GATE 2018 Standardization & Calculation of Gate Score

Last Updated - March 10 2018

What is GATE 2018 Standardization Procedure?

As GATE being a major exam, lacs of aspirants appear for it and it is not possible to organize the exam in one go. So, GATE exam is held in multiple sessions. To arbitrate the difference in the level of difficulty, the Standardization procedure is applied. Therefore, it is quite natural that there will be variations in the level of difficulty of the exam.

The following papers in GATE 2018 will be organized in multiple sessions-

• Electrical Engineering (EE)
• Electronic Engineering (EC)
• Computer Science (CS)

• Mechanical Engineering (ME)
• Civil Engineering (CE)

The basic presumption behind the concept of Standardization is that "in all multi-session GATE papers, the distribution of abilities of aspirants is the same across all the sessions". Now, based on the following parameters, and taking into account different Standardization methods, the committee has arrived at the following formula for calculating the normalized marks for the multi-session papers.

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Standardization mark of Jth aspirant in the ith session ij

Here,

Mij: is the actual marks secured by the jth aspirant in the ith sessions

gt: is the average marks of the top 0.1% of the aspirants considering all sessions

Mgq: is the sum of mean and standard deviation marks of the aspirants in the paper considering all sessions

ti: is the average marks of the top 0.1% of the aspirants in the ith sessions

Miq: is the sum of the mean marks and standard deviation of the ith session

Further, the graph can help understand the relationship between the two scores better.

Graph showing the linear relationship between "actual marks" and "normalized marks" of an aspirant, in a multiple-session subject (CE, EE, ME, EE or CE) of GATE.

Once the responses are checked and assessed, the normalized marks of an aspirant will be calculated corresponding to the actual marks secured by the aspirant in the examination and the GATE 2018 Scorecard will be devised based on the normalized marks. The papers which are actually organized in a single session only, the actual marks secured by the aspirant will be used for calculating the GATE 2018 score.

Calculation of GATE Scores for all Papers

After the Standardization of marks, the final GATE 2018 score will be calculated using the following formula;

In this formulae,

M: marks secured by the aspirant in the personal papers (actual marks for single session papers and normalized marks for multi-session papers)

Mq: is the qualifying marks for general category aspirant in the paper

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

Sq: 350, is the score assigned to Mq

St: 350, is the score assigned to t

In the GATE 2018 formula, Mq is normally 25 marks (out of 100) or +σ, whichever is larger. In this case, μis the mean and σ is the standard deviation of marks of all the aspirants who appeared in the paper.

GATE 2018 scorecards can be downloaded by aspirants from various categories after the declaration of results.

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

Qualifying Marks for GATE 2018

The aspirants should have a brief idea about the qualifying marks for various streams and papers in GATE 2018 on the basis of last year’s cut off (2016).The cut off marks for personal papers have not varied too much for personal papers in last few years.

However, with no.of aspirants varying in every session, there is no fixed format or margin for GATE score. The table below contains separate cut off list for various categories.

As expected the qualifying marks are higher for the general aspirants for all the subjects thus requiring them to score better marks in GATE 2018 in order to get qualified in all the personal subjects.

GATE 2015 Stream and Category Wise Qualifying Cut-Off

GATE Paper CodeNo of Aspirants AppearedGATE PaperQualifying Marks FOR (General)Qualifying Marks FOR (OBC-NCL)Qualifying Marks FOR (SC/ST/PwD)
EC172714Electronics and Communication2522.516.67
CS115425Computer Science and IT2522.516.67
ME22367Mechanical Engineering32.7329.4621.82
EE125851Electrical Engineering2522.516.67
IN185758Instrumentation Engineering25.4522.916.96
CE101429Civil Engineering2522.516.67
CH15844Chemical Engineering27.5224.7718.34
BT10719Biotechnology26.0823.4717.39

GATE 2018 FAQs

Q.1. How is the difficulty level determined for all the papers of the same stream in GATE 2018?

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 via a pre-defined mode to calculate personal score and rank. One can, therefore, say that the calculation of score is genuinely fair irrespective of the difficulty level of various sessions.

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

Ans: Yes, it is true. In case an aspirant 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 in the difficult session then their normalized marks will be greater than the marks they have actually secured. Let’s take an example:

The formula for GATE 2018 Standardization

Mij: is the actual marks secured by the jth aspirant in the ith session

gt: is the average marks of the top 0.1% of the aspirants considering all sessions

Mgq: is the sum of mean and standard deviation marks of the aspirants in the paper considering all sessions

ti: is the average marks of the top 0.1% of the aspirants in the ith session

Miq: is the sum of the mean marks and standard deviation of the ith session

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

Q.3. How much does the difficulty level of question papers vary among various sessions of the GATE 2018 exam?

Ans: The difficulty level does vary to a certain extent and that is the reason some of the students prefer a difficult session as there are chances of getting additional marks for a personal paper after the marks are normalized and vice-versa.

However, the selection of aspirants for a particular slot is done via a computerized random procedure and hence beyond the control of any person.So the candidates should focus on preparation for personal papers rather than worrying about the difficulty level since it is clearly a matter of fate.

Q.4. What will be the range of reduction for a particular paper that would be held during multiple sessions in GATE 2018?

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 offer a rough boundary only.

 GATE Expected Cutoff for NITS GATE Round Wise CCMT Cutoff GATE Paper Wise Cutoff

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

Ans: A aspirant’s percentile indicates the percentage of aspirants scoring lower than that personal aspirant in the GATE exam. It is calculated via this formula:

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

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

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