IITs Cutoff for B.Tech Biotechnology
Updated on Dec 30, 2019 - 7:35 p.m. IST by Saakshi Lama
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IITs Cutoff for B.Tech Biotechnology - The concerned authorities will release the cutoff of IITs for B.Tech Biotechnology in online mode after the counselling session is conducted. IITs cutoff for B.Tech Biotechnology will act as a screening process for admissions. Through the cutoff of IITs for B.Tech Biotechnology, candidates will be able to know the minimum ranks which are required for higher chances of admissions. If the cutoff is not met, the candidates will have less chance of being allotted into the Biotechnology courses offered by IITs. The cutoff will be prepared by the authorities after checking various factors related to JEE Advanced and the institute such as level of difficulty, previous year cutoffs, number of applicants and more. Read the full article to know more about IITs Cutoff for B.Tech Biotechnology.

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IITs Cutoff 2020 for B.Tech Biotechnology

Candidates will be able to check the cutoff of IITs 2020 for B.Tech Biotechnology in this article as soon as it is released by the authorities. The minimum ranks to be secured to have a greater chance of admissions will be present in the cutoff. The candidates will have to note that the cutoff will be different as per the institute and categories that they select.

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Factors Determining IITs Cutoff 2020 for B.Tech Biotechnology

While preparing the cutoff of JEE Advanced 2020 for Biotechnology courses offered by IITs, the following factors will be considered by the authorities:

  • Number of seats available in the institutes.

  • Number of applicants for Biotechnology.

  • Previous years’ cutoff trends

  • Difficulty level of JEE Advanced 2020

Previous Years’ IITs Cutoff for B.Tech Biotechnology

IITs Cutoff 2019 for B.Tech Biotechnology

Name of Institutes

Category

Male

Female

IIT Guwahati

OBC

3971

6350

EWS

1122

1861

General

8643

12991

SC

2408

3271

ST

1028

--

IIT Roorkee

OBC

3358

5688

EWS

732

1780

General

7566

9982

SC

2096

2785

ST

872

--

IITs Cutoff 2018 for B.Tech Biotechnology

S.No.

Name of the institute

Category

Opening Rank

Closing Rank

1

IIT Guwahati

General

5543

8265

OBC

3056

3907

SC

1164

1989

ST

901

455P

2

IIT Roorkee

General

4577

7074

OBC

3030

3346

SC

1372

1741

ST

897

126P

3

IIT Jodhpur

General

7078

11224

OBC

2986

4599

SC

1788

2261

ST

28P

244P

Suffix 'P' indicates that the corresponding rank is from Preparatory Rank List.

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IITs Cutoff 2017 for B.Tech Biotechnology

S.No.

Name of the institutes

Categories

Opening Rank

Closing Rank

1

IIT Guwahati

General

5537

7915

OBC

2451

3285

SC

1638

2027

ST

682

953

2

IIT Roorkee

General

5143

6459

OBC

2154

2896

SC

1651

1899

ST

804

917

IITs Cutoff 2016 for B.Tech Biotechnology

IITs

General Category

General-PWD

OBC-NCL

SC

ST

IIT Guwahati

OR

CR

OR

CR

OR

CR

OR

CR

OR

CR

5388

6323

44

44

2691

3040

1556

1738

15

1116

5394

6713

44

44

2691

3158

1556

1802

54

1116

5394

6713

44

44

2691

3209

1556

1882

54

1116

5394

6848

44

44

2691

3209

1556

1882

54

1116

5394

7088

--

--

2691

3224

1556

1973

54

1116

5394

8000

--

--

2691

3340

1556

1973

782

954

IIT Roorkee

5222

5797

--

--

2527

2884

846

1608

850

973

5301

5912

--

--

2527

2945

1527

1667

850

1026

5301

5912

--

--

2527

2945

1527

1704

850

1026

5301

6001

--

--

2527

2945

1527

1704

850

1084

5417

6098

--

--

2527

2945

1527

1704

850

1084

5417

6255

--

--

2527

2975

1527

1704

850

1084

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What was jee advanced cutoff 2019 for st category

Yastika Kumar 28th Jan, 2020

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This year anyone can tell 80 percentile student of jee mains in obc can qualify for jee advance xams??

Manisha Gupta Expert 28th Jan, 2020

Hello Aspirant,

It's difficult to tell you the exact about the qualifying cutoff & the exact rank. But as per the previous year cutoff you should have decent chance to qualify the exam as OBC category.

You may calculate your expected general rank by this formula


100 - your total percentile score is the percentage  of people ahead of you. 874469 are the number of candidates appeared for Jan exam. So rank formula is (100 - your total score) X 869010 /100. If the number of candidates for April exam increased to 1200000 then your final rank shall be (100 - your total score) X 1200000/100

Using the formula

(100 - your total score) X 869010 /100

Your probable general rank will be 1,73,802.

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https://engineering.careers360.com/jee-main-college-predictor

Good Luck!


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Hi Rejani,

I would like to inform you that you will have to clear JEE Mains paper 1 to be able to qualify for JEE Advance exam. If you clear JEE Advance exam, you need to appear for AAT(Architecture Aptitude Test) exam conducted by the IITs. Then you can get a seat in IIT for B.Arch degree else by appearing in JEE Mains paper 2, you can get into an NIT according to your rank.

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ARNAV SATRUSAL 27th Jan, 2020

Hi Aneena,

I would like to inform tyou that there is no JEE Advance exam being conducted for B.Arch. To appear for JEE Advance exam you have to qualify in the JEE Mains exam i.e, JEE Main B.tech paper. Then you will have to clear JEE Advance exam and score a good AIR. Then you will have to appear for AAT(Architecture Aptitude Test) conducted by IITs to get a seat in B.Arch in IITs.

You can go through this link to predict the different colleges which you can get based upon your rank and percentile.

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