JEE Main 2024 Question Paper With Solutions PDF

JEE Main 2024 Question Paper With Solutions PDF

Team Careers360Updated on 11 Oct 2025, 11:05 AM IST

JEE Main 2024 question paper - NTA has released the JEE Main question paper 2024. Candidates can download the JEE Main 2024 question with solutions from this article. JEE Main ( Joint Entrance Examination) is one of the most common exams of engineering. It was conducted in two sessions, one in January and the other In April. Aspirants can find the direct link to download the JEE Main 2024 question paper with solutions PDF for both sessions. It is necessary to check the JEE Main question paper to cross-check the answers. The authority had released the JEE Main answer key along with the question paper at jeemain.nta.ac.in.

JEE Main 2024 Question Paper With Solutions PDF
JEE Main 2024 Question Paper PDF With Solution

To download the official JEE Mains 2024 question paper pdf, candidates can go through this article and download the corresponding pdf by clicking on the corresponding links provided in the article.

JEE Main Question Paper 2024 Session 1

Below are links to download the 2024 session 1 question paper with the answer key. Using these links you can easily access the question paper and answer key. Below is the link for the JEE Main Question Paper 2024 pdf.

JEE Main 2024 Questions Paper Analysis with Answers

JEE Main 2024 paper was conducted in 2 sessions. Let's have a look at the detailed analysis of both session question papers by experts of careers360. JEE Main 2024 question paper based on the latest syllabus and latest pattern which was changed by NTA.

JEE Main 2024 Questions and Paper Analysis (April Session)

JEE Main 2024 Questions and Paper Analysis (January Session)

How to download the JEE Main 2024 Question Papers With Solutions?

Here are the steps to download the JEE Mains 2024 previous year paper with solutions.

  • Click on the links provided in the tables above.

  • Register using your email ID.

  • The JEE question paper pdf download link will be sent to the registered email ID.

  • Simply download it and view the file.

Also Check:

JEE Main Last 10 Years Question Papers with Solutions

JEE Main Question Paper with Solutions and Answer Keys

JEE Main Syllabus: Subjects & Chapters
Select your preferred subject to view the chapters

Benefits of Solving JEE Main 2024 Question Paper

The previous year's JEE Main question papers are an important resource for exam preparation. The JEE Main 2024 question paper based on the latest syllabus and latest pattern offers the following benefits-

  • Helps in revising and identifying topics that need more practice and conceptual clarity.

  • Students gain a better understanding of the JEE Main exam pattern, frequently appearing topics and more. It acts as an effective preparation strategy for the exam.

  • Candidates can improve their time management skills. They can also use it to determine which sections to concentrate on for better accuracy and speed.

  • Regular practice of JEE Main question paper pdf with solutions can help aspirants gain confidence and reduce anxiety before the exam.

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JEE Main Previous Year Questions 2024

Chemistry

1. Cyclohexene 1721277874250is __________ type of an organic compound.

a)Benzenoid non-aromatic

b)Benzenoid aromatic

c)Alicyclic

d)Acyclic

Solution : 1721277874171 is alicyclic compound

2. Which of the following statements are correct?

A. Glycerol is purified by vacuum distillation because it decomposes at its normal boiling point.

B. Aniline can be purified by steam distillation as aniline is miscible in water.

C. Ethanol can be separated from the ethanol-water mixture by azeotropic distillation because it forms azeotrope.

D. An organic compound is pure, if mixed M.P. remains the same.

Choose the most appropriate answer from the options given below :

Solution: Option (B) is incorrect because aniline is immiscible in water.


Physics

1. By what percentage will the illumination of the lamp decrease if the current drops by 20%?

a)56% b) 26% c) 36% d)46%

Solution: Heat and Power developed in a resistor

Heat developed in a resistor: When a steady current flows through a resistance R for time t, the loss in electric potential energy appears as increased thermal energy(Heat H) of the resistor and H=i^2RtThe power developed = \frac{energy}{time}=i^2R=iR=\frac{V^2}{R} (from Ohm's law)

The unit of heat is the joule (J)

The unit of power is the watt (W)

2. A block of mass 5 kg is there on a smooth inclined plane which is making 30 degrees angle with the horizontal, And a hanging mass of 6 kg is attached with it by a string which passes through a pulley; the acceleration of the system is

  1. 4m/s-2 b)1.8m.s-2 c) 4.5m/s-2 d) 5m/s-2

JEE Main 2026: Preparation Tips & Study Plan
Download the JEE Main 2026 Preparation Tips PDF to boost your exam strategy. Get expert insights on managing study material, focusing on key topics and high-weightage chapters.
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Solution: Let us see the fbd for the system


1721277875916

For acceleration, we can write the equation as


a=\frac{6g-5gsin30^{o}}{6+5}

\Rightarrow a=\frac{60-(50\times \frac{1}{2})}{11}=3.18ms^{-2}

Maths

1. If the mean and variance of five observations are \frac{24}{5} and \frac{194}{25} respectively and the mean of the first four observations is \frac{7}{2}, then the variance of the first four observations in equal to

a)\frac{4}{5} b)\frac{5}{4} c)\frac{3}{5} d)\frac{105}{4}

Solution: MEAN (Arithmetic Mean)

The mean is the sum of the value of each observation in a dataset divided by the number of observations.

Mean of the Ungrouped Data

If n observations in data are x1, x2, x3, ……, xn, then arithmetic mean \mathit{\bar x} is given by

\mathit{\bar x}=\frac{x_1+x_2+x_3+\ldots\dots +x_n }{n}=\frac{1}{n}\sum^{n}_{i=1}x_i

Variance and Standard Deviation

The mean of the squares of the deviations from the mean is called the variance and is denoted by σ2 (read as sigma square).

