JEE Main 2024 Question Paper With Solutions PDF

JEE Main 2024 Question Paper With Solutions PDF

Vikash Kumar VishwakarmaUpdated on 01 Nov 2024, 09:28 AM IST

JEE Main 2024 question paper - Download the 2024 question paper with the solution in pdf format from this article. JEE Main ( Joint Entrance Examination) is one of the most common exams of engineering. JEE Main Exam is conducted in two sessions one in January and the other In April. Herein, in this article, we provide a link to download the JEE Main 2024 question paper with solutions for both sessions. NTA will publish the JEE Main Question Paper 2025 within a week of the JEE Main exam conclusion. The JEE Main answer key along with the question paper will be released 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
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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

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Questions related to JEE Main

On Question asked by student community

Have a question related to JEE Main ?

The approximate annual cost for 11th/12th PCM and JEE coaching is around 1.5 to 3.5 lakhs for institutes, excluding hostel and mess fees. The total fees including hostel as well as mess fees can rise upto 4.5 to 6.5 lakhs and above, depending on location and institute quality.

Hi dear candidate,

You can anytime visit our official website to find the previous 10 years JEE Mains question papers with solutions. Kindly refer to the link attached below to download them in PDF format:

JEE Main Last 10 Years Question Papers with Solutions (2025 to 2015)

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Hello,

For JEE Main and JEE Advanced , the cut-offs are lower for ST category students. Here is a simple idea based on recent trends:

  • JEE Main qualification for ST : Around 50–60 marks is usually enough to qualify for JEE Advanced.

  • JEE Advanced qualification for ST : You just need to clear the JEE Main cut-off, then appear for Advanced.

  • To get good NITs or IITs , you will need higher marks.

    • For NITs (Hyderabad or good branches), try for 120+ marks in JEE Main .

    • For IITs , even with ST quota, you should aim for at least 80–100+ marks in JEE Advanced for decent branches.

Since you are from ST category and Hyderabad , you don’t need 300 marks in JEE Main. Try to score as high as possible to get better branches, but even moderate marks can qualify you.

Hope it helps !

Yes, you can. If you do 11th from CBSE in 2025 and 12th from NIOS in 2026, then you are allowed to attempt JEE. NIOS is a valid board, so you can give JEE in 2026 and again in 2027. Just take admission on time and pass your 12th exams.


For the JEE 2026, priority-wise, high weightage chapters include Integral

1. Calculus, Coordinate Geometry, Vectors, 3D Geometry, Complex Numbers, and Quadratic Equations in Maths

2. Semiconductors, Current Electricity, Work, Power and Energy, and Gravitation in Physics

3. Organic Chemistry Containing Oxygen, Hydrocarbons, and P-Block Elements in Chemistry.