JEE Main Marks vs Percentile vs Rank 2024 - Calculate Percentile using Marks

JEE Main Marks vs Percentile vs Rank 2024 - How to Calculate Percentile using Marks

Edited By Team Careers360 | Updated on May 02, 2024 02:11 PM IST | #JEE Main

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NTA will release the JEE Main Marks vs Percentile Vs Rank 2024 very soon. Careers360 presents the comparison table of expected JEE Mains 2024 marks vs percentile vs rank on this page. The JEE Main marks vs percentile 2024 gives a range of marks required to obtain a particular percentile range in JEE Main 2024. NTA updates the JEE Main 2024 marks vs percentile vs rank along with the JEE exam results at jeemain.nta.ac.in. Students can also check the JEE Main marks vs rank to know the probable range of ranks they may secure based on their scores. Candidates who score 300 marks in the Joint entrance examination Main will achieve a 100 percentile.
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This Story also Contains

Expected JEE Main 2024 Marks vs Rank vs Percentile
What is JEE Main 2024 Percentile Score?
Why IIT JEE Main Result announced in percentile?
JEE Main 2024 Marks vs Percentile Normalisation Formula
How To Calculate Percentile in JEE Main ?
What is a good score in JEE Main?
JEE Main 2024 Marks vs Ranks

JEE Main Marks vs Percentile vs Rank 2024 - How to Calculate Percentile using Marks

Expected JEE Main 2024 Marks vs Rank vs Percentile

Marks out of 300	Rank	Percentile
286- 292	19-12	99.99826992- 99.99890732
280-284	42-23	99.99617561 - 99.99790569
268- 279	106-64	99.99034797 - 99.99417236
250- 267	524-108	99.95228621- 99.99016586
231-249	1385-546	99.87388626-99.95028296
215-230	2798-1421	99.74522293-99.87060821
200-214	4667-2863	99.57503767- 99.73930423
189-199	6664- 4830	99.39319714- 99.56019541
175-188	10746-7152	99.02150308 - 99.3487614
160-174	16163-11018	98.52824811-98.99673561
149-159	21145-16495	98.07460288-98.49801724
132-148	32826-22238	97.0109678-97.97507774
120-131	43174-33636	96.0687115-96.93721175
110-119	54293-44115	95.05625037-95.983027
102-109	65758-55269	94.01228357-94.96737888
95-101	76260-66999	93.05600452 -93.89928202
89-94	87219-78111	92.05811248 -92.88745828
79-88	109329-90144	90.0448455 -91.79177119
62-87	169542-92303	84.56203931-91.59517945
41-61	326517-173239	70.26839007-84.22540213
1-40	1025009-334080	6.66590786-69.5797271

Also Read: JEE Main exam analysis | JEE Mains question paper with solutions | JEE Main rank predictor

JEE Main Marks vs Percentile 2024 Shift Wise - (Sri Chaitanya)

The table mentions the JEE Mains 2024 marks vs percentile January session below.

Percentile	27th Jan Shift 1	27th Jan Shift 2	29th Jan Shift 1	29th Jan Shift 2	30th Jan Shift 1	30th Jan Shift 2	31st Jan Shift 1	31st Jan Shift 2	1st Feb Shift 1	1st Feb Shift 2
99.50	200	190	193	200	198	185	184	187	196	189
99.00 - 99.49	190	175	177	185	183	168	165	170	180	173
98.00 - 98.99	172	154	158	165	162	149	145	150	160	153
97.00 – 97.99	155	138	142	148	145	132	128	133	144	136
96.00 – 96.99	142	125	128	133	131	118	115	120	130	121
95.00 – 95.99	132	114	117	120	119	107	103	110	118	111
94.00 – 94.99	124	105	107	110	109	100	95	101	108	102
93.00 – 93.99	115	98	99	102	102	95	91	95	100	97
92.00 – 92.99	108	93	93	96	96	89	86	90	95	92
91.00 – 91.99	103	88	89	92	93	85	82	86	91	89
90.00 – 90.99	99	85	86	90	89	80	77	82	88	85

