We all are familiar with Multiple Choice Questions (MCQs). These are the questions which give us some options, usually 4, out of which one, or sometimes more than one, is correct. The Joint Entrance Examination Main, or JEE Main, has 30 Questions per subject out of which 20 are MCQs with only one correct option. Generally in mathematics for some questions, you will find the options in integer values like chapters trigonometry, sequence and series and others for which you can easily get the answer by using the values given in options directly to solve questions and verify it in less time. In this article, we are going to explain some ticks and trips with the help of examples which you can implement in various questions. Let us understand the trick with the help of some examples.
Example 1
If A + B = 45°, then find the value of equals
a)1
b)0
c)2
d)1
Let us first try to understand the question. It means that if A + B = 45°, then the value of is always constant, which is equal to one of the options. So, for all combinations of angles A and B whose sum is 45°, the value of (
will always be the same, which is equal to one of the four options.
So, whether (A = 0° and B = 45°) or (A = 1 ° and B = 44°) or (A = 10° and B = 35°), they all will give the same value of (1+tanA)(1+tanB). So, we can easily replace A and B with some convenient values of angles and get the required value of (1+tanA) * (1+tanB). Let us put A = 0° and B = 45°,
= (1+tan0°)(1+tan45°) = (1 + 0)(1 + 1) = 1*2 = 2
So, option C is correct
Example 2
Now we have to select the option that gives the correct sum for all natural number values of n. So, the correct option has to be true for n = 1, 2, 3, …and all other natural number values of n. So, if any option is giving the wrong sum for n = 1, we can be sure that that option is incorrect.
Also Read,
Let us try to find the options that are incorrect.
For n = 1, the actual sum of the given series is =
Checking what value option A for n = 1: . So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Checking what value option B for n = 1: (. So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option C for n = 1: So, it gives the correct answer for n = 1. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option D for n = 1: So, it gives the wrong answer for n = 1, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Now both options A and D are eliminated, and one of option B or C is correct. Let us now see which of these two is giving the correct answer for n = 2.
For n = 2, the actual sum of the given series is
Checking what value option B for n = 2: . So, it gives the correct answer for n = 2 as well. But this might give wrong answers for higher values of n. So, we will not mark this as correct right now.
Checking what value option C for n = 2: . So, it gives the wrong answer for n = 2, and hence it cannot give the correct answer for all values of n. Hence this option is wrong.
Now we know that 3 out of 4 options are wrong, and only option B is left. So, we can safely mark option B as correct.
Example 3
If the value of the determinant given equals ka3b3c3, then the value of k is
a)1
b)0
c)-1
d)2
This question means that the value of the determinant always equals ka3b3c3 for all sets of values of a, b and c. In such questions, we can substitute some values of a, b and c and check to see the values of the determinant and the value of ka3b3c3. Also try to keep values of a, b and c such that ka3b3c3 does not become 0. So we will keep non-zero values of a, b and c. Let us put a = b = c = 1.
The value of the determinant is
which equals 0 (As two columns are the same) and the value of ka3b3c3 is k. So comparing these, we get k = 0.
Know when to use this trick and when not to. Don't use it for questions with more than one right answer or for multiple-choice questions where 'None of these' or 'All of these' is the only correct option.
On Question asked by student community
In Malda district, West Bengal, the JEE Main 2025 exam centre code is WB20. Candidates from the district can choose Malda as their preferred exam centre during registration. The final allotment of centres depends on availability and preferences submitted by students. The exact number of exam halls may vary each year based on the number of applicants. Students should confirm their allotted centre through the official JEE Main admit card.
Yes, you should apply EWS in JEE Main form if you are eligible. For 2026 attempt, it’s better to make your EWS certificate in advance and keep it ready, because at the time of counseling they ask for a valid certificate.
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.
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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 !
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