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JEE Main: Did These Questions Have Wrong Answers In JEE Main Answer Key?
Updated on May 05, 2022 08:57 AM IST | #JEE Main
The Joint Entrance Examination (JEE) Main is conducted by the National Testing Agency (NTA), and every year, lakhs of students appear for the exam. It is the entrance test – or the preliminary selection round – for the top engineering colleges in the country including the Indian Institutes of Technology (IIT), National Institutes of Technology (NIT), Indian Institutes of Information Technology (IIIT) and many more government institutions.
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Clearly, JEE Main is critical. However, JEE Main 2021 had questions which were conceptually or factually ambiguous and some of them were simply incorrect, going by the JEE Main answer key. Last year, the exam was conducted in four phases – February, March, July and August – across 26 Shifts. The result was declared in September.
Here, Careers360 looks into the questions asked in the July and August shifts, and highlights some of the errors in the JEE Main previous year question paper’s Chemistry section. The necessary changes required for these questions were absent from the NTA JEE Main answer keys, which could have affected the scorecards of many candidates in the JEE Main result.
These questions, along with their respective problems, are listed out below:
JEE Main 2021: Attempt 3
July 20, Shift 1
Question ID : 86435117255
JEE Main Answer Key: 172
The issue:
The unit in which Kp is to be reported has not been specified in the question.
So, a student solving for the unit as (Pa)-1 will get an answer of 0.172 which would be rounded off to 0 while the given answer will be in units of (kPa)-1
____________________________________________________________________________
July 20, Shift 2
Question ID : 86435117528
JEE Main Answer Key: 1
The issue:
There is a conceptual error in the question. A negatively-charged sol cannot be precipitated by a negatively-charged ion.
Also, the coagulating value is defined as the minimum concentration of an electrolyte in millimoles per litre required to cause precipitation of a sol in two hours.
So, there arises another problem: How do you extrapolate data for two hours from the one-hour data?
____________________________________________________________________________
July 22, Shift 2
Question ID : 86435117785
JEE Main Answer Key: 86435159952 (Option 2)
Question and Option IDs
The issue:
Multiple options given in the question (Options 1,2 and 3) are capable of forming tertiary alcohol on reaction with excess of Grignard reagent (CH3MgBr) followed by hydrolysis.
Option 1: Alpha- Beta unsaturated carbonyl compounds show direct addition to the carbonyl group with Grignard reagent. Hence, tertiary alcohols will be obtained from the carbonyl groups
Option 3: Grignard reagent can add to a ketone, forming a tertiary alcohol.
____________________________________________________________________________
July 25, Shift 1
Question ID : 86435118234
JEE Main Answer Key: 86435161299 (Option 1)
The issue:
Lewis acids are not used with Phenol because Phenol itself forms a complex with the Lewis acid, due to which the ring becomes deactivated towards Electrophilic Substitution.
In fact, this reaction of Phenol with a Lewis acid FeCl3 is actually used to detect the presence of Phenol in solution. Phenol forms a complex with neutral FeCl3 and produces a violet colour.
Correct answer should be 86435161298 [Option (3)]
____________________________________________________________________________
July 25, Shift 1
Question ID : 86435118225
JEE Main Answer Key: 86435161264 (Option 3)
The issue:
Be is not an alkaline earth metal as its oxide and hydroxide are Amphoteric in nature.
Here is an excerpt from NCERT textbook’s chapter on the s Block elements confirming it.
Hence, Statement I is also correct in the given question.
Correct answer should be 86435161261 [Option (2)]
____________________________________________________________________________
July 27, Shift 1
Question ID : 86435118870
JEE Main Answer Key: 5
The issue:
Data given is insufficient to solve the problem.
Molality (m) can be calculated from the density of solution only when the Molarity (M) is also given in the question.
In the question, Molarity has not been provided and thus the question cannot be solved.
____________________________________________________________________________
July 27, Shift 2
Question ID : 86435119236
JEE Main Answer Key: 300
The issue :
The answer in the JEE Main final answer key was wrong. The answer should have been 600.