JEE Main Marks vs Percentile vs Rank 2024 (Expected)

Candidates can check the expected JEE Mains Marks vs Percentile vs Rank here. As per JEE Main marks vs percentile vs rank, candidates must score above 250 marks to obtain a percentile of more than 90.
Also Read: JEE Main Syllabus | JEE Main Exam Dates

Knowing marks vs percentile vs rank 2024 of JEE Main can help students understand the process. The JEE Main marks vs percentile are the percentile scores corresponding to the marks in the NTA JEE Main exam. NTA adopts a unique normalisation formula to arrive at JEE Main Marks Vs Percentile Vs Rank 2024.

What is JEE Main 2024 Percentile Score?

NTA will use the unique normalisation formula to arrive at the JEE Main 2024 percentile score in Chemistry, Maths, and physics. Candidates with the top score in each section of IIT JEE Main 2024 will be assigned the 100 percentile. The qualifying percentile for jee mains 2024 will be calculated to 7 decimal places to avoid ties between two candidates who appeared for the exam. The percentile for each candidate is calculated on a scale of 100 to 0 and through this candidates can analyse their JEE Mains marks vs percentile 2024. The NTA score and percentile refer to the same score in the JEE Main 2024 result. Candidates can also check the expected marks vs percentile JEE Mains 2024 and use the JEE Main college predictor to get information on colleges based on marks.

JEE Main 2024 Marks vs Rank vs Percentile

Candidates will be able to check the JEE Main scores/marks which land them at a certain JEE Main percentile calculated after normalisation. The JEE Main marks vs percentile 2024 helps candidates to estimate their JEE Main percentile using their score. Previous year JEE Main 2023 marks vs percentile vs rank are given in the data below. This data is indicative.

Why IIT JEE Main Result announced in percentile?

NTA conducts the JEE Main 2024 on multiple dates, generally in two shifts per day. The candidates will be given different sets of questions per session and it is quite possible that despite all efforts to maintain equivalence among various question papers, the difficulty level of the question papers of different sessions may not be the same. Some of the candidates may end up attempting a relatively tougher set of questions when compared to other sets. The candidates who attempt the comparatively tougher examination are likely to get lower marks as compared to those who attempt the easier one. To overcome such a situation, the “Normalisation procedure based on Percentile Score” is used to ensure that candidates are neither benefited nor disadvantaged due to the difficulty level of the examination. To ensure that a candidate’s true merit is identified and that a level playing field is created in the above context, the Normalisation Procedure, set out below shall be adopted, for compiling the NTA scores for multi-session papers.

The official brochure says, "The process of Normalisation is an established practice for comparing candidate scores across multi-session papers and is similar to those being adopted in other large educational selection tests conducted in India. For normalisation across sections, NTA shall use the percentile equivalence."

JEE Main 2024 Marks vs Percentile Normalisation Formula

Normalisation is incorporated to equalise the difficulty levels of the exam when it is conducted in two sessions. So NTA will declare the percentile in JEE Mains 2024 obtained after normalisation. Percentile scores indicate the percentage of candidates that have scored equal to or below (same or lower raw scores) that particular percentile in that examination and this is generally from a scale of 100 to 0 for each session of exams. Therefore, the higher scorer of each JEE Main 2024 session will end up securing the same percentile of 100 which is desirable.

NTA will compile the raw marks of the students and normalise the same to arrive at the percentile scores for each subject (Maths, Physics, and Chemistry)as well as the total as a part of the marking scheme of JEE Mains 2024. The highest score of each JEE Main 2024 session will get the same percentile of 100 which is ideal. The marks obtained between the highest and lowest scores will also be converted to percentiles and candidates can get to know their JEE Main rank vs percentile. This percentile score will be used for the preparation of the merit lists. It is to be noted that the percentile scores will be calculated up to 7 decimal places such that the bunching effect can be reduced as can ties amongst candidates with the same scores. To better understand the correlation between marks vs percentile in JEE Mains 2024, we have provided a table showing the marks vs percentile for the previous years of JEE Main.

