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JEE Main Marks vs Rank 2024 - Calculate Rank using Marks

JEE Main Marks vs Rank 2024 - Calculate Rank using Marks

Edited By Team Careers360 | Updated on May 02, 2024 01:15 PM IST | #JEE Main
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JEE Main 2024 Marks vs Rank - Candidates can check expected JEE Mains marks vs rank 2024 here. Careers360 publishes this information based on last year's stats. Aspirants can check the marks vs rank of JEE Main 2024 to know what marks are required for a high rank in the JEE Mains exam. The official JEE Main marks vs rank will be released along with the JEE results. Candidates can obtain a maximum score of 300 marks in JEE Mains 2024. A score of 160+ marks will ensure a 98 percentile. Candidates can use the JEE Main marks vs percentile vs rank to know their percentile as per their NTA JEE scores. Moreover, students can use the JEE 2024 rank predictor to predict their ranks. Aspirants should note that JEE Main 2024 marks vs rank are indicative.

JEE Main Marks Vs Rank 2024- Expected

Marks out of 300Rank
288- 29420-11
280-28444-22
270- 279107-63
252- 268522-106
231-2491385-546
215-2302798-1421
202-2144666-2862
190-2006664- 4830
175-18910746-7151
161-17416163-11018
149-15921145-16495
132-14832826-22238
120-13143174-33636
110-11954293-44115
102-10965758-55269
95-10176260-66999
89-9487219-78111
79-88109329-90144
64-87169542-92303
44-62326517-173239
1-421025009-334080

JEE Main Marks vs Rank 2024

According to the previous year's JEE Mains marks vs rank, the expected marks vs rank are provided here. Candidates can check the expected JEE Main marks vs rank 2024 to have an idea of how many marks to obtain. Moreover, students can also check the JEE Main rank vs percentile 2024 to know what percentile they will get based on their score in JEE Main. The data given is indicative.

Amity University, Noida B.Tech Admissions 2024

Asia's Only University with the Highest US & UK Accreditation

UPES B.Tech Admissions 2024

Ranked #52 among universities in India by NIRF | Highest CTC 50 LPA | 100% Placements | JEE Scores Accepted

Also Read: JEE Main Question Papers | JEE Main Syllabus | JEE Main Dates

JEE Main 2023 Marks Vs Rank

The tentative marks vs rank JEE Mains 2023. NTA doesn't provide JEE Mains marks vs rank. Below mentioned JEE Main marks vs rank cut off were created after evaluation of the JEE Main final result and seat allotment results.

Marks vs Rank JEE Mains 2023

Marks out of 300Rank
286- 29219-12
280-28442-23
268- 279106-64
250- 267524-108
231-2491385-546
215-2302798-1421
200-2144667-2863
189-1996664- 4830
175-18810746-7152
160-17416163-11018
149-15921145-16495
132-14832826-22238
120-13143174-33636
110-11954293-44115
102-10965758-55269
95-10176260-66999
89-9487219-78111
79-88109329-90144
62-87169542-92303
41-61326517-173239
1-401025009-334080
JEE Main College Predictor
Predict your top engineering college admission chances based on your JEE Main All India Rank & NTA Score.
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In 2017, with the removal of the normalization of Class 12th marks, qualifying marks for JEE Advanced are based on the cutoff, and admissions to NITs, IIITs, and GFTIs are based on the JEE Main rank, making it the biggest parameter. Candidates can check here the NTA JEE Main marks vs percentile 2024. NTA will conduct the JEE Main 2024 sessions 1 and 2 in January and April. Read the full article on JEE Main 2024 marks vs percentile vs ranks to know more details. It is to be noted that marks vs ranks of JEE Main 2024 have been compiled by taking students' data from multiple sources. We also suggest that candidates check the best engineering colleges in India.

For more details, Check the JEE Main Percentile Predictor and JEE Main Rank Predictor

How to Calculate JEE Mains Marks vs Rank 2024?

To calculate your percentile rank among your classmates' test scores, you can use the formula: divide the number of scores below your score by the total number of scores, and then multiply by 100. To illustrate, consider that Candidates scored an 88 on the test, and you want to determine your percentile rank among scores ranging from 67 to 95.

Factors For Determining JEE Main Marks Vs Rank 2024

The factors typically considered when determining the relationship between JEE Main marks, rank, and percentile include:

  • Total number of registered candidates

  • Number of questions in the paper

  • Difficulty level of the exam

  • Candidates' overall performance

  • Trends in previous years' JEE Main marks versus ranks data

Amrita Vishwa Vidyapeetham | B.Tech Admissions 2024

Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | JEE Scores Accepted

Lovely Professional University B.Tech Admissions 2024

India's Largest University | 100% Placements Record | Highest CTC 3 Cr PA

How JEE Main Result is Prepared?