____________________________________________________________________________
JEE Main 2021: Attempt 4
August 26, Shift 2
Question ID : 86435120226
JEE Main Answer Key: 50
The issue :
The answer given in the answer key is wrong. The answer should be 0.
The answer is 0.02 which would be rounded off to 0.
For the answer to be 50, the value of (Kc)-1 had to be asked.
____________________________________________________________________________
August 27, Shift 1
Question ID : 86435120586
JEE Main Answer Key: 4
The issue :
The answer given is 4 because the inner 4f electrons have not been counted.
The question does not say that we need to consider only the 5f electrons
____________________________________________________________________________
August 31, Shift 2
Question ID : 86435121290
JEE Main Answer Key: 86435170463 (Option 4)
The issue :
Alcoholic NaOH causes dehydrohalogenation (elimination) and not substitution.
The answer should be 86435170462 (Option 2)
____________________________________________________________________________
August 31, Shift 2
Question ID : 86435121287
JEE Main Answer Key: 86435170452 (Option 2)
The issue :
None of the options is correct.
All the given options contain permanganate ion (MnO4-) which contains Mn in its highest oxidation state (+7) and hence, it cannot show disproportionation reaction
Question should have been awarded as bonus
____________________________________________________________________________
September 1, Shift 2
Question ID : 86435121562
JEE Main Answer Key: 86435171281 (Option 4)
The issue :
Options 1 and 4 can both be correct but only Option 4 has been given as the correct answer.
Acenaphthene (compound in Option 1) is also Aromatic in nature.
____________________________________________________________________________
Apart from the above-mentioned questions, there were also some in the integer section that had very lengthy mathematical calculations and whose answers had to be rounded off to the nearest integers.
These questions are usually solved by approximations and it is very difficult for students to arrive at the correct integer answer. Previously, these questions had a correct range for the answer, which gave students some leeway to make appropriate approximations in calculations.
Now, with negative marking in the numerical response questions, this becomes an even bigger problem. Considering the high-stakes nature of the JEE Mains, even a few such errors can have significant ramifications for applicants and be deeply demoralising.
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Questions related to JEE Main
Have a question related to JEE Main ?
Hello there,
With 44,000 OBC rank in JEE Mains, getting a free (zero tuition fee) seat in top NITs or IIITs is tough, especially for popular branches like CSE, ECE, or IT.
However, you may still get lower-demand branches like Civil, Mechanical, Metallurgy, or Production in newer NITs like NIT Mizoram, NIT Nagaland, NIT Sikkim, or NIT Puducherry. Some IIITs like IIIT Bhagalpur, IIIT Una, or IIIT Kottayam may also offer seats in lesser-demand branches.
Also, tuition fee waivers for OBC students depend on your family income (if it’s below Rs. 1 lakh per year, full waiver; between Rs. 1-5 lakh, 2/3rd waiver).
Apply through JoSAA counselling and fill maximum choices to increase your chances.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
kumardurgesh1802 28 April,2025
Hello Niyati,
With a 314286 rank in JEE Mains 2025 , it will be difficult to get top government colleges in Delhi like DTU, NSUT, IIIT-D, or IGDTUW through JAC Delhi counselling.
However, you can try for admission in private colleges in Delhi NCR such as:
-
Amity University, Noida
-
GNIOT, Greater Noida
-
Galgotias University, Greater Noida
-
Sharda University, Greater Noida
-
Manav Rachna International Institute, Faridabad
These colleges offer direct admission or admission with low JEE Main ranks. You can also apply for spot rounds or management quota seats.
If you prefer government colleges, you can look for other states' counselling where vacant seats are filled with lower ranks.
I would recommend you to use the JEE Mains College Predictor Tool by Careers360 to get the list of possible colleges you can get with your rank.
To get personalised support for JEE Mains counselling, you can contact through WhatsApp on ( +916005956597 ) to get assistance.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
kumardurgesh1802 27 April,2025
Hello Ankit,
With a 141486 CRL and 46953 OBC rank in JEE Mains, you have moderate chances for CSE or CSE specialization in some lower-tier NITs , IIITs , and state government colleges through home state quota or special rounds .
