Panjab University All Questions

Hello Aspirant,

The KCET rank will be calculated by taking 50% weightage of broad marks and 50% of KCET marks.

You have secured a decent rank.

According to this score your rank will be around the range of 10,000 to 20,000 . This is the expected rank.

You can go through the link to get detailed information about previous year KCET marks vs rank :

https://engineering.careers360.com/articles/kcet-marks-vs-rank

Cheack out this link to predict colleges :

https://www.careers360.com/kcet-college-predictor?utm_source=qna&utm_medium=ini-cet_cp

Hope this information will help you.

Best wishes!

Hi there!

With the marks you mentioned your rank would be anywhere in between 10000 to 20000.

Please note: This is not actual rank, these is been predicted by analyzing previous year trend of rank vs marks.

It is not possible to get exact rank. as it depends on various factors like:

1- Difficulty level of examination

2- No. of students appeared in exam

3- Category of an individual

4- Relative marking and many more

But the above predicted rank will give you a rough idea of ranking.

Also if you wish to know about the colleges in which you may get admission corresponding to your score, click on the link given below to use college predictor

https://www.careers360.com/kcet-college-predictor

I hope this info will be useful!

Hello,

It is difficult to predict your actual overall rank as there are various factors that determine the rank of the candidate such as total number of candidates appeared, highest marks in the exam, normalization of marks etc. Based on KCET 2020 Marks vs Rank trend your rank is expected to vary in between 5000-15000 The marks scored in class 12th are also considered to determine the rank so your actual rank is subjected to change. You may know your actual rank after the results are declared. For your reference you may check KCET 2022 Marks vs Rank trend table using the link below

https://engineering.careers360.com/articles/kcet-marks-vs-rank

Hello Aspirant,

As per the marks mentioned by you in the question your rank in KCET 2022 is expected to be in the range of  1500-2000 .  But note that this is not your exact rank because it is not yet released and the rank predicted above is on the basis of previous year's data.

As the rank Vs marks keeps on changing depending upon various factors such as :

** Number of candidates appeared for examination.

** Top score for that particular year.

** Total number of seats available.

And some more factors, therefore your actual rank in KCET 2022 might vary with slight variations.

You can use KCET 2022 College Predictor toll to predict the colleges that you can get based upon your KCET 2022 rank. You can use KCET College predictor from the link given below :-

https://www.careers360.com/kcet-college-predictor?icn=QnA&ici=qna_answer

I hope this information helps you.

Good Luck!!

Hello Aspirant,

Your rank will be somewhere in the range of 20000-40000. Your KCET rank is calculated by taking 50 percent of your board marks and 50 percent of your KCET score. It is not possible to tell your exact rank. This prediction is based on previous year's information . As cutoff varies considering many factors like number of students appeared for the exam, toughness of the exam, performance of the students etc. You can use the below link to get information about KCET rank vs marks:

https://engineering.careers360.com/articles/kcet-marks-vs-rank

If you want to predict colleges based on your KCET score then use the below link:

KCET College Predictor 2022 - Predict Colleges based on your Rank (careers360.com)

Hope this may help you.

Best wishes

Hey there,

Yes you are eligible. The minimum percentage is 75%. Your scored marks is enough to get u a seat. All the best.

Thank you.

hii Sahil , 

According to expected cut off for B.Sc. honour chemistry in Panjab university , you have fare chances of admission , but if not selected in first round then you must need to wait for 2nd or 3rd round counciling sedule . Your admission chances in main campus are fare. You need to continue visit official website for furthur update related to addmission .

hope it help!!!

Hello Aspirant,

The KCET rank will be calculated by taking 50% weightage of broad marks and 50% of KCET marks.

Don't be demotivated, you have secured a decent rank.

According to this score your rank will be around the range of 10,000 to 20,000 . This is the expected rank.

You can go through the link to get detailed information about previous year KCET marks vs rank :

https://engineering.careers360.com/articles/kcet-marks-vs-rank

Cheack out this link to predict colleges :

https://www.careers360.com/kcet-college-predictor?utm_source=qna&utm_medium=ini-cet_cp

Hope this information will help you.

Best wishes!

Hello Aspirant,

Yes, it can happen. You may get these colleges

HKBK College of Engineering

NHCE Bangalore - New Horizon College of Engineering

Cambridge Institute of Technology, K R Puram

Reva University

DSU - Dayananda Sagar University

Sir M. Visvesvaraya Institute of Technology

Acharya Institute of Technology

AMC Engineering College

B.N.M Institute of Technology

Faculty of Engineering and Technology, M.S. Ramaiah University of Applied Sciences

KLE Technological University

Sri Jayachamarajendra College of Engineering (SJCE Mysore), JSS STU

M.V.J. College of Engineering

CMR Institute of Technology, Bangalore

RNS Institute of Technology

MITE - Mangalore Institute of Technology and Engineering

Global Academy of Technology

Sahyadri College of Engineering and Management

PES College of Engineering

NIE Institute of Technology

KLS Gogte Institute of Technology, Belagavi

You can predict the chances of getting a college here

https://www.careers360.com/kcet-college-predictor?utm_source=qna&utm_medium=ini-cet_cp

KCET exam

Karnataka Examinations Authority (KEA) accomplishes Karnataka Common Entrance Test ( KCET) for entrance into Engineering, B. Pharm, and Farming Science programs in govt academies or private financed or private non-aided academies in Karnataka.

Regards.