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 Main 2026 Registration LIVE: NTA publishes FAQs for BTech entrance exam; apply on jeemain.nta.nic.inNov 21, 2025 | 9:06 PM IST

Candidates who are giving improvement exam or re-appearing in one or more subjects to enhance scores in Class 12 will have to choose 'passed' option while filling the JEE Main application form 2025.

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.

Amrita University B.Tech 2026

Recognized as Institute of Eminence by Govt. of India | NAAC ‘A++’ Grade | Upto 75% Scholarships

UPES B.Tech Admissions 2026

Ranked #43 among Engineering colleges in India by NIRF | Highest Package 1.3 CR , 100% Placements

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.
Download EBook

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.

Articles
|
Upcoming Engineering Exams
Ongoing Dates
HITSEEE Application Date

5 Nov'25 - 22 Apr'26 (Online)

Ongoing Dates
SNUSAT Application Date

19 Nov'25 - 31 Mar'26 (Online)

Certifications By Top Providers
Basic Programming using Python
Via Indian Institute of Technology Bombay
Introduction to Aerospace Engineering
Via Indian Institute of Technology Bombay
Fundamental Concepts of Electricity
Via Indian Institute of Technology Bombay
Artificial Intelligence
Via Columbia University, New York
Computer Fundamentals
Via Devi Ahilya Vishwavidyalaya, Indore
B.Sc in Programming and Data Science
Via Indian Institute of Technology Madras
Udemy
 1525 courses
Swayam
 817 courses
NPTEL
 773 courses
Coursera
 697 courses
Edx
 608 courses
Explore Top Universities Across Globe

Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

Hello murali

No, your son is not eligible for OBC NCL for IIT JEE because you fall in the "creamy layer" occupational category, regardless of your current employment status or family income. Students whose family income is less than Rs. 8 lakhs annually and they are not belong to the "creamy layer".

Note -

  • Children of professionals like doctors, lawyers, and engineers, as well as government officials in Group A and Group B services, are generally considered in the "Creamy layer" category.
  • Your current unemployment does not change your occupational status. The eligibility is based on your profession, not your current job status.

Thank You

Hello,

You can fill the JEE Main form even if you are a private candidate

Write the name of the school/board from where you are appearing as a private candidate .
If your Class 12 admit card or registration slip shows a school/centre name, use that exactly.

If your board lists you as a “Private Candidate” under the board name , then write:

CBSE – Private Candidate
(or your board name – Private Candidate)

Use the pin code of the examination centre/school mentioned on your Class 12 private candidate admit card or registration details.

If your board does not give any school address and only shows the regional office address, then use the regional office address pin code given by your board.

Hope it helps !

Hello Aspirant

You should not leave the OBC-NCL certificate ID blank in the JEE Main form it can create problems later.

NTA wants the certificate details while filling the form, not just at counselling. If you can, apply for the OBC-NCL certificate immediately so you get the ID on time.

If you fail to submit the certificate during counselling, your category will shift to General. It’s safer to enter OBC-NCL only if you’re sure you’ll get the certificate before counselling.

Hope it will help you

Hello,

No, X-rays are not deleted for the JEE 2026 exams. X-rays are part of the electromagnetic spectrum, a topic that remains in the syllabus for both JEE Main and JEE Advanced. Here in this article you will find more about the JEE Main syllabus.

I hope it will clear your query!!

Of course Lakshmi, that's why we are here. As you mentioned, you didn't start preparation for JEE Mains Exam. Don't worry start TODAY. Because today is the best day to start.

Okay you can start with the previous years question papers which you can download by Clicking Here .

You also can browse the JEE Mains Syllabus by Clicking Here . These help you to know which will come in the exam.

If you need any strategical need -- Click Here . This will effectively help you to do more in less time