Preparation-of-JEE-Main-Result

JEE Mains Marks Vs Rank of Candidates (JEE Main 2019)

JEE-Main_marks_vs_RankJEE-Main_marks_vs_Rank-1


Marks vs Rank of JEE Main in Figures

How many seats are available through JEE Main?*

Institutes/Seats

NITS

IIITs

CFTIs

Number of Institutes

31

25

28

Total Seats

17967

4023

4683

Open + Open-PwD

9264

2078

2878

OBC-NCL + OBC-NCL-PwD

4858

1089

776

SC + SC-PwD

2762

609

658

ST + ST-PwD

1736

310

391

*Based on JoSAA Seat Matrix

What is JEE Main 2024 Cutoff?

JEE Mains cutoff 2024 is the minimum mark required by the candidates to qualify for the exam. The cutoff of JEE Main 2024 is different for all categories. The JEE Main admission cutoff 2024 is the opening and closing ranks within which admission is offered in institutes. The cutoff of IIT JEE Main 2024 for admission is released by JoSAA after each round of counselling. Candidates will be able to check the opening and closing ranks for admissions into NITs, IIITs, and CFTIs through the JEE Main 2024 Cutoff. Those candidates who manage to secure within the cutoff marks of JEE Main 2024 will have a higher chance of admission into the institute. One important thing to note is that the cutoff will differ according to the course and institute selected by the candidates.

JEE Main Cutoff Trends

Category/Year

General

OBC

SC

ST

202390.778864273.611422751.977602737.2348772
202288.412138367.009029743.082095426.7771328
202187.899224168.023444746.882533834.6728999
202070.243551872.888796950.176024539.0696101
201978.217486974.316655754.012815544.3345172

2018

74

45

29

24

2017

81

49

32

27

2016

100

70

52

48

2015

105

70

50

44

2014

115

74

53

47

2013

113

70

50

45

Frequently Asked Question (FAQs)

1. Is 96 percentile in JEE Mains good?

At 96 JEE Main percentile, candidates' positions in CRL would be in the range of 30k - 45k. It can be considered a good JEE Main score.

2. What is a good percentile in JEE mains?

Good JEE Main percentile can be said in the range of 99. At this JEE percentile, candidates belonging to any category can get admission of his desired NIT, IIITs or GFTI. However, JEE Main score in range of 99 to 95 is also good.

3. Are All India Ranks and Category Ranks same?

No, AIR and CR are different, AIR is the overall rank obtained by the candidate respective of their category and CR is the rank candidates got in their respective category.

4. 99 percentile in jee mains means how many marks?

Candidates having 99 percentile can be having marks between 175 to 292. Candidates having 98 percentile have marks in range of 149-174.

5. How to calculate JEE main rank?

The formula to calculate JEE Main Rank from Percentile Score is ((100 – P)/100)*N + 1, where p denotes the NTA score, and N denotes the number of students. 

6. What is your JEE Main 2024 percentile?

JEE Main 2024 percentile of a candidate is prepared by NTA using a formula.  

7. Is 95 percentile good in JEE mains?

Yes! JEE Main percentile in range of 95 would position the candidate in range of 45K to 55K CRL. It can be considered good JEE Main percentile. 

8. Jee mains marks vs rank 2019?

If we look in to JEE Main Marks vs Rank 2019, the marks of Rank 1 candidate was 331 out of 360, marks of candidate with rank 103 was 287, marks of candidate having rank 5365 was 171, candidate having 50,000 rank range had marks in range of 60. 

9. What is the probable rank for JEE Main 150 marks?

The JEE Main rank is expected to fall between 18000 to 20000 for approximate 150 marks.

10. Jee main rank predictor from percentile

Candidates can use JEE Main rank predictor to get idea about their rank on the basis of percentile they get in exam.

11. How to calculate JEE main rank?

The formula to calculate JEE Main Rank from Percentile Score is ((100 – P)/100)*N + 1, where p denotes the NTA score, and N denotes the number of students. 

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Questions related to JEE Main

Have a question related to JEE Main ?

Hello Student,
Your chances after considering your category as general, at this given score will be there in some Institutes for CS Branch.
Like: Puducherry Technological University, Puducherry , Indian Institute of Information Technology, Manipur

To know your chances in detail, kindly check our College Predictor Tool: https://engineering.careers360.com/jee-main-college-predictor

Hello aspirant,

If you fall into the general category, securing admission through JoSAA Counselling might pose a challenge. However, you can explore alternative options such as applying to state and university counsellings or retaking the exam to improve your score.

You can Target following colleges with this percentile:

1) KIIT University

2) BIT, Ranchi

3) University of Engineering and Management, Jaipur

4) Lovely Professional University, Jalandhar

5) College of Engineering, Pune

6) Jaipur Engineering College, Jaipur

7) Dream Institute of Technology, Kolkata

Thank you

Hope this information helps you.