How To Calculate Percentile in JEE Main ?

Candidates can follow the below formulas to calculate the JEE Main Percentile score for maths, physics and chemistry as well as the complete paper.

Preparation-of-JEE-Main-Result_85g8RsX

The detailed process with examples for the JEE Main normalisation is explained.

JEE Main Marks vs Percentile Vs Rank: Tiebreaker guidelines

Tiebreaker guidelines: These guidelines come into effect when two or more candidates obtain equal percentile scores. The inter-se-merit- is determined in the order given as follows

Candidates obtaining higher Percentile Scores in Mathematics will be ranked higher.
Candidates obtaining higher Percentile Scores in Physics will be given the higher JEE Main rank in 2024.
Candidates obtaining a higher Percentile Score in Chemistry will get a higher rank.
Candidates older in age will be given preference.

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JEE Main 2024 results will be released separately for each of the sessions on jeemain.nta.nic.in. The NTA will announce the results for Paper 1 of Session 2 on its official website. The best of the NTA scores is taken for candidates who attempt all the sessions. The merit list of JEE Main 2024 will be prepared based on the final scores and ranks announced based on the same. Also, candidates may check the marks required for the 90 percentile in JEE Main and get a preparation plan.

Also, check how to calculate your JEE Main rank using your percentile scores

What is a good score in JEE Main?

Candidates looking for answers to questions like "Is 120 a good score in JEE Mains?", "Is 160 a good score in JEE Main?" or "Is 150 a good score in JEE Mains?" are advised to know that the answers to "What is a good score in JEE Main" significantly depends on the college and institute that a student wishes to take admission to.

JEE Main 2024 Marks vs Ranks

Along with JEE marks vs percentile 2024, candidates also check what rank they can get on scoring a particular JEE Main percentile. The following table can be referred to understand JEE Main marks vs rank.

JEE Main Marks vs Ranks 2024

Marks	Rank
286- 292	19-12
280-284	42-23
268- 279	106-64
250- 267	524-108
231-249	1385-546
215-230	2798-1421
200-214	4667-2863
189-199	6664- 4830
175-188	10746-7152
160-174	16163-11018
149-159	21145-16495
132-148	32826-22238
120-131	43174-33636
110-119	54293-44115
102-109	65758-55269
95-101	76260-66999
89-94	87219-78111
79-88	109329-90144
62-87	169542-92303
41-61	326517-173239
1-40	1025009-334080

JEE Main Marks Vs Percentile Vs Rank (Previous Years)

JEE Percentile vs Marks 2021 (shift 1)

Marks	Percentile
286- 292	99.99826992- 99.99890732
280-284	99.99617561 - 99.99790569
268- 279	99.99034797 - 99.99417236
250- 267	99.95228621- 99.99016586
231-249	99.87388626-99.95028296
215-230	99.74522293-99.87060821
200-214	99.57503767- 99.73930423
189-199	99.39319714- 99.56019541
175-188	99.02150308 - 99.3487614
160-174	98.52824811-98.99673561
149-159	98.07460288-98.49801724
132-148	97.0109678-97.97507774
120-131	96.0687115-96.93721175
110-119	95.05625037-95.983027
102-109	94.01228357-94.96737888
95-101	93.05600452 -93.89928202
89-94	92.05811248 -92.88745828
79-88	90.0448455 -91.79177119
62-87	84.56203931-91.59517945
41-61	70.26839007-84.22540213
1-40	6.66590786-69.5797271

JEE Main Marks vs Percentile 2021 (shift 2)

