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How many marks are required to score 90 percentile in JEE Main 2024?

How many marks are required to score 90 percentile in JEE Main 2024?

Edited By Team Careers360 | Updated on Jun 26, 2024 04:48 PM IST | #JEE Main

How many marks are required to score 90 percentile in JEE Main 2024? - A percentile above 90 is considered a good percentile. Every students want to get 90m percentile in the JEE Main exam. NTA uses the JEE Main normalization process for calculating the marks. Students with high percentile scores will be able to cross the JEE Main cutoff. Careers360 brings the understanding of marks for 90 percentile in JEE Mains 2024 exam as explained by Prof. Uma Sankar (All India IIT Coordinator at Sri Chaitanya Educational Institutions).

How many marks are required to score 90 percentile in JEE Main 2024?
How many marks are required to score 90 percentile in JEE Main 2024?

However, the score does not in itself assure that the student will get a college and this depends upon the JEE Main result which decides the marks, rank as well as percentile. Hence, candidates may refer to the article below that will give an idea of how many marks are required to score 90 percentile in JEE Main exam 2024.

Also Read:

How many Marks are required to score 90 percentile in JEE Mains

It can be said based upon the previous trend of a few years that to score 90 percentile, candidates need to have a score of at least 70 to 75 marks. However, this is just the probable range of marks because it varies for particular sessions. It shall depend upon how difficult the question paper comes up to be, how many students appear in the exam, and how they performed in that examination as it varies according to normalization.

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How to score 90 percentile in JEE Main?

Students appearing for the ongoing session of JEE Main 2024 should keep their spirits high and eye on scoring higher. Moreover, candidates often think is 90 percentile good in JEE Mains and how tough it is to score the 90 percentile. Talking about the average score, the 90 percentile is not something very difficult to achieve as suggested by Prof. Uma Sankar. All one has to do is correctly attempt 7-8 questions in each section out of 25. This is not a difficult task because those who prepare for the examination and concentrate on the essence of the questions will break through. Capturing and pocketing such easy questions while attempting the paper will help them achieve the bar of minimum passing marks and therefore head towards other questions keeping regard with negative marking. For this, candidates who thoroughly complete the books for board exams, particularly NCERT books, and then solve previous JEE Main question papers shall prove sufficient.
Also Read:

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JEE Main Marks versus Percentile

Candidates who are looking for the contrasting calculation of JEE Main marks vs percentile may check out the answer to 90 percentile in JEE Main means how many marks. The table below shows the average range of marks and the corresponding range of percentile that is based on the previous trends. However, the ratio is dependent upon various factors like the difficulty level of the question paper, the number of students that appeared in the exam, etc. Checking the previous year marks vs percentile will help in understanding what marks for 90 percentile in JEE Mains.

JEE Main Percentile vs Marks

Candidates can check the table to understand the distribution of marks and percentile. Moreover, they will get an idea of what marks for 90 percentile in JEE Mains is required.

JEE Main marks Vs JEE Main Percentile

JEE Main ScorePercentile
-75 - -200.843517743614459 - 0.843517743614459
-19 - -100.843517743614459 - 0.843517743614459
0 - 100.843517743614459 - 9.69540662201048
11 - 2013.4958497103427 - 33.2291283360524
21 - 3037.6945295632834 - 56.5693109770195
31 - 4058.1514901857346 - 71.3020522957121
41 - 5073.2878087751462 - 80.9821538087469
51 - 6082.0160627661434 - 86.9079446541208
61 - 7087.5122250914779 - 90.7022005707394
71 - 8091.0721283110867 - 93.1529718505396
81 - 9093.4712312797351 - 94.7494792463808
91 - 10094.9985943180054 - 96.0648502433078
101 - 11096.2045500677875 - 96.9782721725982
111 - 12097.1429377776765 - 97.6856721385145
121 - 13097.8112608696124 - 98.2541321080562
131 - 14098.3174149345299 - 98.6669358629096
141 - 15098.7323896268267 - 98.9902969950969
151 - 16099.0286140409721 - 99.2397377073381
161 - 17099.272084675244 - 99.4312143898418
171 - 18099.4569399985455 - 99.573193698637
181 - 19099.5973996511304 - 99.6885790237511
191 - 20099.7108311325455 - 99.7824720681761
201 - 21099.7950635053476 - 99.845212160289
211 - 22099.8516164257469 - 99.8937326121479
221 - 23099.9011137994553 - 99.9289017987302
231 - 24099.9349804235716 - 99.9563641573886
241 - 25099.9601632979145 - 99.9750342194015
250 - 26299.9772051568448 - 99.9888196721667
263 - 27099.9909906096101 - 99.9940299220308
271 - 28099.9946812032638 - 99.997394875068
30099.99989145
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Normalization in JEE Main

For those who are confused about why the question on marks required to score 90 percentile in JEE Main happened to exist at first, the reason is the normalization process that is used by NTA to put the students who appeared in the various sessions of JEE Main on a common platform. It is done to avoid the contention that one session was more difficult than the other causing a sense of injustice with one or the other. The percentile scores indicate the percentage of candidates who scored equal to or below the particular percentile in the exam and are generally on a scale of 100 to 0 for each session.

Also read:

Frequently Asked Question (FAQs)

1. How many marks should I obtain in the JEE Main 2024 exam in order to get 90 percentile?

In order to score 90 percentile, candidates need to have a score of at least 70 to 75 marks.