The requirement for the JEE Main exam can vary depending on the specific AI course and the institution offering it. Some AI courses may require JEE Main scores as part of their admission criteria, especially if they have a strong focus on engineering or computer science fundamentals.

However, many AI courses, particularly at the postgraduate level, may have their own entrance exam or may not require JEE Main score at all. It's essential to check the admission requirements of the specific AI course and institutions you're interested in to determine whether JEE Main is mandatory.

Hello Student,


As per your given rank your chances will be tough to get admission into NITs.
Still you can try for State Counselling where the JEE Mains Score is accepted.

To know in detail kindly check our college predictor tool: https://engineering.careers360.com/jee-main-college-predictor

Hello aspirant,

With your JEE Main score considerably below the previous year's cutoff for general category students, eligibility to apply to NITs, IIITs, or GFTIs through JoSAA counseling is unlikely. Considering the exam's high difficulty level and stringent cutoffs, gaining admission to these top colleges would be challenging at this percentile.

Thank you

Hope this information helps you.

View All

 5 g of Na2SO4 was dissolved in x g of H2O. The change in freezing point was found to be 3.820C.  If Na2SO4 is 81.5% ionised, the value of x (Kf for water=1.860C kg mol−1) is approximately : (molar mass of S=32 g mol−1 and that of Na=23 g mol−1)
Option: 1  15 g
Option: 2  25 g
Option: 3  45 g
Option: 4  65 g  
 

 50 mL of 0.2 M ammonia solution is treated with 25 mL of 0.2 M HCl.  If pKb of ammonia solution is 4.75, the pH of the mixture will be :
Option: 1 3.75
Option: 2 4.75
Option: 3 8.25
Option: 4 9.25
 

CH_3-CH=CH-CH_3+Br_2\overset{CCl_4}{\rightarrow}A

What is A?

Option: 1

CH_3-CH(Br)-CH_2-CH_3


Option: 2

CH_3-CH(Br)-CH(Br)-CH_3


Option: 3

CH_3-CH_2-CH_2-CH_2Br


Option: 4

None


\mathrm{NaNO_{3}} when heated gives a white solid A and two gases B and C. B and C are two important atmospheric gases. What is A, B and C ?

Option: 1

\mathrm{A}: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 2

A: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{N}_2


Option: 3

A: \mathrm{NaNO}_2 \mathrm{~B}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


Option: 4

\mathrm{A}: \mathrm{Na}_2 \mathrm{OB}: \mathrm{O}_2 \mathrm{C}: \mathrm{Cl}_2


C_1+2 C_2+3 C_3+\ldots .n C_n=

Option: 1

2^n


Option: 2

\text { n. } 2^n


Option: 3

\text { n. } 2^{n-1}


Option: 4

n \cdot 2^{n+1}


 

A capacitor is made of two square plates each of side 'a' making a very small angle \alpha between them, as shown in the figure. The capacitance will be close to : 
Option: 1 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{4 d } \right )

Option: 2 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 + \frac{\alpha a }{4 d } \right )

Option: 3 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{\alpha a }{2 d } \right )

Option: 4 \frac{\epsilon _{0}a^{2}}{d}\left ( 1 - \frac{3 \alpha a }{2 d } \right )
 

 Among the following compounds, the increasing order of their basic strength is
Option: 1  (I) < (II) < (IV) < (III)
Option: 2  (I) < (II) < (III) < (IV)
Option: 3  (II) < (I) < (IV) < (III)
Option: 4  (II) < (I) < (III) < (IV)
 

 An ideal gas undergoes a quasi static, reversible process in which its molar heat capacity C remains constant.  If during  this process the relation of pressure P and volume V is given by PVn=constant,  then n is given by (Here CP and CV are molar specific heat at constant pressure and constant volume, respectively)
Option: 1  n=\frac{C_{p}}{C_{v}}


Option: 2  n=\frac{C-C_{p}}{C-C_{v}}


Option: 3 n=\frac{C_{p}-C}{C-C_{v}}

Option: 4  n=\frac{C-C_{v}}{C-C_{p}}
 

As shown in the figure, a battery of emf \epsilon is connected to an inductor L and resistance R in series. The switch is closed at t = 0. The total charge that flows from the battery, between t = 0 and t = tc (tc is the time constant of the circuit ) is : 


Option: 1 \frac{\epsilon L }{R^{2}} \left ( 1 - \frac{1}{e} \right )
Option: 2 \frac{\epsilon L }{R^{2}}


Option: 3 \frac{\epsilon R }{eL^{2}}

Option: 4 \frac{\epsilon L }{eR^{2}}
 

As shown in the figure, a particle of mass 10 kg is placed at a point A. When the particle is slightly displaced to its right, it starts moving and reaches the point B. The speed  of the particle at B is x m/s. (Take g = 10 m/s2 ) The value of 'x' to the nearest is ___________.
Option: 1 10
Option: 2 20
Option: 3 40
Option: 4 15

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