Last Updated -
March 10 2018

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.

Standardization mark of **J ^{th}** aspirant in the ith session

**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 aspirants 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

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.**

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)

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

** M̅**_{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)

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

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

In the **GATE 2018** 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 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.

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 Paper Code | No of Aspirants 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 |

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.

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

** 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 aspirants 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

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:

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.

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.

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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