Marks	JEE main 2021 percentile
300-281	100 – 99.99989145
271 – 280	99.994681 – 99.997394
263 – 270	99.990990 – 99.994029
250 – 262	99.977205 – 99.988819
241 – 250	99.960163 – 99.975034
231 – 240	99.934980 – 99.956364
221 – 230	99.901113 – 99.928901
211 – 220	99.851616 – 99.893732
191 – 200	99.710831 – 99.782472
181 – 190	99.597399 – 99.688579
171 – 180	99.456939 – 99.573193
161 – 170	99.272084 – 99.431214
151 – 160	99.028614 – 99.239737
141 – 150	98.732389 – 98.990296
131 – 140	98.317414 – 98.666935
121 – 130	97.811260 – 98.254132
111 – 120	97.142937 – 97.685672
101 – 110	96.204550 – 96.978272
91 – 100	94.998594 – 96.064850
81 – 90	93.471231 – 94.749479
71 – 80	91.072128 – 93.152971
61 – 70	87.512225 – 90.702200
51 – 60	82.016062 – 86.907944
41 – 50	73.287808 – 80.982153
31 – 40	58.151490 – 71.302052
21 – 30	37.694529 – 56.569310
20 – 11	13.495849 – 33.229128
0 – 10	0.8435177 – 9.6954066

How many students appeared for IIT JEE Main exam 2024?

NTA will release the details on the number of students who appeared in JEE Mains 2024 session 1 exam. However, as per the official notification, a total of 8.6 lakh students applied for the IIT JEE Main 2024 January session exam, out of which, 8.22 lakh candidates appeared for it.

Total students appeared in JEE Mains 2024 (January)

Particular	Details
Total number of students applied for JEE Main 2024 January session	Paper 1 - 8.6 lakh Paper 2 - 0.46 Lakh
Number of females registered for JEE Main 2024 exam session 1	Paper 1 - 2.6 lakh Paper 2 - 21 K
Number of males registered for NTA JEE Main 2024 session 1	Paper 1 - 6 Lakh Paper 2 - 25 K
Total number of students appeared for JEE Main Jan session	8.22 Lakh
Total percentage of candidates appeared	95.79 %

JEE Main Marks Vs Percentile vs Rank: Check last 5 Year's Qualifying Cutoff

JEE Main Percentile Predictor 2024

Candidates who wish to predict their rank based on the probable scores can check JEE Main rank predictor tool 2024 by careers360. The JEE Main percentile predictor 2024 helps candidates predict their JEE Main 2024 Percentile from the Marks out of 300 and Paper / Shift's Difficulty Level.

JEE Main Marks vs Percentile 2024: Frequently Asked Questions

The JEE Main 2024 percentile predictor helps to predict the IIT JEE percentile a student can get in JEE Mains. The final marks and percentile of IIT JEE Main 2024 is announced through JEE Main result. Moreover, candidates can take the help of rank predictor JEE Main 2024 to predict their rank based on the score. JEE Main Marks vs Rank 2024 will help the candidates to know the qualifying marks for JEE Mains 2024. The JEE Main 2024 session 1 exam conducted from January 24 to February 1, 2024 and for session 2 it will be conducted from April 1 to 15, 2024. Candidates will be able to check JEE Main 2024 marks vs rank to get an idea of their all-India rank. JEE Main rank predictor is also a very helpful tool in deciding the rank in NTA JEE Main. Here are some of the most asked questions related to JEE Main 2024 marks vs rank vs percentile.

Q. 80 marks in JEE Mains percentile

Ans. Candidates having 80 marks in JEE Mains will get around 85 to 90 percentile.

Q. 20 marks in JEE Mains percentile

Ans. With 20 marks, aspirants will get around 65 to 70 percentile.

Q. 90 marks in JEE Mains percentile

Ans. 90 marks in NTA JEE Main 2024 exam will fetch a percentile between 89 to 93.

Q. 100 marks in JEE Mains percentile

Ans. With 100 marks in IIT JEE Main exam 2024, students are expected to get a percentile between 93 to 96.

Q. 150 marks in JEE Mains percentile

Ans. 150 marks in NTA JEE Main 2024 exam will fetch a percentile between 98 to 99.

Q. 40 marks in JEE Mains percentile

Ans. 40 marks in the NTA JEE Main 2024 exam will let a student get a percentile around 70.

Q. 30 Marks in JEE Mains percentile

Ans. 30 marks in the NTA JEE Main 2024 exam will fetch a percentile between 86 to 90.

