JEE Main Question Paper 2018 (Available): Check Previous Year Question Paper with Solution PDF (2024-2014)

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Collegedunia Team

Content Curator | Updated 3+ months ago

JEE Main 2018 Question Papers with Solutions are available for both Paper 1 and Paper 2. The previous years’ question papers pdf can also be downloaded from he article below. Candidates appearing for JEE Main can use the links below to download the question papers with solution for free to prepare better for the exam. 

JEE Main 2018 Question Papers with Solution- Download PDFs

JEE Main Question Papers Date Shift/Slot Question Paper with Solution
B.E./B.Tech April 8, 2018 Code A Check Here
B.E./B.Tech April 8, 2018 Code B Check Here
B.E./B.Tech April 8, 2018 Code C Check Here
B.E./B.Tech April 8, 2018 Code D Check Here
B.E./B.Tech April 15, 2018 Shift 1 Check Here
B.E./B.Tech April 15, 2018 Shift 2 Check Here
B.E./B.Tech April 16, 2018 Shift 1 Check Here
JEE Main Mock Test Series

*The article might have information for the previous academic years, which will be updated soon subject to the notification issued by the University/College.

JEE Main 2018 Questions

  • 1.
    In a group of 3 girls and 4 boys, there are two boys \( B_1 \) and \( B_2 \). The number of ways in which these girls and boys can stand in a queue such that all the girls stand together, all the boys stand together, but \( B_1 \) and \( B_2 \) are not adjacent to each other, is:

      • 144
      • 120
      • 72
      • 96

    • 2.
      If \( \alpha + i\beta \) and \( \gamma + i\delta \) are the roots of the equation \( x^2 - (3-2i)x - (2i-2) = 0 \), \( i = \sqrt{-1} \), then \( \alpha\gamma + \beta\delta \) is equal to:

        • 6
        • 2
        • -2
        • -6

      • 3.
        A galvanometer having a coil of resistance 30 \( \Omega \) needs 20 mA of current for full-scale deflection. If a maximum current of 3 A is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be \( X \, \Omega \), where \( X \) is:

          • 447
          • 298
          • 149
          • 596

        • 4.
          The variance of the numbers 8, 21, 34, 47, \dots, 320, is:


            • 5.
              Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). 
              Assertion (A) : If Young’s double slit experiment is performed in an optically denser medium than air, then the consecutive fringes come closer. 
              Reason (R) : The speed of light reduces in an optically denser medium than air while its frequency does not change. 
              In the light of the above statements, choose the most appropriate answer from the options given below :

                • Both (A) and (R) are true and (R) is the correct explanation of (A)
                • (A) is false but (R) is true.
                • Both (A) and (R) are true but (R) is not the correct explanation of (A)
                • (A) is true but (R) is false.

              JEE Main 2025 : 6 Answered Questions

              Ques. What is the expected rank (AIR) for Sastra University Thanjavur for Biotechnology?

              If you are targeting B.Tech in Biotechnology at SASTRA University, Thanjavur , a JEE Main All India Rank of 9,000 to 13,000 (General category) would typically be safe based on past trends and anticipated cutoffs. Quick Details: Criteria Details Expected JEE Main Rank (Gen) 9,000 – 13,000 Eligibility 10+2 with Physics, Chemistry & Maths/Bio Admission Mode Based on JEE Main + Class 12 marks Seat Distribution 70% seats through JEE Main stream Branch Expected Rank Range CSE (Computer Science) 1,500 – 5,000 IT (Information Technology) 3,000 – 7,000 ECE (Electronics & Communication) 4,000 – 8,000 Biotechnology 9,000 – 13,000 So, if your JEE Main ranking is close to 9,000 to 13,000, you have a highly likely chance of securing a seat in Biotech in SASTRA. Ensure your 12th board marks are also good as they're taken into account as well....Read More
              Answer By Muskan Agrahari 27 Jun 25
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              Ques. I got 281441 rank in JEE Mains. Can I take admission to MNIT Jaipur

              You are not eligible to enroll at MNIT Jaipur under the general All India quota in B.Tech because of your JEE Main rank of 281,441. For this reason, depending on the branch, the lowest rank in the General category (AI quota) at MNIT Jaipur in 2025 was approximately 38,000. Metallurgy and civil engineering, two even less competitive fields, close below 80,000–90,000. Your rank is far beyond this range. Here is what can you do: – Search other NITs or GFTIs: Lower cutoffs will be there in New NITs or Government Funded Technical Institutes. If you belong to SC/ST/OBC/EWS, there is hope.  – State-Level Colleges: Apply to state engineering colleges or private universities with higher cutoffs that take JEE Main scores. Examples: RTU, LNCT, VIT, etc.  – Management Quota / Direct Admission Private colleges permit students to accept admissions on the basis of their 12th-grade marks or low JEE ranks.  – Take a Gap Year If you wish to enter NITs or top colleges, take a gap to prepare for JEE Main 2026. For details of different colleges, check out Collegedunia Page...Read More
              Answer By Muskan Agrahari 26 Jun 25
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              Ques. Is there any college I can get into if my JEE Main CRL is 112660 in the Gen category?

              You can have a college of this standard, but the best NITs and core branches such as CSE or ECE might not be available in the open category. But the newer NITs, IIITs, or GFTIs with less competitive streams such as civil, metallurgy, or production provide opportunities, particularly in spot or special rounds. Attempt home state counseling with better cutoffs. State and private engineering colleges consider this rank. Complete a few options in JoSAA and remain flexible. You can get into a good college but may not get the top branch. For detail of different colleges, you can check out Collegedunia’s Page...Read More
              Answer By Muskan Agrahari 26 Jun 25
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              Ques. I scored 121 in my JEE exam. I would like to know if I’m eligible for admission to IIT Madras.

              No, with 121 in JEE Main, you will not be eligible for IIT Madras admission because IITs accept JEE Advanced scores and not JEE Main. If your 121 is JEE Main, check if you are among the first 2.5 lakh students who can appear for JEE Advanced. Only candidates with a good rank in JEE Advanced can get through in IIT Madras. Quick Breakdown: Criteria Details Exam Required JEE Advanced (not JEE Main) What your score means 121 in JEE Main may not qualify you for JEE Advanced cutoff Admission Process Qualify JEE Main ? Appear for JEE Advanced ? Get rank within cutoff Cutoff at IIT Madras (2024) CSE: ~AIR 171 (OPEN), Least competitive: ~AIR 6926...Read More
              Answer By Muskan Agrahari 24 Jun 25
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              Ques. My EWS rank in JEE Main is 26,066. Do I have a chance of getting admission into any NIT? If yes, which ones could I possibly get?

              Since EWS rank 26,066 is quite low, getting admission to top NITs or popular branches like CSE, ECE, or IT is difficult since these usually close below 10,000 EWS rank. Don't lose hope, though! You can still get options in some of the newer NITs or less popular branches, particularly during the last rounds of JoSAA counselling. Possible Options You Can Explore: NITs to Target Likely Available Branches NIT Puducherry, Mizoram, Sikkim Civil, Mechanical, Chemical, Metallurgy NIT Nagaland, Arunachal Pradesh Production, Bio-Tech, Instrumentation, etc. Note: These are tentative and depend on this year’s cutoff trends.  Check detailed cutoffs at NIT Cutoff – ?collegedunia What you should do is fill as many options in JoSAA (all NITs + all branches) as possible, do not leave any counselling round, and keep backup options free like IIITs, GFTIs, or state colleges based on JEE Main score...Read More
              Answer By Muskan Agrahari 25 Jun 25
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