding part 'a' is correct. However, Q81b, 82b, 83b, 84b and 85b will not carry any negative marks.

c. Papers bearing the code AG, CE, CH, CS, EC, EE, IN, IT, ME, MN, MT, PI, TF will contain questions on Engineering Mathematics to the extent of 20 to 25 marks.

d. The multiple choice objective test questions can be of the following type:i. Each choice containing a single stand-alone statement/phrase/data.

Example:Q. The time independent Schrodinger equation of a system represents the conservation of the

A. total binding energy of the system B. total potential energy of the system C. total kinetic energy of the system D. total energy of the system

ii. Each choice containing a combination of option codes.

The question may be accompanied by four options P, Q, R, S and the choices may be a combination of these options. The candidate has to choose the right combination as the correct answer.

Example:Q. The infra-red stretching frequency νco of(P) Mn(CO)6

+ (Q) CO (R) H3B←CO (S) [V(CO)6]- follows the order

A. P>R>S>Q B. S>P>R>Q C. Q>S>P>R D. R>Q>P>S

iii. Assertion[a]/Reason[r] type with the choices stating if [a]/[r] are True/False and/or stating if [r] is correct/incorrect reasoning of [a]

Example:Q. Determine the correctness or otherwise of the following Assertion [a] and the Reason [r]

Assertion: For a fully developed laminar flow in a circular pipe the average velocity is one half of the maximum velocity.

Reason: The velocity for a fully developed laminar flow in a circular pipe varies linearly in the radial direction.

A. Both [a] and [r] are true and [r] is the correct reason for [a] B. Both [a] and [r] are true but [r] is not the correct reason for [a] C. Both [a] and [r] are false D. [a] is true but [r] is false

iv. Match items: Match all items in Column 1 with correct options from those given in Column 2 and choose the correct set of combinations from the choices A, B, C and D.

Example:Q. Match the following and choose the correct combination

Capital State

P. Chennai 1. Andhra Pradesh

Q. Bangalore 2. West Bengal

R. Mumbai 3. Rajasthan

S. Kolkata 4. Karnataka

5. Tamil Nadu

6. Maharashtra

(A) (B) (C) (D)

P - 1 P - 5 P - 5 P - 4

Q - 6 Q - 4 Q - 4 Q - 5

R - 4 R - 3 R - 6 R - 6

S - 5 S - 2 S - 2 S - 2

v. Common data based questions: Multiple questions may be linked to a common problem data, passage and the like. Two or three questions can be formed from the given common problem data. Each question is independent and its solution obtainable from the above problem data/passage directly. (Answer of the previous question is not required to solve the next question). Each question under this group will carry two marks.

Example:Common data for Q. 78,79,80: The gas phase reaction, 2P + 4Q → 2R which is first order in P and first order in Q is to be carried out isothermally in a plug flow reactor. The entering volumetric flow rate is 2.5 dm3/min and the feed is equimolar in P and Q. The entering temperature and pressure are 727oC and 10 atm respectively. The specific reaction rate at this temperature

is 4 dm3/gmol min and the activation energy is 15,000 cal/gmol.

Q.78. What is the volumetric flow rate in dm3/min when the conversion of P is 25%?

(A) 1.88 (B) 5.40 (C) 7.10 (D) 10.80

Q.79. What is the rate of reaction in gmol/(dm3 min) when the conversion of P is 40%

(A) 1.82 x 103 (B) 4.95 x 10-3 (C) 6.2 x 10-3 (D) 9.73 x 103

Q.80. What is the value of the specific reaction rate constant in dm3/gmol min at 1227oC?

(A) 17.68 (B) 22.32 (C) 49.60 (D) 59.75

vi. Linked answers question: The question will consist of a problem statement followed by two sub-questions (a) and (b) based on the problem statement. The solution to part (b) depends upon the answer to part (a). Each part (a) as well as (b) in such linked answer questions will carry two marks.

Example:Statement for linked answer Q. 81a & 81b: A reversible Carnot engine operates between the actual heat input temperature of 1000 K and actual heat rejection temperature of 250 K. The ambient temperature is 200 K.

Q.81a The efficiency of this engine will be

A) 5% (B) 20% (C) 25% (D) 75%

Q.81b The above engine is to provide the power output of 100 kW. The heat input required will be

(A) 133.33 kW (B) 400 kW (C) 500 kW (D) 2000 kW

In the above simplistic example, the calculation of heat input in Q.81b requires the value of efficiency calculated in Q.81a as the first step.

Structure of the XE/XL Paper Sectionsa. XE and XL papers contain a number of sections. Each Section is of 50 marks.

Each Section will be fully objective type and the questions are divided into three groups.

i. Group I: Question Numbers 1 to 10 (10 questions) will carry one mark each.

ii. Group II: Question numbers 11 to 26 (16 questions) will carry two marks each.

iii. Group III: Question Numbers 27a to 28b (4 questions) will carry two marks each. Each number in this series (27, 28) will have two subquestions (a & b). The solution to part 'b' will be linked to the correct answer to part 'a', as described above in (e) (vi).

b. All questions have four choices with only one being correct.c. Wrong answers carry 25% negative marks in Q1 to Q26 and Q27a, 28a. Marks

for correct answers to Q27b, 28b, will be given only if the answer to the

corresponding part 'a' is correct. However, Q27b, 28b will not carry any negative marks.

d. The types of multiple choice questions are the same as in the Main papers as describ

Many students may not be aware that there are several institutions in this country, offering specialized postgraduate programmes in various disciplines. Attractive scholarship / Assistantship for postgraduate courses in engineering / Technology? Architecture /Pharmacy at Engineering colleges / institutes in the country ,are available to those who qualify through GATE. Some Engineering colleges / institutes specify GATE as a mandatory qualification even for admission do students to post graduate programmes. The candidate is required to find the procedure of final selection and award of scholarship / Assistantship from the respective Institution to which the candidate seeks admission. GATE qualified candidates in Engineering subjects) will also be eligible for the award of junior research fellowship in CSIR Laboratories.