Variance is a quantity that leads to a proper measure of dispersion.

The variance of n observations x1 , x2 ,..., xn is given by

\sigma^{2}=\frac{1}{n} \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2}

\bar{X}=\frac{24}{5} ; \sigma^2=\frac{194}{25}

Let the first four observations be \mathrm{x}_1, \mathrm{x}_2, \mathrm{x}_3, \mathrm{x}_4

Here, \frac{x_1+x_2+x_3+x_4+x_5}{5}=\frac{24}{5}

Also, \frac{\mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4}{4}=\frac{7}{2}\Rightarrow \mathrm{x}_1+\mathrm{x}_2+\mathrm{x}_3+\mathrm{x}_4=14

Now from eqn -1

\mathrm{x}_5=10 Now, \sigma^2=\frac{194}{25}

\begin{aligned} & \frac{x_1^2+x_2^2+x_3^2+x_4^2+x_5^2}{5}-\frac{576}{25}=\frac{194}{25} \\ & \Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=54 \end{aligned}

Now, the variance of the first 4 observations

\mathrm{Var}=\frac{\sum_{\mathrm{i}=1}^4 \mathrm{x}_{\mathrm{i}}^2}{4}-\left(\frac{\sum_{\mathrm{i}=1}^4 \mathrm{x}_{\mathrm{i}}}{4}\right)^2=\frac{54}{4}-\frac{49}{4}=\frac{5}{4}

2. If the general term of the first AP = 3n-5 and that of the second AP is 3n+5. Then the 50th term of AP where both the AP are added

a)155 b) 145 c) 300 d)150

Solution: As we learned,

Properties of an AP:

If a_{1},a_{2},a_{3}\: \: and\: b_{1},b_{2},b_{3} are two APs,

then a_{1}+ b_{1},a_{2}+ b_{2,}a_{3}+ b_{3} is also an AP.

In this Question,

T_n=3n+5+3n-5= 6n

T_{50}=6\times 50=300

Frequently Asked Questions (FAQs)

Q: Upto which year should I solve the question paper of JEE Main?
A:

It is advised to candidates that they should solve at least 10 previous year's question papers to know more about the exam pattern.

Q: Is it necessary to attempt all 75 questions in JEE Mains?
A:

Yes, candidates must attempt all the 75 questions in JEE Mains. However, candidates should only attempt the questions if they are confident about the answer.

Q: Is there a negative marking in the JEE Main Exam?
A:

Yes, there is a negative marking in the JEE Main Exam in both sections for each subject.

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Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

Hello Suhana

If you want to prepare for the JEE Mains exam 2026, then you need to manage your study time and try to give very much attention to your studies. JEE Mains is an entrance exam which is conducted by NTA(National Test Agency), and only science stream students give this exam to get admission in top engineering colleges to get admission in B.Tech or B.E courses.

Students give this exam after completing or appearing in the 12th class. This entrance exam is conducted at the undergraduate level. After clearing the JEE Mains exam, you become eligible to give the JEE Advanced, which helps to get admission in top IIT & NIT colleges of our country.

Here are some tips which help you at the time of your preparation-

  1. Cover the whole syllabus of JEE Mains
  2. Try to manage the study time
  3. Use the NCERT book for better preparation
  4. Give more mock papers and solve the previous question paper
  5. Manage the time & give some time for revision

I hope this information he;lps you. You also do your preparation on our official website, careers360.

All the best for your exam.


Hello,

Yes, you are eligible to attempt JEE Main 2026, as eligibility is based on appearing for or passing the class 12 exam in 2024, 2025, or 2026. Since you passed Class 12 in 2024, you are within the eligibility window to appear in 2026. Here in this article you will find more about the JEE Main eligibility criteria.

I hope it will clear your query!!

Hello,

If you have already passed Class 12 earlier and you are only taking the improvement exam in 2026, then you must choose “Passed” in the JEE Main 2026 form .

JEE Main always considers the first year in which you passed Class 12 as your official passing year. Improvement does not change that. So your status is not “appeared” and not “appearing.” It stays “Passed.”

You should enter:

  • Pass Status: Passed

  • Year of Passing: Your original Class 12 pass year

  • Marks: Your original marks (you can update later only if NTA allows it)

If you are giving Class 12 for the first time in 2026, then you must choose “Appearing.”

This is the correct and safe option for your JEE Main 2026 registration .

Hope it helps !

Hi Hriday

If your father's name is different on your Aadhar card and 10th marksheet, you need to change your father's name in the Aadhar card and match it with the spelling of the 10th marksheet, because when you take admission in college through JEE Mains score. They match your name, your father's name, and your mother's name with your 10th class marksheet, which serves as a real identity for document verification.

What college typically accepts-

  • National Testing Agency(NTA), JEE Mains- NTA requires applicants to match their application name to official identity documents. These official identity documents are the PAN card, Aadhar Card, Driving License, Passport, Voter ID card and School ID card.
  • Colleges Admission Authority- At the time of admission, the admission authority requires to match the application details be matched with the 10th and 12th marksheet as well as official identity documents.

I hope you will understand my point. These are some major issues which many students face during the time of admission in college.

Thank you.


Hello,

Yes, you can apply for JEE Main even if you do not have a caste certificate right now.

You only need the caste certificate if you want to claim a reserved category (SC/ST/OBC/EWS).
If you do not have it at the time of filling the form, you can apply as General or General-EWS/OBC-NCL without certificate , and later you can upload the valid certificate during counselling or document verification.

So yes, you can fill the JEE Main form without a caste certificate.

Hope it helps !