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JEE Main 2024 Question Paper With Solutions PDF

JEE Main 2024 Question Paper With Solutions PDF

#JEE Main
Team Careers360Updated on 08 Dec 2025, 12:21 PM IST

JEE Main 2024 Question Paper - The National Testing Agency has released the JEE Main question paper 2024 on the official website, jeemain.nta.nic.in. Candidates can download the JEE Main 2024 question paper along with solutions on this page. Solving the previous year's JEE Mains paper helps students improve time management, boost confidence and make their preparation more structured focused and efficient. These JEE Main previous year question PDFs allow applicants to analyze questions, compare answers, and understand the exam pattern and difficulty level. 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.

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

The authority had released the JEE Main answer key along with the question paper at jeemain.nta.ac.in. 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 2024 Question Paper Session 1

Below are links to download the JEE Main question paper pdf download 2024. 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 Question Paper 2024 Session 2

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)

JEE Main 2024 Papers: Questions and Analysis for 27, 29,30,31 January and 1 February

JEE Main 2024 Paper: Questions and Analysis of 27th January Morning Shift

JEE Main 2024 Paper: Questions and Analysis of 27th January Evening Shift

JEE Main 2024 Paper: Questions and Analysis of 29th January Morning Shift

JEE Main 2024 Paper: Questions and Analysis of 29th January Evening Shift

JEE Main 2024 Paper: Questions and Analysis of 30th January Morning Shift

JEE Main 2024 Paper: Questions and Analysis of 30th January Evening Shift

JEE Main 2024 Paper: Questions and Analysis of 31st January Morning Shift

JEE Main 2024 Paper: Questions and Analysis of 31st January Evening Shift

JEE Main 2024 Paper: Questions and Analysis of 1st February Morning Shift

JEE Main 2024 Paper: Questions and Analysis of 1st February Evening Shift

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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

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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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Have a question related to JEE Main ?
Can you give me jee previous year question papers in Tamil

Hello,
Since the NTA conducts exams in Tamil, these official papers will have a Tamil language option. Kindly check the following link to get the question papers.
https://engineering.careers360.com/articles/jee-main-question-papers
I hope this helps you.

Read Complete Answer
A
AKKSHEETHA M
16 Dec'25
Jee mains previous year questions on periodic table

Hello there,

Understanding and solving different question papers is one of the best practice for the preparation specially when it comes to JEE mains. It gives you proper understanding of the exam pattern, important topics to cover and marking scheme.

Here is the link attached from the official website of Careers360 which will provide you with the link of previous year question papers on chemistry in PDF format. Hope it helps!

https://engineering.careers360.com/articles/jee-mains-chemistry-questions-in-last-year-exam-premium

thank you!

Read Complete Answer
R
Ritika
16 Dec'25
i have passed 12th with pcb & now i have added math with nios. so now i have 2 marksheets , am i eligible for JEE mains or MHT-CET??

Hello,

Yes, you can be eligible , but it depends on how you passed Mathematics.

JEE Main
You are eligible if:

  • You passed Class 12 with Physics and Mathematics.

  • Mathematics was passed as a full subject from NIOS.

  • NIOS is a recognized board.

  • Having two marksheets is allowed.

You are not eligible if:

  • Mathematics was taken only as an improvement or additional without passing it as a full subject.

MHT-CET
You are eligible if:

  • You passed Class 12 with Physics and Mathematics.

  • Mathematics from NIOS is shown as a passed subject.

  • NIOS is recognized for Maharashtra admissions.

  • Mathematics was passed before the admission year.

You are not eligible if:

  • Mathematics is not shown as a passed subject.

Important

  • Mathematics must be a separate and passed subject.

  • Keep both marksheets during counselling.

Always check the current year information brochure before applying.

Hope it helps !

Read Complete Answer
P
Prachi Kumari
15 Dec'25
how many minimum marks are required to get 99+ percentile in jan attempt in jee

The marks needed for a 99+ percentile in the JEE Main January attempt depend on the difficulty of the paper and the total number of candidates. Generally, you need roughly *180–200* marks out of 300 to hit the 99+ percentile. The exact cutoff varies each session, so checking the official NTA percentile score calculator or previous year cutoffs gives a more precise idea.


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S
Shaily Mishra
14 Dec'25
I have total 68% in my 12th boards PCM , and the forms for improvment are closed by cbse , if i attempt on demand examination exam from NIOS in PCM will jee , BITS , vit , srm , these will accept my result? as i will have 2 marksheets where 3 subjects improved from nios and rest 2 are in cbse.

Hello aspirant

JEE Main accepts NIOS, so you can appear if you meet the basic eligibility.

BITS does not accept marks from two different boards, so this option won’t work for BITS.

VIT and SRM generally accept NIOS, but having two separate mark sheets can be an issue. You should check their official eligibility rules before applying.

Thankyou I hope this help

Read Complete Answer
S
Shaily Mishra
14 Dec'25
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