Vyomworld.com have made every attempt to include what matters for GATE aspirants. We hope that you enjoy exploring the site.

What is GATE ?

The Graduate Aptitude Test in Engineering (GATE) is an all -India Examination conducted by the six Indian Institutes of Technology and Indian Institute of Science, Bangalore, on behalf of the National Coordinating Board - GATE, Ministry of Human Resources Development (MHRD), Government of India.From Freshersworld point of view we have tried our best to give you a clear picture of GATE.

Objective

To identify meritorious and motivated candidates for admission to Post Graduate Programmes in Engineering, Technology, Architecture and Pharmacy at the National level. To serve as benchmark for normalisation of the Undergraduate Engineering Education in the country.

Here is an opportunity for advanced engineering education in India. An M.E or M.Tech degree is a desirable qualification for our young engineers seeking a rewarding professional career. Engineering students, while in the final year of their degree course, spend considerable time in seeking an opening for studies in foreign universities.

The GATE is held every year on the second Sunday of February, across the country in over 100 cities. At present nearly 60,000 students write GATE every year. Candidates can choose a single paper of 3 hours duration to appear in GATE from the discipline papers shown in the following Table.

Agricultural Engineering AG Mathematics MA

Architecture AR Mechanical Engineering ME

Civil Engineering CE Mining Engineering MN

Chemical Engineering CH Metallurgical Engineering MT

Computer Science & Engg. CS Physics PH

Chemistry CY Production & Industrial Engg. PI

Electronics & Comm. Engg. EC Pharmaceutical Sciences PY

Electrical Engineering EE Textile Engg.& Fibre Science TF

Geology & Geophysics GG Engineering Sciences XE

Instrumentation Engineering IN Life Sciences

Papers XE and XL are general in nature and comprise of the following sections: Candidates appearing in XE or XL papers are required to answer

Three Sections, one compulsory as indicated below:

ENGINEERING SCIENCES(XE) CODE LIFE SCIENCES(XL) CODE

Engg. Maths (Compulsory) A Chemistry (Compulsory) I

Computational Science B Biochemistry J

Electrical Sciences C Biotechnology K

Fluid Mechanics D Botany L

Materials Science E Microbiology M

Solid Mechanics F Zoology N

Statistics G

Thermodynamics H

GATE Results

The GATE result is declared every year on 31 st March and the score of the qualified candidates shows their All India Rank and Percentile Score in the discipline paper chosen by

the candidates.

GATE Score Card

a. Score card will be sent only to the qualified candidates. No information will be sent to candidates who are not qualified.

b. The GATE score card is a valuable document. Care should be taken to preserve it. Additional Score Cards, (upto a maximum of two) will be issued on payment basis only once.

c. The Score Card cannot be treated as a proof of category.d. The score card of the Qualified Candidates will include GATE Score, Percentile Score

and Rank.

i. GATE Score

The GATE SCORE of a candidate is a statistical performance index in the range 0 to 1000. It reflects the ability of a candidate, irrespective of the paper or year in which he/she has qualified. Candidates with same GATE SCORE from different disciplines and/or years can be considered to be of equal ability.

where,

m = marks obtained by the candidate.

a = average of marks of all candidates who appeared in the paper mentioned on this scorecard, in the current year.

s = standard deviation of marks of all candidates who appeared in the paper mentioned on this scorecard, in the current year.

K1 and K2 are determined respectively from the mean and standard deviation of marks of all candidates across all papers and years since GATE 2002.

A typical qualitative interpretation of the GATE SCORE, for example, can be as follows:

i. o

GATE Score Range Ability Level

800 to 1000 Outstanding

675 to 800 Excellent

550 to 675 Very good

425 to 550 Good

300 to 425 Above average

100 to 300 Average

Below 100 Below average

oii. Percentile Score

The percentile score is not the same as percentage of marks. The percentile score of a candidate shows what percentage of candidates, who appeared in the same paper in GATE 2005, scored less marks than him/her. It is calculated as follows: Let N be the total number of candidates appearing in that paper and nc be the number of candidates who have the same all India rank c in the same paper (there can be bunching at a given all India rank). Then all the candidates, whose all India rank is r, will have the same percentile score P, where

The percentile score in each paper is calculated as follows: Let N be the total number of candidates appearing in that paper, and nc be the number of candidates who have the same all India rank c in the same paper (there can be bunching at a given all India rank), then all

the candidates, whose all India rank is r, will have the same percentile score P, where

P = {(no. of candidates securing marks less than the candidate concerned)/N}x100

The evaluation of the ORS is carried out by a computerized process using scanning machines, with utmost care. Requests for revaluation of the answer script and re-totaling of marks will not be entertained.

The GATE result and particulars of the qualified candidates will be made available to interested organizations (educational institutions, R and D laboratories, industries etc.) in India and abroad based on written request by the organization and on payment. Details can be obtained from GATE Chairmen of IITs / IISc.

graduate aptitude test in engineering ( gate )

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