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In **GATE 2017**, there will be some exams **(for CE, CS, EC, EE and ME papers)** which will be conducted in multiple sessions.The **GATE **organizing committee will therefore apply a suitable normalization method in consideration of any variation in the difficulty levels of the question papers across different sessions.The basic presumption behind the concept of normalization is that "in all multi-session **GATE papers**,the distribution of abilities of candidates is the same across all the sessions".

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Now,based on the following parameters,and taking into account various normalization methods,the committee has arrived at the following formula for calculating the normalized marks for the multi session papers.

**READ MORE ABOUT GATE Calculation of Normalization Marks**

Normalization mark of **J ^{th}** candidate in the

**Here,**

**M _{ij }**: is the actual marks secured by the

**M̅ ^{g}_{t} **: is the average marks of the top

**M ^{g}_{q}** : is the sum of mean and standard deviation marks of the candidates in the paper considering all sessions

**M̅ _{ti}** : is the average marks of the top

**M _{iq}** : is the sum of the mean marks and standard deviation of the

**READ MORE ABOUT GATE Score Percentile Conversion**

Once the answers are checked and assessed,the normalized marks of a candidate will be calculated corresponding to the actual marks secured by the candidate in the examination and the **GATE 2017 Score card** will be devised based on the normalized marks.The papers which are actually conducted in a single session only,the actual marks secured by the candidate will be used for calculating the **GATE 2017 **score**.**

After the normalization of marks,the final **GATE 2017 score** will be calculated using another formula;

**READ ABOUT MORE GATE Score Calculation**

** M **: marks secured by the candidate in the individual papers (actual marks for single session papers and normalized marks for multi-session papers)

** M**_{q }: is the qualifying marks for general category candidate in the paper

** M̅**_{t : }is the mean of marks of top **0.1% or top 10** (whichever is larger) of the candidates who appeared in the paper (in case of multi-session papers including all sessions)

** S _{q}**: 350, is the score assigned to

** S _{t}**: 350, is the score assigned to

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In the **GATE 2017** formula, **M**_{q }is normally 25 marks (out of 100) or **+ ****σ** ,whichever is larger. In this case,**μ**is the mean and **σ**** **is the standard deviation of marks of all the candidates who appeared in the paper.

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**GATE 2017 score cards can be downloaded by candidates from different categories after the declaration of results.**

- All
**SC/ST/PwD**candidates whose marks are greater than or equal to the qualifying mark designated for**SC/ST/PwD**candidates in their respective papers, - All other candidates whose marks are greater than or equal to the qualifying mark designated for
**OBC (NCL)**candidates in their respective papers.

The candidates should have a brief idea about the qualifying marks for different streams and papers in **GATE 2017** on the basis of last year’s cut off (2016).The cut off marks for individual papers have not varied too much for individual papers in last few years.However with no.of candidates varying in every session there is no fixed pattern or margin for** GATE score** in 2017.The table below contains separate cut off list for different categories.As expected the qualifying marks are higher for the general candidates for all the subjects thus requiring them to score better marks in **GATE 2017** in order to get qualified in all the individual subjects.

**READ MORE ABOUT GATE Qualifying Marks**

GATE Paper Code | No of Candidates Appeared | GATE Paper | Qualifying Marks FOR (General) | Qualifying Marks FOR (OBC-NCL) | Qualifying Marks FOR (SC/ST/PwD) |
---|---|---|---|---|---|

EC | 172714 | Electronics and Communication | 25 | 22.5 | 16.67 |

CS | 115425 | Computer Science and IT | 25 | 22.5 | 16.67 |

ME | 22367 | Mechanical Engineering | 32.73 | 29.46 | 21.82 |

EE | 125851 | Electrical Engineering | 25 | 22.5 | 16.67 |

IN | 185758 | Instrumentation Engineering | 25.45 | 22.9 | 16.96 |

CE | 101429 | Civil Engineering | 25 | 22.5 | 16.67 |

CH | 15844 | Chemical Engineering | 27.52 | 24.77 | 18.34 |

BT | 10719 | Biotechnology | 26.08 | 23.47 | 17.39 |

Q.1. How is the difficulty level determined for all the papers of the same stream in GATE 2017?

**ANS: The difficulty level for all the papers is determined by analyzing density functions of that particular stream.So if we assume that the distribution of marks is denser in the region 50-60 for session 1 for a certain paper and in the same paper for session 2 it is 40-50,then in that case the session two will be considered easier.However, the students should not worry as the marks get normalized through a pre-defined method to calculate individual score and rank.One can therefore say that the calculation of score is genuinely fair irrespective of the difficulty level of different sessions.**

**READ MORE ABOUT GATE Frequently Asked Question**

Q.2. Is the normalized marks always greater than the actual mark obtained for a relatively difficult paper when compared to other sessions of same stream?

**ANS: Yes, it is true.In case a candidate has appeared for the easier session then their normalized marks will be lesser than the actual marks they had secured.Similarly,If they have appeared for the difficult session then their normalized marks will be greater than the marks they have actually secured.Let’s take an example: **

** M _{ij }**: is the actual marks secured by the

** M̅ ^{g}_{t }**:is the average marks of the top

** M ^{g}_{q }**:is the sum of mean and standard deviation marks of the candidates in the paper considering all sessions

** M̅ _{ti }**:is the average marks of the top

** M _{iq }**:is the sum of the mean marks and standard deviation of the

**READ MORE ABOUT GATE Avaibility of Score Card**

Average marks of all the candidates in all the sessions (of a particular subject,say CSE) are 30,average marks of top 0.1% is 80,assuming that the candidate has appeared in the easier session,average marks of top 0.1% in that session is 85,average marks of all candidates in your session is 35 and if candidate has secured 70 marks, then:

Q.3. How much does the difficulty level of question papers vary among different sessions of the GATE 2017 exam?

**ANS: The difficulty level does vary to a certain extend and that is the reason some of the students prefer a difficult session as there are chances of getting additional marks for an individual paper after the marks are normalized and vice-versa.However, the selection of candidates for a particular slot is done through a computerized random process and hence beyond the control of any individual.So the aspirants should focus on preparation for individual papers rather than worrying about the difficulty level since it is completely a matter of fate.**

**READ MORE ABOUT GATE Exam Paper Analysis**

Q.4. What will be the range of reduction for a particular paper that would be held during multiple sessions in GATE 2017?

**ANS: Min limit: (0.9)*your marks; Max limit: (1.15)*your marks **

**However there is no predefined limit.Therefore one cannot presume that if someone is getting an easier set,then their marks are going to be decreased by 10%.It also depends on the other factors/variables in that section.The above formulas are meant to provide a rough boundary only.**

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Q.5. What is GATE Percentile? How is it calculated?

**ANS: A candidate’s percentile indicates the percentage of candidates scoring lower than that individual candidate in the GATE exam. It is calculated through this formula:**

**AIR RANK= It is the all India rank of a particular candidate.**

**N= It is the total no.candidates appearing in that particular stream.**

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