Q. 60 marks in JEE Mains percentile

Ans. 60 marks in the NTA JEE Main 2024 exam will let a student get percentile around 82 to 86.

Q. How much percentile is required for NIT?

Ans. Candidates must have a percentile between 86-96 to be eligible for admission to NITs.

Frequently Asked Question (FAQs)

1. How to calculate JEE main percentile with marks?

NTA uses a formula to arrive at the percentile score of candidates with respect to the raw score. Percentile scores are normalised scores as the exam is conducted in multiple session.

2. I score 280 marks what, will be my JEE Main percentile score

With 280 marks, percentile would be in range of 99

3. What is a good percentile in JEE Mains 2024?

NTA percentile score in range of 99 or within 98 can be considered good percentile in JEE Main 2024

4. Is 70 percentile good in JEE Mains?

With 70 percentile, candidates CRL would be in range of 40,000. Candidates belonging to reserved category can get good colleges in this CRL.

5. What is Highest Percentile in JEE Mains?

The highest score within each session is assigned a perfect 100 percentile.

6. When will JEE Main qualifying marks be declared?

Qualifying JEE Main 2024 cutoff will be announced depicting the minimum marks required by candidates.

7. How to Check JEE Mains Percentile ?

The JEE Main percentile calculation formula is – 100 x (The number of students who secured a score equal to or less than the student) / (The total number of students who appeared in a particular examination session).

8. What is the percentile of 70 marks in mains?

Around 80 percentile can be obtained with 70 marks in JEE Mains.

9. Is 250 marks in JEE Main considered as a good score?

With 250 marks in JEE Mains, candidates is expected to score a percentile of above 95. Therefore, 250 marks in JEE Main can be considered as a good score.

10. What is the percentile for 50 marks?

With 50 marks in JEE Main exam, aspirants will get a percentile somewhere between 80 to 85.

11. What is the percentile for 130 marks?

If a candidate has secured 130 marks then the probable obtained percentile is around 95 to 96.

12. What is a good score in JEE Main?

A score of 250 marks is deemed favorable in the JEE Main exam. Candidates achieving 250 or more in the exams will obtain a percentile score ranging from 85 to 95.

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Questions related to JEE Main

Have a question related to JEE Main ?

What was JEE rank for general category to get admission in BIT Sindri

Hello aspirant,

Based on past year's trends, I think you need to be at least in the 93 percentile to get into ECE, ME, or EE, or maybe Mining or Metallurgical, to get admission in BIT Sindri. You roughly need ranking under 80,000 in JEE Mains for admission in this college.

Thank you

Hope this information helps you.

Read Complete Answer

Sajal Trivedi 04 May,2024

I have paid 1600 for JEE advance 2024, but the page shows an error, but the money has been debited from my account. Despite emailing their email address and calling them, I have not received a response. I have emailed the transaction details. Could anyone please help me?

You need pay again and raise dispute with bank in case od error

Read Complete Answer

nathgabhilash 05 May,2024

sc category rank 34000 in jee mains any colleges?

Jee mains sc category 83.7 percentile rank 13161 hai konsi nit milega

Read Complete Answer

chandansarkar781 04 May,2024

i have made my payment for jee advanced registration and it is debited from my account but still not showing.....i have not yet got the confirmation message on email

You can do two things either wait for few hours or till tomorrow or you can do payment again because in that case previous money will be refunded shortly

Read Complete Answer

shikhartiwari1206 05 May,2024

hlo sir , i am so much tensed from 2-3 days , i have sc category certificate which is made from uttar pradesh and not central certificate it is state certificate but the format is same as per josaa annexure, can i submit it in jee advance form or i have to make central caste certificate, plzz reply

I am also a parent looking for admission process...the information I got regarding caste certificate is that we have get a fresh certificate and central caste certificate is different from the state one

Read Complete Answer

atmajasaraikar 04 May,2024

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Popular JEE Main Questions

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 (K $f$ 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 pK_b 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 C_P and C_V 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 = t_c (t_c 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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