2. Is 75 a good percentile in JEE Mains?

One cannot apply for admission in an NIT or IIIT with a JEE Main percentile score between 70 and 80

3. Can I get IIIT with 75 percentile?

A safe score to secure admission in NIT or IIIT for SC category candidate is 94-96 percentile or above.

4. Can I get a govt college with 80 percentile in JEE mains?

With this percentile you will get rank between 1,50,000 to 3,00,000. This means that you have the opportunity to get admission in NITs and IIITs for BTech program.

5. Is 90 percentile good in JEE Mains?

Yes, 90 Percentile score in JEE Mains 2024 is good enough to get admission into top institutions like IITs, IIITs, NITs, or IISc Bangalore. A 90 percentile score in JEE Main secures a rank between 1,00,000 and 1,50,000.

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Questions related to JEE Main

Have a question related to JEE Main ?

Hello,


Yes, even after clearing JEE, you need to pay fees to study in a college. The fee structure varies depending on the college and course you choose. Here's a general idea:


1. Government Colleges: Fees are relatively low, around 5,000 to 50,000 per year.

2. Government-Aided Colleges: Fees are moderate, around 50,000 to 1,50,000 per year.

3. Private Colleges: Fees can be higher, ranging from 1,50,000 to 5,00,000 or more per year.

4. IITs (Indian Institutes of Technology): Fees are around 2,00,000 to 3,00,000 per year.

5. NITs (National Institutes of Technology): Fees are around 1,50,000 to 2,50,000 per year.


Additionally, you may need to pay for:


· Hostel fees (if you opt for on-campus accommodation)

· Mess fees (for food)

· Other expenses (books, materials, etc.)


Hope this helps,

Thank you

Hi there,

Congratulations on getting a percentile of 96.21 in the jee mains examination.

At this percentile you will be certainly in the top list of candidates participating in the exam and the counselling. You can also check about the previous year cutoffs and the opening and closing ranks by visiting the official website of the nta and then checking about the score required to get an admission in the nits.

Also, you can take the help of our college predictor which will definitely help you in finding the exact list of the colleges which are suitable for admission based upon your jee mains rank and score.

The link of the predictor is mentioned below. You can take  the help from there.

https://engineering.careers360.com/jee-main-college-predictor?utm_source=qna&utm_medium=jee_cp

Hope this resolves your query.

Yes you can participate in Centralized Admission Process (CAP) conducted by the State Common Entrance Test Cell, Maharashtra with your JEE mains score under 15% jee quota.some private universities such as Pict ,Spit, vit ,rocem take admission under this quota but Vjti,Coep state government dont accept JEE marks.for counselling you need to report at the time of counselling and get your documents verified then the merit list will be displayed and admission process is done by the authority.for more refer link

https://engineering.careers360.com/articles/mht-cet-counselling-through-jee-main

Hello,


The syllabus for JEE Main 2026 is based on the Class 11 and 12 curriculum of CBSE. Here's a general overview of the syllabus and most concentrated parts:


Physics:


· Mechanics (30-35%): Kinematics, dynamics, work, energy, power, rotational motion, gravitation

· Thermodynamics (15-20%): Laws of thermodynamics, heat transfer, kinetic theory of gases

· Electromagnetism (20-25%): Electric charges, fields, potential, circuits, magnetism, electromagnetic induction

· Optics (10-15%): Reflection, refraction, lenses, wave optics

· Modern Physics (10-15%): Atomic structure, dual nature of matter, nuclear physics


Chemistry:


· Physical Chemistry (30-35%): Atomic structure, chemical bonding, thermodynamics, kinetics, equilibrium

· Inorganic Chemistry (25-30%): Periodic table, s-block, p-block, d-block, f-block elements

· Organic Chemistry (30-35%): Hydrocarbons, functional groups, reactions, stereochemistry


Mathematics:


· Algebra (30-35%): Sets, relations, functions, matrices, determinants, complex numbers

· Calculus (25-30%): Limits, derivatives, integrals, differential equations

· Coordinate Geometry (20-25%): Lines, circles, parabolas, ellipses, hyperbolas

· Trigonometry (10-15%): Triangles, trigonometric functions, identities


Most Concentrated Parts:


· Physics: Mechanics, Electromagnetism, and Thermodynamics

· Chemistry: Physical Chemistry, Inorganic Chemistry, and Organic Chemistry

· Mathematics: Algebra, Calculus, and Coordinate Geometry


Important Topics:


· Physics: Newton's laws, work-energy theorem, electric circuits, magnetic fields, optics

· Chemistry: Atomic structure, chemical bonding, acid-base equilibria, redox reactions, organic reactions

· Mathematics: Quadratic equations, functions, limits, derivatives, integrals, matrices


hope this helps,

Thank you

Hello,


Yes, admission to the Indian Statistical Institute (ISI) is possible through JEE Advanced score. ISI considers JEE Advanced scores for admission to its undergraduate programs, including:


1. Bachelor of Statistics (Honours) [B.Stat (Hons)]

2. Bachelor of Mathematics (Honours) [B.Math (Hons)]


To be eligible, you need to:


1. Appear for JEE Advanced and score a minimum qualifying mark.

2. Apply to ISI separately, mentioning your JEE Advanced roll number and score.

3. Meet the cutoff score set by ISI for admission.


hope this helps,

Thank 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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