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.

LiveJEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabusDec 3, 2025 | 7:46 AM IST

The marking scheme for JEE Mains 2026 paper 1 and paper 2 is similar. JEE Main total mark is 400 for Paper 2. The total marks in JEE Mains 2026 paper 1 is 300. Each incorrect answer will lead to a deduction of 1 mark, and for each correct response, candidates will get +4 marks.

  • +4 marks for each correct answer and -1 for each wrong answer.
  • Drawing Test: Two questions are evaluated out of 100 marks. There is no negative marking in the drawing test.
Read More
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

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Have a question related to JEE Main ?

Hello,
It is possible to prepare for the JEE session in a short time, but you will need a focused and realistic approach. Instead of trying to cover everything, concentrate on the important chapters and strengthen the topics you already know. Solve previous questions and take practice tests to improve speed and accuracy. Manage your time well and revise regularly. With consistent effort and smart preparation, you can still aim for a good performance even with limited time.
Hope this helps you.

Hello there,

Studying important topics is very essential. It will give you an advantage in examination specially in exams like JEE mains.

Here is the link attached from the official website of Careers360 which will give you the list of all the important topics from all the subjects of JEE mains that is Physics, Chemistry and math. Hope it helps!

https://engineering.careers360.com/articles/most-important-chapters-of-jee-main

thank you!


Yes, you can correct the annual income in the correction window. The correction window opens after the deadline of the application form, where you can correct your wrong details by logging into your account. If the application window is closed right now, and the correction window may be open now, please confirm the date and fill in the write information using the correction window. And if there is a situation where the correction window gets closed, then you need to submit your corrected income certificate at the time of admission. If you need more information related to the JEE Mains Form Correction 2026, then you can read the article JEE Mains Form Correction 2026 on our official website.

Thank you.

Hello,
If you scored one hundred and thirty in the exam, eligibility for IIT depends on the qualifying marks for that year. To get into an IIT, you must first qualify for the next level and then secure a high enough rank. Admission finally depends on the qualifying cutoff and your performance in the next step of the process.
Hope this helps you.

Hello,
If you are not able to open the links given in the lecture plan and you have not started preparing for JEE yet, begin with the basic textbooks you already have. Focus on understanding the main concepts in Physics, Chemistry, and Maths. Make a simple study plan and cover important chapters step by step. Use whatever study material is available to you without depending on the links. With regular practice and revision, you can start your preparation smoothly even without those resources.
Hope this helps you.