Here are some possible options:
-
NIT Mizoram, NIT Nagaland, NIT Meghalaya, NIT Arunachal Pradesh (CSE or related branches in later rounds)
-
IIIT Bhagalpur, IIIT Manipur, IIIT Ranchi (CSE or specializations)
-
State Government Colleges like GECs in Gujarat, Odisha, Haryana (if you apply through state quota)
-
Special Round CSAB (Vacancy round after JoSAA) — you can get better chances here for CSE or allied branches.
Tip: Participate in CSAB Special Rounds seriously as cutoffs drop.
I would recommend you to use the JEE Mains College Predictor Tool by Careers360 to get the list of possible colleges you can get with your rank.
To get personalised support for JEE Mains counselling, you can contact through WhatsApp on ( +916005956597 ) to get assistance.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
kumardurgesh1802 27 April,2025
Hello there,
With a
24 percentile in JEE Mains
, and considering your
12th score of 76%
and
10th score of 82%
, it may be challenging to secure a seat in top-tier colleges for
B.Tech in IT
or
ECE
. However, you can still explore
lower-tier government
and
private colleges
with
affordable fees
. Here are a few options to consider:
Government Colleges:
-
State Government Colleges (State Quota)
-
Engineering Colleges in UP, MP, Odisha, Rajasthan
-
Govt. Engineering Colleges in Bihar (like BCE Patna, MIT Muzaffarpur)
-
Govt. Engineering Colleges in Tamil Nadu (if applying through the state quota)
-
-
Lower Tier NITs or IIITs
-
NIT Manipur
-
NIT Mizoram
-
NIT Nagaland
Private Colleges:
-
-
Shiv Nadar University (Greater Noida)
-
KIIT University (Bhubaneswar)
-
UPES (Dehradun)
-
SRM University (Chennai)
-
Shri Mata Vaishno Devi University (Katra)
-
Amity University (Noida)
-
VIT Bhopal
-
Graphic Era University (Dehradun)
These colleges offer good infrastructure and affordable fees. Private colleges might have moderate fees but offer scholarship programs that can reduce the cost.
I would recommend you to use the JEE Mains College Predictor Tool by Careers360 to get the list of possible colleges you can get with your rank/percentile.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
kumardurgesh1802 27 April,2025
Hello there,
With a JEE Mains percentile of
92.55
and an
OBC NCL rank of 35,798
, your chances of securing a seat in
Computer Science Engineering (CSE)
in the following institutions depend on various factors like the number of seats, category rank, and previous year's cutoffs:
1.
NITs (National Institutes of Technology)
-
Cutoff Range for CSE : For NITs, the cutoff for CSE typically ranges from around 10,000 to 40,000 in the general category, but this may vary depending on the institution and state.
-
Chances : You may have a chance at NITs in lesser-known or newer NITs in states with lower demand for CSE, especially if you opt for home state quota.
2. Government IITs
-
Cutoff Range for CSE : The cutoffs for IITs are generally much higher, often in the 500-1,500 range in the open category for CSE.
-
Chances : It’s very unlikely that you will get CSE in any of the older or established IITs with a 35,798 rank.
3. GFTIs (Government Funded Technical Institutes)
-
Cutoff Range for CSE : For GFTIs, the cutoff for CSE is generally 30,000 to 45,000 for OBC NCL category in some institutes.
-
Chances : You may have a chance in some of the GFTIs for CSE, as these institutes have relatively higher cutoff ranks compared to IITs and NITs.
4. Other Good Institutes
-
Private Institutes : Institutes like SRM , VIT , and Bennett University offer CSE in their B.Tech programs, though their admission is typically through entrance exams specific to the institute.
-
Chances : You will likely have good chances of admission into private engineering colleges that offer good placement support and infrastructure.
Summary
-
NITs : Possible in lesser-known ones, especially through home state quota.
-
IITs : Unlikely for CSE due to high cutoff.
-
GFTIs : Moderate chances for CSE.
-
Private Institutes : Good chances for admission.
For better chances, consider opting for branches like IT or Electronics in NITs or GFTIs, as CSE may be more competitive.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
kumardurgesh1802 27 April,2025
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